The 54th William Lowell Putnam Mathematical Competition

Saturday, December 4, 1993

A–1 The horizontal line y = c intersects the curve y = 2x − B–1 Find the smallest positive integer n such that for every
3x3 in the first quadrant as in the figure. Find c so that integer m with 0 < m < 1993, there exists an integer k
the areas of the two shaded regions are equal. [Figure for which
not included. The first region is bounded by the y-axis,
the line y = c and the curve; the other lies under the m k m+1
< < .
curve and above the line y = c between their two points 1993 n 1994
B–2 Consider the following game played with a deck of 2n
of intersection.] cards numbered from 1 to 2n. The deck is randomly
A–2 Let (xn )n≥0 be a sequence of nonzero real numbers such shuffled and n cards are dealt to each of two players.
that xn2 − xn−1 xn+1 = 1 for n = 1, 2, 3, . . . . Prove there Beginning with A, the players take turns discarding one
exists a real number a such that xn+1 = axn − xn−1 for of their remaining cards and announcing its number.
all n ≥ 1. The game ends as soon as the sum of the numbers on the
discarded cards is divisible by 2n + 1. The last person
A–3 Let Pn be the set of subsets of {1, 2, . . . , n}. Let c(n, m) to discard wins the game. Assuming optimal strategy
be the number of functions f : Pn → {1, 2, . . . , m} such by both A and B, what is the probability that A wins?
that f (A ∩ B) = min{ f (A), f (B)}. Prove that
B–3 Two real numbers x and y are chosen at random in the
m interval (0,1) with respect to the uniform distribution.
c(n, m) = ∑ jn . What is the probability that the closest integer to x/y is
j=1
even? Express the answer in the form r + sπ, where r
and s are rational numbers.
A–4 Let x1 , x2 , . . . , x19 be positive integers each of which is
less than or equal to 93. Let y1 , y2 , . . . , y93 be positive B–4 The function K(x, y) is positive and continuous for 0 ≤
integers each of which is less than or equal to 19. Prove x ≤ 1, 0 ≤ y ≤ 1, and the functions f (x) and g(x) are
that there exists a (nonempty) sum of some xi ’s equal to positive and continuous for 0 ≤ x ≤ 1. Suppose that for
a sum of some y j ’s. all x, 0 ≤ x ≤ 1,
Z 1
A–5 Show that
f (y)K(x, y) dy = g(x)
2 0
x2 − x
Z −10 
3
dx+ and
−100 x − 3x + 1
Z 1  2 Z 1
11 x2 − x
dx+ g(y)K(x, y) dy = f (x).
1
101
x3 − 3x + 1 0
Z 11  2
10 x2 − x Show that f (x) = g(x) for 0 ≤ x ≤ 1.
dx
101 x3 − 3x + 1
100 B–5 Show there do not exist four points in the Euclidean
is a rational number. plane such that the pairwise distances between the
points are all odd integers.
A–6 The infinite sequence of 2’s and 3’s
B–6 Let S be a set of three, not necessarily distinct, posi-
2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 2, 3, 3, tive integers. Show that one can transform S into a set
containing 0 by a finite number of applications of the
3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 2, 3, 3, 3, 2, . . .
following rule: Select two of the three integers, say x
has the property that, if one forms a second sequence and y, where x ≤ y and replace them with 2x and y − x.
that records the number of 3’s between successive 2’s,
the result is identical to the given sequence. Show that
there exists a real number r such that, for any n, the nth
term of the sequence is 2 if and only if n = 1 + brmc
for some nonnegative integer m. (Note: bxc denotes the
largest integer less than or equal to x.)
 The 55th William Lowell Putnam Mathematical Competition
Saturday, December 3, 1994

A–1 Suppose that a sequence a1 , a2 , a3 , . . . satisfies 0 < an ≤ of integers, then at most 512 of the functions in F map
a2n + a2n+1 for all n ≥ 1. Prove that the series ∑∞ n=1 an A to itself.
diverges.
B–1 Find all positive integers n that are within 250 of exactly
A–2 Let A be the area of the region in the first quadrant 15 perfect squares.
bounded by the line y = 12 x, the x-axis, and the ellipse
1 2 2 B–2 For which real numbers c is there a straight line that
9 x + y = 1. Find the positive number m such that A intersects the curve
is equal to the area of the region in the first quadrant
bounded by the line y = mx, the y-axis, and the ellipse x4 + 9x3 + cx2 + 9x + 4
1 2 2
9 x + y = 1.

A–3 Show that if the points of an isosceles right triangle of in four distinct points?
side length 1 are each colored with one of four colors, B–3 Find the set of all real numbers k with the following
then there must be two points
√ of the same color whch property: For any positive, differentiable function f that
are at least a distance 2 − 2 apart. satisfies f 0 (x) > f (x) for all x, there is some number N
A–4 Let A and B be 2 × 2 matrices with integer entries such such that f (x) > ekx for all x > N.
that A, A + B, A + 2B, A + 3B, and A + 4B are all invert- B–4 For n ≥ 1, let dn be the greatest common divisor of the
ible matrices whose inverses have integer entries. Show entries of An − I, where
that A + 5B is invertible and that its inverse has integer 
3 2
 
1 0

entries. A= and I = .
4 3 0 1
A–5 Let (rn )n≥0 be a sequence of positive real numbers such
that limn→∞ rn = 0. Let S be the set of numbers repre- Show that limn→∞ dn = ∞.
sentable as a sum B–5 For any real number α, define the function fα (x) =
bαxc. Let n be a positive integer. Show that there exists
ri1 + ri2 + · · · + ri1994 , an α such that for 1 ≤ k ≤ n,
with i1 < i2 < · · · < i1994 . Show that every nonempty in- fαk (n2 ) = n2 − k = fα k (n2 ).
terval (a, b) contains a nonempty subinterval (c, d) that
does not intersect S.
B–6 For any integer n, set
A–6 Let f1 , . . . , f10 be bijections of the set of integers such
that for each integer n, there is some composition fi1 ◦ na = 101a − 100 · 2a .
fi2 ◦ · · · ◦ fim of these functions (allowing repetitions)
which maps 0 to n. Consider the set of 1024 functions Show that for 0 ≤ a, b, c, d ≤ 99, na + nb ≡ nc + nd
(mod 10100) implies {a, b} = {c, d}.
e
F = { f1e1 ◦ f2e2 ◦ · · · ◦ f1010 },

ei = 0 or 1 for 1 ≤ i ≤ 10. ( fi0 is the identity function

and fi1 = fi .) Show that if A is any nonempty finite set
 The 56th William Lowell Putnam Mathematical Competition
Saturday, December 2, 1995
A–1 Let S be a set of real numbers which is closed under row sums a, b, c of the resulting matrix be rearranged
multiplication (that is, if a and b are in S, then so is ab). (if necessary) so that a ≤ b ≤ c. Show that for some
Let T and U be disjoint subsets of S whose union is n ≥ 1995, it is at least four times as likely that both
S. Given that the product of any three (not necessarily b = a + 1 and c = a + 2 as that a = b = c.
distinct) elements of T is in T and that the product of B–1 For a partition π of {1, 2, 3, 4, 5, 6, 7, 8, 9}, let π(x) be
any three elements of U is in U, show that at least one the number of elements in the part containing x. Prove
of the two subsets T,U is closed under multiplication. that for any two partitions π and π 0 , there are two dis-
tinct numbers x and y in {1, 2, 3, 4, 5, 6, 7, 8, 9} such that
A–2 For what pairs (a, b) of positive real numbers does the π(x) = π(y) and π 0 (x) = π 0 (y). [A partition of a set S
improper integral is a collection of disjoint subsets (parts) whose union is
Z ∞ q√ √ S.]
√ √
q 
x+a− x− x − x − b dx
b B–2 An ellipse, whose semi-axes have lengths a
and b, rolls
without slipping on the curve y = c sin ax . How are
converge? a, b, c related, given that the ellipse completes one rev-
olution when it traverses one period of the curve?
A–3 The number d1 d2 . . . d9 has nine (not necessarily dis-
tinct) decimal digits. The number e1 e2 . . . e9 is such that B–3 To each positive integer with n2 decimal digits, we asso-
each of the nine 9-digit numbers formed by replacing ciate the determinant of the matrix obtained by writing
just one of the digits di is d1 d2 . . . d9 by the correspond- the digits in order across the rows. For example, for n =
ing digit ei (1 ≤ i ≤ 9) is divisible by 7. The number

8 6
f1 f2 . . . f9 is related to e1 e2 . . . e9 is the same way: that 2, to the integer 8617 we associate det = 50.
1 7
is, each of the nine numbers formed by replacing one of Find, as a function of n, the sum of all the determinants
the ei by the corresponding fi is divisible by 7. Show associated with n2 -digit integers. (Leading digits are as-
that, for each i, di − fi is divisible by 7. [For example, sumed to be nonzero; for example, for n = 2, there are
if d1 d2 . . . d9 = 199501996, then e6 may be 2 or 9, since 9000 determinants.)
199502996 and 199509996 are multiples of 7.]
B–4 Evaluate
A–4 Suppose we have a necklace of n beads. Each bead is s
labeled with an integer and the sum of all these labels 1
is n − 1. Prove that we can cut the necklace to form a
8 2207 − 1
.
2207 − 2207−...
string whose consecutive labels x1 , x2 , . . . , xn satisfy
√
a+b c
k Express your answer in the form d , where a, b, c, d
∑ xi ≤ k − 1 for k = 1, 2, . . . , n. are integers.
i=1
B–5 A game starts with four heaps of beans, containing 3,4,5
A–5 Let x1 , x2 , . . . , xn be differentiable (real-valued) func- and 6 beans. The two players move alternately. A move
tions of a single variable f which satisfy consists of taking either

dx1 a) one bean from a heap, provided at least two beans

= a11 x1 + a12 x2 + · · · + a1n xn are left behind in that heap, or
dt
dx2 b) a complete heap of two or three beans.
= a21 x1 + a22 x2 + · · · + a2n xn
dt
.. .. The player who takes the last heap wins. To win the
. . game, do you want to move first or second? Give a
dxn winning strategy.
= an1 x1 + an2 x2 + · · · + ann xn
dt B–6 For a positive real number α, define
for some constants ai j > 0. Suppose that for all i,
S(α) = {bnαc : n = 1, 2, 3, . . . }.
xi (t) → 0 as t → ∞. Are the functions x1 , x2 , . . . , xn nec-
essarily linearly dependent? Prove that {1, 2, 3, . . . } cannot be expressed as the dis-
A–6 Suppose that each of n people writes down the numbers joint union of three sets S(α), S(β ) and S(γ). [As usual,
1,2,3 in random order in one column of a 3 × n matrix, bxc is the greatest integer ≤ x.]
with all orders equally likely and with the orders for
different columns independent of each other. Let the
 Solutions to the 56th William Lowell Putnam Mathematical Competition
Saturday, December 2, 1995

Kiran Kedlaya

A–1 Suppose on the contrary that there exist t1 ,t2 ∈ T A–5 Everyone (presumably) knows that the set of solutions
with t1t2 ∈ U and u1 , u2 ∈ U with u1 u2 ∈ T . Then of a system of linear first-order differential equations
(t1t2 )u1 u2 ∈ U while t1t2 (u1 u2 ) ∈ T , contradiction. with constant coefficients is n-dimensional, with ba-
sis vectors of the form fi (t)~vi (i.e. a function times
A–2 The integral converges iff a = b. The easiest proof uses a constant vector), where the ~vi are linearly indepen-
“big-O” notation and the fact that (1+x)1/2 = 1+x/2+ dent. In particular, our solution ~x(t) can be written as
O(x2 ) for |x| < 1. (Here O(x2 ) means bounded by a ∑ni=1 ci fi (t)~v1 .
constant times x2 .)
Choose a vector ~w orthogonal to~v2 , . . . ,~vn but not to~v1 .
So Since ~x(t) → 0 as t → ∞, the same is true of ~w ·~x; but
√ √ p that is simply (~w ·~v1 )c1 f1 (t). In other words, if ci 6= 0,
x + a − x = x1/2 ( 1 + a/x − 1) then fi (t) must also go to 0.
= x1/2 (1 + a/2x + O(x−2 )), However, it is easy to exhibit a solution which does
not go to 0. The sum of the eigenvalues of the matrix
hence A = (ai j ), also known as the trace of A, being the sum
q√ of the diagonal entries of A, is nonnegative, so A has
√
x + a − x = x1/4 (a/4x + O(x−2 )) an eigenvalue λ with nonnegative real part, and a cor-
responding eigenvector ~v. Then eλt~v is a solution that
and similarly does not go to 0. (If λ is not real, add this solution to
q
√ √ its complex conjugate to get a real solution, which still
x − x − b = x1/4 (b/4x + O(x−2 )). doesn’t go to 0.)
Hence one of the ci , say c1 , is zero, in which case ~x(t) ·
Hence the integral we’re looking at is
~w = 0 for all t.
Z ∞
x1/4 ((a − b)/4x + O(x−2 )) dx. A–6 View this as a random walk/Markov process with states
b (i, j, k) the triples of integers with sum 0, correspond-
ing to the difference between the first, second and third
The term x1/4 O(x−2 ) is bounded by a constant times
rows with their average (twice the number of columns).
x−7/4 , whose integral converges. Thus we only have to Adding a new column adds on a random permutation
decide whether x−3/4 (a − b)/4 converges. But x−3/4 of the vector (1, 0, −1). I prefer to identify the triple
has divergent integral, so we get convergence if and (i, j, k) with the point (i − j) + ( j − k)ω + (k − i)ω 2 in
only if a = b (in which case the integral telescopes any- the plane, where ω is a cube root of unity. Then adding
way). a new column corresponds to moving to one of the six
A–3 Let D and E be the numbers d1 . . . d9 and e1 . . . e9 , re- neighbors of the current position in a triangular lattice.
spectively. We are given that (ei − di )109−i + D ≡ 0 What we’d like to argue is that for large enough n, the
(mod 7) and ( fi − ei )109−i + E ≡ 0 (mod 7) for i = ratio of the probabilities of being in any two particular
1, . . . , 9. Sum the first relation over i = 1, . . . , 9 and we states goes to 1. Then in fact, we’ll see that eventually,
get E − D + 9D ≡ 0 (mod 7), or E + D ≡ 0 (mod 7). about six times as many matrices have a = b − 1, b =
Now add the first and second relations for any partic- c − 1 than a = b = c. This is a pain to prove, though,
ular value of i and we get ( fi − di )109−i + E + D ≡ 0 and in fact is way more than we actually need.
(mod 7). But we know E + D is divisible by 7, and 10 Let Cn and An be the probability that we are at the ori-
is coprime to 7, so di − fi ≡ 0 (mod 7). gin, or at a particular point adjacent to the origin, re-
A–4 Let sk = x1 + · · · + xk − k(n − 1)/n, so that sn = s0 = 0. spectively. Then Cn+1 = An . (In fact, Cn+1 is 1/6 times
These form a cyclic sequence that doesn’t change when the sum of the probabilities of being at each neighbor of
you rotate the necklace, except that the entire sequence the origin at time n, but these are all An .) So the desired
gets translated by a constant. In particular, it makes result, which is that Cn /An ≥ 2/3 for some large n, is
sense to choose xi for which si is maximum and make equivalent to An+1 /An ≥ 2/3.
that one xn ; this way si ≤ 0 for all i, which gives x1 + Suppose on the contrary that this is not the case; then
· · · + xi ≤ i(n − 1)/n, but the right side may be replaced An < c(2/3)n for some constant n. However, if n = 6m,
by i − 1 since the left side is an integer. the probability that we chose each of the six types
of moves m times is already (6m)!/[m!6 66m ], which
 2

by Stirling’s approximation is asymptotic to a constant Now how to compute the eighth root of L? Notice that
times m−5/2 . This term alone is bigger than c(2/3)n , so if x satisfies the quadratic x2 − ax + 1 = 0, then we have
we must have An+1 /An ≥ 2/3 for some n. (In fact, we
must have An+1 /An ≥ 1 − ε for any ε > 0.) 0 = (x2 − ax + 1)(x2 + ax + 1)

B–1 For a given π, no more than three different values of = x4 − (a2 − 2)x2 + 1.
π(x) are possible (four would require one part each of
Clearly, then, the positive square roots of the quadratic
size at least 1,2,3,4, and that’s already more than 9 el-
x2 − bx + 1 satisfy the quadratic x2 − (b2 + 2)1/2 x + 1 =
ements). If no such x, y exist, each pair (π(x), π 0 (x))
occurs for at most 1 element of x, and since there are 0. Thus we compute that L1/2 is the greater root of x2 −
only 3 × 3 possible pairs, each must occur exactly once. 47x + 1 = 0, L1/4 is the greater root of x2 − 7x + 1 = 0,
In particular, each value of π(x) must occur 3 times. and L1/8 is the greater
√ root of x2 − 3x + 1 = 0, otherwise
However, clearly any given value of π(x) occurs kπ(x) known as (3 + 5)/2.
times, where k is the number of distinct partitions of that B–5 This problem is dumb if you know the Sprague-
size. Thus π(x) can occur 3 times only if it equals 1 or Grundy theory of normal impartial games (see Conway,
3, but we have three distinct values for which it occurs, Berlekamp and Guy, Winning Ways, for details). I’ll de-
contradiction. scribe how it applies here. To each position you assign
B–2 For those who haven’t taken enough physics, “rolling a nim-value as follows. A position with no moves (in
without slipping” means that the perimeter of the ellipse which case the person to move has just lost) takes value
and the curve pass at the same rate, so all we’re saying 0. Any other position is assigned the smallest number
is that the perimeter of the ellipse equals the length of not assigned to a valid move from that position.
one period of the sine curve. So set up the integrals: For a single pile, one sees that an empty pile has value
0, a pile of 2 has value 1, a pile of 3 has value 2, a pile
Z 2π q of 4 has value 0, a pile of 5 has value 1, and a pile of 6
(−a sin θ )2 + (b cos θ )2 dθ has value 0.
0
Z 2πa q You add piles just like in standard Nim: the nim-value
= 1 + (c/a cos x/a)2 dx. of the composite of two games (where at every turn you
0
pick a game and make a move there) is the “base 2 ad-
Let θ = x/a in the second integral and write 1 as dition without carries” (i.e. exclusive OR) of the nim-
sin2 θ + cos2 θ and you get values of the constituents. So our starting position, with
piles of 3, 4, 5, 6, has nim-value 2 ⊕ 0 ⊕ 1 ⊕ 0 = 3.
Z 2π p
a2 sin2 θ + b2 cos2 θ dθ A position is a win for the player to move if and only if
0 it has a nonzero value, in which case the winning strat-
Z 2π q
egy is to always move to a 0 position. (This is always
= a2 sin2 θ + (a2 + c2 ) cos2 θ dθ . possible from a nonzero position and never from a zero
0
position, which is precisely the condition that defines
Since the left side is increasing as a function of b, we the set of winning positions.) In this case, the winning
have equality if and only if b2 = a2 + c2 . move is to reduce the pile of 3 down to 2, and you can
easily describe the entire strategy if you so desire.
B–3 For n = 1 we obviously get 45, while for n = 3 the
answer is 0 because it both changes sign (because de- B–6 Obviously α, β , γ have to be greater than 1, and no two
terminants are alternating) and remains unchanged (by can both be rational, so without loss of generality as-
symmetry) when you switch any two rows other than sume that α and β are irrational. Let {x} = x − bxc
the first one. So only n = 2 is left. By the multilinear- denote the fractional part of x. Then m ∈ S(α) if and
ity of the determinant, the answer is the determinant of only if f (m/α) ∈ (1 − 1/α, 1) ∪ {0}. In particular, this
the matrix whose first (resp. second) row is the sum of means that S(α) ∩ {1, . . . , n} contains d(n + 1)/αe − 1
all possible first (resp. second) rows. There are 90 first elements, and similarly. Hence for every integer n,
rows whose sum is the vector (450, 405), and 100 sec-      
ond rows whose sum is (450, 450). Thus the answer is n+1 n+1 n+1
n= + + − 3.
450 × 450 − 450 × 405 = 45 × 450 = 20250. α β γ
B–4 The infinite continued fraction is defined as the limit Dividing through by n and taking the limit as n → ∞
of the sequence L0 = 2207, Ln+1 = 2207 − 1/Ln . No- shows that 1/α + 1/β + 1/γ = 1. That in turn implies
tice that the sequence is strictly decreasing (by induc- that for all n,
tion) and thus indeed has a limit L, which satisfies L =
2207 − 1/L, or rewriting, L2 − 2207L + 1 = 0. More-
     
n+1 n+1 n+1
− + − + − = 2.
over, we want the greater of the two roots. α β γ

Our desired contradiction is equivalent to showing that

the left side actually takes the value 1 for some n. Since
 3

the left side is an integer, it suffices to show that {−(n + does hold. Since α and β are irrational, by the
1)/α} + {−(n + 1)/β } < 1 for some n. one-dimensional Weil theorem, the set of points
A result in ergodic theory (the two-dimensional version ({−n/α}, {−n/β } is dense in the set of (x, y) in the
of the Weil equidistribution theorem) states that if 1, r, s unit square such that ax + by is an integer. It is simple
are linearly independent over the rationals, then the enough to show that this set meets the region {(x, y) ∈
set of points ({nr}, {ns} is dense (and in fact equidis- [0, 1]2 : x + y < 1} unless a + b is an integer, and that
tributed) in the unit square. In particular, our claim def- would imply that 1/α + 1/β , a quantity between 0 and
initely holds unless a/α + b/β = c for some integers 1, is an integer. We have our desired contradiction.
a, b, c.
On the other hand, suppose that such a relation
 The 57th William Lowell Putnam Mathematical Competition
Saturday, December 7, 1996

A–1 Find the least number A such that for any two squares of minimal selfish sets, that is, selfish sets none of whose
combined area 1, a rectangle of area A exists such that proper subsets is selfish.
the two squares can be packed in the rectangle (without
interior overlap). You may assume that the sides of the B–2 Show that for every positive integer n,
squares are parallel to the sides of the rectangle.   2n−1   2n+1
2n − 1 2 2n + 1 2
A–2 Let C1 and C2 be circles whose centers are 10 units < 1 · 3 · 5 · · · (2n − 1) < .
e e
apart, and whose radii are 1 and 3. Find, with proof,
the locus of all points M for which there exists points X B–3 Given that {x1 , x2 , . . . , xn } = {1, 2, . . . , n}, find, with
on C1 and Y on C2 such that M is the midpoint of the proof, the largest possible value, as a function of n (with
line segment XY . n ≥ 2), of

A–3 Suppose that each of 20 students has made a choice of x1 x2 + x2 x3 + · · · + xn−1 xn + xn x1 .

anywhere from 0 to 6 courses from a total of 6 courses
offered. Prove or disprove: there are 5 students and 2 B–4 For any square matrix A, we can define sin A by the
courses such that all 5 have chosen both courses or all 5 usual power series:
have chosen neither course.
∞
(−1)n
A–4 Let S be the set of ordered triples (a, b, c) of distinct sin A = ∑ (2n + 1)! A2n+1 .
n=0
elements of a finite set A. Suppose that
Prove or disprove: there exists a 2 × 2 matrix A with
1. (a, b, c) ∈ S if and only if (b, c, a) ∈ S;
real entries such that
2. (a, b, c) ∈ S if and only if (c, b, a) ∈
/ S;  
1 1996
3. (a, b, c) and (c, d, a) are both in S if and only if sin A = .
0 1
(b, c, d) and (d, a, b) are both in S.
Prove that there exists a one-to-one function g from A B–5 Given a finite string S of symbols X and O, we write
to R such that g(a) < g(b) < g(c) implies (a, b, c) ∈ S. ∆(S) for the number of X’s in S minus the number of
Note: R is the set of real numbers. O’s. For example, ∆(XOOXOOX) = −1. We call a
string S balanced if every substring T of (consecutive
A–5 If p is a prime number greater than 3 and k = b2p/3c, symbols of) S has −2 ≤ ∆(T ) ≤ 2. Thus, XOOXOOX
prove that the sum is not balanced, since it contains the substring OOXOO.
   
p p
 
p Find, with proof, the number of balanced strings of
+ +···+ length n.
1 2 k
B–6 Let (a1 , b1 ), (a2 , b2 ), . . . , (an , bn ) be the vertices of a
of binomial coefficients is divisible by p2 . convex polygon which contains the origin in its inte-
rior. Prove that there exist positive real numbers x and y
A–6 Let c > 0 be a constant. Give a complete description,
such that
with proof, of the set of all continuous functions f :
R → R such that f (x) = f (x2 + c) for all x ∈ R. Note (a1 , b1 )xa1 yb1 + (a2 , b2 )xa2 yb2 + · · ·
that R denotes the set of real numbers.
+(an , bn )xan ybn = (0, 0).
B–1 Define a selfish set to be a set which has its own cardi-
nality (number of elements) as an element. Find, with
proof, the number of subsets of {1, 2, . . . , n} which are
 Solutions to the 57th William Lowell Putnam Mathematical Competition
Saturday, December 7, 1996
Manjul Bhargava and Kiran Kedlaya

A–1 If x and y are the sides of two squares with combined pairs (s, p). As f (k) is minimized for k =
3, it follows
area 1, then x2 + y2 = 1. Suppose without loss of gen- that every student occurs in at least 6 = 32 + 32 such


erality that x ≥ y. Then the shorter side of a rectangle pairs (s, p). Hence there can be at most 120/6 = 20 stu-
containing both squares without overlap must be at least dents, with equality only if each student takes 3 courses,
x, and the longer side must be at least x + y. Hence the and for each set of two courses, there are exactly 4 stu-
desired value of A is the maximum of x(x + y). dents who take both and exactly 4 who take neither.
To find this maximum, we let x = cos θ , y = sin θ with Since there are only 4 ways to complete a given pair
θ ∈ [0, π/4]. Then we are to maximize of courses to a set of 3, and only 4 ways to choose 3
courses not containing the given pair, the only way for
1 there to be 20 students (under our hypotheses) is if all
cos2 θ + sin θ cos θ = (1 + cos 2θ + sin 2θ ) sets of 3 courses are in fact taken. This is the desired
2 √
1 2 conclusion.
= + cos(2θ − π/4)
2 √2 However, Robin Chapman has pointed out that the so-
1+ 2 lution is not unique in the problem as stated, because a
≤ , given selection of courses may be made by more than
2
one student. One alternate solution is to identify the 6
with equality for θ = π/8. Hence this value is the de- courses with pairs of antipodal vertices of an icosahe-
sired value of A. dron, and have each student pick a different face and
choose the three vertices touching that face. In this ex-
A–2 Let O1 and O2 be the centers of C1 and C2 , respectively. ample, each of 10 selections is made by a pair of stu-
(We are assuming C1 has radius 1 and C2 has radius dents.
3.) Then the desired locus is an annulus centered at the
midpoint of O1 O2 , with inner radius 1 and outer radius A–4 In fact, we will show that such a function g exists with
2. the property that (a, b, c) ∈ S if and only if g(d) <
For a fixed point Q on C2 , the locus of the midpoints of g(e) < g( f ) for some cyclic permutation (d, e, f ) of
the segments PQ for P lying on C1 is the image of C1 (a, b, c). We proceed by induction on the number of
under a homothety centered at Q of radius 1/2, which elements in A. If A = {a, b, c} and (a, b, c) ∈ S, then
is a circle of radius 1/2. As Q varies, the center of this choose g with g(a) < g(b) < g(c), otherwise choose g
smaller circle traces out a circle C3 of radius 3/2 (again with g(a) > g(b) > g(c).
by homothety). By considering the two positions of Q Now let z be an element of A and B = A − {z}. Let
on the line of centers of the circles, one sees that C3 is a1 , . . . , an be the elements of B labeled such that g(a1 ) <
centered at the midpoint of O1 O2 , and the locus is now g(a2 ) < · · · < g(an ). We claim that there exists a unique
clearly the specified annulus. i ∈ {1, . . . , n} such that (ai , z, ai+1 ) ∈ S, where hereafter
an+k = ak .
A–3 The claim is false. There are 63 = 20 ways to choose


We show existence first. Suppose no such i exists; then
3 of the 6 courses; have each student choose a different
for all i, k ∈ {1, . . . , n}, we have (ai+k , z, ai ) ∈
/ S. This
set of 3 courses. Then each pair of courses is chosen by
holds by property 1 for k = 1 and by induction on k in
4 students (corresponding to the four ways to complete
general, noting that
this pair to a set of 3 courses) and is not chosen by 4
students (corresponding to the 3-element subsets of the (ai+k+1 , z, ai+k ),(ai+k , z, ai ) ∈ S
remaining 4 courses).
⇒ (ai+k , ai+k+1 , z), (z, ai , ai+k ) ∈ S
Note: Assuming that no two students choose the same
⇒ (ai+k+1 , z, ai ) ∈ S.
courses, the above counterexample is unique (up to per-
muting students). This may be seen as follows: Given a Applying this when k = n, we get (ai−1 , z, ai ) ∈ S, con-
group of students, suppose that for any pair of courses tradicting the fact that (ai , z, ai−1 ) ∈ S. Hence existence
(among the six) there are at most 4 students taking both, follows.
and at most 4 taking neither. Then there are at most
6

120 = (4 + 4) 2 pairs (s, p), where s is a student, and Now we show uniqueness. Suppose (ai , z, ai+1 ) ∈
p is a set of two courses of which s is taking either both S; then for any j 6= i − 1, i, i + 1, we have
or none. On the other hand, if a student
s is taking k (ai , ai+1 , a j ), (a j , a j+1 , ai ) ∈ S by the assumption on G.
courses, then he/she occurs in f (k) = 2k + 6−k


2 such
 2
√
Therefore define x0 = x and xn+1 = xn − c for each positive in-
teger x. By induction on n, r2 < xn+1 < xn for all n, so
(ai , z, ai+1 ), (ai+1 , a j , ai ) ∈ S ⇒ (a j , ai , z) ∈ S the sequence {xn } tends to a limit L which is a root of
(ai , z, a j ), (a j , a j+1 , ai ) ∈ S ⇒ (z, a j , a j+1 ), x2 + c = x not less than r2 . Of course this means L = r2 .
Since f (x) = f (xn ) for all n and xn → r2 , we conclude
so (a j , z, a j+1 ) ∈
/ S. The case j = i + 1 is ruled out by f (x) = f (r2 ), so f is constant on x ≥ r2 .
If r1 < x < r2 and xn is defined as before, then by in-
(ai , z, ai+1 ), (ai+1 , ai+2 , ai ) ∈ S ⇒ (z, ai+1 , ai+2 ) ∈ S
duction, xn < xn+1 < r2 . Note that the sequence can be
and the case j = i − 1 is similar. defined because r1 > c; the latter follows by noting that
the polynomial x2 − x + c is positive at x = c and has
Finally, we put g(z) in (g(an ), +∞) if i = n, and its minimum at 1/2 > c, so both roots are greater than
(g(ai ), g(ai+1 )) otherwise; an analysis similar to that c. In any case, we deduce that f (x) is also constant on
above shows that g has the desired property. r1 ≤ x ≤ r2 .
A–5 (due to Lenny Ng) For 1 ≤ n ≤ p − 1, p divides np and


Finally, suppose x < r1 . Now define x0 = x, xn+1 =
  xn2 + c. Given that xn < r1 , we have xn+1 > xn . Thus if
1 p 1 p−1 p−2 p−n+1 we had xn < r1 for all n, by the same argument as in the
= ··· first case we deduce xn → r1 and so f (x) = f (r1 ). Actu-
p n n 1 2 n−1
ally, this doesn’t happen; eventually we have xn > r1 , in
(−1)n−1
≡ (mod p), which case f (x) = f (xn ) = f (r1 ) by what we have al-
n ready shown. We conclude that f is a constant function.
where the congruence x ≡ y (mod p) means that x − y is (Thanks to Marshall Buck for catching an inaccuracy in
a rational number whose numerator, in reduced form, is a previous version of this solution.)
divisible by p. Hence it suffices to show that Now suppose c > 1/4. Then the sequence xn defined
by x0 = 0 and xn+1 = xn2 + c is strictly increasing and
k
(−1)n−1 has no limit point. Thus if we define f on [x0 , x1 ] as any
∑ ≡ 0 (mod p). continuous function with equal values on the endpoints,
n=1 n
and extend the definition from [xn , xn+1 ] to [xn+1 , xn+2 ]
We distinguish two cases based on p (mod 6). First sup- by the relation f (x) = f (x2 + c), and extend the defini-
pose p = 6r + 1, so that k = 4r. Then tion further to x < 0 by the relation f (x) = f (−x), the
resulting function has the desired property. Moreover,
4r
(−1)n−1 4r
1 2r
1 any function with that property clearly has this form.
∑ n = −
∑ n ∑ 2n2
n=1 n=1 n=1 B–1 Let [n] denote the set {1, 2, . . . , n}, and let fn denote
2r 
1 1
 3r 
1 1
 the number of minimal selfish subsets of [n]. Then the
=∑ − + ∑ + number of minimal selfish subsets of [n] not containing
n=1 n n n=2r+1 n 6r + 1 − n
n is equal to fn−1 . On the other hand, for any mini-
3r mal selfish subset of [n] containing n, by subtracting 1
p
= ∑ ≡ 0 (mod p), from each element, and then taking away the element
n=2r+1 n(p − n)
n − 1 from the set, we obtain a minimal selfish subset
since p = 6r + 1. of [n − 2] (since 1 and n cannot both occur in a selfish
set). Conversely, any minimal selfish subset of [n − 2]
Now suppose p = 6r + 5, so that k = 4r + 3. A similar gives rise to a minimal selfish subset of [n] containing
argument gives n by the inverse procedure. Hence the number of mini-
4r+3 mal selfish subsets of [n] containing n is fn−2 . Thus we
(−1)n−1 4r+3 1 2r+1
1 obtain fn = fn−1 + fn−2 . Since f1 = f2 = 1, we have
∑ = ∑ +2 ∑
n=1 n n=1 n n=1 2n fn = Fn , where Fn denotes the nth term of the Fibonacci
2r+1 
1 1
 3r+2 
1 1
 sequence.
= ∑ − + ∑ +
n=1 n n n=2r+2 n 6r + 5 − n B–2 By estimating the area under the graph of ln x using up-
3r+2 per and lower rectangles of width 2, we get
p
= ∑ ≡ 0 (mod p).
n=2r+2 n(p − n)
Z 2n−1
ln x dx ≤ 2(ln(3) + · · · + ln(2n − 1))
1
Z 2n+1
A–6 We first consider the case c ≤ 1/4; we shall show in this
case f must be constant. The relation ≤ ln x dx.
3

f (x) = f (x2 + c) = f ((−x)2 + c) = f (−x)

R
Since ln x dx = x ln x − x +C, we have, upon exponen-

proves that f is an even function. Let r1 ≤ r2 be the

roots of x2 + c − x, both of which are real. If x > r2 ,
 3

tiating and taking square roots, when x and y equal n − 1 and n − 2. Fortunately, this
yields precisely the difference between the claimed up-
 2n−1
per bound for n and the assumed upper bound for n − 1,

2n − 1 2 2n−1
< (2n − 1) 2 e−n+1 completing the induction.
e
≤ 1 · 3 · · · (2n − 1) B–4 Suppose such a matrix A exists. If the eigenvalues of
2n+1 e
−n+1 A (over the complex numbers) are distinct, then there
≤ (2n + 1) 2 exists a complex matrix C such that B = CAC−1 is
33/2
 2n+1 diagonal. Consequently, sin B is diagonal. But then
sin A = C−1 (sin B)C must be diagonalizable, a contra-

2n + 1 2
< ,
e diction. Hence the eigenvalues of A are the same, and A
has a conjugate B = CAC−1 over the complex numbers
using the fact that 1 < e < 3. of the form
 
B–3 View x1 , . . . , xn as an arrangement of the numbers x y
1, 2, . . . , n on a circle. We prove that the optimal ar- .
0 x
rangement is
A direct computation shows that
. . . , n − 4, n − 2, n, n − 1, n − 3, . . .  
sin x y · cos x
To show this, note that if a, b is a pair of adjacent num- sin B = .
0 sin x
bers and c, d is another pair (read in the same order
around the circle) with a < d and b > c, then the seg- Since sin A and sin B are conjugate, their eigenvalues
ment from b to c can be reversed, increasing the sum must be the same, and so we must have sin x = 1. This
by implies cos x = 0, so that sin B is the identity matrix, as
must be sin A, a contradiction. Thus A cannot exist.
ac + bd − ab − cd = (d − a)(b − c) > 0.
Alternate solution (due to Craig Helfgott and Alex
Now relabel the numbers so they appear in order as fol- Popa): Define both sin A and cos A by the usual power
lows: series. Since A commutes with itself, the power series
identity
. . . , an−4 , an−2 , an = n, an−1 , an−3 , . . .
sin2 A + cos2 A = I
where without loss of generality we assume an−1 >
an−2 . By considering the pairs an−2 , an and an−1 , an−3 holds. But if sin A is the given
 matrix, then 
by the above
and using the trivial fact an > an−1 , we deduce an−2 > 0 −2 · 1996
identity, cos2 A must equal which is a
an−3 . We then compare the pairs an−4 , an−2 and 0 0
an−1 , an−3 , and using that an−1 > an−2 , we deduce nilpotent matrix. Thus cos A is also nilpotent. How-
an−3 > an−4 . Continuing in this fashion, we prove that ever, the square of any 2 × 2 nilpotent matrix must be
an > an−1 > · · · > a1 and so ak = k for k = 1, 2, . . . , n, zero (e.g., by the Cayley-Hamilton theorem). This is a
i.e. that the optimal arrangement is as claimed. In par- contradiction.
ticular, the maximum value of the sum is B–5 Consider a 1 × n checkerboard, in which we write an
n-letter string, one letter per square. If the string is
1 · 2 + (n − 1) · n + 1 · 3 + 2 · 4 + · · · + (n − 2) · n balanced, we can cover each pair of adjacent squares
= 2 + n2 − n + (12 − 1) + · · · + [(n − 1)2 − 1] containing the same letter with a 1 × 2 domino, and
these will not overlap (because no three in a row can
(n − 1)n(2n − 1)
= n2 − n + 2 − (n − 1) + be the same). Moreover, any domino is separated from
6 the next by an even number of squares, since they must
2n3 + 3n2 − 11n + 18 cover opposite letters, and the sequence must alternate
= .
6 in between.
Conversely, any arrangement of dominoes where ad-
Alternate solution: We prove by induction that the value
jacent dominoes are separated by an even number of
given above is an upper bound; it is clearly a lower
squares corresponds to a unique balanced string, once
bound because of the arrangement given above. As-
we choose whether the string starts with X or O. In
sume this is the case for n − 1. The optimal arrange-
other words, the number of balanced strings is twice
ment for n is obtained from some arrangement for n − 1
the number of acceptable domino arrangements.
by inserting n between some pair x, y of adjacent terms.
This operation increases the sum by nx + ny − xy = We count these arrangements by numbering the squares
n2 − (n − x)(n − y), which is an increasing function of 0, 1, . . . , n − 1 and distinguishing whether the dominoes
both x and y. In particular, this difference is maximal start on even or odd numbers. Once this is decided, one
 4

simply chooses whether or not to put a domino in each Note that this maximum is positive for (u, v) 6= (0, 0): if
eligible position. Thus we have 2bn/2c arrangements in we had ai u + bi v < 0 for all i, then the subset ur + vs < 0
the first case and 2b(n−1)/2c in the second, but note that of the rs-plane would be a half-plane containing all of
the case of no dominoes has been counted twice. Hence the points (ai , bi ), whose convex hull would then not
the number of balanced strings is contain the origin, a contradiction.
The function maxi (ai u + bi v) is clearly continuous on
2b(n+2)/2c + 2b(n+1)/2c − 2. the unit circle u2 + v2 = 1, which is compact. Hence it
has a global minimum M > 0, and so for all u, v,
B–6 We will prove the claim assuming only that the convex p
hull of the points (ai , bi ) contains the origin in its inte- max(ai u + bi v) ≥ M u2 + v2 .
rior. (Thanks to Marshall Buck for pointing out that the i
last three words are necessary in the previous sentence!)
Let u = log x, v = log y so that the left-hand side of the p particular, f ≥ n + 1 on the disk of radius
In
(n + 1)/M. Since f (0, 0) = n, the infimum of f is the
given equation is
same over the entire uv-plane as over this disk, which
again is compact. Hence f attains its infimal value at
(a1 , b1 ) exp(a1 u + b1 v) + (a2 , b2 ) exp(a2 u + b2 v)+ some point in the disk, which is the desired global min-
· · · + (an , bn ) exp(an u + bn v). (1) imum.
Noam Elkies has suggested an alternate solution as fol-
Now note that (1) is the gradient of the function
lows: for r > 0, draw the loop traced by (1) as (u, v)
f (u, v) = exp(a1 u + b1 v) + exp(a2 u + b2 v)+ travels counterclockwise around the circle u2 + v2 = r2 .
For r = 0, this of course has winding number 0 about
· · · + exp(an u + bn v), any point, but for r large, one can show this loop has
winding number 1 about the origin, so somewhere in
and so it suffices to show f has a critical point. We will
between the loop must pass through the origin. (Prov-
in fact show f has a global minimum.
ing this latter fact is a little tricky.)
Clearly we have
 
f (u, v) ≥ exp max(ai u + bi v) .
i
 The 58th William Lowell Putnam Mathematical Competition
Saturday, December 6, 1997

A–1 A rectangle, HOMF, has sides HO = 11 and OM = 5. B–1 Let {x} denote the distance between the real number
A triangle ABC has H as the intersection of the altitudes, x and the nearest integer. For each positive integer n,
O the center of the circumscribed circle, M the midpoint evaluate
of BC, and F the foot of the altitude from A. What is
6n−1
the length of BC? m m
Fn = ∑ min({ }, { }).
m=1 6n 3n
A–2 Players 1, 2, 3, . . . , n are seated around a table, and each
has a single penny. Player 1 passes a penny to player (Here min(a, b) denotes the minimum of a and b.)
2, who then passes two pennies to player 3. Player 3 B–2 Let f be a twice-differentiable real-valued function sat-
then passes one penny to Player 4, who passes two pen- isfying
nies to Player 5, and so on, players alternately passing
one penny or two to the next player who still has some f (x) + f 00 (x) = −xg(x) f 0 (x),
pennies. A player who runs out of pennies drops out
of the game and leaves the table. Find an infinite set where g(x) ≥ 0 for all real x. Prove that | f (x)| is
of numbers n for which some player ends up with all n bounded.
pennies.
B–3 For each positive integer n, write the sum ∑nm=1 1/m in
A–3 Evaluate the form pn /qn , where pn and qn are relatively prime
positive integers. Determine all n such that 5 does not
x3 x5 x7
Z ∞ 
x− + − +··· divide qn .
0 2 2·4 2·4·6
2 B–4 Let am,n denote the coefficient of xn in the expansion of
x4 x6
 
x
1 + 2 + 2 2 + 2 2 2 + · · · dx. (1 + x + x2 )m . Prove that for all [integers] k ≥ 0,
2 2 ·4 2 ·4 ·6
b 2k
3 c
A–4 Let G be a group with identity e and φ : G → G a func-
tion such that
0≤ ∑ (−1)i ak−i,i ≤ 1.
i=0

φ (g1 )φ (g2 )φ (g3 ) = φ (h1 )φ (h2 )φ (h3 ) B–5 Prove that for n ≥ 2,
whenever g1 g2 g3 = e = h1 h2 h3 . Prove that there exists n terms n − 1 terms
an element a ∈ G such that ψ(x) = aφ (x) is a homo- z}|{
2
z}|{
2··· ···2
morphism (i.e. ψ(xy) = ψ(x)ψ(y) for all x, y ∈ G). 2 ≡ 22 (mod n).
A–5 Let Nn denote the number of ordered n-tuples of pos-
itive integers (a1 , a2 , . . . , an ) such that 1/a1 + 1/a2 + B–6 The dissection of the 3–4–5 triangle shown below (into
. . . + 1/an = 1. Determine whether N10 is even or odd. four congruent right triangles similar to the original) has
diameter 5/2. Find the least diameter of a dissection of
A–6 For a positive integer n and any real number c, define xk this triangle into four parts. (The diameter of a dissec-
recursively by x0 = 0, x1 = 1, and for k ≥ 0, tion is the least upper bound of the distances between
pairs of points belonging to the same part.)
cxk+1 − (n − k)xk
xk+2 = .
k+1
Fix n and then take c to be the largest value for which
xn+1 = 0. Find xk in terms of n and k, 1 ≤ k ≤ n.
 Solutions to the 58th William Lowell Putnam Mathematical Competition
Saturday, December 6, 1997
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A–1 The centroid G of the triangle is collinear with H and O A–3 Note that the series on the left is simply x exp(−x2 /2).
(Euler line), and the centroid lies two-thirds of the way By integration by parts,
from A to M. Therefore H is also two-thirds of the way Z ∞ Z ∞
2 /2 2 /2
from A to F, so AF = 15. Since the triangles BFH and x2n+1 e−x dx = 2n x2n−1 e−x dx
AFC are similar (they’re right triangles and 0 0

∠HBC = π/2 − ∠C = ∠CAF), and so by induction,

Z ∞
2 /2
we have x2n+1 e−x dx = 2 × 4 × · · · × 2n.
0
BF/FH = AF/FC Thus the desired integral is simply
or ∞
1 √
∑ 2n n! = e.
BF · FC = FH · AF = 75. n=0

Now A–4 In order to have ψ(x) = aφ (x) for all x, we must in par-
ticular have this for x = e, and so we take a = φ (e)−1 .
BC2 = (BF + FC)2 = (BF − FC)2 + 4BF · FC, We first note that
but φ (g)φ (e)φ (g−1 ) = φ (e)φ (g)φ (g−1 )
BF − FC = BM + MF − (MC − MF) = 2MF = 22, and so φ (g) commutes with φ (e) for all g. Next, we
note that
so
p √ φ (x)φ (y)φ (y−1 x−1 ) = φ (e)φ (xy)φ (y−1 x−1 )
BC = 222 + 4 · 75 = 784 = 28.
and using the commutativity of φ (e), we deduce
A–2 We show more precisely that the game terminates with
one player holding all of the pennies if and only if n = φ (e)−1 φ (x)φ (e)−1 φ (y) = φ (e)−1 φ (xy)
2m +1 or n = 2m +2 for some m. First suppose we are in
the following situation for some k ≥ 2. (Note: for us, a or ψ(xy) = ψ(x)ψ(y), as desired.
“move” consists of two turns, starting with a one-penny
pass.) A–5 We may discard any solutions for which a1 6= a2 , since
those come in pairs; so assume a1 = a2 . Similarly, we
– Except for the player to move, each player has k may assume that a3 = a4 , a5 = a6 , a7 = a8 , a9 = a10 .
pennies; Thus we get the equation
– The player to move has at least k pennies. 2/a1 + 2/a3 + 2/a5 + 2/a7 + 2/a9 = 1.
We claim then that the game terminates if and only if
the number of players is a power of 2. First suppose Again, we may assume a1 = a3 and a5 = a7 , so we get
the number of players is even; then after m complete 4/a1 + 4/a5 + 2/a9 = 1; and a1 = a5 , so 8/a1 + 2/a9 =
rounds, every other player, starting with the player who 1. This implies that (a1 − 8)(a9 − 2) = 16, which by
moved first, will have m more pennies than initially, and counting has 5 solutions. Thus N10 is odd.
the others will all have 0. Thus we are reduced to the A–6 Clearly xn+1 is a polynomial in c of degree n, so it suf-
situation with half as many players; by this process, we fices to identify n values of c for which xn+1 = 0. We
eventually reduce to the case where the number of play- claim these are c = n − 1 − 2r for r = 0, 1, . . . , n − 1;
ers is odd. However, if there is more than one player, in this case, xk is the coefficient of t k−1 in the polyno-
after two complete rounds everyone has as many pen- mial f (t) = (1 − t)r (1 + t)n−1−r . This can be verified
nies as they did before (here we need m ≥ 2), so the by noticing that f satisfies the differential equation
game fails to terminate. This verifies the claim.
Returning to the original game, note that after one com- f 0 (t) n − 1 − r r
= −
plete round, b n−1
2 c players remain, each with 2 pennies
f (t) 1+t 1−t
except for the player to move, who has either 3 or 4 pen-
nies. Thus by the above argument, the game terminates
if and only if b n−1
2 c is a power of 2, that is, if and only
if n = 2m + 1 or n = 2m + 2 for some m.
 2

(by logarithmic differentiation) or equivalently, It is immediate (e.g., by induction) that In ≡

1, −1, 1, 0, 0 (mod 5) for n ≡ 1, 2, 3, 4, 5 (mod 5) respec-
(1 − t 2 ) f 0 (t) = f (t)[(n − 1 − r)(1 − t) − r(1 + t)] tively, and moreover, we have the equality
= f (t)[(n − 1 − 2r) − (n − 1)t]
k
1
Hn = ∑ I m ,
m bn/5 c
and then taking the coefficient of tk on both sides: m=0 5

(k + 1)xk+2 − (k − 1)xk = where k = k(n) denotes the largest integer such that
(n − 1 − 2r)xk+1 − (n − 1)xk . 5k ≤ n. We wish to determine those n such that the
above sum has nonnegative 5–valuation. (By the 5–
n−1
valuation of a number a we mean the largest integer v
In particular, the largest such c is n − 1, and xk = k−1
for k = 1, 2, . . . , n. such that a/5v is an integer.)
Greg Kuperberg has suggested an alternate approach to If bn/5k c ≤ 3, then the last term in the above sum
show directly that c = n − 1 is the largest root, without has 5–valuation −k, since I1 , I2 , I3 each have valua-
computing the others. Note that the condition xn+1 = 0 tion 0; on the other hand, all other terms must have 5–
states that (x1 , . . . , xn ) is an eigenvector of the matrix valuation strictly larger than −k. It follows that Hn has
 5–valuation exactly −k; in particular, Hn has nonneg-
 i j = i+1 ative 5–valuation in this case if and only if k = 0, i.e.,
Ai j = n − j j = i − 1 n = 1, 2, or 3.
 0 otherwise
Suppose now that bn/5k c = 4. Then we must also have
20 ≤ bn/5k−1 c ≤ 24. The former condition implies that
with eigenvalue c. By the Perron-Frobenius theorem, the last term of the above sum is I4 /5k = 1/(12 · 5k−2 ),
A has a unique eigenvector with positive entries, whose which has 5–valuation −(k − 2).
eigenvalue has modulus greater than or equal to that of
any other eigenvalue, which proves the claim. It is clear that I20 ≡ I24 ≡ 0 (mod 25); hence if bn/5k−1 c
equals 20 or 24, then the second–to–last term of the
m m m
B–1 It is trivial to check that 6n = { 6n } ≤ { 3n } for 1 ≤ above sum (if it exists) has valuation at least −(k −
m m m
m ≤ 2n, that 1 − 3n = { 3n } ≤ { 6n } for 2n ≤ m ≤ 3n, 3). The third–to–last term (if it exists) is of the form
m m m
that 3n − 1 = { 3n } ≤ { 6n } for 3n ≤ m ≤ 4n, and that Ir /5k−2 , so that the sum of the last term and the third to
m m m
1 − 6n = { 6n } ≤ { 3n } for 4n ≤ m ≤ 6n. Therefore the last term takes the form (Ir + 1/12)/5k−2 . Since Ir can
desired sum is be congruent only to 0,1, or -1 (mod 5), and 1/12 ≡ 3
(mod 5), we conclude that the sum of the last term
2n−1
m 3n−1  m and third–to–last term has valuation −(k − 2), while
∑ + ∑ 1−
m=1 6n m=2n 3n all other terms have valuation strictly higher. Hence
4n−1   6n−1  Hn has nonnegative 5–valuation in this case only when
m m k ≤ 2, leading to the values n = 4 (arising from k = 0),
+ ∑ −1 + ∑ 1− = n.
m=3n 3n m=4n 6n 20,24 (arising from k = 1 and bn/5k−1 c = 20 and 24
resp.), 101, 102, 103, and 104 (arising from k = 2,
B–2 It suffices to show that | f (x)| is bounded for x ≥ 0, since bn/5k−1 c = 20) and 120, 121, 122, 123, and 124 (aris-
f (−x) satisfies the same equation as f (x). But then ing from k = 2, bn/5k−1 c = 24).
Finally, suppose bn/5k c = 4 and bn/5k−1 c = 21, 22,
d
( f (x))2 + ( f 0 (x))2 = 2 f 0 (x)( f (x) + f 00 (x)) or 23. Then as before, the first condition implies that


dx the last term of the sum in (*) has valuation −(k − 2),
= −2xg(x)( f 0 (x))2 ≤ 0, while the second condition implies that the second–to–
last term in the same sum has valuation −(k −1). Hence
so that ( f (x))2 ≤ ( f (0))2 + ( f 0 (0))2 for x ≥ 0. all terms in the sum (*) have 5–valuation strictly higher
than −(k − 1), except for the second–to–last term, and
B–3 The only such n are the numbers 1–4, 20–24, 100–104, therefore Hn has 5–valuation −(k − 1) in this case. In
and 120–124. For the proof let particular, Hn is integral (mod 5) in this case if and only
n if k ≤ 1, which gives the additional values n = 21, 22,
1
Hn = ∑ and 23.
m=1 m
B–4 Let sk = ∑i (−1)i ak−1,i be the given sum (note that ak−1,i
and introduce the auxiliary function is nonzero precisely for i = 0, . . . , b 2k
3 c). Since
1 am+1,n = am,n + am,n−1 + am,n−2 ,
In = ∑ .
1≤m≤n,(m,5)=1
m
 3

we have and that AD < CD. In any dissection of the triangle

into four parts, some two of A, B,C, D, E must belong to
sk − sk−1 + sk+2 = ∑(−1)i (an−i,i + an−i,i+1 + an−i,i+2 ) the same part, forcing the least diameter to be at least
i 25/13.
= ∑(−1)i an−i+1,i+2 = sk+3 . We now exhibit a dissection with least diameter 25/13.
i
(Some variations of this dissection are possible.) Put
By computing s0 = 1, s1 = 1, s2 = 0, we may easily F = (15/13, 19/13), G = (15/13, 0), H = (0, 19/13),
verify by induction that s4 j = s4 j+1 = 1 and s4 j+2 = J = (32/15, 15/13), and divide ABC into the convex
s4 j+3 = 0 for all j ≥ 0. (Alternate solution suggested by polygonal regions ADFH, BEJ, CGFH, DFGEJ. To
John Rickert: write S(x, y) = ∑∞ 2 2 3 i check that this dissection has least diameter 25/13, it
i=0 (y + xy + x y ) , and
note note that sk is the coefficient of yk in S(−1, y) = suffices (by the following remark) to check that the dis-
(1 + y)/(1 − y4 ).) tances

B–5 Define the sequence x1 = 2, xn = 2xn−1 for n > 1. It suf- AD, AF, AH, BE, BJ, DE,CF,CG,CH,
fices to show that for every n, xm ≡ xm+1 ≡ · · · (mod n) DF, DG, DH, DJ, EF, EG, EJ, FG, FH, FJ, GJ
for some m < n. We do this by induction on n, with
n = 2 being obvious. are all at most 25/13. This can be checked by a long
Write n = 2a b,where b is odd. It suffices to show that numerical calculation, which we omit in favor of some
xm ≡ · · · modulo 2a and modulo b, for some m < n. shortcuts: note that ADFH and BEJ are contained in
For the former, we only need xn−1 ≥ a, but clearly circular sectors centered at A and B, respectively, of ra-
xn−1 ≥ n by induction on n. For the latter, note that dius 25/13 and angle less than π/3, while CGFH is a
xm ≡ xm+1 ≡ · · · (mod b) as long as xm−1 ≡ xm ≡ · · · rectangle with diameter CF < 25/13.
(mod φ (b)), where φ (n) is the Euler totient function. Remark. The preceding argument uses implicitly the
By hypothesis, this occurs for some m < φ (b) + 1 ≤ n. fact that for P a simple closed polygon in the plane, if
(Thanks to Anoop Kulkarni for catching a lethal typo in we let S denote the set of points on or within P, then the
an earlier version.) maximum distance between two points of S occurs be-
tween some pair of vertices of P. This is an immediate
B–6 The answer is 25/13. Place the triangle on the cartesian consequence of the compactness of S (which guarantees
plane so that its vertices are at C = (0, 0), A = (0, 3), B = the existence of a maximum) and the convexity of the
(4, 0). Define also the points D = (20/13, 24/13), and function taking (x, y) ∈ S × S to the squared distance be-
E = (27/13, 0). We then compute that tween x and y (which is obvious in terms of Cartesian
25 coordinates).
= AD = BE = DE
13
27
= BC −CE = BE < BC
13
39 p
= AC < AC2 +CE 2 = AE
13
40
= AB − AD = BD < AB
13
 The 59th William Lowell Putnam Mathematical Competition
Saturday, December 5, 1998

A–1 A right circular cone has base of radius 1 and height 3. B–1 Find the minimum value of
A cube is inscribed in the cone so that one face of the
cube is contained in the base of the cone. What is the (x + 1/x)6 − (x6 + 1/x6 ) − 2
side-length of the cube? (x + 1/x)3 + (x3 + 1/x3 )

A–2 Let s be any arc of the unit circle lying entirely in the for x > 0.
first quadrant. Let A be the area of the region lying be-
low s and above the x-axis and let B be the area of the B–2 Given a point (a, b) with 0 < b < a, determine the min-
region lying to the right of the y-axis and to the left of imum perimeter of a triangle with one vertex at (a, b),
s. Prove that A + B depends only on the arc length, and one on the x-axis, and one on the line y = x. You may
not on the position, of s. assume that a triangle of minimum perimeter exists.

A–3 Let f be a real function on the real line with continuous B–3 let H be the unit hemisphere {(x, y, z) : x2 + y2 + z2 =
third derivative. Prove that there exists a point a such 1, z ≥ 0}, C the unit circle {(x, y, 0) : x2 + y2 = 1}, and
that P the regular pentagon inscribed in C. Determine the
surface area of that portion of H lying over the pla-
f (a) · f 0 (a) · f 00 (a) · f 000 (a) ≥ 0. nar region inside P, and write your answer in the form
A sin α + B cos β , where A, B, α, β are real numbers.
A–4 Let A1 = 0 and A2 = 1. For n > 2, the number An is de- B–4 Find necessary and sufficient conditions on positive in-
fined by concatenating the decimal expansions of An−1 tegers m and n so that
and An−2 from left to right. For example A3 = A2 A1 = mn−1
10, A4 = A3 A2 = 101, A5 = A4 A3 = 10110, and so forth. ∑ (−1)bi/mc+bi/nc = 0.
Determine all n such that 11 divides An . i=0

A–5 Let F be a finite collection of open discs in R2 whose B–5 Let N be the positive integer with 1998 decimal digits,
union contains a set E ⊆ R2 . Show that there is a pair- all of them 1; that is,
wise disjoint subcollection D1 , . . . , Dn in F such that
N = 1111 · · · 11.
E ⊆ ∪nj=1 3D j .
√
Find the thousandth digit after the decimal point of N.
Here, if D is the disc of radius r and center P, then 3D
is the disc of radius 3r and center P. B–6 Prove that, for any √
integers a, b, c, there exists a positive
integer n such that n3 + an2 + bn + c is not an integer.
A–6 Let A, B,C denote distinct points with integer coordi-
nates in R2 . Prove that if

(|AB| + |BC|)2 < 8 · [ABC] + 1

then A, B,C are three vertices of a square. Here |XY |

is the length of segment XY and [ABC] is the area of
triangle ABC.
 Solutions to the 59th William Lowell Putnam Mathematical Competition
Saturday, December 5, 1998
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A–1 Consider the plane containing both the axis of the cone for all x. Similarly, f 0 (x) > 0 and f 00 (x) > 0 imply that
and two opposite vertices of the cube’s bottom face. f (x) > 0 for all x. Therefore f (x) f 0 (x) f 00 (x) f 000 (x) > 0
The cross section of the cone and the cube
√ in this plane for all x, and we are done.
consists of a rectangle of sides s and s 2 inscribed in
an isosceles triangle of base 2 and √
height 3, where s is A–4 The number of digits in the decimal expansion of An is
the side-length of the cube. (The s 2 side of the rect- the Fibonacci number Fn , where F1 = 1, F2 = 1, and
angle lies on the base Fn = Fn−1 + Fn−2 for n > 2. It follows that the se-
√ of the triangle.)√Similar triangles quence {An }, modulo 11, satisfies the recursion An =
yield s/3 = (1 − s 2/2)/1, or s = (9 2 − 6)/7.
(−1)Fn−2 An−1 + An−2 . (Notice that the recursion for An
A–2 First solution: to fix notation, let A be the area of re- depends only on the value of Fn−2 modulo 2.) Using
gion DEFG, and B be the area of DEIH; further let C these recursions, we find that A7 ≡ 0 and A8 ≡ 1 modulo
denote the area of sector ODE, which only depends on 11, and that F7 ≡ 1 and F8 ≡ 1 modulo 2. It follows that
the arc length of s. If [XY Z] denotes the area of trian- An ≡ An+6 (mod 11) for all n ≥ 1. We find that among
gle [XY Z], then we have A = C + [OEG] − [ODF] and A1 , A2 , A3 , A4 , A5 , and A6 , only A1 vanishes modulo 11.
B = C + [ODH] − [OEI]. But clearly [OEG] = [OEI] Thus 11 divides An if and only if n = 6k + 1 for some
and [ODF] = [ODH], and so A + B = 2C. nonnegative integer k.
A–5 Define the sequence Di by the following greedy algo-
D rithm: let D1 be the disc of largest radius (breaking ties
H arbitrarily), let D2 be the disc of largest radius not meet-
E ing D1 , let D3 be the disc of largest radius not meeting
I
D1 or D2 , and so on, up to some final disc Dn . To see
that E ⊆ ∪nj=1 3D j , consider a point in E; if it lies in one
of the Di , we are done. Otherwise, it lies in a disc D of
O F G
radius r, which meets one of the Di having radius s ≥ r
(this is the only reason a disc can be skipped in our al-
gorithm). Thus the centers lie at a distance t < s+r, and
so every point at distance less than r from the center of
D lies at distance at most r + t < 3s from the center of
the corresponding Di .
Second solution: We may parametrize a point in s by
any of x, y, or θ = tan−1 (y/x). Then A and B are just the A–6 Recall the inequalities |AB|2 +|BC|2 ≥ 2|AB||BC| (AM-
integrals of y dx and x dy over the appropriate intervals; GM) and |AB||BC| ≥ 2[ABC] (Law of Sines). Also re-
thus A + B is the integral of x dy − y dx (minus because call that the area of a triangle with integer coordinates
the limits of integration are reversed). But dθ = x dy − is half an integer (if its vertices lie at (0, 0), (p, q), (r, s),
y dx, and so A + B = ∆θ is precisely the radian measure the area is |ps − qr|/2), and that if A and B have integer
of s. (Of course, one can perfectly well do this problem coordinates, then |AB|2 is an integer (Pythagoras). Now
by computing the two integrals separately. But what’s observe that
the fun in that?)
8[ABC] ≤ |AB|2 + |BC|2 + 4[ABC]
A–3 If at least one of f (a), f 0 (a), f 00 (a),
or f 000 (a)
vanishes
≤ |AB|2 + |BC|2 + 2|AB||BC|
at some point a, then we are done. Hence we may as-
sume each of f (x), f 0 (x), f 00 (x), and f 000 (x) is either < 8[ABC] + 1,
strictly positive or strictly negative on the real line. By
replacing f (x) by − f (x) if necessary, we may assume and that the first and second expressions are both in-
f 00 (x) > 0; by replacing f (x) by f (−x) if necessary, tegers. We conclude that 8[ABC] = |AB|2 + |BC|2 +
we may assume f 000 (x) > 0. (Notice that these substitu- 4[ABC], and so |AB|2 + |BC|2 = 2|AB||BC| = 4[ABC];
tions do not change the sign of f (x) f 0 (x) f 00 (x) f 000 (x).) that is, B is a right angle and AB = BC, as desired.
Now f 00 (x) > 0 implies that f 0 (x) is increasing, and B–1 Notice that
f 000 (x) > 0 implies that f 0 (x) is convex, so that f 0 (x +
a) > f 0 (x) + a f 00 (x) for all x and a. By letting a increase (x + 1/x)6 − (x6 + 1/x6 ) − 2
in the latter inequality, we see that f 0 (x + a) must be =
(x + 1/x)3 + (x3 + 1/x3 )
positive for sufficiently large a; it follows that f 0 (x) > 0
(x + 1/x)3 − (x3 + 1/x3 ) = 3(x + 1/x)
 2

(difference of squares). The latter is easily seen (e.g., by where r < 10−2000 . Now the digits after the decimal
AM-GM) to have minimum value 6 (achieved at x = 1). point of 10999 /3 are given by .3333 . . ., while the dig-
its after the decimal point of 61 10−999 are given by
B–2 Consider a triangle as described by the problem; label .00000
its vertices A, B,C so that A = (a, b), B lies on the x-axis, √. . . 1666666 . . .. It follows that the first 1000 dig-
its of N are given by .33333 . . . 3331; in particular, the
and C lies on the line y = x. Further let D = (a, −b) be thousandth digit is 1.
the reflection of A in the x-axis, and let E = (b, a) be
the reflection of A in the line y = x. Then AB = DB and B–6 First solution: Write p(n) = n3 + an2 + bn + c. Note
AC = CE, and pso the perimeter of ABC √ is DB + BC + that p(n) and p(n + 2) have the same parity, and recall
CE ≥ DE = (a − b)2 + (a + b)2 = 2a2 + 2b2 . It is that any perfect square is congruent to 0 or 1 (mod 4).
clear that this lower bound can be achieved; just set B Thus if p(n) and p(n + 2) are perfect squares, they are
(resp. C) to be the intersection between the segment congruent mod 4. But p(n + 2) − p(n) ≡ 2n2 + 2b (mod
4), which is not divisible by 4 if n and b have opposite
√ line x = y); thus the minimum
DE and the x-axis (resp.
perimeter is in fact 2a2 + 2b2 . parity.
Second solution: We prove more generally that for any
B–3 We use the well-known result that the surface area of
polynomial P(z) with integer coefficients which is not
the “sphere cap” {(x, y, z) | x2 + y2 + z2 = 1, z ≥ z0 } is
a perfect square, there exists a positive integer n such
simply 2π(1 − z0 ). (This result is easily verified using
that P(n) is not a perfect square. Of course it suffices
calculus; we omit the derivation here.) Now the desired
to assume P(z) has no repeated factors, which is to say
surface area is just 2π minus the surface areas of five
P(z) and its derivative P0 (z) are relatively prime.
identical halves of sphere caps; these caps, up to isome-
try, correspond to z0 being the distance from the center In particular, if we carry out the Euclidean algorithm
of the pentagon to any of its sides, i.e., z0 = cos π5 . Thus on P(z) and P0 (z) without dividing, we get an integer
the desired area is 2π − 25 2π(1 − cos π5 ) = 5π cos π5 −

D (the discriminant of P) such that the greatest com-
3π (i.e., B = π/2). mon divisor of P(n) and P0 (n) divides D for any n.
Now there exist infinitely many primes p such that p
B–4 For convenience, define fm,n (i) = b mi c + b ni c, so that divides P(n) for some n: if there were only finitely
the given sum is S(m, n) = ∑mn−1 fm,n (i) . If m and many, say, p1 , . . . , pk , then for any n divisible by m =
i=0 (−1)
n are both odd, then S(m, n) is the sum of an odd num- P(0)p1 p2 · · · pk , we have P(n) ≡ P(0) (mod m), that is,
ber of ±1’s, and thus cannot be zero. Now consider P(n)/P(0) is not divisible by p1 , . . . , pk , so must be ±1,
the case where m and n have opposite parity. Note that but then P takes some value infinitely many times, con-
b mi c + bk − i+1 tradiction. In particular, we can choose some such p not
m c = k − 1 for all integers i, k, m. Thus
b mi c + b mn−i−1 c = n − 1 and b ni c + b mn−i−1 c = m − 1; dividing D, and choose n such that p divides P(n). Then
m n
this implies that fm,n (i)+ fm,n (mn−i−1) = m+n−2 is P(n + kp) ≡ P(n) + kpP0 (n)(mod p) (write out the Tay-
lor series of the left side); in particular, since p does not
odd, and so (−1) fm,n (i) = −(−1) fm,n (mn−i−1) for all i. It
divide P0 (n), we can find some k such that P(n + kp)
follows that S(m, n) = 0 if m and n have opposite parity.
is divisible by p but not by p2 , and so is not a perfect
Now suppose that m = 2k and n = 2l are both even. square.
2j
Then b 2m c = b 22m
j+1
c for all j, so S can be computed as Third solution: (from David Rusin, David Savitt, and
twice the sum over only even indices: Richard Stanley independently) Assume that n3 +an2 +
2kl−1
bn + c is a square for all n > 0. For sufficiently large n,
S(2k, 2l) = 2 ∑ (−1) fk,l (i) = S(k, l)(1 + (−1)k+l ).
1
i=0 (n3/2 + an1/2 − 1)2 < n3 + an2 + bn + c
2
Thus S(2k, 2l) vanishes if and only if S(k, l) vanishes (if 1
< (n3/2 + an1/2 + 1)2 ;
1 + (−1)k+l = 0, then k and l have opposite parity and 2
so S(k, l) also vanishes).
thus if n is a large even perfect square, we have n3 +
Piecing our various cases together, we easily deduce
an2 + bn + c = (n3/2 + 21 an1/2 )2 . We conclude this is an
that S(m, n) = 0 if and only if the highest powers of 2
equality of polynomials, but the right-hand side is not a
dividing m and n are different.
perfect square for n an even non-square, contradiction.
B–5 Write N = (101998 − 1)/9. Then (The reader might try generalizing this approach to ar-
bitrary polynomials.√ A related argument, due to Greg
√ 10999 p 3 2
Kuperberg: write n + an + bn + c as n 3/2 times a
N= 1 − 10−1998 power series in 1/n and take two finite differences to
3
10999 1 get an expression which tends to 0 as n → ∞, contradic-
= (1 − 10−1998 + r), tion.)
3 2
Note: in case n3 + an2 + bn + c has no repeated factors,
it is a square for only finitely many n, by a theorem
 3

of Siegel; work of Baker gives an explicit (but large)

bound on such n. (I don’t know whether the graders
will accept this as a solution, though.)
 The 60th William Lowell Putnam Mathematical Competition
Saturday, December 4, 1999

A–1 Find polynomials f (x),g(x), and h(x), if they exist, such B–1 Right triangle ABC has right angle at C and ∠BAC = θ ;
that for all x, the point D is chosen on AB so that |AC| = |AD| = 1;
 the point E is chosen on BC so that ∠CDE = θ . The
−1
 if x < −1 perpendicular to BC at E meets AB at F. Evaluate
| f (x)| − |g(x)| + h(x) = 3x + 2 if −1 ≤ x ≤ 0 limθ →0 |EF|.

−2x + 2 if x > 0.
B–2 Let P(x) be a polynomial of degree n such that P(x) =
Q(x)P00 (x), where Q(x) is a quadratic polynomial and
A–2 Let p(x) be a polynomial that is nonnegative for all P00 (x) is the second derivative of P(x). Show that if
real x. Prove that for some k, there are polynomials P(x) has at least two distinct roots then it must have n
f1 (x), . . . , fk (x) such that distinct roots.
B–3 Let A = {(x, y) : 0 ≤ x, y < 1}. For (x, y) ∈ A, let
k
p(x) = ∑ ( f j (x))2 . S(x, y) = ∑ x m yn ,
j=1 1 ≤ m ≤2
2 n

A–3 Consider the power series expansion where the sum ranges over all pairs (m, n) of positive
∞ integers satisfying the indicated inequalities. Evaluate
1
= ∑ an xn .
1 − 2x − x 2
n=0 lim (1 − xy2 )(1 − x2 y)S(x, y).
(x,y)→(1,1),(x,y)∈A
Prove that, for each integer n ≥ 0, there is an integer m
such that B–4 Let f be a real function with a continuous third deriva-
tive such that f (x), f 0 (x), f 00 (x), f 000 (x) are positive for
a2n + a2n+1 = am . all x. Suppose that f 000 (x) ≤ f (x) for all x. Show that
f 0 (x) < 2 f (x) for all x.
A–4 Sum the series
B–5 For an integer n ≥ 3, let θ = 2π/n. Evaluate the de-
∞ ∞
m2 n terminant of the n × n matrix I + A, where I is the
∑ ∑ 3m (n3m + m3n ) . n × n identity matrix and A = (a jk ) has entries a jk =
m=1 n=1 cos( jθ + kθ ) for all j, k.

A–5 Prove that there is a constant C such that, if p(x) is a B–6 Let S be a finite set of integers, each greater than 1.
polynomial of degree 1999, then Suppose that for each integer n there is some s ∈ S such
Z 1 that gcd(s, n) = 1 or gcd(s, n) = s. Show that there exist
s,t ∈ S such that gcd(s,t) is prime.
|p(0)| ≤ C |p(x)| dx.
−1

A–6 The sequence (an )n≥1 is defined by a1 = 1, a2 = 2, a3 =

24, and, for n ≥ 4,

6a2n−1 an−3 − 8an−1 a2n−2

an = .
an−2 an−3
Show that, for all n, an is an integer multiple of n.
 Solutions to the 60th William Lowell Putnam Mathematical Competition
Saturday, December 4, 1999
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A–1 Note that if r(x) and s(x) are any two functions, then an (an−1 + an+1 ). Then

max(r, s) = (r + s + |r − s|)/2. 2b2n+1 + b2n = 2an an+1 + 2an−1 an + a2n−1 + a2n

Therefore, if F(x) is the given function, we have = 2an an+1 + an−1 an+1 + a2n
= a2n+1 + a2n = b2n+2 ,
F(x) = max{−3x − 3, 0} − max{5x, 0} + 3x + 2
= (−3x − 3 + |3x + 3|)/2 and similarly 2b2n + b2n−1 = b2n+1 , so that {bn } satis-
fies the same recurrence as {an }. Since further b0 =
− (5x + |5x|)/2 + 3x + 2
1, b1 = 2 (where we use the recurrence for {an } to cal-
1 culate a−1 = 0), we deduce that bn = an for all n. In
= |(3x + 3)/2| − |5x/2| − x + ,
2 particular, a2n + a2n+1 = b2n+2 = a2n+2 .
so we may set f (x) = (3x + 3)/2, g(x) = 5x/2, and Second solution: Note that
h(x) = −x + 12 .
1
A–2 First solution: First factor p(x) = q(x)r(x), where q has 1 − 2x − x2
all real roots and r has all complex roots. Notice that √ √ !
1 2+1 2−1
each root of q has even multiplicity, otherwise p would = √ √ + √
have a sign change at that root. Thus q(x) has a square 2 2 1 − (1 + 2)x 1 − (1 − 2)x
root s(x).
and that
Now write r(x) = ∏kj=1 (x − a j )(x − a j ) (possible be-
cause r has roots in complex conjugate pairs). Write 1 ∞ √
∏kj=1 (x − a j ) = t(x) + iu(x) with t, x having real coeffi-
√ = ∑ (1 ± 2)n xn ,
1 + (1 ± 2)x n=0
cients. Then for x real,
so that
p(x) = q(x)r(x)
1 √ √ 
= s(x)2 (t(x) + iu(x))(t(x) + iu(x)) an = √ ( 2 + 1)n+1 − (1 − 2)n+1 .
2 2
= (s(x)t(x))2 + (s(x)u(x))2 .
A simple computation (omitted here) now shows that
(Alternatively, one can factor r(x) as a product of a2n + a2n+1 = a2n+2 .
quadratic polynomials with real coefficients, write each
as a sum of squares, then multiply together to get a sum Third
 solution (by Richard Stanley): Let A be the matrix

of many squares.) 0 1
. A simple induction argument shows that
1 2
Second solution: We proceed by induction on the de-
gree of p, with base case where p has degree 0. As in  
n+2 an an+1
the first solution, we may reduce to a smaller degree in A = .
an+1 an+2
case p has any real roots, so assume it has none. Then
p(x) > 0 for all real x, and since p(x) → ∞ for x → ±∞,
The desired result now follows from comparing the top
p has a minimum value c. Now p(x) − c has real roots,
left corner entries of the equality An+2 An+2 = A2n+4 .
so as above, we deduce that p(x)−c is√ a sum of squares.
Now add one more square, namely ( c)2 , to get p(x)
A–4 Denote the series by S, and let an = 3n /n. Note that
as a sum of squares.
∞ ∞
A–3 First solution: Computing the coefficient of xn+1 in the 1
S= ∑ ∑ am (am + an )
identity (1 − 2x − x2 ) ∑∞ m
m=0 am x = 1 yields the recur- m=1 n=1
rence an+1 = 2an + an−1 ; the sequence {an } is then ∞ ∞
1
characterized by this recurrence and the initial condi- = ∑ ∑ an (am + an ) ,
m=1 n=1
tions a0 = 1, a1 = 2.
Define the sequence {bn } by b2n = a2n−1 + a2n , b2n+1 = where the second equality follows by interchanging m
 2

and n. Thus and Q(x)(R(x) − kx) has more roots in [0, 1] than does P

1 1
 (and has the same value at 0). Repeating this argument
2S = ∑ ∑ + shows that 01 |P(x)| dx is greater than the correspond-
R
m n am (am + an ) a n (a m + an )
ing integral for some polynomial with all of its roots in
1 [0, 1].
= ∑∑
m n m an
a
Under this assumption, we have
!2
∞
n 1999
= ∑ n .
n=1 3 P(x) = c ∏ (x − ri )
i=1
But
∞
n 3 for some ri ∈ (0, 1]. Since
∑ 3n = 4
n=1 P(0) = −c ∏ ri = 1,
since, e.g., it’s f 0 (1), where
we have
xn
∞
3
f (x) = ∑ n = , |c| ≥ ∏ |ri−1 | ≥ 1.
n=0 3 3 − x
and we conclude that S = 9/32. Thus it suffices to prove that if Q(x) is a monic polyno-
A–5 First solution: (by Reid Barton) Let r1 , . . . , r1999 be the mial of degree 1999 with all of its roots in [0, 1], then
R1
roots of P. Draw a disc of radius ε around each ri , 0 |Q(x)|Rdx ≥ D for some constant D > 0. But the in-
where ε < 1/3998; this disc covers a subinterval of tegral of 01 ∏1999
i=1 |x − ri | dx is a continuous function for
[−1/2, 1/2] of length at most 2ε, and so of the 2000 (or ri ∈ [0, 1]. The product of all of these intervals is com-
fewer) uncovered intervals in [−1/2, 1/2], one, which pact, so the integral achieves a minimum value for some
we call I, has length at least δ = (1 − 3998ε)/2000 > 0. ri . This minimum is the desired D.
We will exhibit an explicit lower bound for the integral Third solution (by Abe Kunin): It suffices to prove the
of |P(x)|/P(0) over this interval, which will yield such stronger inequality
a bound for the entire integral.
Z 1
Note that
sup |P(x)| ≤ C |P(x)| dx
−1
|P(x)| 1999 |x − ri | x∈[−1,1]
= .
|P(0)| ∏i=1 |ri | holds for some C. But this follows immediately from
Also note that by construction, |x − ri | ≥ ε for each x ∈ the following standard fact: any two norms on a finite-
dimensional vector space (here the polynomials of de-
I. If |ri | ≤ 1, then we have |x−r i|
|ri | ≥ ε. If |ri | > 1, then gree at most 1999) are equivalent. (The proof of this
|x − ri | statement is also a compactness argument: C can be
= |1 − x/ri | ≥ 1 − |x/ri | ≥= 1/2 > ε. taken to be the maximum of the L1-norm divided by
|ri |
R the sup norm over the set of polynomials with L1-norm
We conclude that I |P(x)/P(0)| dx ≥ δ ε, independent 1.)
of P.
Note: combining the first two approaches gives a con-
Second solution: It will be a bit more convenient to structive solution with a constant that is better than that
assume P(0) = 1 (which we may achieve by rescal- given by the first solution, but is still far from optimal. I
ing unless P(0) = 0, in which case there is nothing to don’t know offhand whether it is even known what the
prove) and to prove that there exists D > 0 such that optimal constant and/or the polynomials achieving that
R1 R1
−1 |P(x)| dx ≥ D, or even such that 0 |P(x)| dx ≥ D. constant are.
We first reduce to the case where P has all of its roots
A–6 Rearranging the given equation yields the much more
in [0, 1]. If this is not the case, we can factor P(x) as
tractable equation
Q(x)R(x), where Q has all roots in the interval and R has
none. Then R is either always positive or always neg- an an−1 an−2
ative on [0, 1]; assume the former. Let k be the largest =6 −8 .
an−1 an−2 an−3
positive real number such that R(x) − kx ≥ 0 on [0, 1];
then Let bn = an /an−1 ; with the initial conditions b2 =
Z 1 Z 1 2, b3 = 12, one easily obtains bn = 2n−1 (2n−2 − 1), and
|P(x)| dx = |Q(x)R(x)| dx so
−1 −1
Z 1 n−1
> |Q(x)(R(x) − kx)| dx, an = 2n(n−1)/2 ∏ (2i − 1).
−1 i=1
 3

To see that n divides an , factor n as 2k m, with m odd. P. Thus we also have that no roots of P0 lie on the sides
Then note that k ≤ n ≤ n(n − 1)/2, and that there exists of the convex hull of P, unless they are also roots of P.
i ≤ m − 1 such that m divides 2i − 1, namely i = φ (m) From this we conclude that if r is a root of P which is
(Euler’s totient function: the number of integers in a vertex of the convex hull of the roots, and which is
{1, . . . , m} relatively prime to m). not also a root of P0 , then f has a single pole at r (as r
cannot be a root of P00 ). On the other hand, if r is a root
B–1 The answer is 1/3. Let G be the point obtained by re-
of P which is also a root of P0 , it is a multiple root, and
flecting C about the line AB. Since ∠ADC = π−θ2 , we
π−θ then f has a double pole at r.
find that ∠BDE = π − θ − ∠ADC = 2 = ∠ADC =
π − ∠BDC = π − ∠BDG, so that E, D, G are collinear. If P has roots not all equal, the convex hull of its roots
Hence has at least two vertices.

|BE| |BE| sin(θ /2) B–3 We first note that

|EF| = = = , xy
|BC| |BG| sin(3θ /2) ∑ xm yn = .
m,n>0 (1 − x)(1 − y)
where we have used the law of sines in 4BDG. But by
l’Hôpital’s Rule, Subtracting S from this gives two sums, one of which is

sin(θ /2) cos(θ /2) x2n+1 x3 y

lim = lim = 1/3. ∑ xm yn = ∑ yn =
θ →0 sin(3θ /2) θ →0 3 cos(3θ /2) m≥2n+1 n 1−x (1 − x)(1 − x2 y)

and the other of which sums to xy3 /[(1 − y)(1 − xy2 )].
B–2 First solution: Suppose that P does not have n dis-
Therefore
tinct roots; then it has a root of multiplicity at least 2,
which we may assume is x = 0 without loss of general- xy x3 y
ity. Let xk be the greatest power of x dividing P(x), so S(x, y) = −
(1 − x)(1 − y) (1 − x)(1 − x2 y)
that P(x) = xk R(x) with R(0) 6= 0; a simple computation
yields xy3
−
(1 − y)(1 − xy2 )
P00 (x) = (k2 − k)xk−2 R(x) + 2kxk−1 R0 (x) + xk R00 (x). xy(1 + x + y + xy − x2 y2 )
=
(1 − x2 y)(1 − xy2 )
Since R(0) 6= 0 and k ≥ 2, we conclude that the great-
est power of x dividing P00 (x) is xk−2 . But P(x) = and the desired limit is
Q(x)P00 (x), and so x2 divides Q(x). We deduce (since Q
is quadratic) that Q(x) is a constant C times x2 ; in fact, lim xy(1 + x + y + xy − x2 y2 ) = 3.
(x,y)→(1,1)
C = 1/(n(n − 1)) by inspection of the leading-degree
terms of P(x) and P00 (x). B–4 (based on work by Daniel Stronger) We make repeated
Now if P(x) = ∑nj=0 a j x j , then the relation P(x) = use of the following fact: if f is a differentiable function
Cx2 P00 (x) implies that a j = C j( j − 1)a j for all j; hence on all of R, limx→−∞ f (x) ≥ 0, and f 0 (x) > 0 for all
a j = 0 for j ≤ n − 1, and we conclude that P(x) = an xn , x ∈ R, then f (x) > 0 for all x ∈ R. (Proof: if f (y) < 0
which has all identical roots. for some x, then f (x) < f (y) for all x < y since f 0 > 0,
but then limx→−∞ f (x) ≤ f (y) < 0.)
Second solution (by Greg Kuperberg): Let f (x) =
P00 (x)/P(x) = 1/Q(x). By hypothesis, f has at most From the inequality f 000 (x) ≤ f (x) we obtain
two poles (counting multiplicity).
f 00 f 000 (x) ≤ f 00 (x) f (x) < f 00 (x) f (x) + f 0 (x)2
Recall that for any complex polynomial P, the roots of
P0 lie within the convex hull of P. To show this, it suf- since f 0 (x) is positive. Applying the fact to the differ-
fices to show that if the roots of P lie on one side of a ence between the right and left sides, we get
line, say on the positive side of the imaginary axis, then 1 00
P0 has no roots on the other side. That follows because ( f (x))2 < f (x) f 0 (x). (1)
if r1 , . . . , rn are the roots of P, 2

On the other hand, since f (x) and f 000 (x) are both posi-
P0 (z) n
1
=∑ tive for all x, we have
P(z) i=1 z − ri
2 f 0 (x) f 00 (x) < 2 f 0 (x) f 00 (x) + 2 f (x) f 000 (x).
and if z has negative real part, so does 1/(z − ri ) for
i = 1, . . . , n, so the sum is nonzero. Applying the fact to the difference between the sides
yields
The above argument also carries through if z lies on the
imaginary axis, provided that z is not equal to a root of f 0 (x)2 ≤ 2 f (x) f 00 (x). (2)
 4

Combining (1) and (2), we obtain Second solution (by Mohamed Omar): Set x = eiθ and
2 write
1 f 0 (x)2

1
< ( f 00 (x))2 1 T 1 T 1 T T
u
 
2 2 f (x) 2 A = u u+ v v = u v
< f (x) f 0 (x), 2 2 2 v

or ( f 0 (x))3 < 8 f (x)3 . We conclude f 0 (x) < 2 f (x), as for

desired.
u = x x2 · · · xn , v = x−1 x−2 · · · xn .



Note: one can actually prove the result with a smaller
constant in place of 2, as follows. Adding 21 f 0 (x) f 000 (x)
to both sides of (1) and again invoking the original We now use the fact that for R an n × m matrix and S an
bound f 000 (x) ≤ f (x), we get m × n matrix,

1 0 1 det(In + RS) = det(Im + SR).

[ f (x) f 000 (x) + ( f 00 (x))2 ] < f (x) f 0 (x) + f 0 (x) f 000 (x)
2 2
3 This yields
≤ f (x) f 0 (x).
2
Applying the fact again, we get det(IN + A)
  
1
u
1 0 3 = det In + uT vT
f (x) f 00 (x) < f (x)2 . 2 v
2 4    
1 u
Multiplying both sides by f 0 (x) and applying the fact uT vT


= det I2 +
once more, we get 2 v
2 + uuT uvT
 
1
1 0 1 = det
( f (x))3 < f (x)3 . 4 vuT 2 + vvT
6 4
2 + (x2 + · · · + x2n )
 
1 n
From this we deduce f 0 (x) < (3/2)1/3 f (x) < 2 f (x), as = det
4 n 2 + (x−2 + · · · + x−2n )
desired.
n2
 
1 2 n
I don’t know what the best constant is, except that it = det = 1− .
4 n 2 4
is not less than 1 (because f (x) = ex satisfies the given
conditions).
B–6 First solution: Choose a sequence p1 , p2 , . . . of primes
B–5 First solution: We claim that the eigenvalues of A are as follows. Let p1 be any prime dividing an element of
0 with multiplicity n − 2, and n/2 and −n/2, each with S. To define p j+1 given p1 , . . . , p j , choose an integer
multiplicity 1. To prove this claim, define vectors v(m) , N j ∈ S relatively prime to p1 · · · p j and let p j+1 be a
0 ≤ m ≤ n − 1, componentwise by (v(m) )k = eikmθ , and prime divisor of N j , or stop if no such N j exists.
note that the v(m) form a basis for Cn . (If we arrange the
v(m) into an n × n matrix, then the determinant of this Since S is finite, the above algorithm eventually termi-
matrix is a Vandermonde product which is nonzero.) nates in a finite sequence p1 , . . . , pk . Let m be the small-
Now note that est integer such that p1 · · · pm has a divisor in S. (By the
n
assumption on S with n = p1 · · · pk , m = k has this prop-
(Av(m) ) j = erty, so m is well-defined.) If m = 1, then p1 ∈ S, and we
∑ cos( jθ + kθ )eikmθ are done, so assume m ≥ 2. Any divisor d of p1 · · · pm
k=1
in S must be a multiple of pm , or else it would also be
ei jθ n
e−i jθ n
=
2 ∑ eik(m+1)θ + 2 ∑ eik(m−1)θ . a divisor of p1 · · · pm−1 , contradicting the choice of m.
k=1 k=1 But now gcd(d, Nm−1 ) = pm , as desired.
Since ∑nk=1 eik`θ = 0 for integer ` unless n | `, we Second solution (from sci.math): Let n be the small-
conclude that Av(m) = 0 for m = 0 or for 2 ≤ est integer such that gcd(s, n) > 1 for all s in n; note
m ≤ n − 1. In addition, we find that (Av(1) ) j = that n obviously has no repeated prime factors. By the
n −i jθ
= n2 (v(n−1) ) j and (Av(n−1) ) j = 2n ei jθ = n2 (v(1) ) j , condition on S, there exists s ∈ S which divides n.
2e
so that A(v(1) ± v(n−1) ) = ± 2n (v(1) ± v(n−1) ). Thus On the other hand, if p is a prime divisor of s, then by
{v(0) , v(2) , v(3) , . . . , v(n−2) , v(1) + v(n−1) , v(1) − v(n−1) } is the choice of n, n/p is relatively prime to some ele-
a basis for Cn of eigenvectors of A with the claimed ment t of S. Since n cannot be relatively prime to t, t
eigenvalues. is divisible by p, but not by any other prime divisor of
Finally, the determinant of I + A is the product of (1 + n (as those primes divide n/p). Thus gcd(s,t) = p, as
λ ) over all eigenvalues λ of A; in this case, det(I + A) = desired.
(1 + n/2)(1 − n/2) = 1 − n2 /4.
 Solutions to the 61st William Lowell Putnam Mathematical Competition
Saturday, December 2, 2000
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A–1 The possible values comprise the interval (0, A2 ). Pythagorean Theorem √ then shows
√ that the rectangle
To see that the values must lie in this interval, note that P2 P4 P6 P8 has sides 2 and 2 2. For notational ease,
denote the area of a polygon by putting brackets around
!2
m m the name of the polygon.
∑ xj = ∑ x2j + ∑ 2x j xk , By symmetry, the area of the octagon can be expressed
j=0 j=0 0≤ j<k≤m as
so ∑mj=0 x2j ≤ A2 − 2x0 x1 . Letting m → ∞, we have [P2 P4 P6 P8 ] + 2[P2 P3 P4 ] + 2[P4 P5 P6 ].
∑∞j=0 x2j ≤ A2 − 2x0 x1 < A2 . √
Note that [P2 P3 P4 ] is 2 times the distance from P3 to
To show that all values in (0, A2 ) can be obtained, we
P2 P4 , which is maximized when P3√ lies on the midpoint
use geometric progressions with x1 /x0 = x2 /x1 = · · · =
of arc P2 P4 ; similarly, [P4 P5 P6 ] is 2/2 times the dis-
d for variable d. Then ∑∞j=0 x j = x0 /(1 − d) and
tance from P5 to P4 P6 , which is maximized when P5
!2 lies on the midpoint of arc P4 P6 . Thus the area of
∞
x2 1−d ∞
the octagon is maximized when P3 is the midpoint of
∑ x2j = 1 −0d 2 = 1 + d ∑ xj .
arc P2 P4 and P5 is the midpoint of arc P4√ P6 . In this
j=0 j=0
case, it is easy
√ to calculate that [P2 P3 P4 ] = 5 − 1 and
As d increases from 0 to 1, (1 − d)/(1 + d) decreases 4 P5 P6 ] = 5/2 − 1, and so the area of the octagon is
[P√
from 1 to 0. Thus if we take geometric progressions 3 5.
with ∑∞j=0 x j = A, ∑∞j=0 x2j ranges from 0 to A2 . Thus the
A–4 To avoid some improper integrals at 0, we may as well
possible values are indeed those in the interval (0, A2 ), replace the left endpoint of integration by some ε > 0.
as claimed. We now use integration by parts:
A–2 First solution: Let a be an even integer such that a2 + 1 Z B Z B
sin x
is not prime. (For example, choose a ≡ 2 (mod 5), so sin x sin x2 dx = sin x2 (2x dx)
that a2 +1 is divisible by 5.) Then we can write a2 +1 as ε ε2x
B
a difference of squares x2 − b2 , by factoring a2 + 1 as rs sin x
with r ≥ s > 1, and setting x = (r + s)/2, b = (r − s)/2. =− cos x2
2x ε
Finally, put n = x2 − 1, so that n = a2 + b2 , n + 1 = x2 , Z B 
cos x sin x
n + 2 = x2 + 1. + − 2 cos x2 dx.
ε 2x 2x
Second solution: It is well-known that the equation
x2 − 2y2 = 1 has infinitely many solutions (the so- Now sin x 2
2x cos x tends to 0 as B → ∞, and the integral
called “Pell” equation). Thus setting n = 2y2 (so that
of sin x
cos x2 converges absolutely by comparison with
n = y2 + y2 , n + 1 = x2 + 02 , n + 2 = x2 + 12 ) yields 2x2
2
infinitely many n with the desired property. 1/x . Thus it suffices to note that
Third solution: As in the first solution, it suffices to ex- Z B
cos x
Z B
cos x
hibit x such that x2 − 1 is the sum of two squares. We cos x2 dx = cos x2 (2x dx)
n ε 2x 4x2
ε
will take x = 32 , and show that x2 − 1 is the sum of two cos x B
n
squares by induction on n: if 32 − 1 = a2 + b2 , then = 2
sin x2
4x ε
Z B
n+1 n n 2x cos x − sin x
(32 − 1) = (32 − 1)(32 + 1) − sin x2 dx,
n−1 n−1 ε 4x3
= (32 a + b)2 + (a − 32 b)2 .
and that the final integral converges absolutely by com-
parison to 1/x3 .
Fourth solution (by Jonathan Weinstein): Let n = 4k4 +
4k2 = (2k2 )2 + (2k)2 for any integer k. Then n + 1 = An alternate approach is to first rewrite sin x sin x2 as
1 2 2
(2k2 + 1)2 + 02 and n + 2 = (2k2 + 1)2 + 12 . 2 (cos(x − x) − cos(x + x)). Then
√ B
A–3 The maximum area is 3 5. sin(x2 + x)
Z B
cos(x2 + x) dx = −
We deduce frompthe area of P1 P3 P5 P7 that the radius ε 2x + 1 ε
of the circle is 5/2. An easy calculation using the Z B
2 sin(x2 + x)
− dx
ε (2x + 1)2
 2
RB
converges absolutely, and 0 cos(x2 − x) can be treated Therefore, if 0 ≤ a0 ≤ a1 ≤ · · · ≤ ar < 1 are the roots
similarly. of fk in [0, 1), then fk+1 has a root in each of the in-
tervals (a0 , a1 ), (a1 , a2 ), . . . , (ar−1 , ar ), so long as we
A–5 Let a, b, c be the distances between the points. Then the adopt the convention that the empty interval (t,t) ac-
area of the triangle with the three points as vertices is tually contains the point t itself. There is also a root in
abc/4r. On the other hand, the area of a triangle whose the “wraparound” interval (ar , a0 ). Thus Nk+1 ≥ Nk .
vertices have integer coordinates is at least 1/2 (for ex-
ample, by Pick’s Theorem). Thus abc/4r ≥ 1/2, and Next, note that if we set z = e2πit ; then
so N
1
max{a, b, c} ≥ (abc) 1/3
≥ (2r) 1/3
>r 1/3
.
f4k (t) =
2i ∑ j4k a j (z j − z− j )
j=1

A–6 Recall that if f (x) is a polynomial with integer coeffi- is equal to z−N times a polynomial of degree 2N. Hence
cients, then m − n divides f (m) − f (n) for any integers as a function of z, it has at most 2N roots; therefore fk (t)
m and n. In particular, if we put bn = an+1 − an , then bn has at most 2N roots in [0, 1]. That is, Nk ≤ 2N for all
divides bn+1 for all n. On the other hand, we are given N.
that a0 = am = 0, which implies that a1 = am+1 and so To establish that Nk → 2N, we make precise the obser-
b0 = bm . If b0 = 0, then a0 = a1 = · · · = am and we are vation that
done. Otherwise, |b0 | = |b1 | = |b2 | = · · · , so bn = ±b0
for all n. N

Now b0 + · · · + bm−1 = am − a0 = 0, so half of the inte-

fk (t) = ∑ j4k a j sin(2π jt)
j=1
gers b0 , . . . , bm−1 are positive and half are negative. In
particular, there exists an integer 0 < k < m such that is dominated by the term with j = N. At the points
bk−1 = −bk , which is to say, ak−1 = ak+1 . From this it t = (2i + 1)/(2N) for i = 0, 1, . . . , N − 1, we have
follows that an = an+2 for all n ≥ k − 1; in particular, N 4k aN sin(2πNt) = ±N 4k aN . If k is chosen large
for m = n, we have enough so that

a0 = am = am+2 = f ( f (a0 )) = a2 . |aN |N 4k > |a1 |14k + · · · + |aN−1 |(N − 1)4k ,

B–1 Consider the seven triples (a, b, c) with a, b, c ∈ {0, 1} then fk ((2i + 1)/2N) has the same sign as
not all zero. Notice that if r j , s j ,t j are not all even, then aN sin(2πNat), which is to say, the sequence
four of the sums ar j + bs j + ct j with a, b, c ∈ {0, 1} are fk (1/2N), fk (3/2N), . . . alternates in sign. Thus be-
even and four are odd. Of course the sum with a = b = tween these points (again including the “wraparound”
c = 0 is even, so at least four of the seven triples with interval) we find 2N sign changes of fk . Therefore
a, b, c not all zero yield an odd sum. In other words, at limk→∞ Nk = 2N.
least 4N of the tuples (a, b, c, j) yield odd sums. By the
B–4 For t real and not a multiple of π, write g(t) = f (cost)
sint .
pigeonhole principle, there is a triple (a, b, c) for which
Then g(t + π) = g(t); furthermore, the given equation
at least 4N/7 of the sums are odd.
implies that
B–2 Since gcd(m, n) is an integer linear combination of m
and n, it follows that f (2 cos2 t − 1) 2(cost) f (cost)
g(2t) = = = g(t).
  sin(2t) sin(2t)
gcd(m, n) n
n m In particular, for any integer n and k, we have

is an integer linear combination of the integers g(1 + nπ/2k ) = g(2k + nπ) = g(2k ) = g(1).

Since f is continuous, g is continuous where it is de-

       
m n n−1 n n n
= and = fined; but the set {1 + nπ/2k |n, k ∈ Z} is dense in the
n m m−1 n m m
reals, and so g must be constant on its domain. Since
and hence is itself an integer. g(−t) = −g(t) for all t, we must have g(t) = 0 when t
is not a multiple of π. Hence f (x) = 0 for x ∈ (−1, 1).
k
B–3 Put fk (t) = ddtfk . Recall Rolle’s theorem: if f (t) is dif- Finally, setting x = 0 and x = 1 in the given equation
ferentiable, then between any two zeroes of f (t) there yields f (−1) = f (1) = 0.
exists a zero of f 0 (t). This also applies when the zeroes
B–5 We claim that all integers N of the form 2k , with k a
are not all distinct: if f has a zero of multiplicity m at
positive integer and N > max{S0 }, satisfy the desired
t = x, then f 0 has a zero of multiplicity at least m − 1
conditions.
there.
 3

It follows from the definition of Sn , and induction on n, and SN = S0 ∪ {N + a : a ∈ S0 }, as desired.

that
B–6 For each point P in B, let SP be the set of points with
∑ x j ≡ (1 + x) ∑ xj all coordinates equal to ±1 which differ from P in ex-
j∈Sn j∈Sn−1 actly one coordinate. Since there are more than 2n+1 /n
points in B, and each SP has n elements, the cardinali-
≡ (1 + x)n ∑ xj (mod 2).
j∈S0
ties of the sets SP add up to more than 2n+1 , which is to
say, more than twice the total number of points. By the
From the identity (x + y)2 ≡ x2 + y2 (mod 2) and in- pigeonhole principle, there must be a point in three of
n n n
duction on n, we have (x + y)2 ≡ x2 + y2 (mod 2). the sets, say SP , SQ , SR . But then any two of P, Q, R dif-
Hence if we choose N to be a power of 2 greater than fer in exactly two coordinates, so PQR is an equilateral
max{S0 }, then triangle, as desired.

∑ ≡ (1 + xN ) ∑ xj
j∈Sn j∈S0
 The 62nd William Lowell Putnam Mathematical Competition
Saturday, December 1, 2001

A1 Consider a set S and a binary operation ∗, i.e., for each numbers on the red squares is equal to the sum of the
a, b ∈ S, a ∗ b ∈ S. Assume (a ∗ b) ∗ a = b for all a, b ∈ S. numbers on the black squares.
Prove that a ∗ (b ∗ a) = b for all a, b ∈ S.
B2 Find all pairs of real numbers (x, y) satisfying the sys-
A2 You have coins C1 ,C2 , . . . ,Cn . For each k, Ck is biased tem of equations
so that, when tossed, it has probability 1/(2k + 1) of
falling heads. If the n coins are tossed, what is the 1 1
+ = (x2 + 3y2 )(3x2 + y2 )
probability that the number of heads is odd? Express x 2y
the answer as a rational function of n. 1 1
− = 2(y4 − x4 ).
x 2y
A3 For each integer m, consider the polynomial

Pm (x) = x4 − (2m + 4)x2 + (m − 2)2 . positive integer n, let hni denote the closest in-
B3 For any √
teger to n. Evaluate
For what values of m is Pm (x) the product of two non-
constant polynomials with integer coefficients?
∞
2hni + 2−hni
∑ .
n=1 2n
A4 Triangle ABC has an area 1. Points E, F, G lie, respec- B4 Let S denote the set of rational numbers different from
tively, on sides BC, CA, AB such that AE bisects BF at {−1, 0, 1}. Define f : S → S by f (x) = x − 1/x. Prove
point R, BF bisects CG at point S, and CG bisects AE or disprove that
at point T . Find the area of the triangle RST .
∞
f (n) (S) = 0,
\
A5 Prove that there are unique positive integers a, n such /
that an+1 − (a + 1)n = 2001. n=1

A6 Can an arc of a parabola inside a circle of radius 1 have where f (n) denotes f composed with itself n times.
a length greater than 4?
B5 Let a and b be real numbers in the interval (0, 1/2),
B1 Let n be an even positive integer. Write the numbers and let g be a continuous real-valued function such that
1, 2, . . . , n2 in the squares of an n × n grid so that the g(g(x)) = ag(x) + bx for all real x. Prove that g(x) = cx
k-th row, from left to right, is for some constant c.
(k − 1)n + 1, (k − 1)n + 2, . . . , (k − 1)n + n. B6 Assume that (an )n≥1 is an increasing sequence of pos-
itive real numbers such that lim an /n = 0. Must there
Color the squares of the grid so that half of the squares exist infinitely many positive integers n such that an−i +
in each row and in each column are red and the other an+i < 2an for i = 1, 2, . . . , n − 1?
half are black (a checkerboard coloring is one possi-
bility). Prove that for each coloring, the sum of the
 Solutions to the 62nd William Lowell Putnam Mathematical Competition
Saturday, December 1, 2001
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A–1 The hypothesis implies ((b ∗ a) ∗ b) ∗ (b ∗ a) = b for all is neither√a square nor√twice a square, then the number
a, b ∈ S (by replacing a by b ∗ a), and hence a ∗ (b ∗ a) = fields Q( m) and Q( 2) are distinct quadratic fields,
b for all a, b ∈ S (using (b ∗ a) ∗ b = a). so their compositum is a number field of√degree √ 4,
whose Galois group acts transitively on {± m ± 2}.
A–2 Let Pn denote the desired probability. Then P1 = 1/3, Thus Pm is irreducible.
and, for n > 1,
    A–4 Choose r, s,t so that EC = rBC, FA = sCA, GB = tCB,
2n 1 and let [XY Z] denote the area of triangle XY Z. Then
Pn = Pn−1 + (1 − Pn−1 )
2n + 1 2n + 1 [ABE] = [AFE] since the triangles have the same alti-
tude and base. Also [ABE] = (BE/BC)[ABC] = 1 − r,
 
2n − 1 1
= Pn−1 + . and [ECF] = (EC/BC)(CF/CA)[ABC] = r(1 − s) (e.g.,
2n + 1 2n + 1
by the law of sines). Adding this all up yields
The recurrence yields P2 = 2/5, P3 = 3/7, and by a sim-
ple induction, one then checks that for general n one has 1 = [ABE] + [ABF] + [ECF]
Pn = n/(2n + 1). = 2(1 − r) + r(1 − s) = 2 − r − rs
Note: Richard Stanley points out the following nonin-
or r(1 + s) = 1. Similarly s(1 + t) = t(1 + r) = 1.
ductive argument. Put f (x) = ∏nk=1 (x + 2k)/(2k + 1);
then the coefficient of xi in f (x) is the probability of Let f : [0, ∞) → [0, ∞) be the function given by f (x) =
getting exactly i heads. Thus the desired number is 1/(1 + x); then f ( f ( f (r))) = r. However, f (x) is
( f (1) − f (−1))/2, and both values of f can be com- strictly decreasing in x, so f ( f (x)) is increasing and
puted directly: f (1) = 1, and f ( f ( f (x))) is decreasing. Thus there is at most one x
such that f ( f ( f (x))) = x; in fact, since
√ the equation
1 3 2n − 1 1 f (z) = z has a positive root z = (−1 + 5)/2, we must
f (−1) = × ×···× = .
3 5 2n + 1 2n + 1 have r = s = t = z.
We now compute [ABF] = (AF/AC)[ABC] = z, [ABR] =
2
√ the quadratic formula, if Pm (x) = 0, then x = m ±
A–3 By (BR/BF)[ABF] = z/2, analogously [BCS] = [CAT ] =
2 2m +√2, and√hence the four roots of Pm are given by z/2, and [RST ] = |[ABC] − [ABR] − [BCS] − [CAT ]| =
√
S = {± m ± 2}. If Pm factors into two nonconstant 7−3 5
|1 − 3z/2| = 4 .
polynomials over the integers, then some subset of S
consisting of one or two elements form the roots of a Note: the key relation r(1 + s) = 1 can also be derived
polynomial with integer coefficients. by computing using homogeneous coordinates or vec-
tors.
First
√ suppose
√ this subset has a single element, say
m ± 2; this
√ element must be√ a rational number. A–5 Suppose an+1 − (a + 1)n = 2001. Notice that an+1 +
√
Then ( m ± 2)2 = 2 + m ± 2 2m is an integer, so [(a + 1)n − 1] is a multiple of a; thus a divides 2002 =
2 2 × 7 × 11 × 13.
√ a perfect√square, say m = 2n . But then
m is twice
√
m ± 2 = (n ± 1) 2 is only rational if n = ±1, i.e., Since 2001 is divisible by 3, we must have a ≡ 1
if m = 2. (mod 3), otherwise one of an+1 and (a + 1)n is a multi-
Next, suppose that the subset contains ple of 3 and the other is not, so their difference cannot
√ two √ elements;
√
then we can take it√to be one of { m ± 2}, { 2 ± be divisible by 3. Now an+1 ≡ 1 (mod 3), so we must
√ √
m} or {±( m + 2)}. In all cases, the sum and the have (a + 1)n ≡ 1 (mod 3), which forces n to be even,
product of the elements of the subset must and in particular at least 2.
√ be a rational
number. In the first case, this means 2 m ∈ Q,√so m is If a is even, then an+1 − (a + 1)n ≡ −(a + 1)n (mod 4).
√ 2 √2 ∈ Q,
a perfect square. In the second case, we have Since n is even, −(a + 1)n ≡ −1 (mod 4). Since
contradiction. In√the third case, we have ( m + 2)2 ∈ 2001 ≡ 1 (mod 4), this is impossible. Thus a is odd,
Q, or m + 2 + 2 2m ∈ Q, which means that m is twice and so must divide 1001 = 7 × 11 × 13. Moreover,
a perfect square. an+1 − (a + 1)n ≡ a (mod 4), so a ≡ 1 (mod 4).
We conclude that Pm (x) factors into two nonconstant Of the divisors of 7 × 11 × 13, those congruent to 1 mod
polynomials over the integers if and only if m is either 3 are precisely those not divisible by 11 (since 7 and 13
a square or twice a square. are both congruent to 1 mod 3). Thus a divides 7 × 13.
Note: a more sophisticated interpretation of this argu- Now a ≡ 1 (mod 4) is only possible if a divides 13.
ment can be given using Galois theory. Namely, if m
 2

We cannot have a = 1, since 1 − 2n 6= 2001 for any n. It follows that

Thus the only possibility is a = 13. One easily checks
that a = 13, n = 2 is a solution; all that remains is to ∑ f (s)n + g(s) + 1 = ∑ f (s)n + g(s) + 1,
check that no other n works. In fact, if n > 2, then s∈R s∈B
13n+1 ≡ 2001 ≡ 1 (mod 8). But 13n+1 ≡ 13 (mod 8) as desired.
since n is even, contradiction. Thus a = 13, n = 2 is the
unique solution. Note: Richard Stanley points out a theorem of Ryser
(see Ryser, Combinatorial Mathematics, Theorem 3.1)
Note: once one has that n is even, one can use that that can also be applied. Namely, if A and B are 0 −
2002 = an+1 + 1 − (a + 1)n is divisible by a + 1 to rule 1 matrices with the same row and column sums, then
out cases. there is a sequence of operations on 2 × 2 matrices of
A–6 The answer is yes. Consider the arc of the parabola the form
y = Ax2 inside the circle x2 + (y − 1)2 = 1, where we 
0 1
 
1 0

initially assume that A > 1/2.√This intersects the circle →
1 0 0 1
in three points, (0, 0) and (± 2A − 1/A, (2A − 1)/A).
We claim that for A sufficiently large,
√ the length L of or vice versa, which transforms A into B. If we iden-
the parabolic arc between (0, 0) and ( 2A − 1/A, (2A− tify 0 and 1 with red and black, then the given color-
1)/A) is greater than 2, which implies the desired result ing and the checkerboard coloring both satisfy the sum
by symmetry. We express L using the usual formula for condition. Since the desired result is clearly true for the
arclength: checkerboard coloring, and performing the matrix op-
Z √2A−1/A q erations does not affect this, the desired result follows
L= 1 + (2Ax)2 dx in general.
0
Z 2√2A−1 p B–2 By adding and subtracting the two given equations, we
1
= 1 + x2 dx obtain the equivalent pair of equations
2A 0
Z 2√2A−1 p
!
1 2/x = x4 + 10x2 y2 + 5y4
= 2+ ( 1 + x2 − x) dx − 2 ,
2A 0 1/y = 5x4 + 10x2 y2 + y4 .

where we have artificially introduced −x into the inte- Multiplying the former by x and the latter by y, then
grand in the last step. Now, for x ≥ 0, adding and subtracting the two resulting equations, we
obtain another pair of equations equivalent to the given
p 1 1 1 ones,
1 + x2 − x = √ > √ ≥ ;
1 + x2 + x 2 1 + x2 2(x + 1)
R √ 3 = (x + y)5 , 1 = (x − y)5 .
since 0∞ dx/(2(x + 1)) diverges, so does 0∞ ( 1 + x2 −
R

x) dx. Hence, for sufficiently large A, we have It follows that x = (31/5 + 1)/2 and y = (31/5 − 1)/2 is
R 2√2A−1 √ the unique solution satisfying the given equations.
0 ( 1 + x2 − x) dx > 2, and hence L > 2.
Note: a numerical computation shows that one must B–3 Since (k − 1/2)2 = k2 − k + 1/4 and (k + 1/2)2 = k2 +
take A > 34.7 to obtain L > 2, and that the maximum k + 1/4, we have that hni = k if and only if k2 − k + 1 ≤
value of L is about 4.0027, achieved for A ≈ 94.1. n ≤ k2 + k. Hence
B–1 Let R (resp. B) denote the set of red (resp. black) squares ∞
2hni + 2−hni ∞
2hni + 2−hni
in such a coloring, and for s ∈ R ∪ B, let f (s)n + g(s) + 1 ∑ = ∑ ∑
n=1 2n k=1 n,hni=k 2n
denote the number written in square s, where 0 ≤
f (s), g(s) ≤ n − 1. Then it is clear that the value of ∞ k2 +k
2k + 2−k
f (s) depends only on the row of s, while the value of = ∑ ∑
k=1 n=k2 −k+1 2n
g(s) depends only on the column of s. Since every row
contains exactly n/2 elements of R and n/2 elements of ∞
2 2
B, = ∑ (2k + 2−k )(2−k +k − 2−k −k )
k=1
∞
∑ f (s) = ∑ f (s). = ∑ (2−k(k−2) − 2−k(k+2) )
s∈R s∈B
k=1
∞ ∞
Similarly, because every column contains exactly n/2
elements of R and n/2 elements of B,
= ∑ 2−k(k−2) − ∑ 2−k(k−2)
k=1 k=3
= 3.
∑ g(s) = ∑ g(s).
s∈R s∈B
 3

Alternate solution: rewrite the sum as ∑∞ −(n+hni) + Suppose g is strictly increasing. If c2 6= 0 for some
n=1 2
∞
∑n=1 2 −(n−hni) . Note that hni =
6 hn + 1i if and only if choice of x0 , then xn is dominated by r2n for n suffi-
n = m2 + m for some m. Thus n + hni and n − hni each ciently negative. But taking xn and xn+2 for n suffi-
increase by 1 except at n = m2 + m, where the former ciently negative of the right parity, we get 0 < xn < xn+2
skips from m2 +2m to m2 +2m+2 and the latter repeats but g(xn ) > g(xn+2 ), contradiction. Thus c2 = 0; since
the value m2 . Thus the sums are x0 = c1 and x1 = c1 r1 , we have g(x) = r1 x for all x.
Analogously, if g is strictly decreasing, then c2 = 0 or
∞ ∞ ∞ ∞
2 2 else xn is dominated by r1n for n sufficiently positive.
∑ 2−n − ∑ 2−m + ∑ 2−n + ∑ 2−m = 2 + 1 = 3.
But taking xn and xn+2 for n sufficiently positive of the
n=1 m=1 n=0 m=1
right parity, we get 0 < xn+2 < xn but g(xn+2 ) < g(xn ),
B–4 For a rational number p/q expressed in lowest terms, contradiction. Thus in that case, g(x) = r2 x for all x.
define its height H(p/q) to be |p| + |q|. Then for B–6 Yes, there must exist infinitely many such n. Let S be
any p/q ∈ S expressed in lowest terms, we have the convex hull of the set of points (n, an ) for n ≥ 0. Ge-
H( f (p/q)) = |q2 − p2 | + |pq|; since by assumption p ometrically, S is the intersection of all convex sets (or
and q are nonzero integers with |p| 6= |q|, we have even all halfplanes) containing the points (n, an ); alge-
braically, S is the set of points (x, y) which can be writ-
H( f (p/q)) − H(p/q) = |q2 − p2 | + |pq| − |p| − |q| ten as c1 (n1 , an1 ) + · · · + ck (nk , ank ) for some c1 , . . . , ck
≥ 3 + |pq| − |p| − |q| which are nonnegative of sum 1.
= (|p| − 1)(|q| − 1) + 2 ≥ 2. We prove that for infinitely many n, (n, an ) is a vertex
on the upper boundary of S, and that these n satisfy the
It follows that f (n) (S) consists solely of numbers of given condition. The condition that (n, an ) is a vertex on
height strictly larger than 2n + 2, and hence the upper boundary of S is equivalent to the existence of
(n)
a line passing through (n, an ) with all other points of S
∩∞
n=1 f (S) = 0.
/ below it. That is, there should exist m > 0 such that

Note: many choices for the height function are possible: ak < an + m(k − n) ∀k ≥ 1. (1)
one can take H(p/q) = max |p|, |q|, or H(p/q) equal to
the total number of prime factors of p and q, and so We first show that n = 1 satisfies (1). The condition
on. The key properties of the height function are that ak /k → 0 as k → ∞ implies that (ak − a1 )/(k − 1) → 0
on one hand, there are only finitely many rationals with as well. Thus the set {(ak − a1 )/(k − 1)} has an upper
height below any finite bound, and on the other hand, bound m, and now ak ≤ a1 + m(k − 1), as desired.
the height function is a sufficiently “algebraic” function Next, we show that given one n satisfying (1), there ex-
of its argument that one can relate the heights of p/q ists a larger one also satisfying (1). Again, the condition
and f (p/q). ak /k → 0 as k → ∞ implies that (ak − an )/(k − n) → 0
B–5 Note that g(x) = g(y) implies that g(g(x)) = g(g(y)) as k → ∞. Thus the sequence {(ak − an )/(k − n)}k>n
and hence x = y from the given equation. That is, g is has a maximum element; suppose k = r is the largest
injective. Since g is also continuous, g is either strictly value that achieves this maximum, and put m = (ar −
increasing or strictly decreasing. Moreover, g cannot an )/(r − n). Then the line through (r, ar ) of slope m lies
tend to a finite limit L as x → +∞, or else we’d have strictly above (k, ak ) for k > r and passes through or lies
g(g(x)) − ag(x) = bx, with the left side bounded and above (k, ak ) for k < r. Thus (1) holds for n = r with m
the right side unbounded. Similarly, g cannot tend to replaced by m − ε for suitably small ε > 0.
a finite limit as x → −∞. Together with monotonicity, By induction, we have that (1) holds for infinitely many
this yields that g is also surjective. n. For any such n there exists m > 0 such that for i =
Pick x0 arbitrary, and define xn for all n ∈ Z recursively 1, . . . , n − 1, the points (n − i, an−i ) and (n + i, an+i ) lie
by xn+1 = g(xn√ ) for n > 0, and xn−1 = g−1 (x√n ) for n < 0. below the line through (n, an ) of slope m. That means
Let r1 = (a + a2 + 4b)/2 and r2 = (a − a2 + 4b)/2 an+i < an +mi and an−i < an −mi; adding these together
and r2 be the roots of x2 −ax −b = 0, so that r1 > 0 > r2 gives an−i + an+i < 2an , as desired.
and 1 > |r1 | > |r2 |. Then there exist c1 , c2 ∈ R such that
xn = c1 r1n + c2 r2n for all n ∈ Z.
 The 63rd William Lowell Putnam Mathematical Competition
Saturday, December 7, 2002

A1 Let k be a fixed positive integer. The n-th derivative of Each player, in turn, signs his or her
n (x)
1
xk −1
has the form (xkP−1) n+1 where Pn (x) is a polynomial.
name on a previously unsigned face. The
winner is the player who first succeeds in
Find Pn (1).
signing three faces that share a common
A2 Given any five points on a sphere, show that some four vertex.
of them must lie on a closed hemisphere.
Show that the player who signs first will always win by
A3 Let n ≥ 2 be an integer and Tn be the number of non- playing as well as possible.
empty subsets S of {1, 2, 3, . . . , n} with the property that B3 Show that, for all integers n > 1,
the average of the elements of S is an integer. Prove that
1 n
 
Tn − n is always even. 1 1 1
< − 1− < .
2ne e n ne
A4 In Determinant Tic-Tac-Toe, Player 1 enters a 1 in an
empty 3 × 3 matrix. Player 0 counters with a 0 in a va- B4 An integer n, unknown to you, has been randomly
cant position, and play continues in turn until the 3 × 3 chosen in the interval [1, 2002] with uniform probabil-
matrix is completed with five 1’s and four 0’s. Player ity. Your objective is to select n in an odd number of
0 wins if the determinant is 0 and player 1 wins other- guesses. After each incorrect guess, you are informed
wise. Assuming both players pursue optimal strategies, whether n is higher or lower, and you must guess an in-
who will win and how? teger on your next turn among the numbers that are still
A5 Define a sequence by a0 = 1, together with the rules feasibly correct. Show that you have a strategy so that
a2n+1 = an and a2n+2 = an +an+1 for each integer n ≥ 0. the chance of winning is greater than 2/3.
Prove that every positive rational number appears in the B5 A palindrome in base b is a positive integer whose base-
set b digits read the same backwards and forwards; for ex-

an−1
 
1 1 2 1 3
 ample, 2002 is a 4-digit palindrome in base 10. Note
:n≥1 = , , , , ,... . that 200 is not a palindrome in base 10, but it is the 3-
an 1 2 1 3 2
digit palindrome 242 in base 9, and 404 in base 7. Prove
that there is an integer which is a 3-digit palindrome in
A6 Fix an integer b ≥ 2. Let f (1) = 1, f (2) = 2, and for base b for at least 2002 different values of b.
each n ≥ 3, define f (n) = n f (d), where d is the number
of base-b digits of n. For which values of b does B6 Let p be a prime number. Prove that the determinant of
the matrix
∞
1
∑ f (n)

x y z

n=1  xp yp zp 
2 2 2
converge? xp yp zp

B1 Shanille O’Keal shoots free throws on a basketball is congruent modulo p to a product of polynomials of
court. She hits the first and misses the second, and the form ax + by + cz, where a, b, c are integers. (We
thereafter the probability that she hits the next shot is say two integer polynomials are congruent modulo p if
equal to the proportion of shots she has hit so far. What corresponding coefficients are congruent modulo p.)
is the probability she hits exactly 50 of her first 100
shots?
B2 Consider a polyhedron with at least five faces such that
exactly three edges emerge from each of its vertices.
Two players play the following game:
 Solutions to the 63rd William Lowell Putnam Mathematical Competition
Saturday, December 7, 2002
Kiran Kedlaya and Lenny Ng

A–1 By differentiating Pn (x)/(xk − 1)n+1 , we find that For i, j = 1, 2, 3, let Ai j denote the position in row i and
Pn+1 (x) = (xk − 1)Pn0 (x) − (n + 1)kxk−1 Pn (x); substi- column j. Without loss of generality, we may assume
tuting x = 1 yields Pn+1 (1) = −(n + 1)kPn (1). Since that Player 1’s first move is at A11 . Player 0 then plays
P0 (1) = 1, an easy induction gives Pn (1) = (−k)n n! for at A22 :
all n ≥ 0.  
1 ∗ ∗
Note: one can also argue by expanding in Taylor series ∗ 0 ∗
around 1. Namely, we have
∗ ∗ ∗
1 1 1
= = (x − 1)−1 + · · · , After Player 1’s second move, at least one of A23 and
xk − 1 k(x − 1) + · · · k
A32 remains vacant. Without loss of generality, assume
so A23 remains vacant; Player 0 then plays there.
dn 1 (−1)n n! After Player 1’s third move, Player 0 wins by playing
= at A21 if that position is unoccupied. So assume instead
dxn xk − 1 k(x − 1)−n−1
that Player 1 has played there. Thus of Player 1’s three
and moves so far, two are at A11 and A21 . Hence for i equal
to one of 1 or 3, and for j equal to one of 2 or 3, the
dn 1 following are both true:
Pn (x) = (xk − 1)n+1
dxn xk − 1

(−1)n n!
 (a) The 2 × 2 submatrix formed by rows 2 and i and
= (k(x − 1) + · · · )n+1 (x − 1)−n−1 + · · · by columns 2 and 3 contains two zeroes and two
k empty positions.
n
= (−k) n! + · · · .
(b) Column j contains one zero and two empty posi-
tions.
A–2 Draw a great circle through two of the points. There are
two closed hemispheres with this great circle as bound- Player 0 next plays at Ai j . To prevent a zero column,
ary, and each of the other three points lies in one of Player 1 must play in column j, upon which Player 0
them. By the pigeonhole principle, two of those three completes the 2 × 2 submatrix in (a) for the win.
points lie in the same hemisphere, and that hemisphere Note: one can also solve this problem directly by mak-
thus contains four of the five given points. ing a tree of possible play sequences. This tree can be
Note: by a similar argument, one can prove that among considerably collapsed using symmetries: the symme-
any n+3 points on an n-dimensional sphere, some n+2 try between rows and columns, the invariance of the
of them lie on a closed hemisphere. (One cannot get by outcome under reordering of rows or columns, and the
with only n + 2 points: put them at the vertices of a reg- fact that the scenario after a sequence of moves does
ular simplex.) Namely, any n of the points lie on a great not depend on the order of the moves (sometimes called
sphere, which forms the boundary of two hemispheres; “transposition invariance”).
of the remaining three points, some two lie in the same Note (due to Paul Cheng): one can reduce Determi-
hemisphere. nant Tic-Tac-Toe to a variant of ordinary tic-tac-toe.
A–3 Note that each of the sets {1}, {2}, . . . , {n} has the de- Namely, consider a tic-tac-toe grid labeled as follows:
sired property. Moreover, for each set S with inte- A11 A22 A33
ger average m that does not contain m, S ∪ {m} also
has average m, while for each set T of more than A23 A31 A12
one element with integer average m that contains m, A32 A13 A21
T \ {m} also has average m. Thus the subsets other than
{1}, {2}, . . . , {n} can be grouped in pairs, so Tn − n is Then each term in the expansion of the determinant oc-
even. curs in a row or column of the grid. Suppose Player
1 first plays in the top left. Player 0 wins by playing
A–4 (partly due to David Savitt) Player 0 wins with opti- first in the top row, and second in the left column. Then
mal play. In fact, we prove that Player 1 cannot prevent there are only one row and column left for Player 1 to
Player 0 from creating a row of all zeroes, a column of threaten, and Player 1 cannot already threaten both on
all zeroes, or a 2 × 2 submatrix of all zeroes. Each of the third move, so Player 0 has time to block both.
these forces the determinant of the matrix to be zero.
 2

A–5 It suffices to prove that for any relatively prime positive that for d ≥ 3,
integers r, s, there exists an integer n with an = r and
an+1 = s. We prove this by induction on r + s, the case 2d −1 Z 2d
1 1 1 dx
r +s = 2 following from the fact that a0 = a1 = 1. Given ∑ < d−1 − d +
n=2d−1
n 2 2 2d−1 x
r and s not both 1 with gcd(r, s) = 1, we must have r 6=
s. If r > s, then by the induction hypothesis we have 1
= + log 2
an = r − s and an+1 = s for some n; then a2n+2 = r and 2d
a2n+3 = s. If r < s, then we have an = r and an+1 = s −r 1
≤ + log 2 < 0.125 + 0.7 < 1.
for some n; then a2n+1 = r and a2n+2 = s. 8
Note: a related problem is as follows. Starting with the Put c = 81 + log 2 and L = 1 + 12 + 6(1−c)
1
. Then we can
sequence i
−1 1
prove that ∑2n=1 f (n) < L for all i ≥ 2 by induction on i.
0 1 The case i = 2 is clear. For the induction, note that by
, ,
1 0 (2),
repeat the following operation: insert between each pair 2i −1 i
a c a+c 1 1 1 1
b and d the pair b+d . Prove that each positive rational ∑ < 1+ + +c ∑
number eventually appears. n=1 f (n) 2 6 d=3 f (d)
Observe that by induction, if ba and dc are consecutive 1 1 1
< 1+ + +c
terms in the sequence, then bc−ad = 1. The same holds 2 6 6(1 − c)
for consecutive terms of the n-th Farey sequence, the 1 1
sequence of rational numbers in [0, 1] with denominator = 1+ + = L,
2 6(1 − c)
(in lowest terms) at most n.
1
A–6 The sum converges for b = 2 and diverges for b ≥ 3. We as desired. We conclude that ∑∞n=1 f (n) converges to a
first consider b ≥ 3. Suppose the sum converges; then limit less than or equal to L.
the fact that f (n) = n f (d) whenever bd−1 ≤ n ≤ bd − 1 Note: the above argument proves that the sum for b = 2
yields is at most L < 2.417. One can also obtain a lower
bound by the same technique, namely 1 + 21 + 6(1−c 1
0)
d
∞
1 ∞
1 b −1 1 with c0 = log 2. This bound exceeds 2.043. (By con-
∑ =∑ ∑ n. (1)
n=1 f (n) d=1 f (d) n=bd−1 trast, summing the first 100000 terms of the series only
yields a lower bound of 1.906.) Repeating the same ar-
However, by comparing the integral of 1/x with a Rie- guments with d ≥ 4 as the cutoff yields the upper bound
mann sum, we see that 2.185 and the lower bound 2.079.

bd −1 Z bd B–1 The probability is 1/99. In fact, we show by induction

1 dx
∑ n> on n that after n shots, the probability of having made
n=bd−1
bd−1 x any number of shots from 1 to n − 1 is equal to 1/(n −
= log(bd ) − log(bd−1 ) = log b, 1). This is evident for n = 2. Given the result for n,
we see that the probability of making i shots after n + 1
where log denotes the natural logarithm. Thus (1) yields attempts is
 
∞
1 ∞
1 i−1 1 i 1 (i − 1) + (n − i)
∑ > (log b) ∑ , + 1− =
f (n) n n−1 n n−1 n(n − 1)
n=1 n=1 f (n)
1
a contradiction since log b > 1 for b ≥ 3. Therefore the = ,
n
sum diverges.
as claimed.
For b = 2, we have a slightly different identity because
f (2) 6= 2 f (2). Instead, for any positive integer i, we B–2 (Note: the problem statement assumes that all polyhe-
have dra are connected and that no two edges share more than
one face, so we will do likewise. In particular, these are
2i −1 d
1 1 1 i
1 2 −1 1 true for all convex polyhedra.) We show that in fact
∑ = 1+ + + ∑ ∑ n. (2)
n=1 f (n) 2 6 d=3 f (d) n=2 d−1
the first player can win on the third move. Suppose the
polyhedron has a face A with at least four edges. If the
Again comparing an integral to a Riemann sum, we see first player plays there first, after the second player’s
first move there will be three consecutive faces B,C, D
adjacent to A which are all unoccupied. The first player
wins by playing in C; after the second player’s second
 3

move, at least one of B and D remains unoccupied, and than one of these guesses. If n ≡ 1 (mod 3), it will be
either is a winning move for the first player. guessed on an odd turn. If n ≡ 0 (mod 3), it will be
It remains to show that the polyhedron has a face with guessed on an even turn. If n ≡ 2 (mod 3), then n + 1
at least four edges. (Thanks to Russ Mann for suggest- will be guessed on an even turn, forcing a guess of n on
ing the following argument.) Suppose on the contrary the next turn. Thus the probability of success with this
that each face has only three edges. Starting with any strategy is 1335/2002 > 2/3.
face F1 with vertices v1 , v2 , v3 , let v4 be the other end- Note: for any positive integer m, this strategy wins
point of the third edge out of v1 . Then the faces ad- when the number is being guessed from [1, m] with
jacent to F1 must have vertices v1 , v2 , v4 ; v1 , v3 , v4 ; and probability m1 b 2m+1
3 c. We can prove that this is best
v2 , v3 , v4 . Thus v1 , v2 , v3 , v4 form a polyhedron by them- possible as follows. Let am denote m times the proba-
selves, contradicting the fact that the given polyhedron bility of winning when playing optimally. Also, let bm
is connected and has at least five vertices. (One can also denote m times the corresponding probability of win-
deduce this using Euler’s formula V − E + F = 2 − 2g, ning if the objective is to select the number in an even
where V, E, F are the numbers of vertices, edges and number of guesses instead. (For definiteness, extend the
faces, respectively, and g is the genus of the polyhe- definitions to incorporate a0 = 0 and b0 = 0.)
dron. For a convex polyhedron, g = 0 and you get the We first claim that am = 1 + max1≤k≤m {bk−1 + bm−k }
“usual” Euler’s formula.) and bm = max1≤k≤m {ak−1 + am−k } for m ≥ 1. To
Note: Walter Stromquist points out the following coun- establish the first recursive identity, suppose that our
terexample if one relaxes the assumption that a pair of first guess is some integer k. We automatically win if
faces may not share multiple edges. Take a tetrahedron n = k, with probability 1/m. If n < k, with probability
and remove a smaller tetrahedron from the center of an (k − 1)/m, then we wish to guess an integer in [1, k − 1]
edge; this creates two small triangular faces and turns in an even number of guesses; the probability of success
two of the original faces into hexagons. Then the sec- when playing optimally is bk−1 /(k − 1), by assumption.
ond player can draw by signing one of the hexagons, Similarly, if n < k, with probability (m − k)/m, then the
one of the large triangles, and one of the small trian- subsequent probability of winning is bm−k /(m − k). In
gles. (He does this by “mirroring”: wherever the first sum, the overall probability of winning if k is our first
player signs, the second player signs the other face of guess is (1 + bk−1 + bm−k )/m. For optimal strategy, we
the same type.) choose k such that this quantity is maximized. (Note
that this argument still holds if k = 1 or k = m, by our
B–3 The desired inequalities can be rewritten as definitions of a0 and b0 .) The first recursion follows,
   and the second recursion is established similarly.
1 1 1
1 − < exp 1 + n log 1 − < 1− . We now prove by induction that am = b(2m + 1)/3c and
n n 2n
bm = b2m/3c for m ≥ 0. The inductive step relies on the
By taking logarithms, we can rewrite the desired in- inequality bxc + byc ≤ bx + yc, with equality when one
equalities as of x, y is an integer. Now suppose that ai = b(2i + 1)/3c
    and bi = b2i/3c for i < m. Then
1 1
− log 1 − < −1 − n log 1 − 
2(k − 1)
 
2(m − k)

2n n 1 + bk−1 + bm−k = 1 + +

1
 3 3
< − log 1 − .  
2m
n ≤
3
Rewriting these in terms of the Taylor expansion of
− log(1 − x), we see that the desired result is also equiv- and similarly ak−1 + am−k ≤ b(2m + 1)/3c, with equal-
alent to ity in both cases attained, e.g., when k = 1. The induc-
∞ ∞ ∞
tive formula for am and bm follows.
1 1 1
∑ i2i ni < ∑ (i + 1)ni < ∑ ini , B–5 (due to Dan Bernstein) Put N = 2002!. Then for d =
i=1 i=1 i=1
1, . . . , 2002, the number N 2 written in base b = N/d − 1
which is evident because the inequalities hold term by has digits d 2 , 2d 2 , d 2 . (Note that these really are digits
term. because 2(2002)2 < (2002!)2 /2002 − 1.)
Note: David Savitt points out that the upper bound can Note: one can also produce an integer N which has base
be improved from 1/(ne) to 2/(3ne) with a slightly b digits 1, ∗, 1 for n different values of b, as follows.
more complicated argument. (In fact, for any c > 1/2, Choose c with 0 < c < 21/n . For m a large positive in-
one has an upper bound of c/(ne), but only for n above teger, put N = 1 + (m + 1) · · · (m + n)bcmcn−2 . For m
a certain bound depending on c.) sufficiently large, the bases

B–4 Use the following strategy: guess 1, 3, 4, 6, 7, 9, . . . until N −1

b= = (m + j)
the target number n is revealed to be equal to or lower (m + i)mn−2 ∏
j6=i
 4

for i = 1, . . . , n will have the properties that N ≡ 1 which is precisely the desired determinant.
(mod b) and b2 < N < 2b2 for m sufficiently large. Note: a simpler conceptual proof is as follows. (Every-
Note (due to Russ Mann): one can also give a “noncon- thing in this paragraph will be modulo p.) Note that
structive” argument. Let N be a large positive integer. for any integers a, b, c, the column vector [ax + by +
For b ∈ (N 2 , N 3 ), the number of 3-digit base-b palin- 2
cz, (ax + by + cz) p , (ax + by + cz) p ] is a linear com-
dromes in the range [b2 , N 6 − 1] is at least bination of the columns of the given matrix. Thus
 6 ax + by + cz divides the determinant. In particular, all
N − b2 N6

− 1 ≥ 2 − b − 2, of the factors of (3) divide the determinant; since both
b b (3) and the determinant have degree p2 + p + 1, they
agree up to a scalar multiple. Moreover, they have the
since there is a palindrome in each interval [kb, (k + 2
same coefficient of z p y p x (since this term only appears
1)b − 1] for k = b, . . . , b2 − 1. Thus the average num- in the expansion of (3) when you choose the first term
ber of bases for which a number in [1, N 6 − 1] is at least in each factor). Thus the determinant is congruent to
3 (3), as desired.
1 N −1 N 6
 
∑ − b − 2 ≥ log(N) − c Either argument can be used to generalize to a corre-
N 6 b=N 2 +1 b sponding n × n determinant, called a Moore determi-
nant; we leave the precise formulation to the reader.
for some constant c > 0. Take N so that the right side Note the similarity with the classical Vandermonde de-
exceeds 2002; then at least one number in [1, N 6 − 1] is
terminant: if A is the n × n matrix with Ai j = xij for
a base-b palindrome for at least 2002 values of b.
i, j = 0, . . . , n − 1, then
B–6 We prove that the determinant is congruent modulo p to
det(A) = ∏ (x j − xi ).
p−1 p−1 1≤i< j≤n
x ∏ (y + ix) ∏ (z + ix + jy). (3)
i=0 i, j=0

We first check that

p−1
∏ (y + ix) ≡ y p − x p−1 y (mod p). (4)
i=0

Since both sides are homogeneous as polynomials in x

and y, it suffices to check (4) for x = 1, as a congruence
between polynomials. Now note that the right side has
0, 1, . . . , p − 1 as roots modulo p, as does the left side.
Moreover, both sides have the same leading coefficient.
Since they both have degree only p, they must then co-
incide.
We thus have

p−1 p−1
x ∏ (y + ix) ∏ (z + ix + jy)
i=0 i, j=0
p−1
≡ x(y p − x p−1 y) ∏ ((z + jy) p − x p−1 (z + jy))
j=0
p−1
≡ (xy p − x p y) ∏ (z p − x p−1 z + jy p − jx p−1 y)
j=0

≡ (xy − x y)((z p − x p−1 z) p

p p

− (y p − x p−1 y) p−1 (z p − x p−1 z))

2 2 −p
≡ (xy p − x p y)(z p − x p zp)
− x(y p − x p−1 y) p (z p − x p−1 z)
2 2 2 −p+1 2
≡ xy p z p − x p yz p − x p y p z p + x p yz p
2 2 −p+1 2 2
− xy p z p + x p ypzp + xpyp z − xp ypz
2 2 2
≡ xy p z p + yz p x p + zx p y p
2 2 2
− xz p y p − yx p z p − zy p x p ,
 The 64th William Lowell Putnam Mathematical Competition
Saturday, December 6, 2003

A1 Let n be a fixed positive integer. How many ways are B2 Let n be a positive integer. Starting with the sequence
there to write n as a sum of positive integers, 1, 21 , 31 , . . . , 1n , form a new sequence of n − 1 entries
3 5 2n−1
4 , 12 , . . . , 2n(n−1) by taking the averages of two consec-
n = a1 + a2 + · · · + ak ,
utive entries in the first sequence. Repeat the averaging
with k an arbitrary positive integer and a1 ≤ a2 ≤ · · · ≤ of neighbors on the second sequence to obtain a third
ak ≤ a1 + 1? For example, with n = 4 there are four sequence of n−2 entries, and continue until the final se-
ways: 4, 2+2, 1+1+2, 1+1+1+1. quence produced consists of a single number xn . Show
that xn < 2/n.
A2 Let a1 , a2 , . . . , an and b1 , b2 , . . . , bn be nonnegative real
numbers. Show that B3 Show that for each positive integer n,
n
(a1 a2 · · · an )1/n + (b1 b2 · · · bn )1/n n! = ∏ lcm{1, 2, . . . , bn/ic}.
i=1
≤ [(a1 + b1 )(a2 + b2 ) · · · (an + bn )]1/n .
(Here lcm denotes the least common multiple, and bxc
A3 Find the minimum value of denotes the greatest integer ≤ x.)

| sin x + cos x + tan x + cot x + sec x + csc x| B4 Let f (z) = az4 + bz3 + cz2 + dz + e = a(z − r1 )(z −
r2 )(z − r3 )(z − r4 ) where a, b, c, d, e are integers, a 6= 0.
for real numbers x. Show that if r1 + r2 is a rational number and r1 + r2 6=
r3 + r4 , then r1 r2 is a rational number.
A4 Suppose that a, b, c, A, B,C are real numbers, a 6= 0 and
A 6= 0, such that B5 Let A, B, and C be equidistant points on the circumfer-
ence of a circle of unit radius centered at O, and let P
|ax2 + bx + c| ≤ |Ax2 + Bx +C| be any point in the circle’s interior. Let a, b, c be the
distance from P to A, B,C, respectively. Show that there
for all real numbers x. Show that is a triangle with side lengths a, b, c, and that the area of
this triangle depends only on the distance from P to O.
|b2 − 4ac| ≤ |B2 − 4AC|.
B6 Let f (x) be a continuous real-valued function defined
A5 A Dyck n-path is a lattice path of n upsteps (1, 1) and n on the interval [0, 1]. Show that
downsteps (1, −1) that starts at the origin O and never Z 1Z 1 Z 1
dips below the x-axis. A return is a maximal sequence | f (x) + f (y)| dx dy ≥ | f (x)| dx.
of contiguous downsteps that terminates on the x-axis. 0 0 0
For example, the Dyck 5-path illustrated has two re-
turns, of length 3 and 1 respectively.

Show that there is a one-to-one correspondence be-

tween the Dyck n-paths with no return of even length
and the Dyck (n − 1)-paths.
A6 For a set S of nonnegative integers, let rS (n) denote the
number of ordered pairs (s1 , s2 ) such that s1 ∈ S, s2 ∈ S,
s1 6= s2 , and s1 + s2 = n. Is it possible to partition the
nonnegative integers into two sets A and B in such a way
that rA (n) = rB (n) for all n ?
B1 Do there exist polynomials a(x), b(x), c(y), d(y) such
that

1 + xy + x2 y2 = a(x)c(y) + b(x)d(y)

holds identically?
 Solutions to the 64th William Lowell Putnam Mathematical Competition
Saturday, December 6, 2003
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A1 There are n such sums. More precisely, there is exactly Third solution: Since both sides are continuous in each
one such sum with k terms for each of k = 1, . . . , n (and ai , it is sufficient to prove the claim with a1 , . . . , an all
clearly no others). To see this, note that if n = a1 + a2 + positive (the general case follows by taking limits as
· · · + ak with a1 ≤ a2 ≤ · · · ≤ ak ≤ a1 + 1, then some of the ai tend to zero). Put ri = bi /ai ; then the
given inequality is equivalent to
ka1 = a1 + a1 + · · · + a1
≤ n ≤ a1 + (a1 + 1) + · · · + (a1 + 1) (1 + r1 )1/n · · · (1 + rn )1/n ≥ 1 + (r1 · · · rn )1/n .
= ka1 + k − 1.
In terms of the function
However, there is a unique integer a1 satisfying these
f (x) = log(1 + ex )
inequalities, namely a1 = bn/kc. Moreover, once a1 is
fixed, there are k different possibilities for the sum a1 + and the quantities si = log ri , we can rewrite the desired
a2 + · · · + ak : if i is the last integer such that ai = a1 , inequality as
then the sum equals ka1 + (i − 1). The possible values
of i are 1, . . . , k, and exactly one of these sums comes
 
1 s1 + · · · + sn
out equal to n, proving our claim. ( f (s1 ) + · · · + f (sn )) ≥ f .
n n
Note: In summary, there is a unique partition of n with
k terms that is “as equally spaced as possible”. One This will follow from Jensen’s inequality if we can ver-
can also obtain essentially the same construction induc- ify that f is a convex function; it is enough to check that
tively: except for the all-ones sum, each partition of n is f 00 (x) > 0 for all x. In fact,
obtained by “augmenting” a unique partition of n − 1.
ex 1
f 0 (x) = = 1−
A2 First solution: Assume without loss of generality that 1 + ex 1 + ex
ai +bi > 0 for each i (otherwise both sides of the desired
inequality are zero). Then the AM-GM inequality gives is an increasing function of x, so f 00 (x) > 0 and Jensen’s
inequality thus yields the desired result. (As long as the
 1/n ai are all positive, equality holds when s1 = · · · = sn ,
a1 · · · an i.e., when the vectors (a1 , . . . , an ) and (b1 , . . . , bn ). Of
(a1 + b1 ) · · · (an + bn ) course other equality cases crop up if some of the ai
 
1 a1 an vanish, i.e., if a1 = b1 = 0.)
≤ +···+ ,
n a1 + b1 an + bn Fourth solution: We apply induction on n, the case n =
1 being evident. First we verify the auxiliary inequality
and likewise with the roles of a and b reversed. Adding
these two inequalities and clearing denominators yields (an + bn )(cn + d n )n−1 ≥ (acn−1 + bd n−1 )n
the desired result.
Second solution: Write the desired inequality in the for a, b, c, d ≥ 0. The left side can be written as
form
an cn(n−1) + bn d n(n−1)
1/n 1/n n
(a1 + b1 ) · · · (an + bn ) ≥ [(a1 · · · an ) + (b1 · · · bn ) ] , n−1  
n − 1 n ni n(n−1−i)
+∑ b c d
expand both sides, and compare the terms on both sides i=1 i
in which k of the terms are among the ai . On the n−1  
n − 1 n n(i−1) n(n−i)
left, one has the product of each k-element subset of +∑ a c d .
{1, . . . , n}; on the right, one has i=1 i − 1

  Applying the weighted AM-GM inequality between

n
(a1 · · · an )k/n · · · (b1 . . . bn )(n−k)/n , matching terms in the two sums yields
k

which is precisely nk times the geometric mean of

(an + bn )(cn + d n )n−1 ≥ an cn(n−1) + bn d n(n−1)
the terms on the left. Thus AM-GM shows that the n−1  
n i n−i (n−1)i (n−1)(n−i)
terms under consideration on the left exceed those on +∑ ab c d ,
i=1 i
the right; adding these inequalities over all k yields the
desired result.
 2

proving the auxiliary inequality. Alternate derivation (due to Zuming Feng): We can also
Now given the auxiliary inequality and the n − 1 case minimize |c + 2/(c − 1)| without calculus (or worrying
of the desired inequality, we apply the auxiliary in- about boundary conditions). For c > 1, we have
1/n 1/n
equality with a = a1 , b = b1 , c = (a2 · · · an )1/n(n−1) , 2 √
d = (b2 . . . bn )1/n(n−1) . The right side will be the n-th 1 + (c − 1) + ≥ 1+2 2
c−1
power of the desired inequality. The left side comes out
to by AM-GM
√ on the last two terms, with equality for c −
1 = 2 (which is out of range). For c < 1, we similarly
(a1 + b1 )((a2 · · · an )1/(n−1) + (b2 · · · bn )1/(n−1) )n−1 , have
2 √
and by the induction hypothesis, the second factor is −1 + 1 − c + ≥ −1 + 2 2,
less than (a2 + b2 ) · · · (an + bn ). This yields the desired 1−c
result. √
here with equality for 1 − c = 2.
Note: Equality holds if and only if ai = bi = 0 for
some i or if the vectors (a1 , . . . , an ) and (b1 , . . . , bn ) Second solution: Write
are proportional. As pointed out by Naoki Sato, the 1 a+b
problem also appeared on the 1992 Irish Mathematical f (a, b) = a + b + + .
ab ab
Olympiad. It is also a special case of a classical in-
equality, known as Hölder’s inequality, which general- Then the problem is to minimize | f (a, b)| subject to
izes the Cauchy-Schwarz inequality (this is visible from the constraint a2 + b2 − 1 = 0. Since the constraint re-
the n = 2 case); the first solution above is adapted from gion has no boundary, it is enough to check the value
the standard proof of Hölder’s inequality. We don’t at each critical point and each potential discontinuity
know whether the declaration “Apply Hölder’s inequal- (i.e., where ab = 0) and select the smallest value (after
ity” by itself is considered an acceptable solution to this checking that f has no sign crossings).
problem. We locate the critical points using the Lagrange mul-
A3 First solution: Write tiplier condition: the gradient of f should be parallel
to that of the constraint, which is to say, to the vector
f (x) = sin x + cos x + tan x + cot x + sec x + csc x (a, b). Since
1 sin x + cos x ∂f 1 1
= sin x + cos x + + . = 1− 2 − 2
sin x cos x sin x cos x ∂a a b a
√
We can write sin x + cos x = 2 cos(π/4 − x); this sug- and similarly for b, the proportionality yields
gests making the substitution y = π/4 − x. In this new
coordinate, a2 b3 − a3 b2 + a3 − b3 + a2 − b2 = 0.
1 1 The irreducible factors of the left side are 1 + a, 1 + b,
sin x cos x = sin 2x = cos 2y,
2 2 a − b, and ab − a − b. So we must check what happens
√ when any of those factors, or a or b, vanishes.
and writing c = 2 cos y, we have
If 1 + a = 0, then b = 0, and the singularity of f be-
comes removable when restricted to the circle. Namely,
 
2
f (y) = (1 + c) 1 + 2 −1 we have
c −1
2 1 b+1
= c+ . f = a+b+ +
c−1 a ab
We√ must
√ analyze this function √ of c in √ the range and a2 +b2 −1 = 0 implies (1+b)/a = a/(1−b). Thus
[− 2, 2]. √ Its value at c√= − 2 is 2 − 3 2 < −2.24, we have f = −2; the same occurs when 1 + b = 0.
and at c = 2 is 2 + 3 2 > 6.24. Its derivative is √
If√a − b = 0, then a = b =
ñ 2/2 and either f = 2 +
1 − 2/(c − 1)2 , √which vanishes when (c − 1)2 √= 2, i.e.,
3 2 > 6.24, or f = 2 − 3 2 < −2.24.
where c = 1 ± 2. Only the value c =√ 1 − 2 is in
bounds, at which the value of f is 1 − 2 2 > −1.83. If a = 0, then either b = −1 as discussed above, or b =
As for the pole at c = 1, we observe that f decreases as 1. In the latter case, f blows up as one approaches this
c approaches from below (so takes negative values for point, so there cannot be a global minimum there.
all c < 1) and increases as c approaches from above (so Finally, if ab − a − b = 0, then
takes positive values for all c > 1); from the data col-
lected so far, we see that f has no sign crossings, so the a2 b2 = (a + b)2 = 2ab + 1
minimum of | f | is achieved at a critical
√ point of f . We
conclude that the minimum of | f | is 2 2 − 1.
 3
√
and so ab = 1 ± 2. √ The plus sign is impossible since A5 First solution: We represent a Dyck n-path by a
|ab| ≤ 1, so ab = 1 − 2 and sequence a1 · · · a2n , where each ai is either (1, 1) or
(1, −1).
1
f (a, b) = ab + +1 Given an (n − 1)-path P = a1 · · · a2n−2 , we distinguish
ab
√ two cases. If P has no returns of even-length, then let
= 1 − 2 2 > −1.83. f (P) denote the n-path (1, 1)(1, −1)P. Otherwise, let
ai ai+1 · · · a j denote the rightmost even-length return in
This yields the smallest value of | f | in the√list (and in- P, and let f (P) = (1, 1)a1 a2 · · · a j (1, −1)a j+1 · · · a2n−2 .
deed no sign crossings are possible), so 2 2 − 1 is the Then f clearly maps the set of Dyck (n − 1)-paths to the
desired minimum of | f |. set of Dyck n-paths having no even return.
Note: Instead of using the geometry of the graph of f We claim that f is bijective; to see this, we simply
to rule out sign crossings, one can verify explicitly that construct the inverse mapping. Given an n-path P, let
f cannot take the value 0. In the first solution, note that R = ai ai+1 ...a j denote the leftmost return in P, and let
c + 2/(c − 1) = 0 implies c2 − c + 2 = 0, which has no g(P) denote the path obtained by removing a1 and a j
real roots. In the second solution, we would have from P. Then evidently f ◦g and g◦ f are identity maps,
proving the claim.
a2 b + ab2 + a + b = −1.
Second solution: (by Dan Bernstein) Let Cn be the
Squaring both sides and simplifying yields number of Dyck paths of length n, let On be the number
of Dyck paths whose final return has odd length, and let
2a3 b3 + 5a2 b2 + 4ab = 0, Xn be the number of Dyck paths with no return of even
length.
whose only real root is ab = 0. But the cases with ab =
We first exhibit a recursion for On ; note that O0 = 0.
0 do not yield f = 0, as verified above.
Given a Dyck n-path whose final return has odd length,
A4 We split into three cases. Note first that |A| ≥ |a|, by split it just after its next-to-last return. For some k (pos-
applying the condition for large x. sibly zero), this yields a Dyck k-path, an upstep, a Dyck
(n − k − 1)-path whose odd return has even length, and
Case 1: B2 −4AC > 0. In this case Ax2 +Bx+C has two a downstep. Thus for n ≥ 1,
distinct real roots r1 and r2 . The condition implies that
ax2 + bx + c also vanishes at r1 and r2 , so b2 − 4ac > 0. n−1
Now On = ∑ Ck (Cn−k−1 − On−k−1 ).
k=0
B2 − 4AC = A2 (r1 − r2 )2
≥ a2 (r1 − r2 )2 We next exhibit a similar recursion for Xn ; note that
X0 = 1. Given a Dyck n-path with no even return, split-
= b2 − 4ac. ting as above yields for some k a Dyck k-path with no
even return, an upstep, a Dyck (n − k − 1)-path whose
Case 2: B2 − 4AC ≤ 0 and b2 − 4ac ≤ 0. Assume with- final return has even length, then a downstep. Thus for
out loss of generality that A ≥ a > 0, and that B = 0 (by n ≥ 1,
shifting x). Then Ax2 + Bx + C ≥ ax2 + bx + c ≥ 0 for n−1
all x; in particular, C ≥ c ≥ 0. Thus Xn = ∑ Xk (Cn−k−1 − On−k−1 ).
2 k=0
4AC − B = 4AC
≥ 4ac To conclude, we verify that Xn = Cn−1 for n ≥ 1, by
2
≥ 4ac − b . induction on n. This is clear for n = 1 since X1 = C0 = 1.
Given Xk = Ck−1 for k < n, we have
Alternate derivation (due to Robin Chapman): the el-
n−1
lipse Ax2 +Bxy+Cy2 = 1 is contained within the ellipse
ax2 + bxy + cy2 = 1, and their respective enclosed areas
Xn = ∑ Xk (Cn−k−1 − On−k−1 )
k=0
are π/(4AC − B2 ) and π/(4ac − b2 ). n−1
Case 3: B2 − 4AC ≤ 0 and b2 − 4ac > 0. Since Ax2 + = Cn−1 − On−1 + ∑ Ck−1 (Cn−k−1 − On−k−1 )
Bx +C has a graph not crossing the x-axis, so do (Ax2 + k=1
Bx +C) ± (ax2 + bx + c). Thus = Cn−1 − On−1 + On−1
= Cn−1 ,
(B − b)2 − 4(A − a)(C − c) ≤ 0,
(B + b)2 − 4(A + a)(C + c) ≤ 0 as desired.
Note: Since the problem only asked about the existence
and adding these together yields of a one-to-one correspondence, we believe that any

2(B2 − 4AC) + 2(b2 − 4ac) ≤ 0.

Hence b2 − 4ac ≤ 4AC − B2 , as desired.

 4

proof, bijective or not, that the two sets have the same g(x2 )). From the evident identities
cardinality is an acceptable solution. (Indeed, it would
be highly unusual to insist on using or not using a spe- 1
= f (x) + g(x)
cific proof technique!) The second solution above can 1−x
also be phrased in terms of generating functions. Also, f (x) = f (x2 ) + xg(x2 )
the Cn
are well-known to equal the Catalan numbers
1 2n g(x) = g(x2 ) + x f (x2 ),
n+1 n ; the problem at hand is part of a famous exer-
cise in Richard Stanley’s Enumerative Combinatorics, we have
Volume 1 giving 66 combinatorial interpretations of the
Catalan numbers. f (x) − g(x) = f (x2 ) − g(x2 ) + xg(x2 ) − x f (x2 )
A6 First solution: Yes, such a partition is possible. To = (1 − x)( f (x2 ) − g(x2 ))
achieve it, place each integer into A if it has an even f (x2 ) − g(x2 )
number of 1s in its binary representation, and into B if = .
f (x) + g(x)
it has an odd number. (One discovers this by simply
attempting to place the first few numbers by hand and
We deduce that f (x)2 − g(x)2 = f (x2 ) − g(x2 ), yielding
noticing the resulting pattern.)
the desired equality.
To show that rA (n) = rB (n), we exhibit a bijection be-
Note: This partition is actually unique, up to inter-
tween the pairs (a1 , a2 ) of distinct elements of A with
changing A and B. More precisely, the condition that
a1 + a2 = n and the pairs (b1 , b2 ) of distinct elements
0 ∈ A and rA (n) = rB (n) for n = 1, . . . , m uniquely deter-
of B with b1 + b2 = n. Namely, given a pair (a1 , a2 )
mines the positions of 0, . . . , m. We see this by induction
with a1 + a2 = n, write both numbers in binary and
on m: given the result for m − 1, switching the location
find the lowest-order place in which they differ (such
of m changes rA (m) by one and does not change rB (m),
a place exists because a1 6= a2 ). Change both numbers
so it is not possible for both positions to work. Robin
in that place and call the resulting numbers b1 , b2 . Then
Chapman points out this problem is solved in D.J. New-
a1 + a2 = b1 + b2 = n, but the parity of the number of
man’s Analytic Number Theory (Springer, 1998); in that
1s in b1 is opposite that of a1 , and likewise between b2
solution, one uses generating functions to find the par-
and a2 . This yields the desired bijection.
tition and establish its uniqueness, not just verify it.
Second solution: (by Micah Smukler) Write b(n) for
the number of 1s in the base 2 expansion of n, and B1 No, there do not.
f (n) = (−1)b(n) . Then the desired partition can be First solution: Suppose the contrary. By setting y =
described as A = f −1 (1) and B = f −1 (−1). Since −1, 0, 1 in succession, we see that the polynomials
f (2n) + f (2n + 1) = 0, we have 1 − x + x2 , 1, 1 + x + x2 are linear combinations of a(x)
( and b(x). But these three polynomials are linearly inde-
n
0 n odd pendent, so cannot all be written as linear combinations
∑ f (n) = f (n) n even. of two other polynomials, contradiction.
i=0
Alternate formulation: the given equation expresses a
If p, q are both in A, then f (p) + f (q) = 2; if p, q are diagonal matrix with 1, 1, 1 and zeroes on the diagonal,
both in B, then f (p) + f (q) = −2; if p, q are in different which has rank 3, as the sum of two matrices of rank 1.
sets, then f (p) + f (q) = 0. In other words, But the rank of a sum of matrices is at most the sum of
the ranks of the individual matrices.
2(rA (n) − rB (n)) = ∑ ( f (p) + f (q))
p+q=n,p<q Second solution: It is equivalent (by relabeling and
rescaling) to show that 1 + xy + x2 y2 cannot be writ-
and it suffices to show that the sum on the right is always ten as a(x)d(y) − b(x)c(y). Write a(x) = ∑ ai xi , b(x) =
zero. If n is odd, that sum is visibly ∑ni=0 f (i) = 0. If n ∑ bi xi , c(y) = ∑ c j y j , d(y) = ∑ d j y j . We now start com-
is even, the sum equals paring coefficients of 1 + xy + x2 y2 . By comparing co-
! efficients of 1 + xy + x2 y2 and a(x)d(y) − b(x)c(y), we
n
get
∑ f (i) − f (n/2) = f (n) − f (n/2) = 0.
i=0
1 = ai di − bi ci (i = 0, 1, 2)
This yields the desired result. 0 = ai d j − bi c j (i 6= j).
Third solution: (by Dan Bernstein) Put f (x) = ∑n∈A xn
The first equation says that ai and bi cannot both vanish,
and g(x) = ∑n∈B xn ; then the value of rA (n) (resp. rB (n))
and ci and di cannot both vanish. The second equation
is the coefficient of xn in f (x)2 − f (x2 ) (resp. g(x)2 −
says that ai /bi = c j /d j when i 6= j, where both sides
should be viewed in R ∪ {∞} (and neither is undeter-
mined if i, j ∈ {0, 1, 2}). But then

a0 /b0 = c1 /d1 = a2 /b2 = c0 /d0

 5

contradicting the equation a0 d0 − b0 c0 = 1. sides is the same. On the left side, it is well-known that
Third solution: We work over the complex numbers, the exponent of p in the prime factorization of n! is
in which we have a primitive cube root ω of 1. We also n  
n
use without further comment unique factorization for ∑ pi .
polynomials in two variables over a field. And we keep i=1
the relabeling of the second solution.
(To see this, note that the i-th term counts the multiples
Suppose the contrary. Since 1 + xy + x2 y2 = (1 − of pi among 1, . . . , n, so that a number divisible exactly
xy/ω)(1 − xy/ω 2 ), the rational function a(ω/y)d(y) − by pi gets counted exactly i times.) This number can
b(ω/y)c(y) must vanish identically (that is, coefficient be reinterpreted as the cardinality of the set S of points
by coefficient). If one of the polynomials, say a, van- in the plane with positive integer coordinates lying on
ished identically, then one of b or c would also, and the or under the curve y = np−x : namely, each summand is
desired inequality could not hold. So none of them van- the number of points of S with x = i.
ish identically, and we can write
On the right side, the exponent of p in the prime
c(y) a(ω/y) factorization of lcm(1, . . . , bn/ic) is blog p bn/icc =
= . blog p (n/i)c. However, this is precisely the number of
d(y) b(ω/y)
points of S with y = i. Thus
Likewise,
n n  
n
c(y) a(ω 2 /y) ∑ pblog bn/icc = ∑ pi ,
= . i=1 i=1
d(y) b(ω 2 /y)
and the desired result follows.
Put f (x) = a(x)/b(x); then we have f (ωx) = f (x)
Second solution: We prove the result by induction on
identically. That is, a(x)b(ωx) = b(x)a(ωx). Since a
n, the case n = 1 being obvious. What we actually show
and b have no common factor (otherwise 1 + xy + x2 y2
is that going from n − 1 to n changes both sides by the
would have a factor divisible only by x, which it doesn’t
same multiplicative factor, that is,
since it doesn’t vanish identically for any particular x),
a(x) divides a(ωx). Since they have the same degree, n−1
lcm{1, 2, . . . , bn/ic}
they are equal up to scalars. It follows that one of n=∏ .
a(x), xa(x), x2 a(x) is a polynomial in x3 alone, and like- i=1 lcm{1, 2, . . . , b(n − 1)/ic}
wise for b (with the same power of x).
Note that the i-th term in the product is equal to 1 if n/i
If xa(x) and xb(x), or x2 a(x) and x2 b(x), are polyno- is not an integer, i.e., if n/i is not a divisor of n. It is
mials in x3 , then a and b are divisible by x, but we also equal to 1 if n/i is a divisor of n but not a prime
know a and b have no common factor. Hence a(x) power, since any composite number divides the lcm of
and b(x) are polynomials in x3 . Likewise, c(y) and all smaller numbers. However, if n/i is a power of p,
d(y) are polynomials in y3 . But then 1 + xy + x2 y2 = then the i-th term is equal to p.
a(x)d(y) − b(x)c(y) is a polynomial in x3 and y3 , con-
tradiction. Since n/i runs over all proper divisors of n, the product
on the right side includes one factor of the prime p for
Note: The third solution only works over fields of char- each factor of p in the prime factorization of n. Thus
acteristic not equal to 3, whereas the other two work the whole product is indeed equal to n, completing the
over arbitrary fields. (In the first solution, one must re- induction.
place −1 by another value if working in characteristic
2.) B4 First solution: Put g = r1 + r2 , h = r3 + r4 , u = r1 r2 ,
v = r3 r4 . We are given that g is rational. The following
B2 It is easy to see by induction that the j-th entry are also rational:
of the k-th sequence
(where the original sequence is
k = 1) is ∑ki=1 k−1 k−1 (i + j − 1)), and so x = −b
i−1 /(2 n = g+h
1 n n−1
n−1
n

/i. Now i−1 /i = i /n; hence a
2n−1 ∑i=1 i−1 c
= gh + u + v
1 n  
n 2n − 1 a
xn = n−1 ∑ = n−1 < 2/n, −d
n2 i=1 i n2 = gv + hu
a
as desired. From the first line, h is rational. From the second line,
u + v is rational. From the third line, g(u + v) − (gv +
B3 First solution: It is enough to show that for each prime
hu) = (g − h)u is rational. Since g 6= h, u is rational, as
p, the exponent of p in the prime factorization of both
desired.
Second solution: This solution uses some basic Galois
theory. We may assume r1 6= r2 , since otherwise they
are both rational and so then is r1 r2 .
 6

Let τ be an automorphism of the field of algebraic num- The triangle 4A0 B0C0 has area four times that of 4ABC.
bers; then τ maps each ri to another one, and fixes We may dissect it into twelve triangles by first split-
the rational number r1 + r2 . If τ(r1 ) equals one of r1 ting it into three quadrilaterals PAC0 B, PBC0 A, PCA0 B,
or r2 , then τ(r2 ) must equal the other one, and vice then splitting each of these in four around the respec-
versa. Thus τ either fixes the set {r1 , r2 } or moves it tive interior points PB , PC , PA . Of the resulting twelve
to {r3 , r4 }. But if the latter happened, we would have triangles, three have side lengths a, b, c, while three are
r1 + r2 = r3 + r4 , contrary to hypothesis. Thus τ fixes equilateral triangles of respective sides lengths a, b, c.
the set {r1 , r2 } and in particular the number r1 r2 . Since The other six are isomorphic to two copies each of
this is true for any τ, r1 r2 must be rational. 4PAB, 4PBC, 4PCA, so their total area is twice that
Note: The conclusion fails if we allow r1 + r2 = r3 + r4 . of 4ABC.
For instance, take the polynomial √ x4 − 2 and label its It thus suffices to compute a2 + b2 + c2 in terms of the
2
roots so that√(x − r1 )(x − r2 ) = x − 2 and (x − r3 )(x − radius of the circle and the distance OP. This can be
r4 ) = x2 + 2. done readily in terms of OP using vectors, Cartesian
coordinates, or complex numbers as in the first solution.
B5 First solution: Place the unit circle on the complex
plane so that A, B,C correspond to the complex num- B6 First solution: (composite of solutions by Feng Xie
bers 1, ω, ω 2 , where ω = e2πi/3 , and let P correspond and David Pritchard) Let µ denote Lebesgue measure
to the complex number x. The distances a, b, c are then on [0, 1]. Define
|x − 1|, |x − ω|, |x − ω 2 |. Now the identity
E+ = {x ∈ [0, 1] : f (x) ≥ 0}
2 2
(x − 1) + ω(x − ω) + ω (x − ω ) = 0 E− = {x ∈ [0, 1] : f (x) < 0};

implies that there is a triangle whose sides, as vec- then E+ , E− are measurable and µ(E+ ) + µ(E− ) = 1.
tors, correspond to the complex numbers x − 1, ω(x − Write µ+ and µ− for µ(E+ ) and µ(E− ). Also define
ω), ω 2 (x − ω 2 ); this triangle has sides of length a, b, c. Z
To calculate the area of this triangle, we first note a more I+ = | f (x)| dx
E+
general formula. If a triangle in the plane has vertices Z
at 0, v1 = s1 + it1 , v2 = s2 + it2 , then it is well known I− = | f (x)| dx,
that the area of the triangle is |s1t2 − s2t1 |/2 = |v1 v2 − E−
v2 v1 |/4. In our case, we have v1 = x − 1 and v2 = ω(x − R1
ω); then so that 0 | f (x)| dx = I+ + I− .
√ From the triangle inequality |a + b| ≥ ±(|a| − |b|), we
v1 v2 − v2 v1 = (ω 2 − ω)(xx − 1) = i 3(|x|2 − 1). have the inequality
√
Hence the area of the triangle is 3(1 − |x|2 )/4, which
ZZ
| f (x) + f (y)| dx dy
depends only on the distance |x| from P to O. E+ ×E−
Second solution: (by Florian Herzig) Let A0 , B0 , C0
ZZ
≥± (| f (x)| − | f (y)|) dx dy
be the points obtained by intersecting the lines AP, E+ ×E−
BP, CP with the unit circle. Let d denote OP. Then = ±(µ− I+ − µ+ I− ),
A0 P = (1 − d 2 )/a, etc., by using the power of the point
P. As triangles A0 B0 P and √ BAP are similar, we get and likewise with + and − switched. Adding these in-
that A0 B0 = AB · A0 P/b = 3(1 − d 2 )/(ab). It follows equalities together and allowing all possible choices of
that triangle A0√B0C0 has sides proportional to a, b, c, the signs, we get
by a factor of 3(1 − d 2 )/(abc). In particular, there ZZ
is a triangle with√sides a, b, c, and it has circumra- | f (x) + f (y)| dx dy
dius R = (abc)/( 3(1 − d 2 )). Its area is abc/(4R) =
√ (E+ ×E− )∪(E− ×E+ )
3(1 − d 2 )/4. ≥ max {0, 2(µ− I+ − µ+ I− ), 2(µ+ I− − µ− I+ )} .
Third solution: (by Samuel Li) Consider the rotation
by the angle π/3 around A carrying B to C, and let PA To this inequality, we add the equalities
be the image of P; define PB , PC similarly. Let A0 be the ZZ
intersection of the tangents to the circle at B,C; define | f (x) + f (y)| dx dy = 2µ+ I+
E+ ×E+
B0 ,C0 , similarly. Put ` = AB = BC = CA; we then have ZZ
| f (x) + f (y)| dx dy = 2µ− I−
AB0 = AC0 = BC0 = BA0 = CA0 = CB0 = ` E− ×E−
Z 1
PA = PPA = PA A = PBC0 = PC A0 = a − | f (x)| dx = −(µ+ + µ− )(I+ + I− )
PB = PPB = PB B = PC A0 = PAC0 = b 0

PC = PPC = PCC = PA B0 = PB A0 = c.
 7

to obtain tending to zero converges to the integral. (The domain

Z 1Z 1 Z 1
is compact, so the function is uniformly continuous.
Hence for any ε > 0 there is a cutoff below which any
| f (x) + f (y)| dx dy − | f (x)| dx
0 0 0 mesh size forces the discrepancy between the Riemann
≥ max{(µ+ − µ− )(I+ + I− ) + 2µ− (I− − I+ ), sum and the integral to be less than ε.)
(µ+ − µ− )(I+ − I− ), Alternate derivation (based on a solution by Dan Bern-
(µ− − µ+ )(I+ + I− ) + 2µ+ (I+ − I− )}. stein): from the discrete analogue, we have

Now simply note that for each of the possible compar- n−2 n
∑ | f (xi ) + f (x j )| ≥ ∑ | f (xi )|,
isons between µ+ and µ− , and between I+ and I− , one 1≤i< j≤n 2 i=1
of the three terms above is manifestly nonnegative. This
yields the desired result. for all x1 , . . . , xn ∈ [0, 1]. Integrating both sides as
(x1 , . . . , xn ) runs over [0, 1]n yields
Second solution: We will show at the end that it is
enough to prove a discrete analogue: if x1 , . . . , xn are n(n − 1) 1 1
Z Z
real numbers, then | f (x) + f (y)| dy dx
2 0 0
1 n 1 n
Z 1
n(n − 2)
∑ |x i + x j | ≥ ∑ |xi |. ≥ | f (x)| dx,
n2 i, j=1 n i=1 2 0

or
In the meantime, we concentrate on this assertion.
Z 1Z 1 Z 1
Let f (x1 , . . . , xn ) denote the difference between the two n−2
| f (x) + f (y)| dy dx ≥ | f (x)| dx.
sides. We induct on the number of nonzero values of 0 0 n−1 0
|xi |. We leave for later the base case, where there is at Taking the limit as n → ∞ now yields the desired result.
most one such value. Suppose instead for now that there
are two or more. Let s be the smallest, and suppose Third solution: (by David Savitt) We give an argument
without loss of generality that x1 = · · · = xa = s, xa+1 = which yields the following improved result. Let µ p and
· · · = xa+b = −s, and for i > a + b, either xi = 0 or |xi | > µn be the measure of the sets {x : f (x) > 0} and {x :
s. (One of a, b might be zero.) f (x) < 0} respectively, and let µ ≤ 1/2 be min(µ p , µn ).
Then
Now consider
Z 1Z 1
a terms b terms | f (x) + f (y)| dx dy
z }| { z }| { 0 0
f ( t, · · · ,t , −t, · · · , −t, xa+b+1 , · · · , xn ) Z 1
≥ (1 + (1 − 2µ)2 ) | f (x)| dx.
as a function of t. It is piecewise linear near s; in fact, 0
it is linear between 0 and the smallest nonzero value Note that the constant can be seen to be best possible
among |xa+b+1 |, . . . , |xn | (which exists by hypothesis). by considering a sequence of functions tending towards
Thus its minimum is achieved by one (or both) of those the step function which is 1 on [0, µ] and −1 on (µ, 1].
two endpoints. In other words, we can reduce the num-
ber of distinct nonzero absolute values among the xi Suppose without loss of generality that µ = µ p . As in
without increasing f . This yields the induction, pending the second solution, it suffices to prove a strengthened
verification of the base case. discrete analogue, namely
As for the base case, suppose that x1 = · · · = xa = s > 0,  ! !
2p 2 1 n

1
xa+1 = · · · = xa+b = −s, and xa+b+1 = · · · = xn = 0. |ai + a j | ≥ 1 + 1 − ∑ |ai | ,
(Here one or even both of a, b could be zero, though the n2 ∑
i, j n n i=1
latter case is trivial.) Then
where p ≤ n/2 is the number of a1 , . . . , an which are
s positive. (We need only make sure to choose meshes so
f (x1 , . . . , xn ) = 2 (2a2 + 2b2 + (a + b)(n − a − b)) that p/n → µ as n → ∞.) An equivalent inequality is
n
s s
− (a + b) = 2 (a2 − 2ab + b2 ) ≥ 0. 
2p2 n

n n
∑ |ai + a j | ≥ n − 1 − 2p + n ∑ |ai |.
This proves the base case of the induction, completing 1≤i< j≤n i=1
the solution of the discrete analogue.
Write ri = |ai |, and assume without loss of generality
To deduce the original statement from the discrete ana-
that ri ≥ ri+1 for each i. Then for i < j, |ai +a j | = ri +r j
logue, approximate both integrals by equally-spaced
if ai and a j have the same sign, and is ri −r j if they have
Riemann sums and take limits. This works because
opposite signs. The left-hand side is therefore equal to
given a continuous function on a product of closed in-
tervals, any sequence of Riemann sums with mesh size n n
∑ (n − i)ri + ∑ r jC j ,
i=1 j=1
 8

where It follows that

n n
C j = #{i < j : sgn(ai ) = sgn(a j )} ∑ |ai + a j | = ∑ (n − i)ri + ∑ r jC j
− #{i < j : sgn(ai ) 6= sgn(a j )}. 1≤i< j≤n i=1 j=1
2p
≥ ∑ (n − i − [i even])ri
Consider the partial sum Pk = ∑kj=1 C j . If exactly pk of i=1
a1 , . . . , ak are positive, then this sum is equal to n
          + ∑ (n − 1 − 2p)ri
pk k − pk k pk k − pk i=2p+1
+ − − − ,
2 2 2 2 2 n
= (n − 1 − 2p) ∑ ri
which expands and simplifies to i=1
  2p
k + ∑ (2p + 1 − i − [i even])ri
−2pk (k − pk ) + . i=1
2
n 2p
For k ≤ 2p even, this partial sum would be minimized ≥ (n − 1 − 2p) ∑ ri + p ∑ ri
with pk = 2k , and would then equal − 2k ; for k < 2p odd, i=1 i=1
n
this partial sum would be minimized with pk = k±1 2p n
2 , and ≥ (n − 1 − 2p) ∑ ri + p ∑ ri ,
would then equal − k−1
2 . Either way, Pk ≥ −b k
2 c. On the i=1 n i=1
other hand, if k > 2p, then
    as desired. The next-to-last and last inequalities each
k k follow from the monotonicity of the ri ’s, the former by
−2pk (k − pk ) + ≥ −2p(k − p) + pairing the ith term with the (2p + 1 − i)th .
2 2
Note: Compare the closely related Problem 6 from the
since pk is at most p. Define Qk to be −b 2k c if k ≤ 2p 2000 USA Mathematical Olympiad: prove that for any
and −2p(k − p) + 2k if k ≥ 2p, so that Pk ≥ Qk . Note


nonnegative real numbers a1 , . . . , an , b1 , . . . , bn , one has
that Q1 = 0.
n n
Partial summation gives ∑ min{ai a j , bi b j } ≤ ∑ min{ai b j , a j bi }.
i, j=1 i, j=1
n n
∑ r jC j = rn Pn + ∑ (r j−1 − r j )Pj−1
j=1 j=2
n
≥ rn Qn + ∑ (r j−1 − r j )Q j−1
j=2
n
= ∑ r j (Q j − Q j−1 )
j=2
n
= −r2 − r4 − · · · − r2p + ∑ ( j − 1 − 2p)r j .
j=2p+1
 The 65th William Lowell Putnam Mathematical Competition
Saturday, December 4, 2004

A1 Basketball star Shanille O’Keal’s team statistician B1 Let P(x) = cn xn + cn−1 xn−1 + · · · + c0 be a polynomial
keeps track of the number, S(N), of successful free with integer coefficients. Suppose that r is a rational
throws she has made in her first N attempts of the sea- number such that P(r) = 0. Show that the n numbers
son. Early in the season, S(N) was less than 80% of N,
but by the end of the season, S(N) was more than 80% cn r, cn r2 + cn−1 r, cn r3 + cn−1 r2 + cn−2 r,
of N. Was there necessarily a moment in between when . . . , cn rn + cn−1 rn−1 + · · · + c1 r
S(N) was exactly 80% of N?
A2 For i = 1, 2 let Ti be a triangle with side lengths ai , bi , ci , are integers.
and area Ai . Suppose that a1 ≤ a2 , b1 ≤ b2 , c1 ≤ c2 , and B2 Let m and n be positive integers. Show that
that T2 is an acute triangle. Does it follow that A1 ≤ A2 ? (m + n)! m! n!
m+n
< m n.
A3 Define a sequence {un }∞
by u0 = u1 = u2 = 1, and
n=0
(m + n) m n
thereafter by the condition that
  B3 Determine all real numbers a > 0 for which there exists
un un+1 a nonnegative continuous function f (x) defined on [0, a]
det = n!
un+2 un+3 with the property that the region

for all n ≥ 0. Show that un is an integer for all n. (By R = {(x, y); 0 ≤ x ≤ a, 0 ≤ y ≤ f (x)}
convention, 0! = 1.)
has perimeter k units and area k square units for some
A4 Show that for any positive integer n there is an integer N real number k.
such that the product x1 x2 · · · xn can be expressed iden-
tically in the form B4 Let n be a positive integer, n ≥ 2, and put θ = 2π/n. De-
fine points Pk = (k, 0) in the xy-plane, for k = 1, 2, . . . , n.
N Let Rk be the map that rotates the plane counterclock-
x1 x2 · · · xn = ∑ ci (ai1 x1 + ai2 x2 + · · · + ain xn )n wise by the angle θ about the point Pk . Let R denote
i=1
the map obtained by applying, in order, R1 , then R2 , . . . ,
where the ci are rational numbers and each ai j is one of then Rn . For an arbitrary point (x, y), find, and simplify,
the numbers −1, 0, 1. the coordinates of R(x, y).

A5 An m × n checkerboard is colored randomly: each B5 Evaluate

square is independently assigned red or black with xn
1 + xn+1
∞ 
probability 1/2. We say that two squares, p and q, are lim .
in the same connected monochromatic region if there is x→1− n=0
∏ 1 + xn
a sequence of squares, all of the same color, starting at
p and ending at q, in which successive squares in the B6 Let A be a non-empty set of positive integers, and let
sequence share a common side. Show that the expected N(x) denote the number of elements of A not exceed-
number of connected monochromatic regions is greater ing x. Let B denote the set of positive integers b that
than mn/8. can be written in the form b = a − a0 with a ∈ A and
a0 ∈ A . Let b1 < b2 < · · · be the members of B, listed
A6 Suppose that f (x, y) is a continuous real-valued func-
in increasing order. Show that if the sequence bi+1 − bi
tion on the unit square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. Show that
is unbounded, then
Z 1 Z 1 2 Z 1 Z 1 2
f (x, y)dx dy + f (x, y)dy dx lim N(x)/x = 0.
x→∞
0 0 0 0
Z 1 Z 1
2 Z 1Z 1
≤ f (x, y)dx dy + [ f (x, y)]2 dx dy.
0 0 0 0
 Solutions to the 65th William Lowell Putnam Mathematical Competition
Saturday, December 4, 2004
Kiran Kedlaya and Lenny Ng

A–1 Yes. Suppose otherwise. Then there would be an N that ∠P1 ≤ ∠P2 . Since the latter is acute (because T2 is
such that S(N) < .8N and S(N + 1) > .8(N + 1); that is, acute), we have sin ∠P1 ≤ sin ∠P2 . By the Law of Sines,
O’Keal’s free throw percentage is under 80% at some
point, and after one subsequent free throw (necessarily 1 1
A1 = b1 c1 sin ∠P1 ≤ b2 c2 sin ∠P2 = A2 .
made), her percentage is over 80%. If she makes m 2 2
of her first N free throws, then m/N < 4/5 and (m +
1)/(N + 1) > 4/5. This means that 5m < 4n < 5m + 1, Remark: Many other solutions are possible; for in-
which is impossible since then 4n is an integer between stance, one uses Heron’s formula for the area of a tri-
the consecutive integers 5m and 5m + 1. angle in terms of its side lengths.
Remark: This same argument works for any fraction A–3 Define a sequence vn by vn = (n − 1)(n − 3) · · · (4)(2) if
of the form (n − 1)/n for some integer n > 1, but not n is odd and vn = (n − 1)(n − 3) · · · (3)(1) if n is even;
for any other real number between 0 and 1. it suffices to prove that un = vn for all n ≥ 2. Now
vn+3 vn = (n + 2)(n)(n − 1)! and vn+2 vn+1 = (n + 1)!,
A–2 First solution: (partly due to Ravi Vakil) Yes, it does
and so vn+3 vn − vn+2 vn+1 = n!. Since we can check that
follow. For i = 1, 2, let Pi , Qi , Ri be the vertices of Ti
un = vn for n = 2, 3, 4, and un and vn satisfy the same
opposide the sides of length ai , bi , ci , respectively.
recurrence, it follows by induction that un = vn for all
We first check the case where a1 = a2 (or b1 = b2 or n ≥ 2, as desired.
c1 = c2 , by the same argument after relabeling). Imag-
ine T2 as being drawn with the base Q2 R2 horizontal A–4 It suffices to verify that
and the point P2 above the line Q2 R2 . We may then po-
1
sition T1 so that Q1 = Q2 , R1 = R2 , and P1 lies above x1 · · · xn = ∑ (e1 · · · en )(e1 x1 + · · · + en xn )n .
2n n! e ∈{−1,1}
the line Q1 R1 = Q2 R2 . Then P1 also lies inside the re- i
gion bounded by the circles through P2 centered at Q2
and R2 . Since ∠Q2 and ∠R2 are acute, the part of this To check this, first note that the right side vanishes iden-
region above the line Q2 R2 lies within T2 . In particu- tically for x1 = 0, because each term cancels the corre-
lar, the distance from P1 to the line Q2 R2 is less than or sponding term with e1 flipped. Hence the right side, as
equal to the distance from P2 to the line Q2 R2 ; hence a polynomial, is divisible by x1 ; similarly it is divisi-
A1 ≤ A2 . ble by x2 , . . . , xn . Thus the right side is equal to x1 · · · xn
times a scalar. (Another way to see this: the right side is
To deduce the general case, put
clearly odd as a polynomial in each individual variable,
r = max{a1 /a2 , b1 /b2 , c1 /c2 }. but the only degree n monomial in x1 , . . . , xn with that
property is x1 · · · xn .) Since each summand contributes
1
Let T3 be the triangle with sides ra2 , rb2 , rc2 , which has 2n x1 · · · xn to the sum, the scalar factor is 1 and we are
area r2 A2 . Applying the special case to T1 and T3 , we done.
deduce that A1 ≤ r2 A2 ; since r ≤ 1 by hypothesis, we Remark: Several variants on the above construction are
have A1 ≤ A2 as desired. possible; for instance,
Remark: Another geometric argument in the case a1 =
1
a2 is that since angles ∠Q2 and ∠R2 are acute, the per- x1 · · · xn = ∑ (−1)n−e1 −···−en (e1 x1 + · · · + en xn )n
n! e ∈{0,1}
pendicular to Q2 R2 through P2 separates Q2 from R2 . i
If A1 > A2 , then P1 lies above the parallel to Q2 R2
through P2 ; if then it lies on or to the left of the vertical by the same argument as above.
line through P2 , we have c1 > c2 because the inequality Remark: These construction work over any field of
holds for both horizontal and vertical components (pos- characteristic greater than n (at least for n > 1). On
sibly with equality for one, but not both). Similarly, if the other hand, no construction is possible over a field
P1 lies to the right of the vertical, then b1 > b2 . of characteristic p ≤ n, since the coefficient of x1 · · · xn
Second solution: (attribution unknown) Retain nota- in the expansion of (e1 x1 + · · · + en xn )n is zero for any
tion as in the first paragraph of the first solution. Since ei .
the angle measures in any triangle add up to π, some Remark: Richard Stanley asks whether one can use
angle of T1 must have measure less than or equal to its fewer than 2n terms, and what the smallest possible
counterpart in T2 . Without loss of generality assume number is.
 2

A–5 First solution: First recall that any graph with n ver- – With probability 1/4, the i-th square is opposite in
tices and e edges has at least n − e connected com- color from the adjacent squares directly above and
ponents (add each edge one at a time, and note that to the left of it. In this case adding the i-th square
it reduces the number of components by at most 1). adds one component.
Now imagine the squares of the checkerboard as a – With probability 1/8, the i-th square is the same
graph, whose vertices are connected if the correspond- in color as the adjacent squares directly above and
ing squares share a side and are the same color. Let A to the left of it, but opposite in color from its diag-
be the number of edges in the graph, and let B be the onal neighbor above and to the left. In this case,
number of 4-cycles (formed by monochromatic 2 × 2 adding the i-th square either removes a component
squares). If we remove the bottom edge of each 4-cycle, or leaves the number unchanged.
the resulting graph has the same number of connected
components as the original one; hence this number is at – In all other cases, the number of components re-
least mains unchanged upon adding the i-th square.
Hence adding the i-th square increases the expected
mn − A + B.
number of components by 1/4 − 1/8 = 1/8.
By the linearity of expectation, the expected number of If the i-th square does abut the left edge of the board,
connected components is at least the situation is even simpler: if the i-th square differs in
color from the square above it, one component is added,
mn − E(A) + E(B). otherwise the number does not change. Hence adding
the i-th square increases the expected number of com-
Moreover, we may compute E(A) by summing over ponents by 1/2; likewise if the i-th square abuts the top
the individual pairs of adjacent squares, and we may edge of the board. Thus the expected number of com-
compute E(B) by summing over the individual 2 × 2 ponents is at least i/8 by induction, as desired.
squares. Thus
Remark: Some solvers attempted to consider adding
1 one row at a time, rather than one square; this must be
E(A) = (m(n − 1) + (m − 1)n), handled with great care, as it is possible that the num-
2
1 ber of components can drop rather precipitously upon
E(B) = (m − 1)(n − 1), adding an entire row.
8
and so the expected number of components is at least A–6 By approximating each integral with a Riemann sum,
we may reduce to proving the discrete analogue: for
1 1 xi j ∈ R for i, j = 1, . . . , n,
mn − (m(n − 1) + (m − 1)n) + (m − 1)(n − 1)
2 8
!2 !2
mn + 3m + 3n + 1 mn n n n n
= > .
8 8 n∑ ∑ xi j +n ∑ ∑ xi j
i=1 j=1 j=1 i=1
Remark: A “dual” approach is to consider the graph n n
!2
n n
whose vertices are the corners of the squares of the ≤ ∑ ∑ xi j + n2 ∑ ∑ xi2j .
checkerboard, with two vertices joined if they are ad- i=1 j=1 i=1 j=1
jacent and the edge between then does not separate two
squares of the same color. In this approach, the 4-cycles The difference between the right side and the left side
become isolated vertices, and the bound on components is
is replaced by a call to Euler’s formula relating the ver-
1 n
tices, edges and faces of a planar figure. (One must be ∑ (xi j + xkl − xil − xk j )2 ,
4 i, j,k,l=1
careful, however, to correctly handle faces which are
not simply connected.)
which is evidently nonnegative. If you prefer not to dis-
Second solution: (by Noam Elkies) Number the cretize, you may rewrite the original inequality as
squares of the checkerboard 1, . . . , mn by numbering the
first row from left to right, then the second row, and so Z 1Z 1Z 1Z 1
on. We prove by induction on i that if we just consider F(x, y, z, w)2 dx dy dz dw ≥ 0
0 0 0 0
the figure formed by the first i squares, its expected
number of monochromatic components is at least i/8. for
For i = 1, this is clear.
F(x, y, z, w) = f (x, y) + f (z, w) − f (x, w) − f (z, y).
Suppose the i-th square does not abut the left edge or
the top row of the board. Then we may divide into three
cases. Remark: (by Po-Ning Chen) The discrete inequality
can be arrived at more systematically by repeatedly ap-
plying the following identity: for any real a1 , . . . , an ,
!2
n n
∑ (xi − x j )2 = n ∑ xi2 − ∑ xi .
1≤i< j≤n i=1 i=1
 3

Remark: (by David Savitt) The discrete inequality can and rewrite the desired inequality as
also be interpreted as follows. For c, d ∈ {1, . . . , n − 1}
and ζn = e2πi/n , put ∏ x ∏ y> ∏ z.
x∈Sm y∈Sn z∈Sm+n
zc,d = ∑ ζnci+d j xi j .
i, j To prove this, it suffices to check that if we sort the
multiplicands on both sides into increasing order, the i-
Then the given inequality is equivalent to th term on the left side is greater than or equal to the i-th
term on the right side. (The equality is strict already for
n−1 i = 1, so you do get a strict inequality above.)
∑ |zc,d |2 ≥ 0.
c,d=1 Another way to say this is that for any i, the number of
factors on the left side which are less than i/(m + n) is
less than i. But since j/m < i/(m + n) is equivalent to
B–1 Let k be an integer, 0 ≤ k ≤ n − 1. Since P(r)/rk = 0,
j < im/(m + n), that number is
we have
   
im in
cn rn−k + cn−1 rn−k+1 + · · · + ck+1 r −1+ −1
m+n m+n
= −(ck + ck−1 r−1 + · · · + c0 r−k ). im in
≤ + − 1 = i − 1.
m+n m+n
Write r = p/q where p and q are relatively prime. Then
the left hand side of the above equation can be written as Third solution: Put f (x) = x(log(x + 1) − log x); then
a fraction with denominator qn−k , while the right hand for x > 0,
side is a fraction with denominator pk . Since p and q
are relatively prime, both sides of the equation must be 1
an integer, and the result follows. f 0 (x) = log(1 + 1/x) −
x+1
Remark: If we write r = a/b in lowest terms, then 1
f 00 (x) = − .
P(x) factors as (bx − a)Q(x), where the polynomial Q x(x + 1)2
has integer coefficients because you can either do the
long division from the left and get denominators divis- Hence f 00 (x) < 0 for all x; since f 0 (x) → 0 as x → ∞,
ible only by primes dividing b, or do it from the right we have f 0 (x) > 0 for x > 0, so f is strictly increasing.
and get denominators divisible only by primes dividing Put g(m) = m log m − log(m!); then g(m + 1) − g(m) =
a. The numbers given in the problem are none other f (m), so g(m + 1) − g(m) increases with m. By induc-
than a times the coefficients of Q. More generally, if tion, g(m + n) − g(m) increases with n for any positive
P(x) is divisible, as a polynomial over the rationals, by integer n, so in particular
a polynomial R(x) with integer coefficients, then P/R
also has integer coefficients; this is known as “Gauss’s g(m + n) − g(m) > g(n) − g(1) + f (m)
lemma” and holds in any unique factorization domain. ≥ g(n)
B–2 First solution: We have
  since g(1) = 0. Exponentiating yields the desired in-
m+n m+n m n equality.
(m + n) > m n
m Fourth solution: (by W.G. Boskoff and Bogdan
Suceavă) We prove the claim by induction on m + n.
because the binomial expansion of (m + n)m+n includes The base case is m = n = 1, in which case the de-
the term on the right as well as some others. Rearrang- sired inequality is obviously true: 2!/22 = 1/2 < 1 =
ing this inequality yields the claim. (1!/11 )(1!/11 ). To prove the induction step, suppose
Remark: One can also interpret this argument combi- m + n > 2; we must then have m > 1 or n > 1 or both.
natorially. Suppose that we choose m + n times (with Because the desired result is symmetric in m and n, we
replacement) uniformly randomly from a set of m + n may as well assume n > 1. By the induction hypothesis,
balls, of which m are red and n are blue. Then the proba- we have
bility of picking each ball exactly once is (m+n)!/(m+
(m + n − 1)! m! (n − 1)!
n)m+n . On the other hand, if p is the probability of pick- < m .
ing exactly m red balls, then p < 1 and the probability (m + n − 1)m+n−1 m (n − 1)n−1
of picking each ball exactly once is p(mm /m!)(nn /n!).
To obtain the desired inequality, it will suffice to check
Second solution: (by David Savitt) Define that

Sk = {i/k : i = 1, . . . , k} (m + n − 1)m+n−1 (m + n)! (n − 1)n−1 n!

m+n
<
(m + n − 1)! (m + n) (n − 1)! (n)n
 4

or in other words z with slope ζ n = 1. It thus equals z+T for some T ∈ C.

m+n−1  Since f1 (1) = 1, we can write 1 + T = Rn ◦ · · · ◦ R2 (1).
1 n−1
 
1 However, we also have
1− < 1− .
m+n n
Rn ◦ · · · ◦ R2 (1) = Rn−1 ◦ R1 (0) + 1
To show this, we check that the function f (x) = (1 −
1/x)x−1 is strictly decreasing for x > 1; while this can by the symmetry in how the Ri are defined. Hence
be achieved using the weighted arithmetic-geometric
mean inequality, we give a simple calculus proof in- Rn (1 + T ) = Rn ◦ R1 (0) + Rn (1) = T + Rn (1);
stead. The derivative of log f (x) is log(1 − 1/x) + 1/x,
so it is enough to check that this is negative for x > 1. that is, Rn (T ) = T . Hence T = n, as desired.
An equivalent statement is that log(1 − x) + x < 0 for B–5 First solution: By taking logarithms, we
0 < x < 1; this in turn holds because the function g(x) = see that the desired limit is exp(L), where
log(1 − x) + x tends to 0 as x → 0+ and has derivative = limx→1− ∑∞ n n+1 ) − ln(1 + xn ) .


1
L n=0 x ln(1 + x
1 − 1−x < 0 for 0 < x < 1. Now
B–3 The answer is {a | a > 2}. If a > 2, then the func- N
∑ xn ln(1 + xn+1 ) − ln(1 + xn )


tion f (x) = 2a/(a − 2) has the desired property; both
perimeter and area of R in this case are 2a2 /(a − 2). n=0
Now suppose that a ≤ 2, and let f (x) be a nonnega- N N
tive continuous function on [0, a]. Let P = (x0 , y0 ) be a = 1/x ∑ xn+1 ln(1 + xn+1 ) − ∑ xn ln(1 + xn )
point on the graph of f (x) with maximal y-coordinate; n=0 n=0
N
then the area of R is at most ay0 since it lies below the
line y = y0 . On the other hand, the points (0, 0), (a, 0), = xN ln(1 + xN+1 ) − ln 2 + (1/x − 1) ∑ xn ln(1 + xn );
n=1
and P divide the boundary of R into three sections. The
length of the section between (0, 0) and P is at least the since limN→∞ (xN ln(1 + xN+1 )) = 0 for 0 < x < 1, we
distance between (0, 0) and P, which is at least y0 ; the conclude that L = − ln 2 + limx→1− f (x), where
length of the section between P and (a, 0) is similarly
at least y0 ; and the length of the section between (0, 0) ∞
and (a, 0) is a. Since a ≤ 2, we have 2y0 + a > ay0 and f (x) = (1/x − 1) ∑ xn ln(1 + xn )
n=1
hence the perimeter of R is strictly greater than the area ∞ ∞
of R. = (1/x − 1) ∑ ∑ (−1)m+1 xn+mn /m.
n=1 m=1
B–4 First solution: Identify the xy-plane with the complex
plane C, so that Pk is the real number k. If z is sent to This final double sum converges absolutely when 0 <
z0 by a counterclockwise rotation by θ about Pk , then x < 1, since
z0 − k = eiθ (z − k); hence the rotation Rk sends z to ζ z +
∞ ∞ ∞
k(1 − ζ ), where ζ = e2πi/n . It follows that R1 followed
by R2 sends z to ζ (ζ z + (1 − ζ )) + 2(1 − ζ ) = ζ 2 z + ∑ ∑ xn+mn /m = ∑ xn (− ln(1 − xn ))
n=1 m=1 n=1
(1 − ζ )(ζ + 2), and so forth; an easy induction shows ∞
that R sends z to < ∑ xn (− ln(1 − x)),
n=1
ζ n z + (1 − ζ )(ζ n−1 + 2ζ n−2 + · · · + (n − 1)ζ + n).
which converges. (Note that − ln(1 − x) and − ln(1 −
Expanding the product (1 − ζ )(ζ n−1 + 2ζ n−2 + · · · + xn ) are positive.) Hence we may interchange the sum-
(n−1)ζ +n) yields −ζ n −ζ n−1 −· · ·−ζ +n = n. Thus mations in f (x) to obtain
R sends z to z + n; in cartesian coordinates, R(x, y) =
(x + n, y).
∞
(−1)m+1 x(m+1)n
∞
f (x) = (1/x − 1) ∑∑
Second solution: (by Andy Lutomirski, via Ravi Vakil) m=1 n=1 m
Imagine a regular n-gon of side length 1 placed with ∞
(−1) m+1
 m
x (1 − x)

its top edge on the x-axis and the left endpoint of that = (1/x − 1) ∑ .
m=1 m 1 − xm+1
edge at the origin. Then the rotations correspond to
rolling this n-gon along the x-axis; after the n rotations, This last sum converges absolutely uniformly in x, so
it clearly ends up in its original rotation and translated it is legitimate to take limits term by term. Since
n units to the right. Hence the whole plane must do so
as well.
Third solution: (attribution unknown) Viewing each
Rk as a function of a complex number z as in the first
solution, the function Rn ◦ Rn−1 ◦ · · · ◦ R1 (z) is linear in
 5

xm 1−x 1
limx→1− 1−xm+1
= m+1 for fixed m, we have where each ei runs over {0, 1}, contains at most one
element of A ; consequently, lim sup N(x)/x ≤ 1/2n .
(−1)m+1
∞
We now produce such bi recursively, starting with b0 =
lim f (x) = ∑
x→1− m=1 m(m + 1) 1 (and both (a) and (b) holding vacuously). Given
∞ 
1 1
 b0 , . . . , bn satisfying (a) and (b), note that b0 + · · · +
= ∑ (−1)m+1 − bn−1 < bn by induction on n. By the hypotheses of the
m=1 m m+1 problem, we can find a set Sn of 6bn consecutive in-
tegers, none of which belongs to B. Let bn+1 be the
!
∞
(−1)m+1
=2 ∑ −1 second-smallest multiple of 2bn in Sn ; then bn+1 + x ∈
m=1 m
Sn for −2bn ≤ x ≤ 0 clearly, and also for 0 ≤ x ≤ 2bn
= 2 ln 2 − 1, because there are most 4bn − 1 elements of Sn preced-
ing bn+1 . In particular, the analogue of (b) with n re-
and hence L = ln 2 − 1 and the desired limit is 2/e. placed by n + 1 holds for en+1 6= 0; of course it holds for
Remark: Note that the last series is not absolutely con- en+1 = 0 because (b) was already known. Since the ana-
vergent, so the recombination must be done without re- logue of (a) holds by construction, we have completed
arranging terms. this step of the construction and the recursion may con-
tinue.
Second solution: (by Greg Price, via Tony Zhang and
Anders Kaseorg) Put tn (x) = ln(1 + xn ); we can then Since we can construct b0 , . . . , bn satisfying (a) and (b)
write xn = exp(tn (x)) − 1, and for any n, we have lim sup N(x)/x ≤ 1/2n for any n,
yielding lim N(x)/x = 0 as desired.
∞
Second solution: (by Paul Pollack) Let S be the set
L = lim ∑ (tn (x) − tn+1 (x))(1 − exp(tn (x))).
x→1− n=0 of possible values of lim sup N(x)/x; since S ⊆ [0, 1]
is bounded, it has a least upper bound L. Suppose by
The expression on the right is a Riemann sum approxi- way of contradiction that L > 0; we can then choose
mating the integral 0ln 2 (1 − et ) dt, over the subdivision A , B satisfying the conditions of the problem such that
R

of [0, ln(2)) given by the tn (x). As x → 1− , the max- lim sup N(x)/x > 3L/4.
imum difference between consecutive tn (x) tends to 0, To begin with, we can certainly find some positive inte-
so the Riemann sum tends to the value of the integral. ger m ∈ / B, so that A is disjoint from A + m = {a + m :
Hence L = 0ln 2 (1 − et ) dt = ln 2 − 1, as desired.
R
a ∈ A }. Put A 0 = A ∪ (A + m) and let N 0 (x) be the
B–6 First solution: (based on a solution of Dan Bernstein) size of A 0 ∩ {1, . . . , x}; then lim sup N 0 (x)/x = 3L/2 >
Note that for any b, the condition that b ∈ / B already L, so A 0 cannot obey the conditions of the problem
forces lim sup N(x)/x to be at most 1/2: pair off 2mb + n statement. That is, if we let B 0 be the set of positive
with (2m+1)b+n for n = 1, . . . , b, and note that at most integers that occur as differences between elements of
one member of each pair may belong to A . The idea of A 0 , then there exists an integer n such that among any
the proof is to do something similar with pairs replaced n consecutive integers, at least one lies in B 0 . But
by larger clumps, using long runs of excluded elements
B 0 ⊆ {b + em : b ∈ B, e ∈ {−1, 0, 1}},
of B.
Suppose we have positive integers b0 = 1, b1 , . . . , bn so among any n + 2m consecutive integers, at least one
with the following properties: lies in B. This contradicts the condition of the problem
statement.
(a) For i = 1, . . . , n, ci = bi /(2bi−1 ) is an integer.
We conclude that it is impossible to have L > 0, so L = 0
/ B.
(b) For ei ∈ {−1, 0, 1}, |e1 b1 + · · · + en bn | ∈ and lim N(x)/x = 0 as desired.
Each nonnegative integer a has a unique “base expan- Remark: A hybrid between these two arguments is
sion” to note that if we can produce c1 , . . . , cn such that
|ci − c j | ∈ / B for i, j = 1, . . . , n, then the translates A +
a = a0 b0 + · · · + an−1 bn−1 + mbn (0 ≤ ai < 2ci ); c1 , . . . , A + cn are disjoint and so lim sup N(x)/x ≤ 1/n.
Given c1 ≤ · · · ≤ cn as above, we can then choose cn+1
if two integers have expansions with the same value to be the largest element of a run of cn + 1 consecutive
of m, and values of ai differing by at most 1 for i = integers, none of which lie in B.
0, . . . , n − 1, then their difference is not in B, so at
most one of them lies in A . In particular, for any
di ∈ {0, . . . , ci − 1}, any m0 ∈ {0, 2c0 − 1} and any mn ,
the set

{m0 b0 + (2d1 + e1 )b0 + · · ·

+ (2dn−1 + en−1 )bn−1 + (2mn + en )bn },
 The 66th William Lowell Putnam Mathematical Competition
Saturday, December 3, 2005

A1 Show that every positive integer is a sum of one or more B2 Find all positive integers n, k1 , . . . , kn such that k1 +· · ·+
numbers of the form 2r 3s , where r and s are nonnegative kn = 5n − 4 and
integers and no summand divides another. (For exam-
ple, 23 = 9 + 8 + 6.) 1 1
+ · · · + = 1.
k1 kn
A2 Let S = {(a, b)|a = 1, 2, . . . , n, b = 1, 2, 3}. A rook tour
of S is a polygonal path made up of line segments con- B3 Find all differentiable functions f : (0, ∞) → (0, ∞) for
necting points p1 , p2 , . . . , p3n in sequence such that which there is a positive real number a such that
(i) pi ∈ S, a x
f0 =
(ii) pi and pi+1 are a unit distance apart, for 1 ≤ i < x f (x)
3n,
for all x > 0.
(iii) for each p ∈ S there is a unique i such that pi = p.
How many rook tours are there that begin at (1, 1) B4 For positive integers m and n, let f (m, n) denote the
and end at (n, 1)? number of n-tuples (x1 , x2 , . . . , xn ) of integers such that
|x1 |+|x2 |+· · ·+|xn | ≤ m. Show that f (m, n) = f (n, m).
(An example of such a rook tour for n = 5 was depicted B5 Let P(x1 , . . . , xn ) denote a polynomial with real coeffi-
in the original.) cients in the variables x1 , . . . , xn , and suppose that
A3 Let p(z) be a polynomial of degree n all of whose zeros  2
∂2

have absolute value 1 in the complex plane. Put g(z) = ∂
+ · · · + 2 P(x1 , . . . , xn ) = 0 (identically)
p(z)/zn/2 . Show that all zeros of g0 (z) = 0 have absolute ∂ x12 ∂ xn
value 1.
and that
A4 Let H be an n × n matrix all of whose entries are ±1
and whose rows are mutually orthogonal. Suppose H x12 + · · · + xn2 divides P(x1 , . . . , xn ).
has an a × b submatrix whose entries are all 1. Show
Show that P = 0 identically.
that ab ≤ n.
R 1 ln(x+1) B6 Let Sn denote the set of all permutations of the numbers
A5 Evaluate dx.
0 x2 +1 1, 2, . . . , n. For π ∈ Sn , let σ (π) = 1 if π is an even
A6 Let n be given, n ≥ 4, and suppose that P1 , P2 , . . . , Pn permutation and σ (π) = −1 if π is an odd permutation.
are n randomly, independently and uniformly, chosen Also, let ν(π) denote the number of fixed points of π.
points on a circle. Consider the convex n-gon whose Show that
vertices are the Pi . What is the probability that at least σ (π) n
one of the vertex angles of this polygon is acute? ∑ = (−1)n+1 .
π∈Sn ν(π) + 1 n + 1
B1 Find a nonzero polynomial P(x, y) such that
P(bac, b2ac) = 0 for all real numbers a. (Note:
bνc is the greatest integer less than or equal to ν.)
 Solutions to the 66th William Lowell Putnam Mathematical Competition
Saturday, December 3, 2005
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A–1 We proceed by induction, with base case 1 = 20 30 . Sup- construction is reversible, lengthening any path in An−1
pose all integers less than n − 1 can be represented. If to a path in An .) In the latter case, P contains the sub-
n is even, then we can take a representation of n/2 and path P20 which joins (n − 1, 3), (n, 3), (n, 2), (n, 1) con-
multiply each term by 2 to obtain a representation of n. secutively; deleting P20 results in a path in Bn−1 , and
If n is odd, put m = blog3 nc, so that 3m ≤ n < 3m+1 . If this construction is also reversible. The desired bijec-
3m = n, we are done. Otherwise, choose a representa- tion follows.
tion (n−3m )/2 = s1 +· · ·+sk in the desired form. Then Similarly, there is a bijection between Bn and An−1 ∪
m Bn−1 for n ≥ 2. It follows by induction that for n ≥ 2,
n = 3 + 2s1 + · · · + 2sk ,
|An | = |Bn | = 2n−2 (|A1 | + |B1 |). But |A1 | = 0 and |B1 | =
and clearly none of the 2si divide each other or 3m . 1, and hence the desired answer is |An | = 2n−2 .
Moreover, since 2si ≤ n − 3m < 3m+1 − 3m , we have Remarks: Other bijective arguments are possible: for
si < 3m , so 3m cannot divide 2si either. Thus n has a instance, Noam Elkies points out that each element
representation of the desired form in all cases, complet- of An ∪ Bn contains a different one of the possible
ing the induction. sets of segments of the form (i, 2), (i + 1, 2) for i =
Remarks: This problem is originally due to Paul Erdős. 1, . . . , n − 1. Richard Stanley provides the reference:
Note that the representations need not be unique: for K.L. Collins and L.B. Krompart, The number of Hamil-
instance, tonian paths in a rectangular grid, Discrete Math. 169
(1997), 29–38. This problem is Theorem 1 of that
11 = 2 + 9 = 3 + 8. paper; the cases of 4 × n and 5 × n grids are also
treated. The paper can also be found online at the URL
kcollins.web.wesleyan.edu/vita.htm.
A–2 We will assume n ≥ 2 hereafter, since the answer is 0
for n = 1. A–3 Note that it is implicit in the problem that p is noncon-
First solution: We show that the set of rook tours from stant, one may take any branch of the square root, and
(1, 1) to (n, 1) is in bijection with the set of subsets of that z = 0 should be ignored.
{1, 2, ..., n} that include n and contain an even number First solution: Write p(z) = c ∏nj=1 (z − r j ), so that
of elements in total. Since the latter set evidently con-
tains 2n−2 elements, so does the former. g0 (z) 1 n
z+rj
= ∑ z−rj .
We now construct the bijection. Given a rook tour P g(z) 2z j=1
from (1, 1) to (n, 1), let S = S(P) denote the set of
all i ∈ {1, 2, . . . , n} for which there is either a directed Now if z 6= r j for all j,then
edge from (i, 1) to (i, 2) or from (i, 3) to (i, 2). It is
clear that this set S includes n and must contain an z+rj (z + r j )(z − r j ) |z|2 − 1 + 2Im(zr j )
even number of elements. Conversely, given a subset = = ,
z−rj |z − r j |2 |z − r j |2
S = {a1 , a2 , . . . , a2r = n} ⊂ {1, 2, . . . , n} of this type with
a1 < a2 < · · · < a2r , we notice that there is a unique path and so
P containing (ai , 2 + (−1)i ), (a1 , 2) for i = 1, 2, . . . , 2r. !
This establishes the desired bijection. zg0 (z) |z|2 − 1 1
Re = ∑ |z − r j |2 .
Second solution: Let An denote the set of rook tours g(z) 2 j
beginning at (1, 1) and ending at (n, 1), and let Bn de-
note the set of rook tours beginning at (1, 1) and ending Since the quantity in parentheses is positive, g0 (z)/g(z)
at (n, 3). can be 0 only if |z| = 1. If on the other hand z = r j for
For n ≥ 2, we construct a bijection between An and some j, then |z| = 1 anyway.
An−1 ∪ Bn−1 . Any path P in An contains either the line Second solution: Write p(z) = c ∏nj=1 (z − r j ), so that
segment P1 between (n−1, 1) and (n, 1), or the line seg-
g0 (z) n  
ment P2 between (n, 2) and (n, 1). In the former case, P 1 1
must also contain the subpath P10 which joins (n − 1, 3), = ∑ − .
g(z) j=1 z − r j 2z
(n, 3), (n, 2), and (n − 1, 2) consecutively; then deleting
P1 and P10 from P and adding the line segment joining We first check that g0 (z) 6= 0 whenever z is real and z >
(n − 1, 3) to (n − 1, 2) results in a path in An−1 . (This 1. In this case, for r j = eiθ j , we have z − r j = (z −
 2

1 1
cos(θ j )) + sin(θ j )i, so the real part of z−r j − 2z is columns. Then the hypothesis implies that the matrix
MM T has n − b’s on the main diagonal and −b’s else-
z − cos(θ j ) 1 z2 − 1 where. Hence the column vector v of length a consist-
− = > 0. ing of all 1’s satisfies MM T v = (n − ab)v, so n − ab is
z2 − 2z cos(θ j ) + 1 2z 2z(z2 − 2z cos(θ j ) + 1)
an eigenvalue of MM T . But MM T is semidefinite, so
Hence g0 (z)/g(z) has positive real part, so g0 (z)/g(z) its eigenvalues are all nonnegative real numbers. Hence
and hence g(z) are nonzero. n − ab ≥ 0.
Applying the same argument after replacing p(z) by Remarks: A matrix as in the problem is called
p(eiθ z), we deduce that g0 cannot have any roots out- a Hadamard matrix, because it meets the equality
side the unit circle. Applying the same argument after condition of Hadamard’s inequality: any n × n matrix
replacing p(z) by zn p(1/z), we also deduce that g0 can- with ±1 entries has absolute determinant at most nn/2 ,
not have any roots inside the unit circle. Hence all roots with equality if and only if the rows are mutually
of g0 have absolute value 1, as desired. orthogonal (from the interpretation of the determinant
as the volume of a paralellepiped whose edges are
Third solution: Write p(z) = c ∏nj=1 (z − r j ) and put parallel to the row vectors). Note that this implies
r j = e2iθ j . Note that g(e2iθ ) is equal to a nonzero con- that the columns are also mutually orthogonal. A
stant times generalization of this problem, with a similar proof,
is known as Lindsey’s lemma: the sum of the entries
n
ei(θ +θ j ) − e−i(θ +θ j ) n
in any√a × b submatrix of a Hadamard matrix is at
h(θ ) = ∏ = ∏ sin(θ + θ j ).
j=1 2i j=1 most abn. Stanley notes that Ryser (1981) asked
for the smallest size of a Hadamard matrix contain-
Since h has at least 2n roots (counting multiplicity) in ing an r × s submatrix of all 1’s, and refers to the
the interval [0, 2π), h0 does also by repeated application URL www3.interscience.wiley.com/cgi-bin/
of Rolle’s theorem. Since g0 (e2iθ ) = 2ie2iθ h0 (θ ), g0 (z2 ) abstract/110550861/ABSTRACT for more informa-
has at least 2n roots on the unit circle. Since g0 (z2 ) is tion.
equal to z−n−1 times a polynomial of degree 2n, g0 (z2 )
has all roots on the unit circle, as then does g0 (z). A–5 First solution: We make the substitution x = tan θ ,
rewriting the desired integral as
Remarks: The second solution imitates the proof of
the Gauss-Lucas theorem: the roots of the derivative of Z π/4
a complex polynomial lie in the convex hull of the roots log(tan(θ ) + 1) dθ .
0
of the original polynomial. The second solution is close
to problem B3 from the 2000 Putnam. A hybrid be- Write
tween the first and third solutions is to check that on the
unit circle, Re(zg0 (z)/g(z)) = 0 while between any two log(tan(θ ) + 1) = log(sin(θ ) + cos(θ )) − log(cos(θ ))
roots of p, Im(zg0 (z)/g(z)) runs from +∞ to −∞ and so √
must have a zero crossing. (This only works when p has and then note that sin(θ ) + cos(θ ) = 2 cos(π/4 − θ ).
distinct roots, but the general case follows by the con- We may thus rewrite the integrand as
tinuity of the roots of a polynomial as functions of the
coefficients.) One can also construct a solution using 1
log(2) + log(cos(π/4 − θ )) − log(cos(θ )).
Rouché’s theorem. 2

A–4 First solution: Choose a set of a rows r1 , . . . , ra con- But over the interval [0, π/4], the integrals of
taining an a × b submatrix whose entries are all 1. Then log(cos(θ )) and log(cos(π/4 − θ )) are equal, so their
for i, j ∈ {1, . . . , a}, we have ri · r j = n if i = j and 0 contributions cancel out. The desired integral is then
otherwise. Hence just the integral of 12 log(2) over the interval [0, π/4],
which is π log(2)/8.
a
∑ ri · r j = an. Second solution: (by Roger Nelsen) Let I denote the
i, j=1 desired integral. We make the substitution x = (1 −
u)/(1 + u) to obtain
On the other hand, the term on the left is the dot product
of r1 + · · · + ra with itself, i.e., its squared length. Since Z 1
(1 + u)2 log(2/(1 + u)) 2 du
this vector has a in each of its first b coordinates, the dot I=
0 2(1 + u2 ) (1 + u)2
product is at least a2 b. Hence an ≥ a2 b, whence n ≥ ab Z 1
as desired. log(2) − log(1 + u)
= du
0 1 + u2
Second solution: (by Richard Stanley) Suppose with- Z 1
du
out loss of generality that the a × b submatrix occupies = log(2) − I,
the first a rows and the first b columns. Let M be the 0 1 + u2
submatrix occupying the first a rows and the last n − b
 3

yielding then J0 = arctan(1) = π

4 and J1 = 12 log(2). Moreover,
Z 1
1 du π log(2) Z 1
1
I= log(2) 2
= . Jn + Jn+2 = xn dx = .
2 0 1+u 8 0 n+1

Third solution: (attributed to Steven Sivek) Define the Write

function m
(−1)i−1
Z 1 Am = ∑
f (t) =
log(xt + 1)
dx i=1 2i − 1
0 x2 + 1 m
(−1)i−1
Bm = ∑ ;
so that f (0) = 0 and the desired integral is f (1). Then i=1 2i
by differentiation under the integral,
then
Z 1
x
f 0 (t) = dx. J2n = (−1)n (J0 − An )
0 (xt + 1)(x2 + 1)
J2n+1 = (−1)n (J1 − Bn ).
By partial fractions, we obtain
Now the 2N-th partial sum of our series equals
x=1
0 2t arctan(x) − 2 log(tx + 1) + log(x2 + 1)
f (t) = N  
2(t 2 + 1) x=0 J2n−1 J2n
∑ −
πt + 2 log(2) − 4 log(t + 1) n=1 2n − 1 2n
= ,
4(t 2 + 1) N
(−1)n−1

(−1)n

=∑ (J1 − Bn−1 ) − (J0 − An )
whence n=1 2n − 1 2n
= AN (J1 − BN−1 ) + BN (J0 − AN ) + AN BN .
log(2) arctan(t) π log(t 2 + 1)
Z t
log(t + 1)
f (t) = + − dt
2 8 0 t2 + 1 As N → ∞, AN → J0 and BN → J1 , so the sum tends to
J0 J1 = π log(2)/8.
and hence
Z 1 Fifth solution: (suggested by Alin Bostan) Note that
π log(2) log(t + 1)
f (1) = − dt. Z 1
x dy
4 0 t2 + 1 log(1 + x) = ,
0 1 + xy
But the integral on the right is again the desired integral
f (1), so we may move it to the left to obtain so the desired integral I may be written as
Z 1Z 1
π log(2) x dy dx
2 f (1) = I= .
4 0 0 (1 + xy)(1 + x2 )
and hence f (1) = π log(2)/8 as desired. We may interchange x and y in this expression, then use
Fourth solution: (by David Rusin) We have Fubini’s theorem to interchange the order of summa-
! tion, to obtain
(−1)n−1 xn
Z 1 Z 1 ∞
log(x + 1) Z 1Z 1
dx = ∑ 2 dx. y dy dx
0 x2 + 1 0 n=1 n(x + 1)
I= .
0 0 (1 + xy)(1 + y2 )
We next justify moving the sum through the integral We then add these expressions to obtain
sign. Note that
Z 1Z 1 
x y dy dx
∞ Z 1
(−1)n−1 xn dx 2I = 2
+ 2
∑ 0 0 1+x 1+y 1 + xy
n=1 0 n(x2 + 1)
x + y + xy2 + x2 y dy dx
Z 1Z 1
=
is an alternating series whose terms strictly decrease to 0 0 (1 + x2 )(1 + y2 ) 1 + xy
Z 1Z 1
zero, so it converges. Moreover, its partial sums alter- (x + y) dy dx
nately bound the previous integral above and below, so = .
0 0 (1 + x2 )(1 + y2 )
the sum of the series coincides with the integral.
Put By another symmetry argument, we have
Z 1Z 1
Z 1 n
x dx x dy dx
Jn = ; 2I = 2 ,
0 x2 + 1 0 0 (1 + x2 )(1 + y2 )
 4

so member of each pair. Let R2 , . . . , Rn be the points of the

Z 1
 Z 1
 pairs which lie in the semicircle, taken in order away
x dx 1 π from Q1 , and let S2 , . . . , Sn be the antipodes of these.
I= = log(2) · .
0 1 + x2 0 1 + y2 8 Then to get the desired situation, we must choose from
the pairs to end up with all but one of the Si , and we can-
Remarks: The first two solutions are related by the not take Rn and the other Si or else ∠Q1 will be acute.
fact that if x = tan(θ ), then 1 − x/(1 + x) = tan(π/4 − That gives us (n − 2) good choices out of 2n−1 ; since
θ ). The strategy of the third solution (introducing we could have chosen Q1 to be any of the n points, the
a parameter then differentiating it) was a favorite of probability is again n(n − 2)2−n+1 .
physics Nobelist (and Putnam Fellow) Richard Feyn-
man. B–1 Take P(x, y) = (y − 2x)(y − 2x − 1). To see that this
R ∞ The fifth2 solution resembles Gauss’s evaluation works, first note that if m = bac, then 2m is an integer
of −∞ exp(−x ) dx. Noam Elkies notes that this inte-
gral is number 2.491#8 in Gradshteyn and Ryzhik, Ta- less than or equal to 2a, so 2m ≤ b2ac. On the other
ble of integrals, series, and products. The Mathemat- hand, m + 1 is an integer strictly greater than a, so 2m +
ica computer algebra system (version 5.2) successfully 2 is an integer strictly greater than 2a, so b2ac ≤ 2m+1.
computes this integral, but we do not know how. B–2 By the arithmetic-harmonic mean inequality or the
A–6 First solution: The angle at a vertex P is acute if and Cauchy-Schwarz inequality,
only if all of the other points lie on an open semicir-  
1 1
cle. We first deduce from this that if there are any (k1 + · · · + kn ) +···+ ≥ n2 .
two acute angles at all, they must occur consecutively. k1 kn
Suppose the contrary; label the vertices Q1 , . . . , Qn in
counterclockwise order (starting anywhere), and sup- We must thus have 5n − 4 ≥ n2 , so n ≤ 4. Without loss
pose that the angles at Q1 and Qi are acute for some of generality, we may suppose that k1 ≤ · · · ≤ kn .
i with 3 ≤ i ≤ n − 1. Then the open semicircle start- If n = 1, we must have k1 = 1, which works. Note that
ing at Q2 and proceeding counterclockwise must con- hereafter we cannot have k1 = 1.
tain all of Q3 , . . . , Qn , while the open semicircle start- If n = 2, we have (k1 , k2 ) ∈ {(2, 4), (3, 3)}, neither of
ing at Qi and proceeding counterclockwise must contain which work.
Qi+1 , . . . , Qn , Q1 , . . . , Qi−1 . Thus two open semicircles
cover the entire circle, contradiction. If n = 3, we have k1 + k2 + k3 = 11, so 2 ≤ k1 ≤ 3.
Hence
It follows that if the polygon has at least one acute an-
gle, then it has either one acute angle or two acute an- (k1 , k2 , k3 ) ∈ {(2, 2, 7), (2, 3, 6), (2, 4, 5), (3, 3, 5), (3, 4, 4)},
gles occurring consecutively. In particular, there is a
unique pair of consecutive vertices Q1 , Q2 in counter- and only (2, 3, 6) works.
clockwise order for which ∠Q2 is acute and ∠Q1 is
If n = 4, we must have equality in the AM-HM inequal-
not acute. Then the remaining points all lie in the arc
ity, which only happens when k1 = k2 = k3 = k4 = 4.
from the antipode of Q1 to Q1 , but Q2 cannot lie in the
arc, and the remaining points cannot all lie in the arc Hence the solutions are n = 1 and k1 = 1, n = 3 and
from the antipode of Q1 to the antipode of Q2 . Given (k1 , k2 , k3 ) is a permutation of (2, 3, 6), and n = 4 and
the choice of Q1 , Q2 , let x be the measure of the coun- (k1 , k2 , k3 , k4 ) = (4, 4, 4, 4).
terclockwise arc from Q1 to Q2 ; then the probability Remark: In the cases n = 2, 3, Greg Kuperberg sug-
that the other points fall into position is 2−n+2 − xn−2 if gests the alternate approach of enumerating the solu-
x ≤ 1/2 and 0 otherwise. tions of 1/k1 + · · · + 1/kn = 1 with k1 ≤ · · · ≤ kn . This is
Hence the probability that the polygon has at least one easily done by proceeding in lexicographic order: one
acute angle with a given choice of which two points will obtains (2, 2) for n = 2, and (2, 3, 6), (2, 4, 4), (3, 3, 3)
act as Q1 and Q2 is for n = 3, and only (2, 3, 6) contributes to the final an-
swer.
Z 1/2
n − 2 −n+1
(2−n+2 − xn−2 ) dx = 2 . B–3 First solution: The functions are precisely f (x) = cxd
0 n−1
for c, d > 0 arbitrary except that we must take c = 1 in
Since there are n(n − 1) choices for which two points case d = 1. To see that these work, note that f 0 (a/x) =
act as Q1 and Q2 , the probability of at least one acute dc(a/x)d−1 and x/ f (x) = 1/(cxd−1 ), so the given equa-
angle is n(n − 2)2−n+1 . tion holds if and only if dc2 ad−1 = 1. If d 6= 1, we may
solve for a no matter what c is; if d = 1, we must have
Second solution: (by Calvin Lin) As in the first solu- c = 1. (Thanks to Brad Rodgers for pointing out the
tion, we may compute the probability that for a particu- d = 1 restriction.)
lar one of the points Q1 , the angle at Q1 is not acute but
the following angle is, and then multiply by n. Imag- To check that these are all solutions, put b = log(a) and
ine picking the points by first choosing Q1 , then pick- y = log(a/x); rewrite the given equation as
ing n − 1 pairs of antipodal points and then picking one
f (eb−y ) f 0 (ey ) = eb−y .
 5

Put corresponds to placing k bars, each in a different posi-

tion behind one of the m fixed stars.
g(y) = log f (ey );
We conclude that
then the given equation rewrites as   
m n
f (m, n, k) = 2k = f (n, m, k);
k k
g(b − y) + log g0 (y) + g(y) − y = b − y,
summing over k gives f (m, n) = f (n, m). (One may also
or
extract easily a bijective interpretation of the equality.)
log g0 (y) = b − g(y) − g(b − y). Second solution: (by Greg Kuperberg) It will be con-
venient to extend the definition of f (m, n) to m, n ≥ 0,
By the symmetry of the right side, we have g0 (b − y) = in which case we have f (0, m) = f (n, 0) = 1.
g0 (y). Hence the function g(y) + g(b − y) has zero
Let Sm,n be the set of n-tuples (x1 , . . . , xn ) of inte-
derivative and so is constant, as then is g0 (y). From this
gers such that |x1 | + · · · + |xn | ≤ m. Then elements of
we deduce that f (x) = cxd for some c, d, both necessar-
Sm,n can be classified into three types. Tuples with
ily positive since f 0 (x) > 0 for all x.
|x1 | + · · · + |xn | < m also belong to Sm−1,n . Tuples with
Second solution: (suggested by several people) Substi- |x1 | + · · · + |xn | = m and xn ≥ 0 correspond to elements
tute a/x for x in the given equation: of Sm,n−1 by dropping xn . Tuples with |x1 | + · · · + |xn | =
a m and xn < 0 correspond to elements of Sm−1,n−1 by
f 0 (x) = . dropping xn . It follows that
x f (a/x)
f (m, n) = f (m − 1, n) + f (m, n − 1) + f (m − 1, n − 1),
Differentiate:
a a2 f 0 (a/x) so f satisfies a symmetric recurrence with symmetric
f 00 (x) = − + . boundary conditions f (0, m) = f (n, 0) = 1. Hence f is
x2 f (a/x) x3 f (a/x)2
symmetric.
Now substitute to eliminate evaluations at a/x: Third solution: (by Greg Martin) As in the second so-
lution, it is convenient to allow f (m, 0) = f (0, n) = 1.
f 0 (x) f 0 (x)2 Define the generating function
f 00 (x) = − + .
x f (x) ∞ ∞

Clear denominators:
G(x, y) = ∑ ∑ f (m, n)xm yn .
m=0 n=0
00 0 0 2
x f (x) f (x) + f (x) f (x) = x f (x) . As equalities of formal power series (or convergent se-
ries on, say, the region |x|, |y| < 13 ), we have
Divide through by f (x)2 and rearrange:

f 0 (x) x f 00 (x) x f 0 (x)2

G(x, y) = ∑ ∑ xm yn ∑ 1
0= + − . m≥0 n≥0 k1 , ..., kn ∈Z
f (x) f (x) f (x)2 |k1 |+···+|kn |≤m

The right side is the derivative of x f 0 (x)/ f (x), so that

= ∑ yn ∑ ∑ xm
n≥0 k1 , ..., kn ∈Z m≥|k1 |+···+|kn |
quantity is constant. That is, for some d,
x|k1 |+···+|kn |
f 0 (x) d
= ∑ yn ∑ 1−x
= . n≥0 k1 , ..., kn ∈Z
f (x) x  n
1 n |k|
= ∑ ∑ y x
Integrating yields f (x) = cxd , as desired. 1 − x n≥0 k∈Z
 n
B–4 First solution: Define f (m, n, k) as the number of n-
1 n 1+x
= ∑ y 1−x
tuples (x1 , x2 , . . . , xn ) of integers such that |x1 | + · · · + 1 − x n≥0
|xn | ≤ m and exactly k of x1 , . . . , xn are nonzero. To 1 1
choose such a tuple, we may choose the k nonzero posi- = ·
1 − x 1 − y(1 + x)/(1 − x)
tions, the signs of those k numbers, and then an ordered
1
k-tuple of positive integers with sum ≤ m. There are = .
n

and 2k for the second. 1 − x − y − xy
k options for the first choice,
m

As for the third, we have k options by a “stars and Since G(x, y) = G(y, x), it follows that f (m, n) =
bars” argument: depict the k-tuple by drawing a num- f (n, m) for all m, n ≥ 0.
ber of stars for each term, separated by bars, and adding
stars at the end to get a total of m stars. Then each tuple
 6

B–5 First solution: Put Q = x12 + · · · + xn2 . Since Q is ho- in characteristic p > 0 because we may take P =
mogeneous, P is divisible by Q if and only if each of (x12 + · · · + xn2 ) p = x12p + · · · + xn2p .) The third solution
the homogeneous components of P is divisible by Q. It can be extended to complex coefficients by replacing
is thus sufficient to solve the problem in case P itself is P(∇) by its complex conjugate, and again the result
homogeneous, say of degree d. may be deduced for any field of characteristic zero.
Suppose that we have a factorization P = Qm R for some Stanley also suggests Section 5 of the arXiv e-print
m > 0, where R is homogeneous of degree d and not math.CO/0502363 for some algebraic background for
divisible by Q; note that the homogeneity implies that this problem.
n
∂R B–6 First solution: Let I be the identity matrix, and let Jx be
∑ xi = dR. the matrix with x’s on the diagonal and 1’s elsewhere.
∂ xi
i=1 Note that Jx − (x − 1)I, being the all 1’s matrix, has rank
1 and trace n, so has n − 1 eigenvalues equal to 0 and
∂2 2
Write ∇2 as shorthand for ∂ x12
+ · · · + ∂∂x2 ; then one equal to n. Hence Jx has n − 1 eigenvalues equal to
n
x − 1 and one equal to x + n − 1, implying
0 = ∇2 P
n
det Jx = (x + n − 1)(x − 1)n−1 .
∂R
= 2mnQm−1 R + Qm ∇2 R + 2 ∑ 2mxi Qm−1
i=1 ∂ xi On the other hand, we may expand the determinant as a
sum indexed by permutations, in which case we get
= Qm ∇2 R + (2mn + 4md)Qm−1 R.

Since m > 0, this forces R to be divisible by Q, contra- det Jx = ∑ sgn(π)xν(π) .

π∈Sn
diction.
Second solution: (by Noam Elkies) Retain notation as Integrating both sides from 0 to 1 (and substituting y =
in the first solution. Let Pd be the set of homogeneous 1 − x) yields
polynomials of degree d, and let Hd be the subset of Z 1
Pd of polynomials killed by ∇2 , which has dimension sgn(π)
∑ ν(π) + 1 = (x + n − 1)(x − 1)n−1 dx
≥ dim(Pd ) − dim(Pd−2 ); the given problem amounts to π∈Sn 0
showing that this inequality is actually an equality. Z 1
Consider the operator Q∇2(i.e., apply ∇2
then multiply = (−1)n+1 (n − y)yn−1 dy
0
by Q) on Pd ; its zero eigenspace is precisely Hd . By the n
calculation from the first solution, if R ∈ Pd , then = (−1)n+1 ,
n+1
∇2 (QR) − Q∇2 R = (2n + 4d)R. as desired.
Second solution: We start by recalling a form of the
Consequently, Q j Hd−2 j is contained in the eigenspace
principle of inclusion-exclusion: if f is a function on
of Q∇2 on Pd of eigenvalue the power set of {1, . . . , n}, then
(2n + 4(d − 2 j)) + · · · + (2n + 4(d − 2)).
f (S) = ∑ (−1)|T |−|S| ∑ f (U).
T ⊇S U⊇T
In particular, the Q j H d−2 j
lie in distinct eigenspaces,
so are linearly independent within Pd . But by dimen- In this case we take f (S) to be the sum of σ (π) over all
sion counting, their total dimension is at least that of Pd . permutations π whose fixed points are exactly S. Then
Hence they exhaust Pd , and the zero eigenspace cannot ∑U⊇T f (U) = 1 if |T | ≥ n − 1 and 0 otherwise (since a
have dimension greater than dim(Pd ) − dim(Pd−2 ), as permutation group on 2 or more symbols has as many
desired. even and odd permutations), so
Third solution: (by Richard Stanley) Write x =
(x1 , . . . , xn ) and ∇ = ( ∂∂x , . . . , ∂∂xn ). Suppose that f (S) = (−1)n−|S| (1 − n + |S|).
1
P(x) = Q(x)(x12 + · · · + xn2 ). Then
The desired sum can thus be written, by grouping over
2 fixed point sets, as
P(∇)P(x) = Q(∇)(∇ )P(x) = 0.
n  
On the other hand, if P(x) = ∑α cα xα (where α = n 1−n+i
(α1 , . . . , αn ) and xα = x1α1 · · · xnαn ), then the constant
∑ i (−1)n−i i + 1
i=0
term of P(∇)P(x) is seen to be ∑α c2α . Hence cα = 0 n   n
n
 
n−i n−i n n
for all α. = ∑ (−1) − ∑ (−1)
i=0 i i=0 i + 1 i
Remarks: The first two solutions apply directly over n  
n n+1
any field of characteristic zero. (The result fails = 0 − ∑ (−1)n−i
i=0 n + 1 i+1
n
= (−1)n+1 .
n+1
 7

Third solution: (by Richard Stanley) The cycle indica- Expanding the right side as a Taylor series and compar-
tor of the symmetric group Sn is defined by ing coefficients yields the desired result.
c (π) c (π) Fourth solution (sketch): (by David Savitt) We prove
Zn (x1 , . . . , xn ) = ∑ x11 · · · xnn , the identity of rational functions
π∈Sn
σ (π) (−1)n+1 n!(x + n − 1)
where ci (π) is the number of cycles of π of length i. Put ∑ =
π∈Sn ν(π) + x x(x + 1) · · · (x + n)
Fn = ∑ σ (π)xν(π) = Zn (x, −1, 1, −1, 1, . . . )
π∈Sn by induction on n, which for x = 1 implies the desired
result. (This can also be deduced as in the other solu-
and tions, but in this argument it is necessary to formulate
Z 1 the strong induction hypothesis.)
σ (π)
f (n) = ∑ ν(π) + 1 = Fn (x) dx. Let R(n, x) be the right hand side of the above equation.
0
π∈Sn It is easy to verify that
A standard argument in enumerative combinatorics (the (−1)n+1
Exponential Formula) gives R(x, n) = R(x + 1, n − 1) + (n − 1)!
x
n−1
∞
tn ∞
tk (n − 1)!
∑ Zn (x1 , . . . , xn ) n! = exp ∑ xk k , + ∑ (−1)l−1 R(x, n − l),
n=0 k=1 l=2 (n − l)!

yielding since the sum telescopes. To prove the desired equality,

it suffices to show that the left hand side satisfies the
tn t2 t3
∞ Z 1  
same recurrence. This follows because we can classify
∑ f (n) = exp xt − + − · · · dx each π ∈ Sn as either fixing n, being an n-cycle, or hav-
n=0 n! 0 2 3
Z 1 ing n in an l-cycle for one of l = 2, . . . , n − 1; writing
= e(x−1)t+log(1+t) dx the sum over these classes gives the desired recurrence.
0
Z 1
= (1 + t)e(x−1)t dx
0
1
= (1 − e−t )(1 + t).
t
 The 67th William Lowell Putnam Mathematical Competition
Saturday, December 2, 2006

A–1 Find the volume of the region of points (x, y, z) such that B–1 Show that the curve x3 + 3xy + y3 = 1 contains only one
set of three distinct points, A, B, and C, which are ver-
(x2 + y2 + z2 + 8)2 ≤ 36(x2 + y2 ). tices of an equilateral triangle, and find its area.
B–2 Prove that, for every set X = {x1 , x2 , . . . , xn } of n real
A–2 Alice and Bob play a game in which they take turns
numbers, there exists a non-empty subset S of X and an
removing stones from a heap that initially has n stones.
integer m such that
The number of stones removed at each turn must be one
less than a prime number. The winner is the player who
takes the last stone. Alice plays first. Prove that there 1
m+ ∑s ≤ .
are infinitely many n such that Bob has a winning strat- s∈S n + 1
egy. (For example, if n = 17, then Alice might take 6 B–3 Let S be a finite set of points in the plane. A linear parti-
leaving 11; then Bob might take 1 leaving 10; then Al- tion of S is an unordered pair {A, B} of subsets of S such
ice can take the remaining stones to win.) that A ∪ B = S, A ∩ B = 0, / and A and B lie on opposite
sides of some straight line disjoint from S (A or B may
A–3 Let 1, 2, 3, . . . , 2005, 2006, 2007, 2009, 2012, 2016, . . . be empty). Let LS be the number of linear partitions of
be a sequence defined by xk = k for k = 1, 2, . . . , 2006 S. For each positive integer n, find the maximum of LS
and xk+1 = xk + xk−2005 for k ≥ 2006. Show that the over all sets S of n points.
sequence has 2005 consecutive terms each divisible by
2006. B–4 Let Z denote the set of points in Rn whose coordinates
are 0 or 1. (Thus Z has 2n elements, which are the ver-
A–4 Let S = {1, 2, . . . , n} for some integer n > 1. Say a per- tices of a unit hypercube in Rn .) Given a vector sub-
mutation π of S has a local maximum at k ∈ S if space V of Rn , let Z(V ) denote the number of members
of Z that lie in V . Let k be given, 0 ≤ k ≤ n. Find the
(i) π(k) > π(k + 1) for k = 1;
maximum, over all vector subspaces V ⊆ Rn of dimen-
(ii) π(k − 1) < π(k) and π(k) > π(k + 1) for 1 < k < sion k, of the number of points in V ∩ Z. [Editorial note:
n; the proposers probably intended to write Z(V ) instead
(iii) π(k − 1) < π(k) for k = n. of “the number of points in V ∩ Z”, but this changes
nothing.]
(For example, if n = 5 and π takes values at 1, 2, 3, 4, 5
of 2, 1, 4, 5, 3, then π has a local maximum of 2 at k = B–5 For each continuous function f : [0, 1] → R, let I( f ) =
R1 2 R1 2
1, and a local maximum of 5 at k = 4.) What is the 0 x f (x) dx and J(x) = 0 x ( f (x)) dx. Find the maxi-
average number of local maxima of a permutation of S, mum value of I( f ) − J( f ) over all such functions f .
averaging over all permutations of S?
B–6 Let k be an integer greater than 1. Suppose a0 > 0, and
A–5 Let n be a positive odd integer and let θ be a real num- define
ber such that θ /π is irrational. Set ak = tan(θ + kπ/n),
1
k = 1, 2, . . . , n. Prove that an+1 = an + √
k a
n
a1 + a2 + · · · + an
a1 a2 · · · an for n > 0. Evaluate

is an integer, and determine its value. ak+1

n
lim .
n→∞ nk
A–6 Four points are chosen uniformly and independently at
random in the interior of a given circle. Find the proba-
bility that they are the vertices of a convex quadrilateral.
 Solutions to the 67th William Lowell Putnam Mathematical Competition
Saturday, December 2, 2006
Kiran Kedlaya and Lenny Ng

p change to cylindrical coordinates, i.e., we put r =

A–1 We where n is fixed and f is a fixed polynomial of n vari-
x2 + y2 . Then the given inequality is equivalent to ables with integer coefficients, for any positive integer
N, the sequence modulo N is eventually periodic. This
r2 + z2 + 8 ≤ 6r, is simply because there are only finitely many possible
sequences of n consecutive values modulo N, and once
or such a sequence is repeated, every subsequent value is
repeated as well.
(r − 3)2 + z2 ≤ 1.
We next observe that if one can rewrite the same recur-
This defines a solid of revolution (a solid torus); the sion as
area being rotated is the disc (x − 3)2 + z2 ≤ 1 in the
xk−n = g(xk−n+1 , . . . , xk ) (k > n),
xz-plane. By Pappus’s theorem, the volume of this
equals the area of this disc, which is π, times the dis- where g is also a polynomial with integer coefficients,
tance through which the center of mass is being rotated, then the sequence extends uniquely to a doubly infi-
which is (2π)3. That is, the total volume is 6π 2 . nite sequence . . . , x−1 , x0 , x1 , . . . which is fully periodic
A–2 Suppose on the contrary that the set B of values of modulo any N. That is the case in the situation at hand,
n for which Bob has a winning strategy is finite; for because we can rewrite the given recursion as
convenience, we include n = 0 in B, and write B =
xk−2005 = xk+1 − xk .
{b1 , . . . , bm }. Then for every nonnegative integer n not
in B, Alice must have some move on a heap of n stones It thus suffices to find 2005 consecutive terms divisible
leading to a position in which the second player wins. by N in the doubly infinite sequence, for any fixed N
That is, every nonnegative integer not in B can be writ- (so in particular for N = 2006). Running the recursion
ten as b+ p−1 for some b ∈ B and some prime p. How- backwards, we easily find
ever, there are numerous ways to show that this cannot
happen. x1 = x0 = · · · = x−2004 = 1
First solution: Let t be any integer bigger than all of x−2005 = · · · = x−4009 = 0,
the b ∈ B. Then it is easy to write down t consecutive
composite integers, e.g., (t +1)!+2, . . . , (t +1)!+t +1. yielding the desired result.
Take n = (t + 1)! + t; then for each b ∈ B, n − b + 1 is
one of the composite integers we just wrote down. A–4 First solution: By the linearity of expectation, the av-
erage number of local maxima is equal to the sum of
Second solution: Let p1 , . . . , p2m be any prime num- the probability of having a local maximum at k over
bers; then by the Chinese remainder theorem, there ex- k = 1, . . . , n. For k = 1, this probability is 1/2: given
ists a positive integer x such that the pair {π(1), π(2)}, it is equally likely that π(1) or
π(2) is bigger. Similarly, for k = n, the probability is
x − b1 ≡ −1 (mod p1 pm+1 )
1/2. For 1 < k < n, the probability is 1/3: given the pair
... {π(k − 1), π(k), π(k + 1)}, it is equally likely that any
x − bn ≡ −1 (mod pm p2m ). of the three is the largest. Thus the average number of
local maxima is
For each b ∈ B, the unique integer p such that x = b +
p − 1 is divisible by at least two primes, and so cannot 1 1 n+1
2· + (n − 2) · = .
itself be prime. 2 3 3
Third solution: (by Catalin Zara) Put b1 = 0, and take
Second solution: Another way to apply the linear-
n = (b2 − 1) · · · (bm − 1); then n is composite because
ity of expectation is to compute the probability that
3, 8 ∈ B, and for any nonzero b ∈ B, n − bi + 1 is di-
i ∈ {1, . . . , n} occurs as a local maximum. The most
visible by but not equal to bi − 1. (One could also take
efficient way to do this is to imagine the permutation as
n = b2 · · · bm − 1, so that n − bi + 1 is divisible by bi .)
consisting of the symbols 1, . . . , n, ∗ written in a circle
A–3 We first observe that given any sequence of integers in some order. The number i occurs as a local maxi-
x1 , x2 , . . . satisfying a recursion mum if the two symbols it is adjacent to both belong to
the set {∗, 1, . . . , i − 1}. There are i(i − 1) pairs of such
xk = f (xk−1 , . . . , xk−n ) (k > n), symbols and n(n − 1) pairs in total, so the probability of
 2

i occurring as a local maximum is i(i − 1)/(n(n − 1)), expected value of a random variable over all choices of
and the average number of local maxima is P, Q, R. Write [XY Z] for the area of triangle XY Z.
n n   If P, Q, R, S are the four points, we may ignore the case
i(i − 1) 2 i
∑ n(n − 1) = n(n − 1) ∑ 2 where three of them are collinear, as this occurs with
i=1 i=1 probability zero. Then the only way they can fail to
form the vertices of a convex quadrilateral is if one of
 
2 n+1
= them lies inside the triangle formed by the other three.
n(n − 1) 3
There are four such configurations, depending on which
n+1
= . point lies inside the triangle, and they are mutually ex-
3 clusive. Hence the desired probability is 1 minus four
One can obtain a similar (if slightly more intricate) so- times the probability that S lies inside triangle PQR.
lution inductively, by removing the known local maxi- That latter probability is simply E([PQR]) divided by
mum n and splitting into two shorter sequences. the area of the disc.
Remark: The usual term for a local maximum in this Let O denote the center of the circle, and let P0 , Q0 , R0
sense is a peak. The complete distribution for the num- be the projections of P, Q, R onto the circle from O. We
ber of peaks is known; Richard Stanley suggests the ref- can write
erence: F. N. David and D. E. Barton, Combinatorial
Chance, Hafner, New York, 1962, p. 162 and subse- [PQR] = ±[OPQ] ± [OQR] ± [ORP]
quent.
for a suitable choice of signs, determined as follows.
A–5 Since the desired expression involves symmetric func- If the points P0 , Q0 , R0 lie on no semicircle, then all of
tions of a1 , . . . , an , we start by finding a polynomial with the signs are positive. If P0 , Q0 , R0 lie on a semicircle
a1 , . . . , an as roots. Note that in that order and Q lies inside the triangle OPR, then
the sign on [OPR] is positive and the others are nega-
1 ± i tan θ = e±iθ sec θ tive. If P0 , Q0 , R0 lie on a semicircle in that order and Q
lies outside the triangle OPR, then the sign on [OPR] is
so that negative and the others are positive.

1 + i tan θ = e2iθ (1 − i tan θ ). We first calculate

Consequently, if we put ω = e2inθ , then the polynomial E([OPQ] + [OQR] + [ORP]) = 3E([OPQ]).

Qn (x) = (1 + ix)n − ω(1 − ix)n Write r1 = OP, r2 = OQ, θ = ∠POQ, so that

1
has among its roots a1 , . . . , an . Since these are distinct [OPQ] = r1 r2 (sin θ ).
and Qn has degree n, these must be exactly the roots. 2
If we write The distribution of r1 is given by 2r1 on [0, 1] (e.g., by
the change of variable formula to polar coordinates),
Qn (x) = cn xn + · · · + c1 x + c0 , and similarly for r2 . The distribution of θ is uniform on
[0, π]. These three distributions are independent; hence
then a1 + · · · + an = −cn−1 /cn and a1 · · · an = −c0 /cn ,
so the ratio we are seeking is cn−1 /c0 . By inspection, E([OPQ])
Z 1 2  Z π 
1 1
cn−1 = nin−1 − ωn(−i)n−1 = nin−1 (1 − ω) = 2r2 dr sin(θ ) dθ
2 0 π 0
c0 = 1 − ω
4
= ,
so 9π
and
(
a1 + · · · + an n n≡1 (mod 4)
=
a1 · · · an −n n ≡ 3 (mod 4). 4
E([OPQ] + [OQR] + [ORP]) = .
3π
Remark: The same argument shows that the ratio be-
tween any two odd elementary symmetric functions of We now treat the case where P0 , Q0 , R0 lie on a semicir-
a1 , . . . , an is independent of θ . cle in that order. Put θ1 = ∠POQ and θ2 = ∠QOR; then
the distribution of θ1 , θ2 is uniform on the region
A–6 First solution: (by Daniel Kane) The probability is
35 0 ≤ θ1 , 0 ≤ θ2 , θ1 + θ2 ≤ π.
1 − 12π 2 . We start with some notation and simplifica-
tions. For simplicity, we assume without loss of gen-
erality that the circle has radius 1. Let E denote the In particular, the distribution on θ = θ1 + θ2 is 2θ
π2
on
[0, π]. Put rP = OP, rQ = OQ, rR = OR. Again, the dis-
tribution on rP is given by 2rP on [0, 1], and similarly
 3

for rQ , rR ; these are independent from each other and the lines PQ, QR, RP, which with probability 1 divide
from the joint distribution of θ1 , θ2 . Write E 0 (X) for the interior of the circle into seven regions. Put a =
the expectation of a random variable X restricted to this [PQR], let b1 , b2 , b3 denote the areas of the three other
part of the domain. regions sharing a side with the triangle, and let c1 , c2 , c3
Let χ be the random variable with value 1 if Q is inside denote the areas of the other three regions. Put A =
triangle OPR and 0 otherwise. We now compute E(a), B = E(b1 ), C = E(c1 ), so that A + 3B + 3C = π.
Note that c1 + c2 + c3 + a is the area of the region in
E 0 ([OPR]) which we can choose a fourth point S so that the quadri-
1
Z 1 2 Z π
2θ
 lateral PQRS fails to be convex. By comparing expecta-
2
= 2r dr 2
sin(θ ) dθ tions, we have 3C + A = 4A, so A = C and 4A + 3B = π.
2 0 0 π
4 We will compute B + 2A = B + 2C, which is the ex-
= pected area of the part of the circle cut off by a chord
9π
through two random points D, E, on the side of the
E 0 (χ[OPR]) chord not containing a third random point F. Let h be
= E 0 (2[OPR]2 /θ ) the distance from the center O of the circle to the line
Z 1 2 Z π  DE. We now determine the distribution of h.
1 2θ −1 2
= 2r3 dr 2
θ sin (θ ) dθ Put r = OD; the distribution of r is 2r on [0, 1]. Without
2 0 0 π
loss of generality, suppose O is the origin and D lies
1 on the positive x-axis. For fixed r, the distribution of
= .
8π h runs over [0, r], and can be computed as the area of
Also recall that given any triangle XY Z, if T is cho- the infinitesimal region in which E can be chosen so the
sen uniformly at random inside XY Z, the expectation chord through DE has distance to O between h and h +
of [T XY ] is the area of triangle bounded by XY and the dh, divided by π. This region splits into two symmetric
centroid of XY Z, namely 13 [XY Z]. pieces, one of which lies between chords making angles
of arcsin(h/r) and arcsin((h + dh)/r) with the x-axis.
Let χ be the random variable with value 1 if Q is inside The angle between these is dθ = dh/(r2 − h2 ). Draw
triangle OPR and 0 otherwise. Then the chord through D at distance h to O, and let L1 , L2 be
the lengths of the parts on opposite sides of D; then the
E 0 ([OPQ] + [OQR] + [ORP] − [PQR])
area we are looking for is 12 (L12 + L22 )dθ . Since
= 2E 0 (χ([OPQ] + [OQR]) + 2E 0 ((1 − χ)[OPR]) p p
2 {L1 , L2 } = 1 − h2 ± r2 − h2 ,
= 2E 0 ( χ[OPR]) + 2E 0 ([OPR]) − 2E 0 (χ[OPR])
3
2 29 the area we are seeking (after doubling) is
0
= 2E ([OPR]) − E 0 (χ[OPR]) = .
3 36π
1 + r2 − 2h2
2 √ .
Finally, note that the case when P0 , Q0 , R0 lie on a semi- r2 − h2
circle in some order occurs with probability 3/4. (The
case where they lie on a semicircle proceeding clock- Dividing by π, then integrating over r, we compute the
wise from P0 to its antipode has probability 1/4; this distribution of h to be
case and its two analogues are exclusive and exhaus-
1 + r2 − 2h2
Z 1
1
tive.) Hence 2 √ 2r dr
π h r2 − h2
E([PQR]) 16
= (1 − h2 )3/2 .
= E([OPQ] + [OQR] + [ORP]) 3π
3
− E 0 ([OPQ] + [OQR] + [ORP] − [PQR]) We now return to computing B + 2A. Let A(h) denote
4
the smaller of the two areas of the disc cut off by a
4 29 35
= − = , chord at distance h. The chance that the third point
3π 48π 48π is in the smaller (resp. larger) portion is A(h)/π (resp.
so the original probability is 1 − A(h)/π), and then the area we are trying to com-
pute is π − A(h) (resp. A(h)). Using the distribution on
4E([PQR]) 35 h, and the fact that
1− = 1− .
π 12π 2 Z 1p
A(h) = 2 1 − h2 dh
Second solution: (by David Savitt) As in the first so- h
lution, it suffices to check that for P, Q, R chosen uni- π p
35
= − arcsin(h) − h 1 − h2 ,
formly at random in the disc, E([PQR]) = 48π . Draw 2
 4
√
we find (−1, −1) to (1/2, 1/2), or 3 2/2.√The area of an equi-
lateral triangle
√ of height h is h2 3/3, so the desired
B + 2A area is 3 3/2.
2 1 16
Z
= A(h)(π − A(h)) (1 − h2 )3/2 dh Remark: The factorization used above is a special case
π 0 3π of the fact that
35 + 24π 2
= . x3 + y3 + z3 − 3xyz
72π
35
= (x + y + z)(x + ωy + ω 2 z)(x + ω 2 y + ωz),
Since 4A + 3B = π, we solve to obtain A = 48π as in the
first solution. where ω denotes a primitive cube root of unity. That
Third solution: (by Noam Elkies) Again, we reduce fact in turn follows from the evaluation of the determi-
to computing the average area of a triangle formed by nant of the circulant matrix
three random points A, B,C inside a unit circle. Let O be  
the center of the circle, and put c = max{OA, OB, OC}; x y z
 z x y
then the probability that c ≤ r is (r2 )3 , so the distribu-
tion of c is 6c5 dc on [0, 1]. y z x

Given c, the expectation of [ABC] is equal to c2 times X, by reading off the eigenvalues of the eigenvectors
the expected area of a triangle formed by two random (1, ω i , ω 2i ) for i = 0, 1, 2.
points P, Q in a circle and a fixed point R on the bound-
ary. We introduce polar coordinates centered at R, in B–2 Let {x} = x − bxc denote the fractional part of x. For
which the circle is given by r = 2 sin θ for θ ∈ [0, π]. i = 0, . . . , n, put si = x1 + · · · + xi (so that s0 = 0). Sort
The distribution of a random point in that circle is the numbers {s0 }, . . . , {sn } into ascending order, and
1 call the result t0 , . . . ,tn . Since 0 = t0 ≤ · · · ≤ tn < 1,
π r dr dθ over θ ∈ [0, π] and r ∈ [0, 2 sin θ ]. If (r, θ )
and (r0 , θ 0 ) are the two random points, then the area is the differences
1 0 0
2 rr sin |θ − θ |.
t1 − t0 , . . . ,tn − tn−1 , 1 − tn
Performing the integrals over r and r0 first, we find
are nonnegative and add up to 1. Hence (as in the pi-
32 π π 3
Z Z
X= sin θ sin3 θ 0 sin |θ − θ 0 | dθ 0 dθ geonhole principle) one of these differences is no more
9π 2 0 0 than 1/(n + 1); if it is anything other than 1 − tn , it
64 π θ 3
Z Z
= 2 sin θ sin3 θ 0 sin(θ − θ 0 ) dθ 0 dθ . equals ±({si } − {s j }) for some 0 ≤ i < j ≤ n. Put
9π 0 0 S = {xi+1 , . . . , x j } and m = bsi c − bs j c; then
This integral is unpleasant but straightforward; it
yields X = 35/(36π), and E([PQR]) = 01 6c7 X dc =
R
m + ∑ s = |m + s j − si |
35/(48π), giving the desired result. s∈S

Remark: This is one of the oldest problems in geo- = |{s j } − {si }|

metric probability; it is an instance of Sylvester’s four- 1
point problem, which nowadays is usually solved us- ≤ ,
n+1
ing a device known as Crofton’s formula. We de-
fer to http://mathworld.wolfram.com/ for further as desired. In case 1 − tn ≤ 1/(n + 1), we take S =
discussion. {x1 , . . . , xn } and m = −dsn e, and again obtain the de-
sired conclusion.
B–1 The “curve” x3 + 3xy + y3 − 1 = 0 is actually reducible,
B–3 The maximum is n2 + 1, achieved for instance by a


because the left side factors as
convex n-gon: besides the trivial partition (in which all
(x + y − 1)(x2 − xy + y2 + x + y + 1). of the points are in one part), each linear partition oc-
curs by drawing a line crossing a unique pair of edges.
Moreover, the second factor is
First solution: We will prove that LS = n2 + 1 in any

1 configuration in which no two of the lines joining points

((x + 1)2 + (y + 1)2 + (x − y)2 ), of S are parallel. This suffices to imply the maximum
2
in all configurations: given a maximal configuration,
so it only vanishes at (−1, −1). Thus the curve in ques- we may vary the points slightly to get another maximal
tion consists of the single point (−1, −1) together with configuration in which our hypothesis is satisfied. For
the line x + y = 1. To form a triangle with three points convenience, we assume n ≥ 3, as the cases n = 1, 2 are
on this curve, one of its vertices must be (−1, −1). easy.
The other two vertices lie on the line x + y = 1, so the Let P be the line at infinity in the real projective plane;
length of the altitude from (−1, −1) is the distance from i.e., P is the set of possible directions of lines in the
 5

plane, viewed as a circle. Remove the directions cor- lines of S0 crossing the segment O0 P are determined by
responding
to lines through two points of S; this leaves which region P lies in.
behind n2 intervals. Thus our original maximum is equal to the maximum
Given a direction in one of the intervals, consider the number of regions into which n−1 lines divide an affine
set of linear partitions achieved by lines parallel to that plane. By
induction on n, this number is easily seen to
direction. Note that the resulting collection of partitions be 1 + n2 .
depends only on the interval. Then note that the collec- Fourth solution: (by Florian Herzig) Say that an S-line
tions associated to adjacent intervals differ in only one is a line that intersects S in at least two points. We claim
element. that the nontrivial linear partitions of S are in natural bi-
The trivial partition that puts all of S on one side is in ev- jection with pairs (`, {X,Y }) consisting of an S-line `
ery such collection. We now observe that for any other and a nontrivial linear partition {X,Y } of ` ∩ S. Since
linear partition {A, B}, the set of intervals to which an S-line ` admits precisely |` ∩ S| − 1 ≤ |`∩S|


2 nontriv-
{A, B} is: ial linear partitions, the claim implies that LS ≤ n2 + 1

(a) a consecutive block of intervals, but with equality iff no three points of S are collinear.

(b) not all of them. Let P be the line at infinity in the real projective plane.
Given any nontrivial linear partition {A, B} of S, the set
For (a), note that if `1 , `2 are nonparallel lines achieving of lines inducing this partition is a proper, open, con-
the same partition, then we can rotate around their point nected subset I of P. (It is proper because it has to omit
of intersection to achieve all of the intermediate direc- directions of S-lines that pass through both parts of the
tions on one side or the other. For (b), the case n = 3 partition and open because we can vary the separating
is evident; to reduce the general case to this case, take line. It is connected because if we have two such lines
points P, Q, R such that P lies on the opposite side of the that aren’t parallel, we can rotate through their point of
partition from Q and R. intersection to get all intermediate directions.) Among
all S-lines that intersect both A and B choose a line `
It follows now that that each linear partition, except for
whose direction is minimal (in the clockwise direction)
the trivial one, occurs in exactly one place as the parti-
with respect to the interval I; also, pick an arbitrary line
tion associated to some interval but not to its immediate
`0 that induces {A, B}. By rotating `0 clockwise to `
counterclockwise neighbor. In other words, the num-
about their point of intersection, we see that the direc-
ber of linear partitions is one more than the number of
tion of ` is the least upper bound of I. (We can’t hit any
intervals, or n2 + 1 as desired.


point of S during the rotation because of the minimality
Second solution: We prove the upper bound by induc- property of `.) The line ` is in fact unique because if the
tion on n. Choose a point P in the convex hull of S. Put (parallel) lines pq and rs are two choices for `, with p,
S0 = S \ {P};

by the induction hypothesis, there are at q ∈ A; r, s ∈ B, then one of the diagonals ps, qr would
most n−12 + 1 linear partitions of S 0 . Note that each
contradict the minimality property of `. To define the
linear partition of S restricts to a linear partition of S0 . above bijection we send {A, B} to (`, {A ∩ `, B ∩ `}).
Moreover, if two linear partitions of S restrict to the
Conversely, suppose that we are given an S-line ` and
same linear partition of S0 , then that partition of S0 is
a nontrivial linear partition {X,Y } of ` ∩ S. Pick any
achieved by a line through P.
point p ∈ ` that induces the partition {X,Y }. If we ro-
By rotating a line through P, we see that there are at tate the line ` about p in the counterclockwise direction
most n − 1 partitions of S0 achieved by lines through by a sufficiently small amount, we get a nontrivial lin-
P: namely, the partition only changes when the rotating ear partitition of S that is independent of all choices. (It
line passes through one of the points of S. This yields is obtained from the partition of S − ` induced by ` by
the desired result. adjoining X to one part and Y to the other.) This defines
Third solution: (by Noam Elkies) We enlarge the plane a map in the other direction.
to a projective plane by adding a line at infinity, then By construction these two maps are inverse to each
apply the polar duality map centered at one of the points other, and this proves the claim.
O ∈ S. This turns the rest of S into a set S0 of n − 1 lines
Remark: Given a finite set S of points in Rn , a non-
in the dual projective plane. Let O0 be the point in the
Radon partition of S is a pair (A, B) of complementary
dual plane corresponding to the original line at infinity;
subsets that can be separated by a hyperplane. Radon’s
it does not lie on any of the lines in S0 .
theorem states that if #S ≥ n + 2, then not every (A, B)
Let ` be a line in the original plane, corresponding to a is a non-Radon partition. The result of this problem has
point P in the dual plane. If we form the linear partition been greatly extended, especially within the context of
induced by `, then the points of S \ {O} lying in the matroid theory and oriented matroid theory. Richard
same part as O correspond to the lines of S0 which cross Stanley suggests the following references: T. H. Bry-
the segment O0 P. If we consider the dual affine plane lawski, A combinatorial perspective on the Radon con-
as being divided into regions by the lines of S0 , then the vexity theorem, Geom. Ded. 5 (1976), 459-466; and T.
 6

Zaslavsky, Extremal arrangements of hyperplanes, Ann. B–5 The answer is 1/16. We have
N. Y. Acad. Sci. 440 (1985), 69-87. Z 1 Z 1
B–4 The maximum is 2k , achieved for instance by the sub- x2 f (x) dx − x f (x)2 dx
0 0
space Z 1
= (x3 /4 − x( f (x) − x/2)2 ) dx
{(x1 , . . . , xn ) ∈ Rn : x1 = · · · = xn−k = 0}. 0
Z 1
≤ x3 /4 dx = 1/16,
First solution: More generally, we show that any affine 0
k-dimensional plane in Rn can contain at most 2k points
in Z. The proof is by induction on k + n; the case k = with equality when f (x) = x/2.
n = 0 is clearly true. B–6 First solution: We start with some easy upper and
Suppose that V is a k-plane in Rn . Denote the hyper- lower bounds on an . We write O( f (n)) and Ω( f (n))
planes {xn = 0} and {xn = 1} by V0 and V1 , respec- for functions g(n) such that f (n)/g(n) and g(n)/ f (n),
tively. If V ∩ V0 and V ∩ V1 are each at most (k − 1)- respectively, are bounded above. Since an is a non-
dimensional, then V ∩ V0 ∩ Z and V ∩ V1 ∩ Z each have decreasing sequence, an+1 − an is bounded above, so
cardinality at most 2k−1 by the induction assumption, −1/k
an = O(n). That means an = Ω(n−1/k ), so
and hence V ∩ Z has at most 2k elements. Otherwise, !
if V ∩ V0 or V ∩ V1 is k-dimensional, then V ⊂ V0 or n
V ⊂ V1 ; now apply the induction hypothesis on V , an = Ω ∑ i−1/k = Ω(n(k−1)/k ).
viewed as a subset of Rn−1 by dropping the last coordi- i=1
nate.
In fact, all we will need is that an → ∞ as n → ∞.
Second solution: Let S be a subset of Z contained in
By Taylor’s theorem with remainder, for 1 < m < 2 and
a k-dimensional subspace of V . This is equivalent to
x > 0,
asking that any t1 , . . . ,tk+1 ∈ S satisfy a nontrivial linear
dependence c1t1 +· · ·+ck+1tk+1 = 0 with c1 , . . . , ck+1 ∈ m(m − 1) 2
R. Since t1 , . . . ,tk+1 ∈ Qn , given such a dependence |(1 + x)m − 1 − mx| ≤ x .
2
we can always find another one with c1 , . . . , ck+1 ∈ Q;
then by clearing denominators, we can find one with −(k+1)/k
Taking m = (k + 1)/k and x = an+1 /an = 1 + an ,
c1 , . . . , ck+1 ∈ Z and not all having a common factor. we obtain
Let F2 denote the field of two elements, and let S ⊆ Fn2
be the reductions modulo 2 of the points of S. Then any (k+1)/k (k+1)/k k+1 k + 1 −(k+1)/k
an+1 − an − ≤ an .
t1 , . . . ,tk+1 ∈ S satisfy a nontrivial linear dependence, k 2k2
because we can take the dependence from the end of
In particular,
the previous paragraph and reduce modulo 2. Hence S
is contained in a k-dimensional subspace of F2n , and the (k+1)/k (k+1)/k k+1
latter has cardinality exactly 2k . Thus S has at most 2k lim an+1 − an = .
n→∞ k
elements, as does S.
Variant (suggested by David Savitt): if S contained k +1 In general, if xn is a sequence with limn→∞ xn = c, then
linearly independent elements, the (k + 1) × n matrix also
formed by these would have a nonvanishing maximal
minor. The lift of that minor back to R would also not 1 n
vanish, so S would contain k + 1 linearly independent
lim
n→∞ n
∑ xi = c
i=1
elements.
by Cesaro’s lemma. Explicitly, for any ε > 0, we can
Third solution: (by Catalin Zara) Let V be a k-
find N such that |xn − c| ≤ ε/2 for n ≥ N, and then
dimensional subspace. Form the matrix whose rows are
the elements of V ∩ Z; by construction, it has row rank
at most k. It thus also has column rank at most k; in 1 n n−N ε N N
c− ∑ xi ≤ n 2 + n ∑ (c − xi ) ;
particular, we can choose k coordinates such that each n i=1 i=1
point of V ∩ Z is determined by those k of its coordi-
nates. Since each coordinate of a point in Z can only for n large, the right side is smaller than ε.
take two values, V ∩ Z can have at most 2k elements. In our case, we deduce that
Remark: The proposers probably did not (k+1)/k
realize that this problem appeared online an k+1
lim =
about three months before the exam, at n→∞ n k
http://www.artofproblemsolving.com/
Forum/viewtopic.php?t=105991. (It may very and so
well have also appeared even earlier.) k
ak+1

k+1
lim n k = ,
n→∞ n k
 7

as desired. and this completes the proof.

Remark: The use of Cesaro’s lemma above is the spe- Third solution: We argue that an → ∞ as in the first
cial case bn = n of the Cesaro-Stolz theorem: if an , bn solution. Write bn = an − Lnk/(k+1) , for a value of L to
are sequences such that bn is positive, strictly increas- be determined later. We have
ing, and unbounded, and
bn+1
an+1 − an −1/k
lim = L, = bn + an − L((n + 1)k/(k+1) − nk/(k+1) )
n→∞ bn+1 − bn
= e1 + e2 ,
then
an where
lim = L.
n→∞ bn −1/k
e1 = bn + an − L−1/k n−1/(k+1)
Second solution: In this solution, rather than applying e2 = L((n + 1)k/(k+1) − nk/(k+1) )
Taylor’s theorem with remainder to (1 + x)m for 1 < − L−1/k n−1/(k+1) .
m < 2 and x > 0, we only apply convexity to deduce
that (1 + x)m ≥ 1 + mx. This gives We first estimate e1 . For −1 < m < 0, by the convexity
of (1 + x)m and (1 + x)1−m , we have
(k+1)/k (k+1)/k k+1
an+1 − an ≥ ,
k 1 + mx ≤ (1 + x)m
and so ≤ 1 + mx(1 + x)m−1 .
(k+1)/k k+1 Hence
an ≥ n+c
k
1
for some c ∈ R. In particular, − L−(k+1)/k n−1 bn ≤ e1 − bn
k
(k+1)/k
1 −(k+1)/k
an k+1 ≤ − bn an .
lim inf ≥ k
n→∞ n k
Note that both bounds have sign opposite to bn ; more-
and so over, by the bound an = Ω(n(k−1)/k ), both bounds have
 k/(k+1) absolutely value strictly less than that of bn for n suffi-
an k+1 ciently large. Consequently, for n large,
lim inf ≥ .
n→∞ nk/(k+1) k
|e1 | ≤ |bn |.
But turning this around, the fact that
We now work on e2 . By Taylor’s theorem with remain-
an+1 − an der applied to (1 + x)m for x > 0 and 0 < m < 1,
−1/k
= an
1 + mx ≥ (1 + x)m
k + 1 −1/(k+1) −1/(k+1)
 
≤ n (1 + o(1)), m(m − 1) 2
k ≥ 1 + mx + x .
2
where o(1) denotes a function tending to 0 as n → ∞, The “main term” of L((n + 1)k/(k+1) − nk/(k+1) )
yields is L k+1k −1/(k+1)
n . To make this coincide with
L −1/k n−1/(k+1) , we take
an
k + 1 −1/(k+1) n −1/(k+1)
   k/(k+1)
k+1
≤ ∑i (1 + o(1)) L= .
k i=1 k
k + 1 k + 1 −1/(k+1) k/(k+1)
 
= n (1 + o(1)) We then find that
k k
|e2 | = O(n−2 ),
k + 1 k/(k+1) k/(k+1)
 
= n (1 + o(1)),
k and because bn+1 = e1 + e2 , we have |bn+1 | ≤ |bn | +
|e2 |. Hence
so
!
 k/(k+1) n
an k+1
lim sup ≤ |bn | = O ∑ i−2 = O(1),
n→∞ nk/(k+1) k i=1
 8

and so termining the asymptotic behavior of sequences of this

k type: replace the given recursion
ak+1

n k+1
lim = Lk+1 = . −1/k
n→∞ nk k an+1 − an = an

Remark: The case k = 2 appeared on the 2004 Roma- by the differential equation
nian Olympiad (district level).
y0 = y−1/k
Remark: One can make a similar argument for any se-
quence given by an+1 = an + f (an ), when f is a de- and determine the asymptotics of the latter.
creasing function.
Remark: Richard Stanley suggests a heuristic for de-
 The 68th William Lowell Putnam Mathematical Competition
Saturday, December 1, 2007

A–1 Find all values of α for which the curves y = αx2 + f ( f (n) + 1) if and only if n = 1. [Editor’s note: one
1 1
αx+ 24 and x = αy2 +αy+ 24 are tangent to each other. must assume f is nonconstant.]
A–2 Find the least possible area of a convex set in the plane B–2 Suppose that f : [0, 1] → R has a continuous derivative
that intersects both branches of the hyperbola xy = 1 and that 01 f (x) dx = 0. Prove that for every α ∈ (0, 1),
R
and both branches of the hyperbola xy = −1. (A set S
in the plane is called convex if for any two points in S 1
Z α
the line segment connecting them is contained in S.) f (x) dx ≤ max | f 0 (x)|.
0 8 0≤x≤1
A–3 Let k be a positive integer. Suppose that the integers √
1, 2, 3, . . . , 3k + 1 are written down in random order. B–3 Let x0 = 1 and for n ≥ 0, let xn+1 = 3xn + bxn 5c. In
What is the probability that at no time during this pro- particular, x1 = 5, x2 = 26, x3 = 136, x4 = 712. Find a
cess, the sum of the integers that have been written up closed-form expression for x2007 . (bac means the largest
to that time is a positive integer divisible by 3? Your integer ≤ a.)
answer should be in closed form, but may include fac-
B–4 Let n be a positive integer. Find the number of pairs
torials.
P, Q of polynomials with real coefficients such that
A–4 A repunit is a positive integer whose digits in base 10
are all ones. Find all polynomials f with real coeffi- (P(X))2 + (Q(X))2 = X 2n + 1
cients such that if n is a repunit, then so is f (n).
and deg P > deg Q.
A–5 Suppose that a finite group has exactly n elements of B–5 Let k be a positive integer. Prove that there exist polyno-
order p, where p is a prime. Prove that either n = 0 or mials P0 (n), P1 (n), . . . , Pk−1 (n) (which may depend on
p divides n + 1. k) such that for any integer n,

A–6 A triangulation T of a polygon P is a finite collection j n kk jnk j n kk−1

= P0 (n) + P1 (n) + · · · + Pk−1 (n) .
of triangles whose union is P, and such that the inter- k k k
section of any two triangles is either empty, or a shared
vertex, or a shared side. Moreover, each side is a side (bac means the largest integer ≤ a.)
of exactly one triangle in T . Say that T is admissible
B–6 For each positive integer n, let f (n) be the number of
if every internal vertex is shared by 6 or more triangles.
ways to make n! cents using an unordered collection of
For example, [figure omitted.] Prove that there is an in-
coins, each worth k! cents for some k, 1 ≤ k ≤ n. Prove
teger Mn , depending only on n, such that any admissible
that for some constant C, independent of n,
triangulation of a polygon P with n sides has at most Mn
triangles. 2 /2−Cn 2 /4 2 /2+Cn 2 /4
nn e−n ≤ f (n) ≤ nn e−n .
B–1 Let f be a polynomial with positive integer coefficients.
Prove that if n is a positive integer, then f (n) divides
 Solutions to the 68th William Lowell Putnam Mathematical Competition
Saturday, December 1, 2007
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

√
A–1 The only such α are 2/3, 3/2, (13 ± 601)/12. Second solution: For any nonzero value of α, the two
First solution: Let C1 and C2 be the curves y = αx2 + conics will intersect in four points in the complex pro-
αx + 241 1
and x = αy2 + αy + 24 , respectively, and let L jective plane P2 (C). To determine the y-coordinates of
be the line y = x. We consider three cases. these intersection points, subtract the two equations to
obtain
If C1 is tangent to L, then the point of tangency (x, x)
satisfies (y − x) = α(x − y)(x + y) + α(x − y).
1
2αx + α = 1, x = αx2 + αx + ; Therefore, at a point of intersection we have either x =
24 y, or x = −1/α − (y + 1). Substituting these two pos-
by symmetry, C2 is tangent to L there, so C1 and C2 are sible linear conditions into the second equation shows
tangent. Writing α = 1/(2x + 1) in the first equation that the y-coordinate of a point of intersection is a root
and substituting into the second, we must have of either Q1 (y) = αy2 + (α − 1)y + 1/24 or Q2 (y) =
αy2 + (α + 1)y + 25/24 + 1/α.
x2 + x 1 If two curves are tangent, then the y-coordinates of at
x= + ,
2x + 1 24 least two of the intersection points will coincide; the
converse is also true because one of the curves is the
which simplifies to 0 = 24x2 − 2x − 1 = (6x + 1)(4x − graph of a function in x. The coincidence occurs pre-
1), or x ∈ {1/4, −1/6}. This yields α = 1/(2x + 1) ∈ cisely when either the discriminant of at least one of
{2/3, 3/2}. Q1 or Q2 is zero, or there is a common root of Q1
If C1 does not intersect L, then C1 and C2 are separated and Q2 . Computing the discriminants of Q1 and Q2
by L and so cannot be tangent. yields (up to constant factors) f1 (α) = 6α 2 − 13α + 6
and f2 (α) = 6α 2 − 13α − 18, respectively. If on the
If C1 intersects L in two distinct points P1 , P2 , then it
other hand Q1 and Q2 have a common root, it must be
is not tangent to L at either point. Suppose at one of
also a root of Q2 (y) − Q1 (y) = 2y + 1 + 1/α, yielding
these points, say P1 , the tangent to C1 is perpendicular
y = −(1 + α)/(2α) and 0 = Q1 (y) = − f2 (α)/(24α).
to L; then by symmetry, the same will be true of C2 , so
C1 and C2 will be tangent at P1 . In this case, the point Thus the values of α for which the two curves are tan-
P1 = (x, x) satisfies gent must be contained in the
√set of zeros of f1 and f2 ,
namely 2/3, 3/2, and (13 ± 601)/12.
1
2αx + α = −1, x = αx2 + αx + ; Remark: The fact that the two conics in P2 (C) meet in
24 four points, counted with multiplicities, is a special case
writing α = −1/(2x + 1) in the first equation and sub- of Bézout’s theorem: two curves in P2 (C) of degrees
stituting into the second, we have m, n and not sharing any common component meet in
exactly mn points when counted with multiplicity.
x2 + x 1 Many solvers were surprised that the proposers chose
x=− + ,
2x + 1 24 the parameter 1/24 to give two rational roots and two
√ nonrational roots. In fact, they had no choice in the
√± 601)/72. This yields α = −1/(2x +
or x = (−23 matter: attempting to make all four roots rational by
1) = (13 ± 601)/12. replacing 1/24 by β amounts to asking for β 2 + β and
If instead the tangents to C1 at P1 , P2 are not perpen- β 2 + β + 1 to be perfect squares. This cannot happen
dicular to L, then we claim there cannot be any point outside of trivial cases (β = 0, −1) ultimately because
where C1 and C2 are tangent. Indeed, if we count in- the elliptic curve 24A1 (in Cremona’s notation) over Q
tersections of C1 and C2 (by using C1 to substitute for y has rank 0. (Thanks to Noam Elkies for providing this
in C2 , then solving for y), we get at most four solutions computation.)
counting multiplicity. Two of these are P1 and P2 , and However, there are choices that make the radical milder,
any point of tangency counts for two more. However, e.g., β = 1/3 gives β 2 + β = 4/9 and β 2 + β + 1 =
off of L, any point of tangency would have a mirror im- 13/9, while β = 3/5 gives β 2 + β = 24/25 and β 2 +
age which is also a point of tangency, and there cannot β + 1 = 49/25.
be six solutions. Hence we have now found all possible
α. A–2 The minimum is 4, achieved by the square with vertices
(±1, ±1).
 2

First solution: To prove that 4 is a lower bound, let S be It follows that the ordering satisfies the given condition
a convex set of the desired form. Choose A, B,C, D ∈ S if and only if the following two conditions hold: the
lying on the branches of the two hyperbolas, with A in first element in the ordering is not divisible by 3, and
the upper right quadrant, B in the upper left, C in the the sequence mod 3 (ignoring zeroes) is of the form
lower left, D in the lower right. Then the area of the 1, 1, −1, 1, −1, . . .. The two conditions are independent,
quadrilateral ABCD is a lower bound for the area of S. and the probability of the first is (2k +1)/(3k +1) while
2k+1

Write A = (a, 1/a), B = (−b, 1/b), C = (−c, −1/c), the probability of the second is 1/ k , since there
are 2k+1


D = (d, −1/d) with a, b, c, d > 0. Then the area of the k ways to order (k + 1) 1’s and k −1’s. Hence
quadrilateral ABCD is the desired probability is the product of these two, or
k!(k+1)!
1 (3k+1)(2k)! .
(a/b + b/c + c/d + d/a + b/a + c/b + d/c + a/d),
2 A–4 Note that n is a repunit if and only if 9n + 1 = 10m for
which by the arithmetic-geometric mean inequality is at some power of 10 greater than 1. Consequently, if we
least 4. put
 
Second solution: Choose A, B,C, D as in the first solu- n−1
tion. Note that both the hyperbolas and the area of the g(n) = 9 f + 1,
9
convex hull of ABCD are invariant under the transfor-
mation (x, y) 7→ (xm, y/m) for any m > 0. For m small, then f takes repunits to repunits if and only if g takes
the counterclockwise angle from the line AC to the line powers of 10 greater than 1 to powers of 10 greater than
BD approaches 0; for m large, this angle approaches 1. We will show that the only such functions g are those
π. By continuity, for some m this angle becomes π/2, of the form g(n) = 10c nd for d ≥ 0, c ≥ 1 − d (all of
that is, AC and BD become perpendicular. The area of which clearly work), which will mean that the desired
ABCD is then AC · BD. polynomials f are those of the form
√
It thus suffices to note that AC ≥ 2 2 (and similarly for 1
BD). This holds because if we draw the tangent lines to f (n) = (10c (9n + 1)d − 1)
the hyperbola xy = 1 at the points (1, 1) and (−1, −1), 9
then A and C lie outside the region between these lines. for the same c, d.
If we project the segment AC orthogonally onto the line
It is convenient to allow “powers of 10” to be of the
x√ = y = 1, the resulting projection has length at least
form 10k for any integer k. With this convention, it suf-
2 2, so AC must as well.
fices to check that the polynomials g taking powers of
Third solution: (by Richard Stanley) Choose A, B,C, D 10 greater than 1 to powers of 10 are of the form 10c nd
as in the first solution. Now fixing A and C, move B and for any integers c, d with d ≥ 0.
D to the points at which the tangents to the curve are
First solution: Suppose that the leading term of g(x)
parallel to the line AC. This does not increase the area
is axd , and note that a > 0. As x → ∞, we have
of the quadrilateral ABCD (even if this quadrilateral is
g(x)/xd → a; however, for x a power of 10 greater than
not convex).
1, g(x)/xd is a power of 10. The set of powers of 10 has
Note that B and D are now diametrically opposite; write no positive limit point, so g(x)/xd must be equal to a
B = (−x, 1/x) and D = (x, −1/x). If we thus repeat the for x = 10k with k sufficiently large, and we must have
procedure, fixing B and D and moving A and C to the a = 10c for some c. The polynomial g(x) − 10c xd has
points where the tangents are parallel to BD, then A and infinitely many roots, so must be identically zero.
C must move to (x, 1/x) and (−x, −1/x), respectively,
Second solution: We proceed by induction on d =
forming a rectangle of area 4.
deg(g). If d = 0, we have g(n) = 10c for some c. Oth-
Remark: Many geometric solutions are possible. An erwise, g has rational coefficients by Lagrange’s inter-
example suggested by David Savitt (due to Chris polation formula (this applies to any polynomial of de-
Brewer): note that AD and BC cross the positive and gree d taking at least d + 1 different rational numbers
negative x-axes, respectively, so the convex hull of to rational numbers), so g(0) = t is rational. More-
ABCD contains O. Then check that the area of trian- over, g takes each value only finitely many times, so the
gle OAB is at least 1, et cetera. sequence g(100 ), g(101 ), . . . includes arbitrarily large
powers of 10. Suppose that t 6= 0; then we can choose
A–3 Assume that we have an ordering of 1, 2, . . . , 3k +1 such
a positive integer h such that the numerator of t is not
that no initial subsequence sums to 0 mod 3. If we omit
divisible by 10h . But for c large enough, g(10c ) − t has
the multiples of 3 from this ordering, then the remain-
numerator divisible by 10b for some b > h, contradic-
ing sequence mod 3 must look like 1, 1, −1, 1, −1, . . .
tion.
or −1, −1, 1, −1, 1, . . .. Since there is one more integer
in the ordering congruent to 1 mod 3 than to −1, the Consequently, t = 0, and we may apply the induction
sequence mod 3 must look like 1, 1, −1, 1, −1, . . .. hypothesis to g(n)/n to deduce the claim.
 3

Remark: The second solution amounts to the fact that face has three edges, and each edge but the n outside
g, being a polynomial with rational coefficients, is con- edges belongs to two faces; hence F = 2E − n. On the
tinuous for the 2-adic and 5-adic topologies on Q. By other hand, each edge has two endpoints, and each of
contrast, the first solution uses the “∞-adic” topology, the V − n internal vertices is an endpoint of at least 6
i.e., the usual real topology. edges; hence a1 + · · · + an + 6(V − n) ≤ 2E. Combining
this inequality with the previous two equations gives
A–5 In all solutions, let G be a finite group of order m.
First solution: By Lagrange’s theorem, if m is not di- a1 + · · · + an ≤ 2E + 6n − 6(1 − F + E)
visible by p, then n = 0. Otherwise, let S be the set of = 4n − 6,
p-tuples (a0 , . . . , a p−1 ) ∈ G p such that a0 · · · a p−1 = e;
then S has cardinality m p−1 , which is divisible by p. as claimed.
Note that this set is invariant under cyclic permutation, Now set A3 = 1 and An = An−1 + 2n − 3 for n ≥ 4; we
that is, if (a0 , . . . , a p−1 ) ∈ S, then (a1 , . . . , a p−1 , a0 ) ∈ S will prove by induction on n that T has at most An trian-
also. The fixed points under this operation are the tuples gles. For n = 3, since a1 +a2 +a3 = 6, a1 = a2 = a3 = 2
(a, . . . , a) with a p = e; all other tuples can be grouped and hence T consists of just one triangle.
into orbits under cyclic permutation, each of which has
Next assume that an admissible triangulation of an
size p. Consequently, the number of a ∈ G with a p = e
(n − 1)-gon has at most An−1 triangles, and let T be
is divisible by p; since that number is n + 1 (only e has
an admissible triangulation of an n-gon. If any ai = 2,
order 1), this proves the claim.
then we can remove the triangle of T containing vertex
Second solution: (by Anand Deopurkar) Assume that vi to obtain an admissible triangulation of an (n − 1)-
n > 0, and let H be any subgroup of G of order p. Let S gon; then the number of triangles in T is at most
be the set of all elements of G \ H of order dividing p, An−1 + 1 < An by induction. Otherwise, all ai ≥ 3. Now
and let H act on G by conjugation. Each orbit has size the average of a1 , . . . , an is less than 4, and thus there
p except for those which consist of individual elements are more ai = 3 than ai ≥ 5. It follows that there is a se-
g which commute with H. For each such g, g and H quence of k consecutive vertices in P whose degrees are
generate an elementary abelian subgroup of G of order 3, 4, 4, . . . , 4, 3 in order, for some k with 2 ≤ k ≤ n − 1
p2 . However, we can group these g into sets of size (possibly k = 2, in which case there are no degree 4
p2 − p based on which subgroup they generate together vertices separating the degree 3 vertices). If we remove
with H. Hence the cardinality of S is divisible by p; from T the 2k − 1 triangles which contain at least one
adding the p − 1 nontrivial elements of H gives n ≡ −1 of these vertices, then we are left with an admissible tri-
(mod p) as desired. angulation of an (n − 1)-gon. It follows that there are at
Third solution: Let S be the set of elements in G hav- most An−1 + 2k − 1 ≤ An−1 + 2n − 3 = An triangles in
ing order dividing p, and let H be an elementary abelian T . This completes the induction step and the proof.
p-group of maximal order in G. If |H| = 1, then we are Remark: We can refine the bound An somewhat. Sup-
done. So assume |H| = pk for some k ≥ 1, and let H posing that ai ≥ 3 for all i, the fact that a1 + · · · + an ≤
act on S by conjugation. Let T ⊂ S denote the set of 4n − 6 implies that there are at least six more indices
fixed points of this action. Then the size of every H- i with ai = 3 than with ai ≥ 5. Thus there exist six
orbit on S divides pk , and so |S| ≡ |T | (mod p). On the sequences with degrees 3, 4, . . . , 4, 3, of total length at
other hand, H ⊂ T , and if T contained an element not most n + 6. We may thus choose a sequence of length
in H, then that would contradict the maximality of H. k ≤ b 6n c + 1, so we may improve the upper bound to
It follows that H = T , and so |S| ≡ |T | = |H| = pk ≡ 0 An = An−1 + 2b 6n c + 1, or asymptotically 61 n2 .
(mod p), i.e., |S| = n + 1 is a multiple of p.
However (as noted by Noam Elkies), a hexagonal
Remark: This result is a theorem of Cauchy; the first swatch of a triangular lattice, with the boundary as close
solution above is due to McKay. A more general (and to regular as possible, achieves asymptotically 16 n2 tri-
more difficult) result was proved by Frobenius: for any angles.
positive integer m, if G is a finite group of order divis-
ible by m, then the number of elements of G of order B–1 The problem fails if f is allowed to be constant, e.g.,
dividing m is a multiple of m. take f (n) = 1. We thus assume that f is nonconstant.
Write f (n) = ∑di=0 ai ni with ai > 0. Then
A–6 For an admissible triangulation T , number the vertices
of P consecutively v1 , . . . , vn , and let ai be the number d
of edges in T emanating from vi ; note that ai ≥ 2 for f ( f (n) + 1) = ∑ ai ( f (n) + 1)i
all i. i=0

We first claim that a1 + · · · + an ≤ 4n − 6. Let V, E, F ≡ f (1) (mod f (n)).

denote the number of vertices, edges, and faces in T .
By Euler’s Formula, (F + 1) − E +V = 2 (one must add If n = 1, then this implies that f ( f (n) + 1) is divisible
1 to the face count for the region exterior to P). Each by f (n). Otherwise, 0 < f (1) < f (n) since f is non-
constant and has positive coefficients, so f ( f (n) + 1)
cannot be divisible by f (n).
 4

B–2 Put B = max0≤x≤1 | f 0 (x)| and g(x) = 0x f (y) dy. Since

R
sequences/). Therein, the sequence is described as
g(0) = g(1) = 0, the maximum value of |g(x)| must oc- the case S(1, 5) of the sequence S(a0 , a1 ) in which an+2
cur at a critical point y ∈ (0, 1) satisfying g0 (y) = f (y) = is the least integer for which an+2 /an+1 > an+1 /an .
0. We may thus take α = y hereafter. Sloane cites D. W. Boyd, Linear recurrence relations
for some generalized Pisot sequences, Advances in
Since 0α f (x) dx = − 01−α f (1 − x) dx, we may assume
R R
Number Theory (Kingston, ON, 1991), Oxford Univ.
that α ≤ 1/2. By then substituting − f (x) for f (x) if
Press, New York, 1993, p. 333–340.
needed, we may assume that 0α f (x) dx ≥ 0. From the
R

inequality f 0 (x) ≥ −B, we deduce f (x) ≤ B(α − x) for B–4 The number of pairs is 2n+1 . The degree condition
0 ≤ x ≤ α, so forces P to have degree n and leading coefficient ±1;
Z α Z α we may count pairs in which P has leading coefficient
f (x) dx ≤ B(α − x) dx 1 as long as we multiply by 2 afterward.
0 0
1 α Factor both sides:
= − B(α − x)2
2 0 (P(X) + Q(X)i)(P(X) − Q(X)i)
α2 1 n−1
= B≤ B = ∏ (X − exp(2πi(2 j + 1)/(4n)))
2 8
j=0
as desired. n−1
· ∏ (X + exp(2πi(2 j + 1)/(4n))).
B–3 First solution: Observing that x2 /2 = 13, x3 /4 = 34, j=0
x4 /8 = 89, we guess that xn = 2n−1 F2n+3 , where Fk is
the k-th Fibonacci number. Thus we claim√ that xn = Then each choice of P, Q corresponds to equating
n−1
2√ 2n+3 − α −(2n+3) ), where α = 1+ 5 , to make the
(α 2
P(X) + Q(X)i with the product of some n factors on
5
2006 the right, in which we choose exactly of the two fac-
answer x2007 = 2√
5
(α 3997 − α −3997 ). tors for each j = 0, . . . , n − 1. (We must take exactly n
We prove the claim by induction; the base case x0 = 1 is factors because as a polynomial in X with complex co-
so it suffices to show that the recursion xn+1 =
true, and √ efficients, P(X) + Q(X)i has degree exactly n. We must
3xn + bxn 5c is satisfied for our formula for xn . Indeed, choose one for each j to ensure that P(X) + Q(X)i and
√
2 3+ 5
since α = 2 , we have P(X) − Q(X)i are complex conjugates, so that P, Q have
real coefficients.) Thus there are 2n such pairs; multi-
√ 2n−1 plying by 2 to allow P to have leading coefficient −1
xn+1 − (3 + 5)xn = √ (2(α 2n+5 − α −(2n+5) ) yields the desired result.
5
√ Remark: If we allow P and Q to have complex coeffi-
− (3 + 5)(α 2n+3 − α −(2n+3) )) cients but still require deg(P)
> deg(Q), then the num-
= 2n α −(2n+3) . ber of pairs increases to 2 2n n , as we may choose any n
2n
of the 2n factors of X +1 to use to form P(X)+Q(X)i.
√ √
Now 2n α −(2n+3) = ( 1−2 5 )3 (3 − 5)n is between −1
B–5 For n an integer, we have nk = n−k j for j the unique
 
and 0; the recursion follows since xn , xn+1 are integers.
integer in {0, . . . , k − 1} congruent to n modulo k; hence
Second solution:√(by Catalin
√ Zara) Since xn is rational,
we have 0 < xn 5 − bxn 5c < 1. We now have the k−1 j 
nk n− j
inequalities ∏ − = 0.
j=0 k k
√
xn+1 − 3xn < xn 5 < xn+1 − 3xn + 1
√ √ By expanding this out, we obtain the desired polynomi-
(3 + 5)xn − 1 < xn+1 < (3 + 5)xn als P0 (n), . . . , Pk−1 (n).
√ √
4xn − (3 − 5) < (3 − 5)xn+1 < 4xn Remark: Variants of this solution are possible that con-
√ √ struct the Pi less explicitly, using Lagrange interpolation
3xn+1 − 4xn < xn+1 5 < 3xn+1 − 4xn + (3 − 5).
or Vandermonde determinants.
√ √
Since 0 < 3 − 5 < 1, this yields bxn+1 5c = 3xn+1 − B–6 (Suggested by Oleg Golberg) Assume n ≥ 2, or else the
4xn , so we can rewrite the recursion as xn+1 = 6xn − problem is trivially false. Throughout this proof, any Ci
4xn−1 for n ≥ 2. It is routine to solve this recursion to will be a positive constant whose exact value is imma-
obtain the same solution as above. terial. As in the proof of Stirling’s approximation, we
Remark: With an initial 1 prepended, this estimate for any fixed c ∈ R,
becomes sequence A018903 in Sloane’s On-
n
Line Encyclopedia of Integer Sequences: 1 1
(http://www.research.att.com/~njas/
∑ (i + c) log i = 2 n2 log n − 4 n2 + O(n log n)
i=1
 5

by comparing the sum to an integral. This gives For a lower bound on f (n), we note that if 0 ≤ ai <
(n − 1)!/i! for i = 2, . . . , n − 1 and an = 0, then 0 ≤
2 /2−C n 2 /4
nn 1 e−n ≤ 11+c 22+c · · · nn+c a2 2! + · · · + an n! ≤ n!, so there is a unique choice of a1
2 /2+C n 2 /4 to complete this to a solution of a1 1! + · · · + an n! = n!.
≤ nn 2 e−n . Hence
We now interpret f (n) as counting the number of n- (n − 1)! (n − 1)!
tuples (a1 , . . . , an ) of nonnegative integers such that f (n) ≥ ···
2! (n − 1)!
a1 1! + · · · + an n! = n!. = 31 42 · · · (n − 1)n−3
2 /2+C n 2 /4
≥ nn 4 e−n .
For an upper bound on f (n), we use the inequalities
0 ≤ ai ≤ n!/i! to deduce that there are at most n!/i! +
1 ≤ 2(n!/i!) choices for ai . Hence

n! n!
f (n) ≤ 2n ···
1! n!
= 2n 21 32 · · · nn−1
2 /2+C n 2 /4
≤ nn 3 e−n .
 The 69th William Lowell Putnam Mathematical Competition
Saturday, December 6, 2008

A1 Let f : R2 → R be a function such that f (x, y)+ f (y, z)+ (The elements of G in the sequence are not required to
f (z, x) = 0 for all real numbers x, y, and z. Prove that be distinct. A subsequence of a sequence is obtained
there exists a function g : R → R such that f (x, y) = by selecting some of the terms, not necessarily consec-
g(x) − g(y) for all real numbers x and y. utive, without reordering them; for example, 4, 4, 2 is a
subsequence of 2, 4, 6, 4, 2, but 2, 2, 4 is not.)
A2 Alan and Barbara play a game in which they take turns
filling entries of an initially empty 2008 × 2008 array. B1 What is the maximum number of rational points that can
Alan plays first. At each turn, a player chooses a real lie on a circle in R2 whose center is not a rational point?
number and places it in a vacant entry. The game ends (A rational point is a point both of whose coordinates
when all the entries are filled. Alan wins if the determi- are rational numbers.)
nant of the resulting matrix is nonzero; Barbara wins if
it is zero. Which player has a winning strategy? B2 RLet F0 (x) = ln x. For n ≥ 0 and x > 0, let Fn+1 (x) =
x
0 Fn (t) dt. Evaluate
A3 Start with a finite sequence a1 , a2 , . . . , an of positive in-
tegers. If possible, choose two indices j < k such that a j n!Fn (1)
lim .
does not divide ak , and replace a j and ak by gcd(a j , ak ) n→∞ ln n
and lcm(a j , ak ), respectively. Prove that if this process B3 What is the largest possible radius of a circle contained
is repeated, it must eventually stop and the final se- in a 4-dimensional hypercube of side length 1?
quence does not depend on the choices made. (Note: B4 Let p be a prime number. Let h(x) be a polynomial with
gcd means greatest common divisor and lcm means integer coefficients such that h(0), h(1), . . . , h(p2 − 1)
least common multiple.) are distinct modulo p2 . Show that h(0), h(1), . . . , h(p3 −
A4 Define f : R → R by 1) are distinct modulo p3 .
( B5 Find all continuously differentiable functions f : R → R
x if x ≤ e such that for every rational number q, the number f (q)
f (x) =
x f (ln x) if x > e. is rational and has the same denominator as q. (The
denominator of a rational number q is the unique posi-
1
Does ∑∞
n=1 f (n) converge? tive integer b such that q = a/b for some integer a with
gcd(a, b) = 1.) (Note: gcd means greatest common di-
A5 Let n ≥ 3 be an integer. Let f (x) and g(x) be poly- visor.)
nomials with real coefficients such that the points
( f (1), g(1)), ( f (2), g(2)), . . . , ( f (n), g(n)) in R2 are the B6 Let n and k be positive integers. Say that a permutation
vertices of a regular n-gon in counterclockwise order. σ of {1, 2, . . . , n} is k-limited if |σ (i) − i| ≤ k for all
Prove that at least one of f (x) and g(x) has degree i. Prove that the number of k-limited permutations of
greater than or equal to n − 1. {1, 2, . . . , n} is odd if and only if n ≡ 0 or 1 (mod 2k +1).

A6 Prove that there exists a constant c > 0 such that in ev-

ery nontrivial finite group G there exists a sequence of
length at most c log |G| with the property that each el-
ement of G equals the product of some subsequence.
 Solutions to the 69th William Lowell Putnam Mathematical Competition
Saturday, December 6, 2008
Kiran Kedlaya and Lenny Ng

A–1 The function g(x) = f (x, 0) works. Substituting a01 , . . . , a0h are not. Repeating this argument for each pair
(x, y, z) = (0, 0, 0) into the given functional equa- (p, m) such that pm divides the initial product a1 , . . . , an ,
tion yields f (0, 0) = 0, whence substituting (x, y, z) = we can determine the exact prime factorization of each
(x, 0, 0) yields f (x, 0) + f (0, x) = 0. Finally, substi- of a01 , . . . , a0n . This proves that the final sequence is
tuting (x, y, z) = (x, y, 0) yields f (x, y) = − f (y, 0) − unique.
f (0, x) = g(x) − g(y). Remark: (by David Savitt and Noam Elkies) Here are
Remark: A similar argument shows that the possible two other ways to prove the termination. One is to ob-
functions g are precisely those of the form f (x, 0) + c serve that ∏ j a jj is strictly increasing at each step, and
for some c. bounded above by (a1 · · · an )n . The other is to notice
that a1 is nonincreasing but always positive, so even-
A–2 Barbara wins using one of the following strategies.
tually becomes constant; then a2 is nonincreasing but
First solution: Pair each entry of the first row with the always positive, and so on.
entry directly below it in the second row. If Alan ever
Reinterpretation: For each p, consider the sequence
writes a number in one of the first two rows, Barbara
consisting of the exponents of p in the prime factoriza-
writes the same number in the other entry in the pair. If
tions of a1 , . . . , an . At each step, we pick two positions
Alan writes a number anywhere other than the first two
i and j such that the exponents of some prime p are in
rows, Barbara does likewise. At the end, the resulting
the wrong order at positions i and j. We then sort these
matrix will have two identical rows, so its determinant
two position into the correct order for every prime p
will be zero.
simultaneously.
Second solution: (by Manjul Bhargava) Whenever
It is clear that this can only terminate with all se-
Alan writes a number x in an entry in some row, Bar-
quences being sorted into the correct order. We must
bara writes −x in some other entry in the same row. At
still check that the process terminates; however, since
the end, the resulting matrix will have all rows summing
all but finitely many of the exponent sequences consist
to zero, so it cannot have full rank.
of all zeroes, and each step makes a nontrivial switch
A–3 We first prove that the process stops. Note first that in at least one of the other exponent sequences, it is
the product a1 · · · an remains constant, because a j ak = enough to check the case of a single exponent sequence.
gcd(a j , ak ) lcm(a j , ak ). Moreover, the last number in This can be done as in the first solution.
the sequence can never decrease, because it is always Remark: Abhinav Kumar suggests the following
proof
replaced by its least common multiple with another that the process always terminates in at most n2 steps.
number. Since it is bounded above (by the product of (This is a variant of the worst-case analysis of the bub-
all of the numbers), the last number must eventually ble sort algorithm.)
reach its maximum value, after which it remains con-
Consider the number of pairs (k, l) with 1 ≤ k < l ≤ n
stant throughout. After this happens, the next-to-last
such that ak does not divide al (call these bad pairs).
number will never decrease, so it eventually becomes
At each step, we find one bad pair (i, j) and eliminate
constant, and so on. After finitely many steps, all of the
it, and we do not touch any pairs that do not involve
numbers will achieve their final values, so no more steps
either i or j. If i < k < j, then neither of the pairs (i, k)
will be possible. This only happens when a j divides ak
and (k, j) can become bad, because ai is replaced by a
for all pairs j < k.
divisor of itself, while a j is replaced by a multiple of
We next check that there is only one possible final se- itself. If k < i, then (k, i) can only become a bad pair if
quence. For p a prime and m a nonnegative integer, we ak divided ai but not a j , in which case (k, j) stops being
claim that the number of integers in the list divisible bad. Similarly, if k > j, then (i, k) and ( j, k) either stay
by pm never changes. To see this, suppose we replace the same or switch status. Hence the number of bad
a j , ak by gcd(a j , ak ), lcm(a j , ak ). If neither of a j , ak is pairs goes

down by at least 1 each time; since it is at
divisible by pm , then neither of gcd(a j , ak ), lcm(a j , ak ) most n2 to begin with, this is an upper bound for the
is either. If exactly one a j , ak is divisible by pm , then number of steps.
lcm(a j , ak ) is divisible by pm but gcd(a j , ak ) is not.
Remark: This problem is closely related to the clas-
gcd(a j , ak ), lcm(a j , ak ) are as well. sification theorem for finite abelian groups. Namely,
If we started out with exactly h numbers not divisible if a1 , . . . , an and a01 , . . . , a0n are the sequences obtained
by pm , then in the final sequence a01 , . . . , a0n , the num- at two different steps in the process, then the abelian
bers a0h+1 , . . . , a0n are divisible by pm while the numbers
 2

groups Z/a1 Z × · · · × Z/an Z and Z/a01 Z × · · · × Z/a0n Z up with 0, ∗, 1 for any of ∗ = a, b, c, whereas for the pen-
are isomorphic. The final sequence gives a canonical tagon we can end up with 0, ∗, 1 for any of ∗ = a, b.
presentation of this group; the terms of this sequence Consequently, the final sequence is determined by the
are called the elementary divisors or invariant factors initial sequence if and only if L is distributive.
of the group.
Remark: (by Tom Belulovich) A lattice is a partially A–4 The sum diverges. From the definition, f (x) = x on
e
ordered set L in which for any two x, y ∈ L, there is a [1, e], x ln x on (e, ee ], x ln x ln ln x on (ee , ee ], and so
unique minimal element z with z ≥ x and z ≥ y, called forth. It follows that on [1, ∞), f is positive, continu-
1
the join and denoted x ∧ y, and there is a unique max- ous, and increasing. Thus ∑∞ n=1 f (n) , if it converges, is
R ∞ dx
imal element z with z ≤ x and z ≤ y, called the meet bounded below by 1 f (x) ; it suffices to prove that the
and denoted x ∨ y. In terms of a lattice L, one can pose integral diverges.
the following generalization of the given problem. Start
Write ln1 x = ln x and lnk x = ln(lnk−1 x)
with a1 , . . . , an ∈ L. If i < j but ai 6≤ a j , it is permit-
for k ≥ 2; similarly write exp1 x = ex and
ted to replace ai , a j by ai ∨ a j , ai ∧ a j , respectively. The k−1
same argument as
above shows that this always termi- expk x = eexp x . If we write y = lnk x, then
nates in at most n2 steps. The question is, under what x = exp y and dx = (expk y)(expk−1 y) · · · (exp1 y)dy =
k

conditions on the lattice L is the final sequence uniquely x(ln1 x) · · · (lnk−1 x)dy. Now on [expk−1 1, expk 1], we
determined by the initial sequence? have f (x) = x(ln1 x) · · · (lnk−1 x), and thus substituting
It turns out that this holds if and only if L is distributive, y = lnk x yields
i.e., for any x, y, z ∈ L, Z expk 1 Z 1
dx
= dy = 1.
x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z). expk−1 1 f (x) 0

k
(This is equivalent to the same axiom with the oper- R ∞ dx ∞ R exp 1 dx
It follows that 1 f (x) = ∑k=1 expk−1 1 f (x) diverges, as
ations interchanged.) For example, if L is a Boolean
desired.
algebra, i.e., the set of subsets of a given set S under in-
clusion, then ∧ is union, ∨ is intersection, and the dis- A–5 Form the polynomial P(z) = f (z) + ig(z) with complex
tributive law holds. Conversely, any finite distributive coefficients. It suffices to prove that P has degree at
lattice is contained in a Boolean algebra by a theorem least n − 1, as then one of f , g must have degree at least
of Birkhoff. The correspondence takes each x ∈ L to n − 1.
the set of y ∈ L such that x ≥ y and y cannot be written
By replacing P(z) with aP(z) + b for suitable a, b ∈
as a join of two elements of L \ {y}. (See for instance
C, we can force the regular n-gon to have vertices
Birkhoff, Lattice Theory, Amer. Math. Soc., 1967.)
ζn , ζn2 , . . . , ζnn for ζn = exp(2πi/n). It thus suffices to
On one hand, if L is distributive, it can be shown that check that there cannot exist a polynomial P(z) of de-
the j-th term of the final sequence is equal to the meet gree at most n − 2 such that P(i) = ζni for i = 1, . . . , n.
of ai1 ∧ · · · ∧ ai j over all sequences 1 ≤ i1 < · · · < i j ≤ n.
We will prove more generally that for any complex
For instance, this can be checked by forming the small-
number t ∈ / {0, 1}, and any integer m ≥ 1, any polyno-
est subset L0 of L containing a1 , . . . , an and closed under
mial Q(z) for which Q(i) = t i for i = 1, . . . , m has degree
meet and join, then embedding L0 into a Boolean alge-
at least m − 1. There are several ways to do this.
bra using Birkhoff’s theorem, then checking the claim
for all Boolean algebras. It can also be checked di- First solution: If Q(z) has degree d and leading coeffi-
rectly (as suggested by Nghi Nguyen) by showing that cient c, then R(z) = Q(z + 1) − tQ(z) has degree d and
for j = 1, . . . , n, the meet of all joins of j-element sub- leading coefficient (1 − t)c. However, by hypothesis,
sets of a1 , . . . , an is invariant at each step. R(z) has the distinct roots 1, 2, . . . , m − 1, so we must
have d ≥ m − 1.
On the other hand, a lattice fails to be distributive if
and only if it contains five elements a, b, c, 0, 1 such that Second solution: We proceed by induction on m. For
either the only relations among them are implied by the base case m = 1, we have Q(1) = t 1 6= 0, so Q must
be nonzero, and so its degree is at least 0. Given the
1 ≥ a, b, c ≥ 0 assertion for m − 1, if Q(i) = t i for i = 1, . . . , m, then
the polynomial R(z) = (t − 1)−1 (Q(z + 1) − Q(z)) has
(this lattice is sometimes called the diamond), or the degree one less than that of Q, and satisfies R(i) = t i
only relations among them are implied by for i = 1, . . . , m − 1. Since R must have degree at least
m − 2 by the induction hypothesis, Q must have degree
1 ≥ a ≥ b ≥ 0, 1≥c≥0 at least m − 1.
(this lattice is sometimes called the pentagon). (For a Third solution: We use the method of finite differ-
proof, see the Birkhoff reference given above.) For each ences (as in the second solution) but without induction.
of these examples, the initial sequence a, b, c fails to de-
termine the final sequence; for the diamond, we can end
 3

Namely, the (m − 1)-st finite difference of P evaluated no greater than

at 1 equals
1 − (2m − m2 /n)/n = s2 .
m−1  
j m−1
∑ (−1) Q(m − j) = t(1 − t)m−1 6= 0, We start out with k = 0 and s = 1 − 1/n; after k steps,
j=0 j k
we have s ≤ (1 − 1/n)2 . It is enough to prove that for
which is impossible if Q has degree less than m − 1. some c > 0, we can always find an integer k ≤ c ln n
such that
Remark: One can also establish the claim by comput-
ing a Vandermonde-type determinant, or by using the k
1 2

1
Lagrange interpolation formula to compute the leading 1− < ,
n n
coefficient of Q.
A–6 For notational convenience, we will interpret the prob- as then we have n − m < 1 and hence H = G.
lem as allowing the empty subsequence, whose product To obtain this last inequality, put
is the identity element of the group. To solve the prob-
lem in the interpretation where the empty subsequence k = b2 log2 nc < (2/ ln 2) ln n,
is not allowed, simply append the identity element to
the sequence given by one of the following solutions. so that 2k+1 ≥ n2 . From the facts that ln n ≤ ln 2 + (n −
2)/2 ≤ n/2 and ln(1 − 1/n) < −1/n for all n ≥ 2, we
First solution: Put n = |G|. We will say that a sequence have
S produces an element g ∈ G if g occurs as the product
n2
 
of some subsequence of S. Let H be the set of elements k 1 n
produced by the sequence S. 2 ln 1 − < − = − < − ln n,
n 2n 2
Start with S equal to the empty sequence. If at any point
the set H −1 H = {h1 h2 : h−1 yielding the desired inequality.
1 , h2 ∈ H} fails to be all of G,
extend S by appending an element g of G not in H −1 H. Remark: An alternate approach in the second solution
Then Hg ∩ H must be empty, otherwise there would be is to distinguish betwen the cases of H small (i.e., m <
an equation of the form h1 g = h2 with h1 , h2 ∈ G, or n1/2 , in which case m can be replaced by a value no
g = h−1
1 h2 , a contradiction. Thus we can extend S by
less than 2m − 1) and H large. This strategy is used in a
one element and double the size of H. number of recent results of Bourgain, Tao, Helfgott, and
others on small doubling or small tripling of subsets of
After k ≤ log2 n steps, we must obtain a sequence S =
finite groups.
a1 , . . . , ak for which H −1 H = G. Then the sequence
a−1 −1
k , . . . , a1 , a1 , . . . , ak produces all of G and has length
In the second solution, if we avoid the rather weak in-
at most (2/ ln 2) ln n. equality ln n ≤ n/2, we instead get sequences of length
log2 (n ln n) = log2 (n) + log2 (ln n). This is close to op-
Second solution:
timal: one cannot use fewer than log2 n terms because
Put m = |H|. We will show that we can append one the number of subsequences must be at least n.
element g to S so that the resulting sequence of k + 1
elements will produce at least 2m − m2 /n elements of B–1 There are at most two such points. For example,
G. To see this, we compute the points (0, 0) and (1, 0) lie on a circle with center
(1/2, x) for any real number x, not necessarily rational.
∑ |H ∪ Hg| = ∑ (|H| + |Hg| − |H ∩ Hg|) On the other hand, suppose P = (a, b), Q = (c, d), R =
g∈G g∈G
(e, f ) are three rational points that lie on a circle. The
= 2mn − ∑ |H ∩ Hg| midpoint M of the side PQ is ((a + c)/2, (b + d)/2),
g∈G which is again rational. Moreover, the slope of the line
= 2mn − |{(g, h) ∈ G2 : h ∈ H ∩ Hg}| PQ is (d −b)/(c−a), so the slope of the line through M
perpendicular to PQ is (a−c)/(b−d), which is rational
= 2mn − ∑ |{g ∈ G : h ∈ Hg}|
h∈H
or infinite.
Similarly, if N is the midpoint of QR, then N is a rational
= 2mn − ∑ |H −1 h| point and the line through N perpendicular to QR has
h∈H
2 rational slope. The center of the circle lies on both of
= 2mn − m .
these lines, so its coordinates (g, h) satisfy two linear
By the pigeonhole principle, we have |H ∪ Hg| ≥ 2m − equations with rational coefficients, say Ag + Bh = C
m2 /n for some choice of g, as claimed. and Dg + Eh = F. Moreover, these equations have a
unique solution. That solution must then be
In other words, by extending the sequence by one ele-
ment, we can replace the ratio s = 1−m/n (i.e., the frac- g = (CE − BD)/(AE − BD)
tion of elements of G not generated by S) by a quantity
h = (AF − BC)/(AE − BD)
 4

(by elementary algebra, or Cramer’s rule), so the center this holds for the unit vector in the same direction as
of the circle is rational. This proves the desired result. (v1i , v2i ), we must have
Remark: The above solution is deliberately more ver-
bose than is really necessary. A shorter way to say this v21i + v22i ≤ 1 (i = 1, . . . , 4).
is that any two distinct rational points determine a ra-
Conversely, if this holds, then the Cauchy-Schwarz in-
tional line (a line of the form ax + by + c = 0 with a, b, c
equality and the above analysis imply that C lies in H.
rational), while any two nonparallel rational lines inter-
sect at a rational point. A similar statement holds with If r is the radius of C, then
the rational numbers replaced by any field. 4 4
Remark: A more explicit argument is to show that 2r2 = ∑ v21i + ∑ v22i
the equation of the circle through the rational points i=1 i=1
(x1 , y1 ), (x2 , y2 ), (x3 , y3 ) is 4

 2 = ∑ (v21i + v22i )
x1 + y21 x1 y1 1

i=1
x2 + y2 x2 y2 1
2 2
≤ 4,
0 = det  x2 + y2 x3 y3 1

3 3 √
x 2 + y2 x y 1 so r ≤ 2. Since this is achieved by the circle through
(1, 1, 0, 0) and (0, 0, 1, 1), it is the desired maximum.
which has the form a(x2 + y2 ) + dx + ey + f = 0 Remark: One may similarly ask for the radius of the
for a, d, e, f rational. The center of this circle is largest k-dimensional ball inside an n-dimensional unit
(−d/(2a), −e/(2a)), which is again a rational point. hypercube; the given problem is the case (n, k) = (4, 2).
Daniel Kane gives the following argument
p to show that
B–2 We claim that Fn (x) = (ln x − an )xn /n!, where an =
the maximum radius in this case is 12 nk . (Thanks for
∑nk=1 1/k. Indeed, temporarily write Gn (x) = (ln x − Noam Elkies for passing this along.)
an )xn /n! for x > 0 and n ≥ 1; then limx→0 Gn (x) = 0
and G0n (x) = (ln x − an + 1/n)xn−1 /(n − 1)! = Gn−1 (x), We again scale up by a factor of 2, so that we are trying
and the claim follows by the Fundamental Theorem of to show that the maximum radius r of a k-dimensional
ball contained in the hypercube [−1, 1]n is nk . Again,
p
Calculus and induction on n.
there is no loss of generality in centering the ball at the
Given the claim, we have Fn (1) = −an /n! and so we
origin. Let T : Rk → Rn be a similitude carrying the
need to evaluate − limn→∞ lnann . But since the function
unit ball to this embedded k-ball. Then there exists a
1/x is strictly decreasingR for x positive, ∑nk=2 1/k = an −
vector vi ∈ Rk such that for e1 , . . . , en the standard basis
1 isRbounded below by 2n dx/x = ln n − ln 2 and above
of Rn , x · vi = T (x) · ei for all x ∈ Rk . The condition of
by 1n dx/x = ln n. It follows that limn→∞ lnann = 1, and
the problem is equivalent to requiring |vi | ≤ 1 for all i,
the desired limit is −1.
√
while the radius r of the embedded ball is determined
B–3 The largest possible radius is 22 . It will be convenient by the fact that for all x ∈ Rk ,
to solve the problem for a hypercube of side length 2 n
instead, in which√ case we are trying to show that the r2 (x · x) = T (x) · T (x) = ∑ x · vi .
largest radius is 2. i=1

Choose coordinates so that the interior of the hypercube Let M be the matrix with columns v1 , . . . , vk ; then
is the set H = [−1, 1]4 in R4 . Let C be a circle centered MM T = r2 Ik , for Ik the k × k identity matrix. We then
at the point P. Then C is contained both in H and its have
reflection across P; these intersect in a rectangular par-
alellepiped each of whose pairs of opposite faces are at kr2 = Trace(r2 Ik ) = Trace(MM T )
most 2 unit apart. Consequently, if we translate C so n
that its center moves to the point O = (0, 0, 0, 0) at the = Trace(M T M) = ∑ |vi |2
center of H, then it remains entirely inside H. i=1

This means that the answer we seek equals the largest ≤ n,

possible radius of a circle C contained in H and pn
centered at O. Let v1 = (v11 , . . . , v14 ) and v2 = yielding the upper bound r ≤ k.
(v21 , . . . , v24 ) be two points on C lying on radii perpen- To show that this bound is optimal, it is enough to show
dicular to each other. Then the points of the circle can that one can find an orthogonal projection of Rn onto Rk
be expressed as v1 cos θ + v2 sin θ for 0 ≤ θ < 2π. Then so that the projections of the ei all have the same norm
C lies in H if and only if for each i, we have (one can then rescale to get the desired configuration
of v1 , . . . , vn ). We construct such a configuration by a
|v1i cos θ + v2i sin θ | ≤ 1 (0 ≤ θ < 2π). “smoothing” argument. Startw with any projection. Let
In geometric terms, the vector (v1i , v2i ) in R2 has
dot product at most 1 with every unit vector. Since
 5

w1 , . . . , wn be the projections of e1 , . . . , en . If the desired equal to some integer c = f 0 ( ab ): f ( an+1 a c

bn ) = f ( b ) + bn .
condition is not achieved, we can choose i, j such that an+1 a
Now c cannot be 0, since otherwise f ( bn ) = f ( b ) for
sufficiently large n has denominator b rather than bn.
1
|wi |2 < (|w1 |2 + · · · + |wn |2 ) < |w j |2 . Similarly, |c| cannot be greater than 1: otherwise if we
n take n = k|c| for k a sufficiently large positive integer,
By precomposing with a suitable rotation that fixes then f ( ab ) + bnc
has denominator bk, contradicting the
an+1
eh for h 6= i, j, we can vary |wi |, |w j | without varying fact that f ( bn ) has denominator bn. It follows that
|wi |2 + |w j |2 or |wh | for h 6= i, j. We can thus choose c = f 0 ( ab ) = ±1.
such a rotation to force one of |wi |2 , |w j |2 to become Thus the derivative of f at any rational number is ±1.
equal to 1n (|w1 |2 + · · · + |wn |2 ). Repeating at most n − 1 Since f is continuously differentiable, we conclude that
times gives the desired configuration. f 0 (x) = 1 for all real x or f 0 (x) = −1 for all real x. Since
f (0) must be an integer (a rational number with denom-
B–4 We use the identity given by Taylor’s theorem: inator 1), f (x) = x+n or f (x) = −x+n for some integer
n.
deg(h) (i)
h (x) Remark: After showing that f 0 (q) is an integer for each
h(x + y) = ∑ yi .
i=0 i! q, one can instead argue that f 0 is a continuous function
from the rationals to the integers, so must be constant.
In this expression, h(i) (x)/i! is a polynomial in x with One can then write f (x) = ax + b and check that b ∈ Z
integer coefficients, so its value at an integer x is an by evaluation at a = 0, and that a = ±1 by evaluation at
integer. x = 1/a.
For x = 0, . . . , p − 1, we deduce that B–6 In all solutions, let Fn,k be the number of k-limited per-
0 2 mutations of {1, . . . , n}.
h(x + p) ≡ h(x) + ph (x) (mod p ).
First solution: (by Jacob Tsimerman) Note that any
(This can also be deduced more directly using the bino- permutation is k-limited if and only if its inverse is k-
mial theorem.) Since we assumed h(x) and h(x + p) limited. Consequently, the number of k-limited per-
are distinct modulo p2 , we conclude that h0 (x) 6≡ 0 mutations of {1, . . . , n} is the same as the number of
(mod p). Since h0 is a polynomial with integer coef- k-limited involutions (permutations equal to their in-
ficients, we have h0 (x) ≡ h0 (x + mp) (mod p) for any verses) of {1, . . . , n}.
integer m, and so h0 (x) 6≡ 0 (mod p) for all integers x. We use the following fact several times: the number of
Now for x = 0, . . . , p2 − 1 and y = 0, . . . , p − 1, we write involutions of {1, . . . , n} is odd if n = 0, 1 and even oth-
erwise. This follows from the fact that non-involutions
h(x + yp2 ) ≡ h(x) + p2 yh0 (x) (mod p3 ). come in pairs, so the number of involutions has the same
parity as the number of permutations, namely n!.
Thus h(x), h(x+ p2 ), . . . , h(x+(p−1)p2 ) run over all of
the residue classes modulo p3 congruent to h(x) mod- For n ≤ k + 1, all involutions are k-limited. By the pre-
ulo p2 . Since the h(x) themselves cover all the residue vious paragraph, Fn,k is odd for n = 0, 1 and even for
classes modulo p2 , this proves that h(0), . . . , h(p3 − 1) n = 2, . . . , k + 1.
are distinct modulo p3 . For n > k + 1, group the k-limited involutions into
Remark: More generally, the same proof shows that classes based on their actions on k + 2, . . . , n. Note
for any integers d, e > 1, h permutes the residue classes that for C a class and σ ∈ C, the set of elements of
modulo pd if and only if it permutes the residue classes A = {1, . . . , k + 1} which map into A under σ depends
modulo pe . The argument used in the proof is related only on C, not on σ . Call this set S(C); then the size of
to a general result in number theory known as Hensel’s C is exactly the number of involutions of S(C). Conse-
lemma. quently, |C| is even unless S(C) has at most one element.
However, the element 1 cannot map out of A because we
B–5 The functions f (x) = x + n and f (x) = −x + n for any are looking at k-limited involutions. Hence if S(C) has
integer n clearly satisfy the condition of the problem; one element and σ ∈ C, we must have σ (1) = 1. Since
we claim that these are the only possible f . σ is k-limited and σ (2) cannot belong to A, we must
Let q = a/b be any rational number with gcd(a, b) = 1 have σ (2) = k + 2. By induction, for i = 3, . . . , k + 1,
and b > 0. For n any positive integer, we have we must have σ (i) = k + i.
If n < 2k + 1, this shows that no class C of odd cardi-
f ( an+1 a
bn ) − f ( b ) nality can exist, so Fn,k must be even. If n ≥ 2k + 1,
 
an + 1 a
1
= bn f − nb f the classes of odd cardinality are in bijection with k-
bn
bn b
limited involutions of {2k + 2, . . . , n}, so Fn,k has the
is an integer by the property of f . Since f is differ- same parity as Fn−2k−1,k . By induction on n, we deduce
entiable at a/b, the left hand side has a limit. It fol- the desired result.
lows that for sufficiently large n, both sides must be
 6

Second solution: (by Yufei Zhao) Let Mn,k be the n × n We next add column 1 to each of columns 2, . . . , k + 1.
matrix with  
( 1 0 0 0 0 0 0 0/
1 |i − j| ≤ k 0 0 0 0 1 0 0 0/ 
(Mn,k )i j =  
0 otherwise. 0

0 0 0 1 1 0 0/ 

 
0 0 0 0 1 1 1 0/ 
Write det(Mn,k ) as the sum over permutations σ of 
0

{1, . . . , n} of (Mn,k )1σ (1) · · · (Mn,k )nσ (n) times the signa-  1 1 1 1 1 1 ? 
ture of σ . Then σ contributes ±1 to det(Mn,k ) if σ is
 
0 0 1 1 1 1 1 ?
k-limited and 0 otherwise. We conclude that
 
0 0 0 1 1 1 1 ?
det(Mn,k ) ≡ Fn,k (mod 2). 0/ 0/ 0/ 0/ ? ? ? ∗

For i = 2, for each of j = i + 1, . . . , 2k + 1 for which the

For the rest of the solution, we interpret Mn,k as a matrix
( j, k + i)-entry is nonzero, add row i to row j.
over the field of two elements. We compute its determi-
nant using linear algebra modulo 2.  
1 0 0 0 0 0 0 0/
We first show that for n ≥ 2k + 1, 0
 0 0 0 1 0 0 0/ 

0 0 0 0 0 1 0 0/ 
 
Fn,k ≡ Fn−2k−1,k (mod 2),  
0 0 0 0 0 1 1 0/ 
provided that we interpret F0,k = 1. We do this by com-
 
0 1 1 1 0 1 1 ?
puting det(Mn,k ) using row and column operations. We 



will verbally describe these operations for general k, 0 0 1 1 0 1 1 ?
 
while illustrating with the example k = 3. 0 0 0 1 0 1 1 ?
To begin with, Mn,k has the following form. 0/ 0/ 0/ 0/ 0/ ? ? ∗

1 1 1 1 0 0 0 0/
 Repeat the previous step for i = 3, . . . , k + 1 in succes-
1 1 1 1 1 0 0 0/  sion.
1 1 1 1 1 1 0 0/   
1 0 0 0 0 0 0 0/
 
1 1 1 1 1 1 1 0/ 
   0 0 0 0 1 0 0 0/ 
0 1 1 1 1 1 1 ?  
 
0 0 1 1 1 1 1 ?  0 0 0 0 0 1 0 0/ 
 
 
0 0 0 1 1 1 1 ?

 0 0 0 0 0 0 1 0/ 

 
0/ 0/ 0/ 0/ ? ? ? ∗ 0 1 1 1 0 0 0 ?
 
 
0 0 1 1 0 0 0 ?
In this presentation, the first 2k + 1 rows and columns  
are shown explicitly; the remaining rows and columns 0 0 0 1 0 0 0 ?
are shown in a compressed format. The symbol 0/ in- 0/ 0/ 0/ 0/ 0/ 0/ 0/ ∗
dicates that the unseen entries are all zeroes, while the
symbol ? indicates that they are not. The symbol ∗ in Repeat the two previous steps with the roles of the rows
the lower right corner represents the matrix Fn−2k−1,k . and columns reversed. That is, for i = 2, . . . , k + 1, for
We will preserve the unseen structure of the matrix by each of j = i+1, . . . , 2k +1 for which the ( j, k +i)-entry
only adding the first k + 1 rows or columns to any of the is nonzero, add row i to row j.
others.  
We first add row 1 to each of rows 2, . . . , k + 1. 1 0 0 0 0 0 0 0/
0 0 0 0 1 0 0 0/ 
   
1 1 1 1 0 0 0 0/ 0 0 0 0 0 1 0 0/ 
 
0  
 0 0 0 1 0 0 0/ 
 0
 0 0 0 0 0 1 0/ 

0 0 0 0 1 1 0 0/ 
  0
 1 0 0 0 0 0 0/ 

 
0 0 0 0 1 1 1 0/ 

0 0 1 0 0 0 0

0/ 
   
0 1 1 1 1 1 1 ? 0 0 0 1 0 0 0 0/ 
 
∗
 
0 0 1 1 1 1 1 ? 0/ 0/ 0/ 0/ 0/ 0/ 0/
 
0 0 0 1 1 1 1 ?
We now have a block diagonal matrix in which the top
0/ 0/ 0/ 0/ ? ? ? ∗ left block is a (2k + 1) × (2k + 1) matrix with nonzero
determinant (it results from reordering the rows of the
 7

identity matrix), the bottom right block is Mn−2k−1,k , We next check that if n ≡ 0, 1 (mod 2k + 1), then Mn,k
and the other two blocks are zero. We conclude that is invertible. Suppose that a1 , . . . , an are scalars such
that a1 r1 + · · · + an rn is the zero vector. The m-th co-
det(Mn,k ) ≡ det(Mn−2k−1,k ) (mod 2), ordinate of this vector equals am−k + · · · + am+k , where
we regard ai as zero if i ∈ / {1, . . . , n}. By comparing
proving the desired congruence. consecutive coordinates, we obtain
To prove the desired result, we must now check that
F0,k , F1,k are odd and F2,k , . . . , F2k,k are even. For n = am−k = am+k+1 (1 ≤ m < n).
0, . . . , k + 1, the matrix Mn,k consists of all ones, so its
determinant is 1 if n = 0, 1 and 0 otherwise. (Alter- In particular, the ai repeat with period 2k + 1. Taking
natively, we have Fn,k = n! for n = 0, . . . , k + 1, since m = 1, . . . , k further yields that
every permutation of {1, . . . , n} is k-limited.) For n =
ak+2 = · · · = a2k+1 = 0
k + 2, . . . , 2k, observe that rows k and k + 1 of Mn,k both
consist of all ones, so det(Mn,k ) = 0 as desired. while taking m = n − k, . . . , n − 1 yields
Third solution: (by Tom Belulovich) Define Mn,k as
in the second solution. We prove det(Mn,k ) is odd for an−2k = · · · = an−1−k = 0.
n ≡ 0, 1 (mod 2k + 1) and even otherwise, by directly
determining whether or not Mn,k is invertible as a matrix For n ≡ 0 (mod 2k + 1), the latter can be rewritten as
over the field of two elements.
a1 = · · · = ak = 0
Let ri denote row i of Mn,k . We first check that if n ≡
2, . . . , 2k (mod 2k + 1), then Mn,k is not invertible. In whereas for n ≡ 1 (mod 2k + 1), it can be rewritten as
this case, we can find integers 0 ≤ a < b ≤ k such that
n + a + b ≡ 0 (mod 2k + 1). Put j = (n + a + b)/(2k + a2 = · · · = ak+1 = 0.
1). We can then write the all-ones vector both as In either case, since we also have
j−1 a1 + · · · + a2k+1 = 0
∑ rk+1−a+(2k+1)i
i=0 from the (k + 1)-st coordinate, we deduce that all of the
ai must be zero, and so Mn,k must be invertible.
and as
Remark: The matrices Mn,k are examples of banded
j−1
matrices, which occur frequently in numerical appli-
∑ rk+1−b+(2k+1)i . cations of linear algebra. They are also examples of
i=0
Toeplitz matrices.
Hence Mn,k is not invertible.
 The 70th William Lowell Putnam Mathematical Competition
Saturday, December 5, 2009

A1 Let f be a real-valued function on the plane such that for B1 Show that every positive rational number can be written
every square ABCD in the plane, f (A) + f (B) + f (C) + as a quotient of products of factorials of (not necessarily
f (D) = 0. Does it follow that f (P) = 0 for all points P distinct) primes. For example,
in the plane?
10 2! · 5!
A2 Functions f , g, h are differentiable on some open inter- = .
9 3! · 3! · 3!
val around 0 and satisfy the equations and initial condi-
tions
B2 A game involves jumping to the right on the real number
1 line. If a and b are real numbers and b > a, the cost of
f 0 = 2 f 2 gh + , f (0) = 1, jumping from a to b is b3 − ab2 . For what real numbers
gh
4 c can one travel from 0 to 1 in a finite number of jumps
0 2
g = f g h + , g(0) = 1, with total cost exactly c?
fh
1 B3 Call a subset S of {1, 2, . . . , n} mediocre if it has the fol-
h0 = 3 f gh2 + , h(0) = 1.
fg lowing property: Whenever a and b are elements of S
whose average is an integer, that average is also an ele-
Find an explicit formula for f (x), valid in some open ment of S. Let A(n) be the number of mediocre subsets
interval around 0. of {1, 2, . . . , n}. [For instance, every subset of {1, 2, 3}
except {1, 3} is mediocre, so A(3) = 7.] Find all posi-
A3 Let dn be the determinant of the n × n matrix whose tive integers n such that A(n + 2) − 2A(n + 1) + A(n) =
entries, from left to right and then from top to bottom, 1.
are cos 1, cos 2, . . . , cos n2 . (For example,
B4 Say that a polynomial with real coefficients in two vari-
cos 1 cos 2 cos 3 ables, x, y, is balanced if the average value of the poly-
d3 = cos 4 cos 5 cos 6 . nomial on each circle centered at the origin is 0. The
cos 7 cos 8 cos 9 balanced polynomials of degree at most 2009 form a
vector space V over R. Find the dimension of V .
The argument of cos is always in radians, not degrees.)
Evaluate limn→∞ dn . B5 Let f : (1, ∞) → R be a differentiable function such that

A4 Let S be a set of rational numbers such that x2 − f (x)2

f 0 (x) = for all x > 1.
(a) 0 ∈ S; x2 ( f (x)2 + 1)
(b) If x ∈ S then x + 1 ∈ S and x − 1 ∈ S; and Prove that limx→∞ f (x) = ∞.
1
(c) If x ∈ S and x 6∈ {0, 1}, then x(x−1) ∈ S.
B6 Prove that for every positive integer n, there is a se-
Must S contain all rational numbers? quence of integers a0 , a1 , . . . , a2009 with a0 = 0 and
a2009 = n such that each term after a0 is either an ear-
A5 Is there a finite abelian group G such that the product of lier term plus 2k for some nonnegative integer k, or of
the orders of all its elements is 22009 ? the form b mod c for some earlier positive terms b and c.
[Here b mod c denotes the remainder when b is divided
A6 Let f : [0, 1]2 → R be a continuous function on the by c, so 0 ≤ (b mod c) < c.]
closed unit square such that ∂∂ xf and ∂∂ yf exist and are
continuous on the interior (0, 1)2 . Let a = 01 f (0, y) dy,
R

b = 01 f (1, y) dy, c = 01 f (x, 0) dx, d = 01 f (x, 1) dx.

R R R

Prove or disprove: There must be a point (x0 , y0 ) in

(0, 1)2 such that

∂f ∂f
(x0 , y0 ) = b − a and (x0 , y0 ) = d − c.
∂x ∂y
 Solutions to the 70th William Lowell Putnam Mathematical Competition
Saturday, December 5, 2009
Kiran Kedlaya and Lenny Ng

A–1 Yes, it does follow. Let P be any point in the plane. Let A–3 The limit is 0; we will show this by checking that
ABCD be any square with center P. Let E, F, G, H be dn = 0 for all n ≥ 3. Starting from the given matrix,
the midpoints of the segments AB, BC,CD, DA, respec- add the third column to the first column; this does not
tively. The function f must satisfy the equations change the determinant. However, thanks to the identity
cos x+cos y = 2 cos x+y x−y
2 cos 2 , the resulting matrix has
0= f (A) + f (B) + f (C) + f (D) the form
0= f (E) + f (F) + f (G) + f (H)
2 cos 2 cos 1 cos 2 ···
 
0= f (A) + f (E) + f (P) + f (H)  2 cos(n + 2) cos 1 cos(n + 2) · · ·
0= f (B) + f (F) + f (P) + f (E) 2 cos(2n + 2) cos 1 2 cos(2n + 2) · · ·
 
0= f (C) + f (G) + f (P) + f (F) .. .. ..
. . .
0= f (D) + f (H) + f (P) + f (G).
with the first column being a multiple of the second.
If we add the last four equations, then subtract the first Hence dn = 0.
equation and twice the second equation, we obtain 0 = Remark. Another way to draw the same conclu-
4 f (P), whence f (P) = 0. sion is to observe that the given matrix is the sum of
Remark. Problem 1 of the 1996 Romanian IMO team the two rank 1 matrices A jk = cos( j − 1)n cos k and
selection exam asks the same question with squares re- B jk = − sin( j − 1)n sin k, and so has rank at most 2.
placed by regular polygons of any (fixed) number of One can also use the matrices A jk = ei(( j−1)n+k) , B jk =
vertices. e−i( j−1)n+k .
A–2 Multiplying the first differential equation by gh, the sec- A–4 The answer is no; indeed, S = Q \ {n + 2/5 | n ∈ Z} sat-
ond by f h, and the third by f g, and summing gives isfies the given conditions. Clearly S satisfies (a) and
(b); we need only check that it satisfies (c). It suffices
( f gh)0 = 6( f gh)2 + 6.
to show that if x = p/q is a fraction with (p, q) = 1 and
Write k(x) = f (x)g(x)h(x); then k0 = 6k2 +6 and k(0) = p > 0, then we cannot have 1/(x(x − 1)) = n + 2/5 for
1. One solution for this differential equation with this an integer n. Suppose otherwise; then
initial condition is k(x) = tan(6x + π/4); by standard
(5n + 2)p(p − q) = 5q2 .
uniqueness, this must necessarily hold for x in some
open interval around 0. Now the first given equation Since p and q are relatively prime, and p divides
becomes 5q2 , we must have p | 5, so p = 1 or p = 5. On the
other hand, p − q and q are also relatively prime, so
f 0 / f = 2k(x) + 1/k(x)
p − q divides 5 as well, and p − q must be ±1 or
= 2 tan(6x + π/4) + cot(6x + π/4); ±5. This leads to eight possibilities for (p, q): (1, 0),
(5, 0), (5, 10), (1, −4), (1, 2), (1, 6), (5, 4), (5, 6). The
integrating both sides gives first three are impossible, while the final five lead to
−2 ln cos(6x + π/4) + ln sin(6x + π/4) 5n + 2 = 16, −20, −36, 16, −36 respectively, none of
ln( f (x)) = + c, which holds for integral n.
6
Remark. More generally, no rational number of the
sin(6x+π/4) 1/6 form m/n, where m, n are relatively prime and neither
 
whence f (x) = ec cos 2 (6x+π/4) . Substituting
of ±m is a quadratic residue mod n, need be in S. If
f (0) = 1 gives ec = 2−1/12 and thus f (x) = x = p/q is in lowest terms and 1/(x(x − 1)) = m/n + k
sin(6x+π/4) 1/6 for some integer k, then p(p − q) is relatively prime to
 
2−1/12 cos 2 (6x+π/4) .
q2 ; q2 /(p(p − q)) = (m + kn)/n then implies that m +
Remark. The answer can be put in alternate forms kn = ±q2 and so ±m must be a quadratic residue mod
using trigonometric identities. One particularly simple n.
one is
A–5 No, there is no such group. By the structure theorem
f (x) = (sec 12x)1/12 (sec 12x + tan 12x)1/4 . for finitely generated abelian groups, G can be written
as a product of cyclic groups. If any of these factors has
odd order, then G has an element of odd order, so the
 2

product of the orders of all of its elements cannot be a Moreover, the partial derivatives
power of 2.
∂f
We may thus consider only abelian 2-groups hereafter. (x0 , y0 ) = 3(1 + y0 )(8x0 − 4)
For such a group G, the product of the orders of all of ∂x
its elements has the form 2k(G) for some nonnegative ∂f
(x0 , y0 ) = 3(2x0 − 1)2 − 1.
integer G, and we must show that it is impossible to ∂y
achieve k(G) = 2009. Again by the structure theorem,
we may write have no common zero in (0, 1)2 . Namely, for the first
partial to vanish, we must have x0 = 1/2 since 1 + y0 is
∞ nowhere zero, but for x0 = 1/2 the second partial cannot
G∼
= ∏(Z/2i Z)ei vanish.
i=1
Remark. This problem amounts to refuting a potential
for some nonnegative integers e1 , e2 , . . . , all but finitely generalization of the Mean Value Theorem to bivariate
many of which are 0. functions. Many counterexamples are possible. Kent
For any nonnegative integer m, the elements of G of Merryfield suggests y sin(2πx), for which all four of the
order at most 2m form a subgroup isomorphic to boundary integrals vanish; here the partial derivatives
are 2πy cos(2πx) and sin(2πx). Catalin Zara suggests
∞
x1/3 y2/3 . Qingchun Ren suggests xy(1 − y).
∏(Z/2min{i,m} Z)ei ,
i=1 B–1 Every positive rational number can be uniquely written
in lowest terms as a/b for a, b positive integers. We
which has 2sm elements for sm = ∑∞
i=1 min{i, m}ei . prove the statement in the problem by induction on the
Hence
largest prime dividing either a or b (where this is con-
∞ sidered to be 1 if a = b = 1). For the base case, we
k(G) = ∑ i(2si − 2si−1 ). can write 1/1 = 2!/2!. For a general a/b, let p be the
i=1
largest prime dividing either a or b; then a/b = pk a0 /b0
Since s1 ≤ s2 ≤ · · · , k(G) + 1 is always divisible by 2s1 . for some k 6= 0 and positive integers a0 , b0 whose largest
In particular, k(G) = 2009 forces s1 ≤ 1. prime factors are strictly less than p. We now have
a0 0
a/b = (p!)k (p−1)! k b0 , and all prime factors of a and
However, the only cases where s1 ≤ 1 are where all of
the ei are 0, in which case k(G) = 0, or where ei = 1 for (p − 1)!k b0 are strictly less than p. By the induction as-
some i and e j = 0 for j 6= i, in which case k(G) = (i − a0
sumption, (p−1)! k b0 can be written as a quotient of prod-
1)2i + 1. The right side is a strictly increasing function a 0
ucts of prime factorials, and so a/b = (p!)k (p−1)! k b0 can
of i which equals 1793 for i = 8 and 4097 for i = 9, so
it can never equal 2009. This proves the claim. as well. This completes the induction.
Remark. One can also arrive at the key congruence Remark. Noam Elkies points out that the representa-
by dividing G into equivalence classes, by declaring tions are unique up to rearranging and canceling com-
two elements to be equivalent if they generate the same mon factors.
cyclic subgroup of G. For h > 0, an element of order
B–2 The desired real numbers c are precisely those for which
2h belongs to an equivalence class of size 2h−1 , so the
1/3 < c ≤ 1. For any positive integer m and any se-
products of the orders of the elements of this equiva-
quence 0 = x0 < x1 < · · · < xm = 1, the cost of jumping
lence class is 2 j for j = h2h−1 . This quantity is divisible
along this sequence is ∑m 2
i=1 (xi − xi−1 )xi . Since
by 4 as long as h > 1; thus to have k(G) ≡ 1 (mod 4),
the number of elements of G of order 2 must be con- m m
gruent to 1 modulo 4. However, there are exactly 2e − 1 1 = ∑ (xi − xi−1 ) ≥ ∑ (xi − xi−1 )xi2
such elements, for e the number of cyclic factors of G. i=1 i=1
m Z xi−1
Hence e = 1, and one concludes as in the given solution.
>∑ t 2 dt
A–6 We disprove the assertion using the example i=1 xi
Z 1
2 1
f (x, y) = 3(1 + y)(2x − 1)2 − y. = t dt = ,
0 3
We have b−a = d −c = 0 because the identity f (x, y) = we can only achieve costs c for which 1/3 < c ≤ 1.
f (1 − x, y) forces a = b, and because It remains to check that any such c can be achieved.
Z 1 Suppose 0 = x0 < · · · < xm = 1 is a sequence with m ≥
c= 3(2x − 1)2 dx = 1, 1. For i = 1, . . . , m, let ci be the cost of the sequence
0 0, xi , xi+1 , . . . , xm . For i > 1 and 0 < y ≤ xi−1 , the cost of
Z 1
d= (6(2x − 1)2 − 1) dx = 1. the sequence 0, y, xi , . . . , xm is
0
ci + y3 + (xi − y)xi2 − xi3 = ci − y(xi2 − y2 ),
 3

which is less than ci but approaches ci as y → 0. By endpoints by subtracting 2k−1 from each element. By
continuity, for i = 2, . . . , m, every value in the interval the induction assumption again, it follows that S must
[ci−1 , ci ) can be achieved, as can cm = 1 by the sequence contain all integers between 2k−1 + 1 and 2k + 1. Thus
0, 1. S = {1, . . . , 2k + 1} and the induction is complete.
To show that all costs c with 1/3 < c ≤ 1 can be Remark. One can also proceed by checking that a
achieved, it now suffices to check that for every ε > 0, nonempty subset of {1, . . . , n} is mediocre if and only if
there exists a sequence with cost at most 1/3 + ε. For it is an arithmetic progression with odd common differ-
instance, if we take xi = i/m for i = 0, . . . , m, the cost ence. Given this fact, the number of mediocre subsets
becomes of {1, . . . , n + 2} containing the endpoints is seen to be
the number of odd factors of n + 1, from which the de-
1 2 (m + 1)(2m + 1) sired result is evident. (The sequence A(n) appears as
(1 + · · · + m2 ) = ,
m3 6m2 sequence A124197 in the Encyclopedia of Integer Se-
which converges to 1/3 as m → +∞. quences.)

Reinterpretation. The cost of jumping along a partic- B–4 Any polynomial P(x, y) of degree at most 2009 can be
ular sequence is an upper Riemann sum of the function written uniquely as a sum ∑2009i=0 Pi (x, y) in which Pi (x, y)
t 2 . The fact that this function admits a Riemann inte- is a homogeneous polynomial of degree i. For r > 0,
gral implies that for any ε > 0, there exists δ0 such that let Cr beHthe path (r cos θ , r sin θ ) for 0 ≤ θ ≤ 2π. Put
the cost of the sequence x0 , . . . , xm is at most 1/3 + ε as λ (Pi ) = C1 Pi ; then for r > 0,
long as maxi {xi − xi−1 } < ε. (The computation of the
integral using the sequence xi = i/m was already known I 2009
to Archimedes.) Cr
P= ∑ ri λ (Pi ).
i=0
B–3 The answer is n = 2k − 1 for some integer k ≥ 1. There
For fixed P, the right side is a polynomial in r, which
is a bijection between mediocre subsets of {1, . . . , n}
vanishes for all r > 0 if and only if its coefficients
and mediocre subsets of {2, . . . , n + 1} given by adding
vanish. In other words, P is balanced if and only if
1 to each element of the subset; thus A(n + 1) − A(n)
λ (Pi ) = 0 for i = 0, . . . , 2009.
is the number of mediocre subsets of {1, . . . , n + 1}
that contain 1. It follows that A(n + 2) − 2A(n + For i odd, we have Pi (−x, −y) = −Pi (x, y). Hence
1) + An = (A(n + 2) − A(n + 1)) − (A(n + 1) − A(n)) λ (Pi ) = 0, e.g., because the contributions to the integral
is the difference between the number of mediocre sub- from θ and θ + π cancel.
sets of {1, . . . , n + 2} containing 1 and the number of For i even, λ (Pi ) is a linear function of the coefficients
mediocre subsets of {1, . . . , n + 1} containing 1. This of Pi . This function is not identically zero, e.g., because
difference is precisely the number of mediocre subsets for Pi = (x2 + y2 )i/2 , the integrand is always positive
of {1, . . . , n + 2} containing both 1 and n + 2, which and so λ (Pi ) > 0. The kernel of λ on the space of ho-
we term “mediocre subsets containing the endpoints.” mogeneous polynomials of degree i is thus a subspace
Since {1, . . . , n + 2} itself is a mediocre subset of itself of codimension 1.
containing the endpoints, it suffices to prove that this
is the only mediocre subset of {1, . . . , n + 2} containing It follows that the dimension of V is
the endpoints if and only if n = 2k − 1 for some k.
(1 + · · · + 2010) − 1005 = (2011 − 1) × 1005 = 2020050.
If n is not of the form 2k − 1, then we can write n + 1 =
2a b for odd b > 1. In this case, the set {1 + mb | 0 ≤ m ≤ B–5 First solution. If f (x) ≥ x for all x > 1, then the desired
2a } is a mediocre subset of {1, . . . , n+2} containing the conclusion clearly holds. We may thus assume hereafter
endpoints: the average of 1 + m1 b and 1 + m2 b, namely that there exists x0 > 1 for which f (x0 ) < x0 .
1 + m1 +m 2
2 b, is an integer if and only if m1 + m2 is even,
in which case this average lies in the set. Rewrite the original differential equation as
It remains to show that if n = 2k − 1, then the only x2 + 1 f (x)2
mediocre subset of {1, . . . , n + 2} containing the end- f 0 (x) = 1 − .
x2 1 + f (x)2
points is itself. This is readily seen by induction on
k. For k = 1, the statement is obvious. For general Put c0 = min{0, f (x0 ) − 1/x0 }. For all x ≥ x0 , we have
k, any mediocre subset S of {1, . . . , n + 2 = 2k + 1} f 0 (x) > −1/x2 and so
containing 1 and 2k + 1 must also contain their aver-
age, 2k−1 + 1. By the induction assumption, the only
Z x
f (x) ≥ f (x0 ) − dt/t 2 > c0 .
mediocre subset of {1, . . . , 2k−1 + 1} containing the x0
endpoints is itself, and so S must contain all integers
between 1 and 2k−1 + 1. Similarly, a mediocre subset In the other direction, we claim that f (x) < x for all
of {2k−1 + 1, . . . , 2k + 1} containing the endpoints gives x ≥ x0 . To see this, suppose the contrary; then by con-
a mediocre subset of {1, . . . , 2k−1 + 1} containing the tinuity, there is a least x ≥ x0 for which f (x) ≥ x, and
 4

this least value satisfies f (x) = x. However, this forces Third solution. (by Noam Elkies) Consider the func-
f 0 (x) = 0 < 1 and so f (x − ε) > x − ε for ε > 0 small, tion g(x) = f (x) + 31 f (x)3 , for which
contradicting the choice of x.
Put x1 = max{x0 , −c0 }. For x ≥ x1 , we have | f (x)| < x f (x)2
g0 (x) = f 0 (x)(1 + f (x)2 ) = 1 −
and so f 0 (x) > 0. In particular, the limit limx→+∞ f (x) = x2
L exists.
for x > 1. Since evidently g0 (x) < 1, g(x)−x is bounded
Suppose that L < +∞; then limx→+∞ f 0 (x) = 1/(1 + above for x large. As in the first solution, f (x) is
L2 ) > 0. Hence for any sufficiently small ε > 0, we can bounded below for x large, so 13 f (x)3 − x is bounded
choose x2 ≥ x1 so that f 0 (x) ≥ ε for x ≥ x2 . But then above by some c > 0. For x ≥ c, we obtain f (x) ≤
f (x) ≥ f (x2 ) + ε(x − x2 ), which contradicts L < +∞. (6x)1/3 .
Hence L = +∞, as desired.
Since f (x)/x → 0 as x → +∞, g0 (x) → 1 and so
Variant. (by Leonid Shteyman) One obtains a similar g(x)/x → 1. Since g(x) tends to +∞, so does f (x).
argument by writing (With a tiny bit of extra work, one shows that in fact
f (x)/(3x)1/3 → 1 as x → +∞.)
1 f (x)2
f 0 (x) = − 2 ,
1 + f (x)2 x (1 + f (x)2 ) B–6 First solution. (based on work of Yufei Zhao) Since
any sequence of the desired form remains of the desired
so that form upon multiplying each term by 2, we may reduce
to the case where n is odd. In this case, take x = 2h for
1 1
− ≤ f 0 (x) − ≤ 0. some positive integer h for which x ≥ n, and set
x2 1 + f (x)2
a0 = 0
Hence f 0 (x) − 1/(1 + f (x)2 ) tends to 0 as x → +∞, so
a1 = 1
f (x) is bounded below,
R and tends to +∞ if and only if
the improper integral dx/(1 + f (x)2 ) diverges. How- a2 = 2x + 1 = a1 + 2x
ever, if the integral were to converge, then as x → +∞ a3 = (x + 1)2 = a2 + x2
we would have 1/(1 + f (x)2 ) → 0; however, since f is
a4 = xn + 1 = a1 + xn
bounded below, this again forces f (x) → +∞.
a5 = n(x + 1) = a4 mod a3
Second solution. (by Catalin Zara) The function g(x) =
f (x) + x satisfies the differential equation a6 = x
a7 = n = a5 mod a6 .
1 − (g(x)/x − 1)2
g0 (x) = 1 + . We may pad the sequence to the desired length by tak-
1 + x2 (g(x)/x − 1)2
ing a8 = · · · = a2009 = n.
This implies that g0 (x) > 0 for all x > 1, so the limit Second solution. (by James Merryfield) Suppose first
L1 = limx→+∞ g(x) exists. In addition, we cannot have that n is not divisible by 3. Recall that since 2 is a primi-
L1 < +∞, or else we would have limx→+∞ g0 (x) = 0 tive root modulo 32 , it is also a primitive root modulo 3h
whereas the differential equation forces this limit to be for any positive integer h. In particular, if we choose h
1. Hence g(x) → +∞ as x → +∞. so that 32h > n, then there exists a positive integer c for
Similarly, the function h(x) = − f (x) + x satisfies the which 2c mod 32h = n. We now take b to be a positive
differential equation integer for which 2b > 32h , and then put

1 − (h(x)/x − 1)2 a0 = 0
h0 (x) = 1 − .
1 + x2 (h(x)/x − 1)2 a1 = 1
a2 = 3 = a1 + 2
This implies that h0 (x) ≥ 0 for all x, so the limit L2 =
limx→+∞ h(x) exists. In addition, we cannot have L2 < a3 = 3 + 2b
+∞, or else we would have limx→+∞ h0 (x) = 0 whereas a4 = 22hb
the differential equation forces this limit to be 1. Hence
h(x) → +∞ as x → +∞. a5 = 32h = a4 mod a3
For some x1 > 1, we must have g(x), h(x) > 0 for all a6 = 2c
x ≥ x1 . For x ≥ x1 , we have | f (x)| < x and hence a7 = n = a6 mod a5 .
f 0 (x) > 0, so the limit L = limx→+∞ f (x) exists. Once
again, we cannot have L < +∞, or else we would If n is divisible by 3, we can force a7 = n − 1 as in the
have limx→+∞ f 0 (x) = 0 whereas the original differen- above construction, then put a8 = a7 + 1 = n. In both
tial equation (e.g., in the form given in the first so- cases, we then pad the sequence as in the first solution.
lution) forces this limit to be 1/(1 + L2 ) > 0. Hence Remark. Hendrik Lenstra, Ronald van Luijk, and
f (x) → +∞ as x → ∞, as desired. Gabriele Della Torre suggest the following variant of
 5

the first solution requiring only 6 steps. For n odd and x It seems unlikely that a shorter solution can be con-
as in the first solution, set structed without relying on any deep number-theoretic
conjectures.
a0 = 0
a1 = 1
a2 = x + 1 = a1 + x
a3 = xn + x + 1 = a2 + xn
a4 = x(n−1)(φ (a3 )−1)
xn + 1
a5 = = a4 mod a3
x+1
a6 = n = a5 mod a2 .
 The 71st William Lowell Putnam Mathematical Competition
Saturday, December 4, 2010

A1 Given a positive integer n, what is the largest k such that B1 Is there an infinite sequence of real numbers
the numbers 1, 2, . . . , n can be put into k boxes so that a1 , a2 , a3 , . . . such that
the sum of the numbers in each box is the same? [When
n = 8, the example {1, 2, 3, 6}, {4, 8}, {5, 7} shows that am m m
1 + a2 + a3 + · · · = m
the largest k is at least 3.]
for every positive integer m?
A2 Find all differentiable functions f : R → R such that
B2 Given that A, B, and C are noncollinear points in the
f (x + n) − f (x) plane with integer coordinates such that the distances
f 0 (x) = AB, AC, and BC are integers, what is the smallest possi-
n
ble value of AB?
for all real numbers x and all positive integers n.
B3 There are 2010 boxes labeled B1 , B2 , . . . , B2010 , and
A3 Suppose that the function h : R2 → R has continuous 2010n balls have been distributed among them, for
partial derivatives and satisfies the equation some positive integer n. You may redistribute the balls
by a sequence of moves, each of which consists of
∂h ∂h choosing an i and moving exactly i balls from box Bi
h(x, y) = a (x, y) + b (x, y)
∂x ∂y into any one other box. For which values of n is it possi-
ble to reach the distribution with exactly n balls in each
for some constants a, b. Prove that if there is a constant
box, regardless of the initial distribution of balls?
M such that |h(x, y)| ≤ M for all (x, y) ∈ R2 , then h is
identically zero. B4 Find all pairs of polynomials p(x) and q(x) with real
coefficients for which
A4 Prove that for each positive integer n, the number
10n n
1010 + 1010 + 10n − 1 is not prime. p(x)q(x + 1) − p(x + 1)q(x) = 1.
A5 Let G be a group, with operation ∗. Suppose that
B5 Is there a strictly increasing function f : R → R such
(i) G is a subset of R3 (but ∗ need not be related to that f 0 (x) = f ( f (x)) for all x?
addition of vectors);
B6 Let A be an n × n matrix of real numbers for some n ≥
(ii) For each a, b ∈ G, either a × b = a ∗ b or a × b = 0 1. For each positive integer k, let A[k] be the matrix
(or both), where × is the usual cross product in obtained by raising each entry to the kth power. Show
R3 . that if Ak = A[k] for k = 1, 2, . . . , n + 1, then Ak = A[k] for
Prove that a × b = 0 for all a, b ∈ G. all k ≥ 1.

A6 Let f : [0, ∞) → R be a strictly decreasing continu-

ous function such that limx→∞ f (x) = 0. Prove that
R ∞ f (x)− f (x+1)
0 f (x) dx diverges.
 Solutions to the 71st William Lowell Putnam Mathematical Competition
Saturday, December 4, 2010
Kiran Kedlaya and Lenny Ng

A–1 The largest such k is b n+1 n

2 c = d 2 e. For n even, this value Since 10n ≥ n ≥ 2m ≥ m + 1, 10n is divisible by 2n and
n n
is achieved by the partition hence by 2m+1 , and similarly 1010 is divisible by 210
and hence by 2 m+1 . It follows that
{1, n}, {2, n − 1}, . . . ;
m
N ≡ 1 + 1 + (−1) + (−1) ≡ 0 (mod 102 + 1).
for n odd, it is achieved by the partition
n m
Since N ≥ 1010 > 10n + 1 ≥ 102 + 1, it follows that N
{n}, {1, n − 1}, {2, n − 2}, . . . . is composite.
One way to see that this is optimal is to note that the A–5 We start with three lemmas.
common sum can never be less than n, since n itself
belongs to one of the boxes. This implies that k ≤ (1 + Lemma 1. If x, y ∈ G are nonzero orthogonal vectors, then
· · · + n)/n = (n + 1)/2. Another argument is that if k > x ∗ x is parallel to y.
(n + 1)/2, then there would have to be two boxes with
one number each (by the pigeonhole principle), but such Proof. Put z = x × y 6= 0, so that x, y, and z = x ∗ y are nonzero
boxes could not have the same sum. and mutually orthogonal. Then w = x × z 6= 0, so w = x ∗ z is
nonzero and orthogonal to x and z. However, if (x∗x)×y 6= 0,
Remark. A much subtler question would be to find
then w = x∗(x∗y) = (x∗x)∗y = (x∗x)×y is also orthogonal
the smallest k (as a function of n) for which no such
to y, a contradiction.
arrangement exists.
Lemma 2. If x ∈ G is nonzero, and there exists y ∈ G nonzero
A–2 The only such functions are those of the form f (x) =
and orthogonal to x, then x ∗ x = 0.
cx + d for some real numbers c, d (for which the prop-
erty is obviously satisfied). To see this, suppose that f
Proof. Lemma 1 implies that x ∗ x is parallel to both y and
has the desired property. Then for any x ∈ R,
x × y, so it must be zero.
2 f 0 (x) = f (x + 2) − f (x) Lemma 3. If x, y ∈ G commute, then x × y = 0.
= ( f (x + 2) − f (x + 1)) + ( f (x + 1) − f (x))
= f 0 (x + 1) + f 0 (x). Proof. If x × y 6= 0, then y × x is nonzero and distinct from
x×y. Consequently, x∗y = x×y and y∗x = y×x 6= x∗y.
Consequently, f 0 (x + 1) = f 0 (x).
We proceed now to the proof. Assume by way of con-
Define the function g : R → R by g(x) = f (x + 1) −
tradiction that there exist a, b ∈ G with a × b 6= 0. Put
f (x), and put c = g(0), d = f (0). For all x ∈ R,
c = a × b = a ∗ b, so that a, b, c are nonzero and lin-
g0 (x) = f 0 (x + 1) − f 0 (x) = 0, so g(x) = c identically,
early independent. Let e be the identity element of G.
and f 0 (x) = f (x+1)− f (x) = g(x) = c, so f (x) = cx+d
Since e commutes with a, b, c, by Lemma 3 we have
identically as desired.
e×a = e×b = e×c = 0. Since a, b, c span R3 , e×x = 0
A–3 If a = b = 0, then the desired result holds trivially, so for all x ∈ R3 , so e = 0.
we assume that at least one of a, b is nonzero. Pick Since b, c, and b × c = b ∗ c are nonzero and mutually
any point (a0 , b0 ) ∈ R2 , and let L be the line given by orthogonal, Lemma 2 implies
the parametric equation L(t) = (a0 , b0 ) + (a, b)t for t ∈
R. By the chain rule and the given equation, we have b ∗ b = c ∗ c = (b ∗ c) ∗ (b ∗ c) = 0 = e.
d
dt (h ◦ L) = h ◦ L. If we write f = h ◦ L : R → R, then
f 0 (t) = f (t) for all t. It follows that f (t) = Cet for some Hence b ∗ c = c ∗ b, contradicting Lemma 3 because b ×
constant C. Since | f (t)| ≤ M for all t, we must have c 6= 0. The desired result follows.
C = 0. It follows that h(a0 , b0 ) = 0; since (a0 , b0 ) was
an arbitrary point, h is identically 0 over all of R2 . A–6 First solution. Note that the hypotheses on f imply
that f (x) > 0 for all x ∈ [0, +∞), so the integrand is a
A–4 Put continuous function of f and the integral makes sense.
10n n
Rewrite the integral as
N = 1010 + 1010 + 10n − 1. Z ∞ 
f (x + 1)
1− dx,
Write n = 2m k with m a nonnegative integer and k a 0 f (x)
positive odd integer. For any nonnegative integer j,
mj m
102 ≡ (−1) j (mod 102 + 1).
 2

and suppose by way of contradiction that it converges If there exist arbitrarily large values of y for which
to a finite limit L. For n ≥ 0, define the Lebesgue mea- f (y) > 2 f (y + 1), we deduce that the original integral is
surable set greater than any multiple of 1/7, and so diverges. Oth-
erwise, for x large we may argue that
f (x + n + 1)
In = {x ∈ [0, 1] : 1 − ≤ 1/2}.
f (x + n) f (x) − f (x + 1) 3 f (x)
> log
1
f (x) 5 f (x + 1)
Then L ≥ ∑∞ n=0 2 (1 − µ(In )), so the latter sum con-
verges. In particular, there exists a nonnegative integer as in the above solution, and again get divergence using
N for which ∑∞ n=N (1 − µ(In )) < 1; the intersection a telescoping sum.
∞ ∞ Second solution. (Communicated by Paul Allen.) Let
b > a be nonnegative integers. Then
[ \
I= In = [0, 1] − ([0, 1] − In )
n=N n=N
Z b b−1 Z 1
f (x) − f (x + 1) f (x + k) − f (x + k + 1)
then has positive Lebesgue measure. dx = ∑ dx
a f (x) k=a 0 f (x + k)
By Taylor’s theorem with remainder, for t ∈ [0, 1/2], Z 1 b−1
f (x + k) − f (x + k + 1)
= ∑ dx
t2 f (x + k)
 
1 0 k=a
− log(1 − t) ≤ t + sup
2 t∈[0,1/2] (1 − t)2 Z 1 b−1
f (x + k) − f (x + k + 1)
≥ ∑ dx
= t + 2t 2 ≤ 2t. 0 k=a f (x + a)
Z 1
f (x + a) − f (x + b)
For each nonnegative integer n ≥ N, we then have = dx.
0 f (x + a)
Z n 
f (x + 1)
L≥ 1− dx Now since f (x) → 0, given a, we can choose an in-
N f (x)
teger l(a) > a for which f (l(a)) < f (a + 1)/2; then
n−1 Z 1  
f (x + i + 1) f (x+a)− f (x+l(a)) (l(a))
≥ 1 − ff (a+1) > 1/2 for all x ∈ [0, 1].
=∑ 1− dx f (x+a)
i=N 0 f (x + i) Thus if we define a sequence of integers an by a0 = 0,
n−1 Z 
f (x + i + 1)
 an+1 = l(an ), then
≥∑ 1− dx
i=N I f (x + i) Z ∞
f (x) − f (x + 1) ∞ Z an+1
f (x) − f (x + 1)
dx = ∑ dx
1 n−1 f (x + i) f (x) f (x)
Z
0 n=0 an
≥ ∑ log dx
2 i=N I f (x + i + 1) ∞ Z 1
! > ∑ (1/2)dx,
n−1
1 f (x + i) n=0 0
Z
= ∑ log f (x + i + 1) dx
2 I i=N and the final sum clearly diverges.
1 f (x + N)
Z
= log dx. Third solution. (By Joshua Rosenberg, communicated
2 I f (x + n) by Catalin Zara.) If the original integral converges, then
on one hand the integrand ( f (x) − f (x + 1))/ f (x) = 1 −
For each x ∈ I, log f (x + N)/ f (x + n) is a strictly f (x + 1)/ f (x) cannot tend to 1 as x → ∞. On the other
increasing unbounded function of n. ByR the mono- hand, for any a ≥ 0,
tone convergence theorem, the integral I log( f (x +
N)/ f (x + n)) dx grows without bound as n → +∞, a f (a + 1)
contradiction. Thus the original integral diverges, as 0<
f (a)
desired. Z a+1
1
Remark. This solution is motivated by the commonly- < f (x) dx
f (a) a
used fact that an infinite product (1 + x1 )(1 + x2 ) · · ·
1
Z ∞
converges absolutely if and only if the sum x1 + x2 + · · · = ( f (x) − f (x + 1)) dx
converges absolutely. The additional measure-theoretic f (a) a
f (x) − f (x + 1)
Z ∞
argument at the beginning is needed because one cannot
≤ dx,
bound − log(1−t) by a fixed multiple of t uniformly for a f (x)
all t ∈ [0, 1).
and the last expression tends to 0 as a → ∞. Hence by
Greg Martin suggests a variant solution that avoids use
the squeeze theorem, f (a + 1)/ f (a) → 0 as a → ∞, a
of Lebesgue measure. √ Note first that if f (y) > 2 f (y +
contradiction.
1),
√ then either f (y) > 2 f (y + 1/2) or f (y + 1/2) >
2 f (y + 1), and in either case we deduce that
Z y+1/2  
f (x) − f (x + 1) 1 1 1
dx > 1− √ > .
y−1/2 f (x) 2 2 7
 3

B–1 First solution. No such sequence exists. If it did, then Remark. Manjul Bhargava points out it is easy to con-
the Cauchy-Schwartz inequality would imply struct sequences of complex numbers with the desired
property if we drop the condition of absolute conver-
8 = (a21 + a22 + · · · )(a41 + a42 + · · · ) gence. Here is an inductive construction (of which sev-
≥ (a31 + a32 + · · · )2 = 9, eral variants are possible). For n = 1, 2, . . . and z ∈ C,
define the finite sequence
contradiction.  
1 2πi j/n
Second solution. (Communicated by Catalin Zara.) sn,z = e : j = 0, . . . , n − 1 .
z
Suppose that such a sequence exists. If a2k ∈ [0, 1] for
all k, then a4k ≤ a2k for all k, and so This sequence has the property that for any positive in-
teger j, the sum of the j-th powers of the terms of sn,z
4 = a41 + a42 + · · · ≤ a21 + a22 + · · · = 2, equals 1/z j if j is divisible by n and 0 otherwise. More-
over, any partial sum of j-th powers is bounded in ab-
contradiction. There thus exists a positive integer k for solute value by n/|z| j .
which a2k ≥ 1. However, in this case, for m large, a2m
k >
The desired sequence will be constructed as follows.
2m and so a12m + a2m
2 + · · · 6
= 2m.
Suppose that we have a finite sequence which has the
Third solution. We generalize the second solution to correct sum of j-th powers for j = 1, . . . , m. (For in-
show that for any positive integer k, it is impossible for stance, for m = 1, we may start with the singleton
a sequence a1 , a2 , . . . of complex numbers to satisfy the sequence 1.) We may then extend it to a new se-
given conditions in case the series ak1 + ak2 + · · · con- quence which has the correct sum of j-th powers for
verges absolutely. This includes the original problem j = 1, . . . , m + 1, by appending k copies of sm+1,z for
by taking k = 2, in which case the series a21 + a22 + · · · suitable choices of a positive integer k and a complex
consists of nonnegative real numbers and so converges number z with |z| < m−2 . This last restriction ensures
absolutely if it converges at all. that the resulting infinite sequence a1 , a2 , . . . is such that
Since the sum ∑∞ k
i=1 |ai | converges by hypothesis, we
for each positive integer m, the series am m
1 + a2 + · · · is
can find a positive integer n such that ∑∞ k
i=n+1 |ai | < 1.
convergent (though not absolutely convergent). Its par-
For each positive integer d, we then have tial sums include a subsequence equal to the constant
value m, so the sum of the series must equal m as de-
n ∞ sired.
kd − ∑ akd
i ≤ ∑ |ai |kd < 1.
i=1 i=n+1 B–2 The smallest distance is 3, achieved by A = (0, 0), B =
(3, 0), C = (0, 4). To check this, it suffices to check
We thus cannot have |a1 |, . . . , |an | ≤ 1, or else the sum that AB cannot equal 1 or 2. (It cannot equal 0 because
∑ni=1 akd
i would be bounded in absolute value by n inde- if two of the points were to coincide, the three points
pendently of d. But if we put r = max{|a1 |, . . . , |an |} > would be collinear.)
1, we obtain another contradiction because for any ε >
The triangle inequality implies that |AC − BC| ≤ AB,
0,
with equality if and only if A, B,C are collinear. If AB =
n 1, we may assume without loss of generality that A =
lim sup(r − ε)−kd ∑ akdi > 0. (0, 0), B = (1, 0). To avoid collinearity, we must have
d→∞ i=1 AC = BC, but this forces C = (1/2, y) for some y ∈ R,
a contradiction. (One can also treat this case by scaling
For instance, this follows from applying the root test to by a factor of 2 to reduce to the case AB = 2, treated in
the rational function the next paragraph.)
!
n
1 ∞ n If AB = 2, then we may assume without loss of gener-
∑ 1 − ak z = ∑ ∑ ai zd ,
kd
ality that A = (0, 0), B = (2, 0). The triangle inequal-
i=1 i d=0 i=1 ity implies |AC − BC| ∈ {0, 1}. Also, for C = (x, y),
AC2 = x2 + y2 and BC2 = (2 − x)2 + y2 have the same
which has a pole within the circle |z| ≤ r−1/k . (An ele- parity; it follows that AC = BC. Hence c = (1, y) for
mentary proof is also possible.) some y ∈ R, so y2 and y2 + 1 = BC2 are consecutive
Fourth solution. (Communicated by Noam Elkies.) perfect squares. This can only happen for y = 0, but
Since ∑k a2k = 2, for each positive integer k we have then A, B,C are collinear, a contradiction again.
a2k ≤ 2 and so a4k ≤ 2a2k , with equality only for a2k ∈ Remark. Manjul Bhargava points out that more gener-
{0, 2}. Thus to have ∑k a4k = 4, there must be a single ally, a Heronian triangle (a triangle with integer sides
index k for which a2k = 2, and the other ak must all equal and rational area) cannot have a side of length 1 or 2
0. But then ∑k a2m m
k = 2 6= 2m for any positive integer
(and again it is enough to treat the case of length 2).
m > 2. The original problem follows from this because a tri-
angle whose vertices have integer coordinates has area
 4

equal to half an integer (by Pick’s formula or the ex- Then r(x) = a, s(x) = c for all x ∈ Z, and hence identi-
plicit formula for the area as a determinant). cally; consequently, p(x) = ax + b, q(x) = cx + d for all
x ∈ Z, and hence identically. For p and q of this form,
B–3 It is possible if and only if n ≥ 1005. Since
p(x)q(x + 1) − p(x + 1)q(x) = bc − ad,
2009 × 2010
1 + · · · + 2009 = = 2010 × 1004.5,
2 so we get a solution if and only if bc − ad = 1, as
for n ≤ 1004, we can start with an initial distribution claimed.
in which each box Bi starts with at most i − 1 balls (so Second solution. (Communicated by Catalin Zara.)
in particular B1 is empty). From such a distribution, no Again, note that p and q must be nonzero. Write
moves are possible, so we cannot reach the desired final
distribution. p(x) = p0 + p1 x + · · · + pm xm
Suppose now that n ≥ 1005. By the pigeonhole prin- q(x) = q0 + q1 x + · · · + qn xn
ciple, at any time, there exists at least one index i for
which the box Bi contains at least i balls. We will de- with pm , qn 6= 0, so that m = deg(p), n = deg(q). It
scribe any such index as being eligible. The following is enough to derive a contradiction assuming that
sequence of operations then has the desired effect. max{m, n} > 1, the remaining cases being treated as in
the first solution.
(a) Find the largest eligible index i. If i = 1, proceed Put R(x) = p(x)q(x+1)− p(x+1)q(x). Since m+n ≥ 2
to (b). Otherwise, move i balls from Bi to B1 , then by assumption, the coefficient of xm+n−1 in R(x) must
repeat (a). vanish. By easy algebra, this coefficient equals (m −
(b) At this point, only the index i = 1 can be eligi- n)pm qn , so we must have m = n > 1.
ble (so it must be). Find the largest index j for For k = 1, . . . , 2m − 2, the coefficient of xk in R(x) is
which B j is nonempty. If j = 1, proceed to (c).
   
Otherwise, move 1 ball from B1 to B j ; in case this j i
makes j eligible, move j balls from B j to B1 . Then ∑ − (pi q j − p j qi )
i+ j>k, j>i k−i k− j
repeat (b).
(c) At this point, all of the balls are in B1 . For i = and must vanish. For k = 2m − 2, the only summand is
2, . . . , 2010, move one ball from B1 to Bi n times. for (i, j) = (m − 1, m), so pm−1 qm = pm qm−1 .
After these operations, we have the desired distribution. Suppose now that h ≥ 1 and that pi q j = p j qi is known to
vanish whenever j > i ≥ h. (By the previous paragraph,
B–4 First solution. The pairs (p, q) satisfying the given we initially have this for h = m − 1.) Take k = m + h − 2
equation are those of the form p(x) = ax + b, q(x) = and note that the conditions i + j > h, j ≤ m force i ≥
cx + d for a, b, c, d ∈ R such that bc − ad = 1. We will h − 1. Using the hypothesis, we see that the only possi-
see later that these indeed give solutions. ble nonzero contribution to the coefficient of xk in R(x)
Suppose p and q satisfy the given equation; note that is from (i, j) = (h − 1, m). Hence ph−1 qm = pm qh−1 ;
neither p nor q can be identically zero. By subtracting since pm , qm 6= 0, this implies ph−1 q j = p j qh−1 when-
the equations ever j > h − 1.
By descending induction, we deduce that pi q j = p j qi
p(x)q(x + 1) − p(x + 1)q(x) = 1 whenever j > i ≥ 0. Consequently, p(x) and q(x) are
p(x − 1)q(x) − p(x)q(x − 1) = 1, scalar multiples of each other, forcing R(x) = 0, a con-
tradiction.
we obtain the equation Third solution. (Communicated by David Feldman.)
As in the second solution, we note that there are no so-
p(x)(q(x + 1) + q(x − 1)) = q(x)(p(x + 1) + p(x − 1)).
lutions where m = deg(p), n = deg(q) are distinct and
The original equation implies that p(x) and q(x) have m+n ≥ 2. Suppose p, q form a solution with m = n ≥ 2.
no common nonconstant factor, so p(x) divides p(x + The desired identity asserts that the matrix
1) + p(x − 1). Since each of p(x + 1) and p(x − 1) has  
p(x) p(x + 1)
the same degree and leading coefficient as p, we must
q(x) q(x + 1)
have
has determinant 1. This condition is preserved by re-
p(x + 1) + p(x − 1) = 2p(x).
placing q(x) with q(x) − t p(x) for any real number t. In
If we define the polynomials r(x) = p(x + 1) − p(x), particular, we can choose t so that deg(q(x) − t p(x)) <
s(x) = q(x + 1) − q(x), we have r(x + 1) = r(x), and m; we then obtain a contradiction.
similarly s(x + 1) = s(x). Put

a = r(0), b = p(0), c = s(0), d = q(0).

 5

B–5 First solution. The answer is no. Suppose otherwise. x = g( f (x)), and we may differentiate to find that
For the condition to make sense, f must be differen- 1 = g0 ( f (x)) f 0 (x) = g0 ( f (x)) f ( f (x)). It follows that
tiable. Since f is strictly increasing, we must have g0 (y) = 1/ f (y) for yR≥ y0 ; since g takes arbitrarily large
f 0 (x) ≥ 0 for all x. Also, the function f 0 (x) is strictly values, the integral y∞0 dy/ f (y) must diverge. One then
increasing: if y > x then f 0 (y) = f ( f (y)) > f ( f (x)) = gets a contradiction from any reasonable lower bound
f 0 (x). In particular, f 0 (y) > 0 for all y ∈ R. on f (y) for y large, e.g., the bound f (x) ≥ αx2 from the
For any x0 ≥ −1, if f (x0 ) = b and f 0 (x0 ) = a > 0, then second solution. (One can also start with a linear lower
f 0 (x) > a for x > x0 and thus f (x) ≥ a(x − x0 ) + b for bound f (x) ≥ β x, then use the integral expression for g
x ≥ x0 . Then either b < x0 or a = f 0 (x0 ) = f ( f (x0 )) = to deduce that g(x) ≤ γ log x, which in turn forces f (x)
f (b) ≥ a(b − x0 ) + b. In the latter case, b ≤ a(x0 + to grow exponentially.)
1)/(a + 1) ≤ x0 + 1. We conclude in either case that B–6 For any polynomial p(x), let [p(x)]A denote the n × n
f (x0 ) ≤ x0 + 1 for all x0 ≥ −1. matrix obtained by replacing each entry Ai j of A by
It must then be the case that f ( f (x)) = f 0 (x) ≤ 1 for p(Ai j ); thus A[k] = [xk ]A. Let P(x) = xn + an−1 xn−1 +
all x, since otherwise f (x) > x + 1 for large x. Now · · · + a0 denote the characteristic polynomial of A. By
by the above reasoning, if f (0) = b0 and f 0 (0) = the Cayley-Hamilton theorem,
a0 > 0, then f (x) > a0 x + b0 for x > 0. Thus for
x > max{0, −b0 /a0 }, we have f (x) > 0 and f ( f (x)) > 0 = A · P(A)
a0 x + b0 . But then f ( f (x)) > 1 for sufficiently large x,
= An+1 + an−1 An + · · · + a0 A
a contradiction.
Second solution. (Communicated by Catalin Zara.) = A[n+1] + an−1 A[n] + · · · + a0 A[1]
Suppose such a function exists. Since f is strictly = [xp(x)]A.
increasing and differentiable, so is f ◦ f = f 0 . In
particular, f is twice differentiable; also, f 00 (x) = Thus each entry of A is a root of the polynomial xp(x).
f 0 ( f (x)) f 0 (x) is the product of two strictly increasing Now suppose m ≥ n + 1. Then
nonnegative functions, so it is also strictly increasing
and nonnegative. In particular, we can choose α > 0 0 = [xm+1−n P(x)]A
and M ∈ R such that f 00 (x) > 4α for all x ≥ M. Then
= A[m+1] + an−1 A[m] + · · · + a0 A[m+1−n]
for all x ≥ M,

f (x) ≥ f (M) + f 0 (M)(x − M) + 2α(x − M)2 . since each entry of A is a root of xm+1−n P(x). On the
other hand,
In particular, for some M 0 > M, we have f (x) ≥ αx2 for
all x ≥ M 0 . 0 = Am+1−n · P(A)

Pick T > 0 so that αT 2 > M 0 . Then for x ≥ T , f (x) > = Am+1 + an−1 Am + · · · + a0 Am+1−n .
M 0 and so f 0 (x) = f ( f (x)) ≥ α f (x)2 . Now
Therefore if Ak = A[k] for m + 1 − n ≤ k ≤ m, then
1 1
Z 2T 0
f (t)
Z 2T Am+1 = A[m+1] . The desired result follows by induction
− = dt ≥ α dt; on m.
f (T ) f (2T ) T f (t)2 T
Remark. David Feldman points out that the result is
however, as T → ∞, the left side of this inequality tends best possible in the following sense: there exist ex-
to 0 while the right side tends to +∞, a contradiction. amples of n × n matrices A for which Ak = A[k] for
Third solution. (Communicated by Noam Elkies.) k = 1, . . . , n but An+1 6= A[n+1] .
Since f is strictly increasing, for some y0 , we can de-
fine the inverse function g(y) of f for y ≥ y0 . Then
 The 72nd William Lowell Putnam Mathematical Competition
Saturday, December 3, 2011

A1 Define a growing spiral in the plane to be a sequence

of points with integer coordinates P0 = (0, 0), P1 , . . . , Pn
1 2
 
such that n ≥ 2 and: 1
lim ∑ Prob(g = x) −
m→∞ b2m n
x∈G
– the directed line segments P0 P1 , P1 P2 , . . . , Pn−1 Pn
are in the successive coordinate directions east is positive and finite.
(for P0 P1 ), north, west, south, east, etc.;
– the lengths of these line segments are positive and B1 Let h and k be positive integers. Prove that for every
strictly increasing. ε > 0, there are positive integers m and n such that
√ √
[Picture omitted.] How many of the points (x, y) with ε < |h m − k n| < 2ε.
integer coordinates 0 ≤ x ≤ 2011, 0 ≤ y ≤ 2011 cannot
be the last point, Pn of any growing spiral? B2 Let S be the set of all ordered triples (p, q, r) of prime
numbers for which at least one rational number x satis-
A2 Let a1 , a2 , . . . and b1 , b2 , . . . be sequences of positive fies px2 + qx + r = 0. Which primes appear in seven or
real numbers such that a1 = b1 = 1 and bn = bn−1 an − 2 more elements of S?
for n = 2, 3, . . . . Assume that the sequence (b j ) is
bounded. Prove that B3 Let f and g be (real-valued) functions defined on an
∞ open interval containing 0, with g nonzero and contin-
1 uous at 0. If f g and f /g are differentiable at 0, must f
S= ∑ a1 ...an
n=1 be differentiable at 0?

converges, and evaluate S. B4 In a tournament, 2011 players meet 2011 times to play
a multiplayer game. Every game is played by all 2011
A3 Find a real number c and a positive number L for which players together and ends with each of the players either
R π/2 r winning or losing. The standings are kept in two 2011×
rc 0 x sin x dx 2011 matrices, T = (Thk ) and W = (Whk ). Initially, T =
lim R π/2 = L.
r→∞ xr cos x dx W = 0. After every game, for every (h, k) (including for
0
h = k), if players h and k tied (that is, both won or both
lost), the entry Thk is increased by 1, while if player h
A4 For which positive integers n is there an n × n matrix
won and player k lost, the entry Whk is increased by 1
with integer entries such that every dot product of a row
and Wkh is decreased by 1.
with itself is even, while every dot product of two dif-
ferent rows is odd? Prove that at the end of the tournament, det(T + iW ) is
a non-negative integer divisible by 22010 .
A5 Let F : R2 → R and g : R → R be twice continuously
differentiable functions with the following properties: B5 Let a1 , a2 , . . . be real numbers. Suppose that there is a
constant A such that for all n,
– F(u, u) = 0 for every u ∈ R;
!2
n
– for every x ∈ R, g(x) > 0 and x2 g(x) ≤ 1; 1
Z ∞
∑ 2
dx ≤ An.
– for every (u, v) ∈ R2 , the vector ∇F(u, v) is either −∞ i=1 1 + (x − ai )
0 or parallel to the vector hg(u), −g(v)i.
Prove there is a constant B > 0 such that for all n,
Prove that there exists a constant C such that for every
n
n ≥ 2 and any x1 , . . . , xn+1 ∈ R, we have
∑ (1 + (ai − a j )2 ) ≥ Bn3 .
C i, j=1
min |F(xi , x j )| ≤ .
i6= j n
B6 Let p be an odd prime. Show that for at least (p + 1)/2
A6 Let G be an abelian group with n elements, and let values of n in {0, 1, 2, . . . , p − 1},
p−1
{g1 = e, g2 , . . . , gk } $ G
∑ k!nk is not divisible by p.
k=0
be a (not necessarily minimal) set of distinct generators
of G. A special die, which randomly selects one of the
elements g1 , g2 , ..., gk with equal probability, is rolled m
times and the selected elements are multiplied to pro-
duce an element g ∈ G. Prove that there exists a real
number b ∈ (0, 1) such that
 Solutions to the 72nd William Lowell Putnam Mathematical Competition
Saturday, December 3, 2011
Kiran Kedlaya and Lenny Ng

bj B
A1 We claim that the set of points with 0 ≤ x ≤ 2011 and Now if (b j ) is bounded above by B, then b j +2 ≤ B+2
0 ≤ y ≤ 2011 that cannot be the last point of a growing B m
for all j, and so 3/2 > Sm ≥ 3/2(1 − ( B+2 ) ). Since
spiral are as follows: (0, y) for 0 ≤ y ≤ 2011; (x, 0) and B
(x, 1) for 1 ≤ x ≤ 2011; (x, 2) for 2 ≤ x ≤ 2011; and B+2 < 1, it follows that the sequence (Sm ) converges to
(x, 3) for 3 ≤ x ≤ 2011. This gives a total of S = 3/2.
A3 We claim that (c, L) = (−1, 2/π) works. Write f (r) =
2012 + 2011 + 2011 + 2010 + 2009 = 10053 R π/2 r
0 x sin x dx. Then
excluded points.
(π/2)r+1
Z π/2
The complement of this set is the set of (x, y) with 0 < f (r) < xr dx =
x < y, along with (x, y) with x ≥ y ≥ 4. Clearly the 0 r+1
former set is achievable as P2 in a growing spiral, while while since sin x ≥ 2x/π for x ≤ π/2,
a point (x, y) in the latter set is P6 in a growing spiral
with successive lengths 1, 2, 3, x + 1, x + 2, and x + y − Z π/2 r+1
2x (π/2)r+1
1. f (r) > dx = .
0 π r+2
We now need to rule out the other cases. Write x1 <
y1 < x2 < y2 < . . . for the lengths of the line segments It follows that
in the spiral in order, so that P1 = (x1 , 0), P2 = (x1 , y1 ),  r+1
P3 = (x1 − x2 , y1 ), and so forth. Any point beyond P0 2
lim r f (r) = 1,
has x-coordinate of the form x1 − x2 + · · · + (−1)n−1 xn r→∞ π
for n ≥ 1; if n is odd, we can write this as x1 + (−x2 +
whence
x3 ) + · · · + (−xn−1 + xn ) > 0, while if n is even, we can
write this as (x1 − x2 ) + · · · + (xn−1 − xn ) < 0. Thus no f (r) r(2/π)r+1 f (r) 2(r + 1) 2
point beyond P0 can have x-coordinate 0, and we have lim = lim · = .
r→∞ f (r + 1) r→∞ (r + 1)(2/π)r+2 f (r + 1) πr π
ruled out (0, y) for 0 ≤ y ≤ 2011.
Next we claim that any point beyond P3 must have Now by integration by parts, we have
y-coordinate either negative or ≥ 4. Indeed, each
such point has y-coordinate of the form y1 − y2 + · · · + Z π/2
1
Z π/2
f (r + 1)
(−1)n−1 yn for n ≥ 2, which we can write as (y1 − y2 ) + xr cos x dx = xr+1 sin x dx = .
0 r+1 0 r+1
· · · + (yn−1 − yn ) < 0 if n is even, and
Thus setting c = −1 in the given limit yields
y1 + (−y2 + y3 ) + · · · + (−yn−1 + yn ) ≥ y1 + 2 ≥ 4
(r + 1) f (r) 2
if n ≥ 3 is odd. Thus to rule out the rest of the forbidden lim = ,
r→∞ r f (r + 1) π
points, it suffices to check that they cannot be P2 or P3
for any growing spiral. But none of them can be P3 = as desired.
(x1 − x2 , y1 ) since x1 − x2 < 0, and none of them can
be P2 = (x1 , y1 ) since they all have y-coordinate at most A4 The answer is n odd. Let I denote the n × n identity
equal to their x-coordinate. matrix, and let A denote the n × n matrix all of whose
entries are 1. If n is odd, then the matrix A − I satisfies
A2 For m ≥ 1, write the conditions of the problem: the dot product of any
  row with itself is n − 1, and the dot product of any two
3 b1 · · · bm distinct rows is n − 2.
Sm = 1− .
2 (b1 + 2) · · · (bm + 2) Conversely, suppose n is even, and suppose that the ma-
trix M satisfied the conditions of the problem. Consider
Then S1 = 1 = 1/a1 and a quick calculation yields
all matrices and vectors mod 2. Since the dot product
b1 · · · bm−1 1 of a row with itself is equal mod 2 to the sum of the en-
Sm − Sm−1 = = tries of the row, we have Mv = 0 where v is the vector
(b2 + 2) · · · (bm + 2) a1 · · · am
(1, 1, . . . , 1), and so M is singular. On the other hand,
for m ≥ 2, since a j = (b j + 2)/b j−1 for j ≥ 2. It follows MM T = A − I; since
that Sm = ∑m n=1 1/(a1 · · · an ).
(A − I)2 = A2 − 2A + I = (n − 2)A + I = I,

we have (det M)2 = det(A − I) = 1 and det M = 1, con-

tradicting the fact that M is singular.
 2

R x Abhinav Kumar) Define G : R → R by G(x) =

A5 (by we show that b ∈ (0, 1) as follows. First suppose b = 0;
0 g(t) dt. By assumption, G is a strictly increasing, then
thrice continuously differentiable function. It is also
k
bounded: for x > 1, we have 1 n
1= ∑ λχ = k ∑ ∑ χ(gi ) = k
Z x Z x χ∈Ĝ i=1 χ∈Ĝ
0 < G(x) − G(1) = g(t) dt ≤ dt/t 2 = 1,
1 1
because ∑χ∈(G) ˆ χ(gi ) equals n for i = 1 and 0 oth-
and similarly, for x < −1, we have 0 > G(x) − G(−1) ≥ erwise. However, this contradicts the hypothesis that
−1. It follows that the image of G is some open interval {g1 , . . . , gk } is not all of G. Hence b > 0. Next suppose
(A, B) and that G−1 : (A, B) → R is also thrice continu- b = 1, and choose χ ∈ Ĝ − {ê} with |λχ | = 1. Since
ously differentiable. each of χ(g1 ), . . . , χ(gk ) is a complex number of norm
Define H : (A, B) × (A, B) → R by H(x, y) = 1, the triangle inequality forces them all to be equal.
F(G−1 (x), G−1 (y)); it is twice continuously dif- Since χ(g1 ) = χ(e) = 1, χ must map each of g1 , . . . , gk
ferentiable since F and G−1 are. By our assumptions to 1, but this is impossible because χ is a nontrivial
about F, character and g1 , . . . , gk form a set of generators of G.
This contradiction yields b < 1.
∂H ∂H ∂ F −1 1 Since v = 1n ∑χ∈Ĝ vχ and Mvχ = λχ vχ , we have
+ = (G (x), G−1 (y)) · −1
∂x ∂y ∂x g(G (x))
1 1
∂ F −1 1 M m v − vê = ∑ λχm vχ .
+ (G (x), G−1 (y)) · = 0. n n
∂y g(G−1 (y)) χ∈Ĝ−{ê}

Therefore H is constant along any line parallel to the Since the vectors vχ are pairwise orthogonal, the limit
vector (1, 1), or equivalently, H(x, y) depends only on we are interested in can be written as
x − y. We may thus write H(x, y) = h(x − y) for some
function h on (−(B − A), B − A), and we then have 1 1 1
lim (M m v − vê ) · (M m v − vê ).
F(x, y) = h(G(x) − G(y)). Since F(u, u) = 0, we have m→∞ b2m n n
h(0) = 0. Also, h is twice continuously differentiable and then rewritten as
(since it can be written as h(x) = H((A + B + x)/2, (A +
B − x)/2)), so |h0 | is bounded on the closed interval 1
lim ∑ |λχ |2m = #{χ ∈ Ĝ : |λχ | = b}.
[−(B − A)/2, (B − A)/2], say by M. m→∞ b2m
χ∈Ĝ−{ê}
Given x1 , . . . , xn+1 ∈ R for some n ≥ 2, the numbers
G(x1 ), . . . , G(xn+1 ) all belong to (A, B), so we can By construction, this last quantity is nonzero and finite.
choose indices i and j so that |G(xi ) − G(x j )| ≤ (B − Remark. It is easy to see that the result fails if we do
A)/n ≤ (B − A)/2. By the mean value theorem, not assume g1 = e: take G = Z/2Z, n = 1, and g1 = 1.
Remark. Harm Derksen points out that a similar ar-
B−A gument applies even if G is not assumed to be abelian,
|F(xi , x j )| = |h(G(xi ) − G(x j ))| ≤ M , provided that the operator g1 + · · · + gk in the group al-
n
gebra Z[G] is normal, i.e., it commutes with the op-
so the claim holds with C = M(B − A). erator g−1 −1
1 + · · · + gk . This includes the cases where
the set {g1 , . . . , gk } is closed under taking inverses and
A6 Choose some ordering h1 , . . . , hn of the elements of G where it is a union of conjugacy classes (which in turn
with h1 = e. Define an n × n matrix M by settting Mi j = includes the case of G abelian).
1/k if h j = hi g for some g ∈ {g1 , . . . , gk } and Mi j = 0
otherwise. Let v denote the column vector (1, 0, . . . , 0). Remark. The matrix M used above has nonnegative
The probability that the product of m random elements entries with row sums equal to 1 (i.e., it corresponds to
of {g1 , . . . , gk } equals hi can then be interpreted as the a Markov chain), and there exists a positive integer m
i-th component of the vector M m v. such that M m has positive entries. For any such matrix,
the Perron-Frobenius theorem implies that the sequence
Let Ĝ denote the dual group of G, i.e., the group of vectors M m v converges to a limit w, and there exists
of complex-valued characters of G. Let ê ∈ Ĝ de- b ∈ [0, 1) such that
note the trivial character. For each χ ∈ Ĝ, the vector
vχ = (χ(hi ))ni=1 is an eigenvector of M with eigenvalue 1 n
λχ = (χ(g1 ) + · · · + χ(gk ))/k. In particular, vê is the lim sup
m→∞
∑ ((Mm v − w)i )2
b2m i=1
all-ones vector and λê = 1. Put
is nonzero and finite. (The intended interpretation in
b = max{|λχ | : χ ∈ Ĝ − {ê}};
case b = 0 is that M m v = w for all large m.) However,
the limit need not exist in general.
 3

B1 Since the rational numbers are dense in the reals, we ( f g) · ( f /g) with the square root function. By the chain
can find positive integers a, b such that rule, f is differentiable at 0.

3ε b 4ε If f (0) = 0, then ( f /g)(0) = 0, so we have

< < .
hk a hk f (x)
( f /g)0 (0) = lim .
By multiplying a and b by a suitably large positive in- x→0 xg(x)
teger, we can also ensure that 3a2 > b. We then have
Since g is continuous at 0, we may multiply limits to
ε b b p deduce that limx→0 f (x)/x exists.
< <√ = a2 + b − a
hk 3a a2 + b + a Second solution. Choose a neighborhood N of 0 on
which g(x) 6= 0. Define the following functions on N \
and
{0}: h1 (x) = f (x)g(x)−x f (0)g(0) ; h2 (x) = f (x)g(0)− f (0)g(x)
xg(0)g(x) ;
p b b ε h3 (x) = g(0)g(x); h4 (x) = g(x)+g(0) 1
. Then by assump-
a2 + b − a = √ ≤ <2 .
a2 + b + a 2a hk tion, h1 , h2 , h3 , h4 all have limits as x → 0. On the other
hand,
We may then take m = k2 (a2 + b), n = h2 a2 .
f (x) − f (0)
B2 Only the primes 2 and 5 appear seven or more times. = (h1 (x) + h2 (x)h3 (x))h4 (x),
The fact that these primes appear is demonstrated by x
the examples f (x)− f (0)
and it follows that limx→0 x exists, as desired.
(2, 5, 2), (2, 5, 3), (2, 7, 5), (2, 11, 5) B4 Number the games 1, . . . , 2011, and let A = (a jk ) be the
2011 × 2011 matrix √ whose jk entry is 1 if player k wins
and their reversals. game j and i = −1 if player k loses game j. Then
It remains to show that if either ` = 3 or ` is a prime ah j a jk is 1 if players h and k tie in game j; i if player h
greater than 5, then ` occurs at most six times as an wins and player k loses in game j; and −i if h loses and
element of a triple in S. Note that (p, q, r) ∈ S if and T
k wins. It follows that T + iW = A A.
only if q2 − 4pr = a2 for some integer a; in particular,
since 4pr ≥ 16, this forces q ≥ 5. In particular, q is odd, Now the determinant of A is unchanged if we subtract
as then is a, and so q2 ≡ a2 ≡ 1 (mod 8); consequently, the first row of A from each of the other rows, pro-
one of p, r must equal 2. If r = 2, then 8p = q2 − a2 = ducing a matrix whose rows, besides the first one, are
(q + a)(q − a); since both factors are of the same sign (1 − i) times a row of integers. Thus we can write
and their sum is the positive number 2q, both factors are det A = (1 − i)2010 (a + bi) for some integers a, b. But
T
positive. Since they are also both even, we have q + a ∈ then det(T + iW ) = det(A A) = 22010 (a2 + b2 ) is a non-
{2, 4, 2p, 4p} and so q ∈ {2p + 1, p + 2}. Similarly, if negative integer multiple of 22010 , as desired.
p = 2, then q ∈ {2r + 1, r + 2}. Consequently, ` occurs
B5 Define the function
at most twice as many times as there are prime numbers
in the list Z ∞
dx
f (y) = 2 )(1 + (x + y)2 )
.
`−1 −∞ (1 + x
2` + 1, ` + 2, , ` − 2.
2 For y ≥ 0, in the range −1 ≤ x ≤ 0, we have
For ` = 3,` − 2 = 1 is not prime. For ` ≥ 7, the numbers
` − 2, `, ` + 2 cannot all be prime, since one of them is (1 + x2 )(1 + (x + y)2 ) ≤ (1 + 1)(1 + (1 + y)2 ) = 2y2 + 4y + 4
always a nontrivial multiple of 3. ≤ 2y2 + 4 + 2(y2 + 1) ≤ 6 + 6y2 .
Remark. The above argument shows that the cases
listed for 5 are the only ones that can occur. By con- We thus have the lower bound
trast, there are infinitely many cases where 2 occurs if 1
either the twin prime conjecture holds or there are in- f (y) ≥ ;
6(1 + y2 )
finitely many Sophie Germain primes (both of which
are expected to be true). the same bound is valid for y ≤ 0 because f (y) = f (−y).
B3 Yes, it follows that f is differentiable. The original hypothesis can be written as
First solution. Note first that at 0, f /g and g are both n
continuous, as then is their product f . If f (0) 6= 0, then ∑ f (ai − a j ) ≤ An
in some neighborhood of 0, f is either always positive i, j=1
or always negative. We can thus choose ε ∈ {±1} so
and thus implies that
that ε f is the composition of the differentiable function
n
1
∑ 2
≤ 6An.
i, j=1 1 + (ai − a j )
 4

By the Cauchy-Schwarz inequality, this implies and note that by Wilson’s theorem again,
n p−1
xk−1
∑ (1 + (ai − a j )2 ) ≥ Bn3 h0 (x) = 1 + ∑ (k − 1)! = x p−1 − 1 + g(x).
i, j=1 k=1

for B = 1/(6A). If z ∈ F p is such that g(z) = 0, then z 6= 0 because g(0) =

Remark. One can also compute explicitly (using par- 1. Therefore, z p−1 = 1, so h(z) = h0 (z) = 0 and so z is
tial fractions, Fourier transforms, or contour integra- at least a double root of h. Since h is a polynomial of
2π
tion) that f (y) = 4+y 2.
degree p, there can be at most (p − 1)/2 zeroes of g in
F p , as desired.
Remark. Praveen Venkataramana points out that the
lower bound can be improved to Bn4 as follows. For Second solution. (By Noam Elkies) Define the polyno-
each z ∈ Z, put Qz,n = {i ∈ {1, . . . , n} : ai ∈ [z, z + 1)} mial f over F p by
and qz,n = #Qz,n . Then ∑z qz,n = n and p−1
n
1 1
f (x) = ∑ k!xk .
6An ≥ ∑ 2
≥ ∑ q2z,n . k=0
i, j=1 1 + (ai − a j ) z∈Z 2
Put t = (p − 1)/2; the problem statement is that f has
If exactly k of the qz,n are nonzero, then ≥ ∑z∈Z q2z,n at most t roots modulo p. Suppose the contrary; since
f (0) = 1, this means that f (x) is nonzero for at most
n2 /k by Jensen’s inequality (or various other methods),
t −1 values of x ∈ F∗p . Denote these values by x1 , . . . , xm ,
so we must have k ≥ n/(6A). Then
where by assumption m < t, and define the polynomial
n k Q over F p by
∑ (1 + (ai − a j )2 ) ≥ n2 + ∑ max{0, (|i − j| − 1)2 }
m t−1
i, j=1 i, j=1

k4 2k3 5k2 k
Q(x) = ∏ (x − xm ) = ∑ Qk x k .
2 k=1 k=0
≥n + − + − .
6 3 6 3
Then we can write
This is bounded below by Bn4 for some B > 0.
P(x)
In the opposite direction, one can weaken the initial up- f (x) = (1 − x p−1 )
Q(x)
per bound to An4/3 and still derive a lower bound of
Bn3 . The argument is similar. where P(x) is some polynomial of degree at most m.
This means that the power series expansions of f (x) and
B6 In order to interpret the problem statement, one must
P(x)/Q(x) coincide modulo x p−1 , so the coefficients of
choose a convention for the value of 00 ; we will take it
xt , . . . , x2t−1 in f (x)Q(x) vanish. In other words, the
to equal 1. (If one takes 00 to be 0, then the problem
product of the square matrix
fails for p = 3.)
First solution. By Wilson’s theorem, A = ((i + j + 1)!)t−1
i, j=0

k!(p − 1 − k)! ≡ (−1)k (p − 1)! ≡ (−1)k+1 (mod p), with the nonzero column vector (Qt−1 , . . . , Q0 ) is zero.
However, by the following lemma, det(A) is nonzero
so we have a congruence of Laurent polynomials modulo p, a contradiction.
p−1 p−1
(−1)k+1 xk Lemma 1. For any nonnegative integer m and any integer n,
∑ k!xk ≡ ∑ (p − 1 − k)! (mod p)
k=0 k=0 m
p−1 det((i + j + n)!)m
i, j=0 = ∏ k!(k + n)!.
(−x)−k
≡ −x p−1 ∑ (mod p). k=0
k=0 k!
Proof. Define the (m+1)×(m+1) matrix Am,n by (Am,n )i, j =
Replacing x with −1/x, we reduce the original problem i+ j+n

; the desired result is then that det(Am,n ) = 1. Note that
i
to showing that the polynomial
(
p−1 k (Am,n )i j i=0
x (Am,n−1 )i j =
g(x) = ∑ k! (Am,n )i j − (Am,n )(i−1) j i > 0;
k=0
that is, Am,n−1 can be obtained from Am,n by elementary row
over F p has at most (p − 1)/2 nonzero roots in F p . To
operations. Therefore, det(Am,n ) = det(Am,n−1 ), so det(Am,n )
see this, write
depends only on m. The claim now follows by observing that
h(x) = x p − x + g(x) A0,0 is the 1 × 1 matrix
 with entry 1 and that Am,−1 has the
1 ∗
block representation .
0 Am−1,0
 5

Remark. Elkies has given a more detailed http://mathoverflow.net/questions/82648.

discussion of the origins of this solution in
the theory of orthogonal polynomials; see
 The 73rd William Lowell Putnam Mathematical Competition
Saturday, December 1, 2012

A1 Let d1 , d2 , . . . , d12 be real numbers in the open interval (i) The functions f1 (x) = ex − 1 and f2 (x) = ln(x + 1)
(1, 12). Show that there exist distinct indices i, j, k such are in S;
that di , d j , dk are the side lengths of an acute triangle. (ii) If f (x) and g(x) are in S, the functions f (x) + g(x)
A2 Let ∗ be a commutative and associative binary operation and f (g(x)) are in S;
on a set S. Assume that for every x and y in S, there (iii) If f (x) and g(x) are in S and f (x) ≥ g(x) for all
exists z in S such that x ∗ z = y. (This z may depend on x ≥ 0, then the function f (x) − g(x) is in S.
x and y.) Show that if a, b, c are in S and a ∗ c = b ∗ c,
then a = b. Prove that if f (x) and g(x) are in S, then the function
f (x)g(x) is also in S.
A3 Let f : [−1, 1] → R be a continuous function such that B2 Let P be a given (non-degenerate) polyhedron. Prove
2
 2  that there is a constant c(P) > 0 with the following
(i) f (x) = 2−x
2 f x
2−x2
for every x in [−1, 1], property: If a collection of n balls whose volumes sum
to V contains the entire surface of P, then n > c(P)/V 2 .
(ii) f (0) = 1, and
√f (x)
B3 A round-robin tournament of 2n teams lasted for 2n − 1
(iii) limx→1− exists and is finite.
1−x days, as follows. On each day, every team played one
game against another team, with one team winning and
Prove that f is unique, and express f (x) in closed form.
one team losing in each of the n games. Over the course
A4 Let q and r be integers with q > 0, and let A and B be of the tournament, each team played every other team
intervals on the real line. Let T be the set of all b + mq exactly once. Can one necessarily choose one winning
where b and m are integers with b in B, and let S be team from each day without choosing any team more
the set of all integers a in A such that ra is in T . Show than once?
that if the product of the lengths of A and B is less than
B4 Suppose that a0 = 1 and that an+1 = an + e−an for n =
q, then S is the intersection of A with some arithmetic
0, 1, 2, . . . . Does an − log n have a finite limit as n → ∞?
progression.
(Here log n = loge n = ln n.)
A5 Let F p denote the field of integers modulo a prime p,
B5 Prove that, for any two bounded functions g1 , g2 : R →
and let n be a positive integer. Let v be a fixed vec-
[1, ∞), there exist functions h1 , h2 : R → R such that, for
tor in Fnp , let M be an n × n matrix with entries of F p ,
every x ∈ R,
and define G : Fnp → Fnp by G(x) = v + Mx. Let G(k)
denote the k-fold composition of G with itself, that is, sup(g1 (s)x g2 (s)) = max(xh1 (t) + h2 (t)).
G(1) (x) = G(x) and G(k+1) (x) = G(G(k) (x)). Determine s∈R t∈R
all pairs p, n for which there exist v and M such that the
pn vectors G(k) (0), k = 1, 2, . . . , pn are distinct. B6 Let p be an odd prime number such that p ≡ 2 (mod 3).
Define a permutation π of the residue classes modulo p
A6 Let f (x, y) be a continuous, real-valued function on R2 . by π(x) ≡ x3 (mod p). Show that π is an even permu-
Suppose that, for every rectangular region R of area 1, tation if and only if p ≡ 3 (mod 4).
the double integral of f (x, y) over R equals 0. Must
f (x, y) be identically 0?
B1 Let S be a class of functions from [0, ∞) to [0, ∞) that
satisfies:
 Solutions to the 73rd William Lowell Putnam Mathematical Competition
Saturday, December 1, 2012
Kiran Kedlaya and Lenny Ng

A1 Without loss of generality, assume d1 ≤ d2 ≤ · · · ≤ d12 . Proof. We may assume #S ≥ 3, as otherwise S is trivially an
2 < d 2 + d 2 for some i ≤ 10, then d , d
If di+2 i i+1 i i+1 , di+2 arithmetic progression. Let a1 , a2 be the smallest and second-
are the side lengths of an acute triangle, since in this smallest elements of S, respectively, and put d = a2 − a1 . Let
case di2 < di+1
2 +d 2 and d 2 < d 2 +d 2 as well. Thus
i+2 i+1 i i+2 m be the smallest positive integer such that a1 + md ∈ / S. Sup-
we may assume di+2 2 ≥ d2 + d2 pose that there exists an integer n contained in S but not in
i i+1 for all i. But then
by induction, di2 ≥ Fi d12 for all i, where Fi is the i-th {a1 , a1 + d, . . . , a1 + (m − 1)d}, and choose the least such n.
Fibonacci number (with F1 = F2 = 1): i = 1 is clear, i = By the hypothesis applied with (a, b, c) = (a1 , a2 , n), we see
2 follows from d2 ≥ d1 , and the induction step follows that n − d also has the property, a contradiction.
from the assumed inequality. Setting i = 12 now gives
2 ≥ 144d 2 , contradicting d > 1 and d < 12.
d12 1 12
We now return to the original problem. By dividing
1
B, q, r by gcd(q, r) if necessary, we may reduce to the
Remark. A materially equivalent problem appeared on case where gcd(q, r) = 1. We may assume #S ≥ 3, as
the 2012 USA Mathematical Olympiad and USA Junior otherwise S is trivially an arithmetic progression. Let
Mathematical Olympiad. a1 , a2 , a3 be any three distinct elements of S, labeled
A2 Write d for a ∗ c = b ∗ c ∈ S. For some e ∈ S, d ∗ e = a, so that a1 < a2 < a3 , and write rai = bi + mi q with
and thus for f = c ∗ e, a ∗ f = a ∗ c ∗ e = d ∗ e = a and bi , mi ∈ Z and bi ∈ B. Note that b1 , b2 , b3 must also be
b ∗ f = b ∗ c ∗ e = d ∗ e = a. Let g ∈ S satisfy g ∗ a = b; distinct, so the differences b2 − b1 , b3 − b1 , b3 − b2 are
then b = g ∗ a = g ∗ (a ∗ f ) = (g ∗ a) ∗ f = b ∗ f = a, as all nonzero; consequently, two of them have the same
desired. sign. If bi − b j and bk − bl have the same sign, then we
must have
Remark. With slightly more work, one can show that
S forms an abelian group with the operation ∗. (ai − a j )(bk − bl ) = (bi − b j )(ak − al )
√
A3 We will prove that f (x) = 1 − x2 for √ all x ∈ [−1, 1]. because both sides are of the same sign, of absolute
Define g : (−1, 1)
√ → R by g(x) = f (x)/ 1 − x2 . Plug- value less than q, and congruent to each other modulo
2
ging f (x) = g(x) 1 − x into equation (i) and simplify- q. In other words, the points (a1 , b1 ), (a2 , b2 ), (a3 , b3 )
ing yields in R2 are collinear. It follows that a4 = a1 + a3 − a2
 2  also belongs to S (by taking b4 = b1 + b3 − b2 ), so S
x satisfies the conditions of the lemma. It is therefore an
g(x) = g (1)
2 − x2 arithmetic progression.
for all x ∈ (−1, 1). Now fix x ∈ (−1, 1) and define a Reinterpretations. One can also interpret this argu-
a2n ment geometrically using cross products (suggested by
sequence {an }∞
n=1 by a1 = x and an+1 = 2−a2n
. Then Noam Elkies), or directly in terms of congruences (sug-
|2
an ∈ (−1, 1) and thus |an+1 | ≤ |an for all n. It follows gested by Karl Mahlburg).
that {|an |} is a decreasing sequence with |an | ≤ |x|n for Remark. The problem phrasing is somewhat confus-
all n, and so limn→∞ an = 0. Since g(an ) = g(x) for ing: to say that “S is the intersection of [the interval]
all n by (1) and g is continuous at 0, we conclude that A with an arithmetic progression” is the same thing as
g(x) = g(0) = f (0) = 1. This holds for all x ∈ (−1, 1) saying that “S is the empty set or an arithmetic progres-
and thus for x = ±1 as well by continuity. The result sion” unless it is implied that arithmetic progressions
follows. are necessarily infinite. Under that interpretation, how-
Remark. As pointed out by Noam Elkies, condition ever, the problem becomes false; for instance, for
(iii) is unnecessary. However, one can use it to derive
a slightly different solution by running the recursion in q = 5, r = 1, A = [1, 3], B = [0, 2],
the opposite direction.
we have
A4 We begin with an easy lemma.
T = {· · · , 0, 1, 2, 5, 6, 7, . . . }, S = {1, 2}.
Lemma. Let S be a finite set of integers with the following
property: for all a, b, c ∈ S with a ≤ b ≤ c, we also have a + A5 The pairs (p, n) with the specified property are those
c − b ∈ S. Then S is an arithmetic progression. pairs with n = 1, together with the single pair (2, 2).
We first check that these do work. For n = 1, it is clear
that taking v = (1) and M = (0) has the desired effect.
 2
 

1 1 by hypothesis, so h(x, y + 1/c) = h(x, y). By the fundamental
For (p, n) = (2, 2), we take v = 0 1 and M =
0 1 theorem of calculus, we may differentiate both sides of this
and then observe that identity with respect to y to deduce that g(x, y + 1/c) = g(x, y).
        Differentiating this new identity with respect to x yields the
(k) 0 0 1 1 desired equality.
G (0) = , , , , k = 0, 1, 2, 3.
0 1 0 1 √
Lemma 2. Let C be a circle whose diameter d is at least 2,
We next check that no other pairs work, keeping in mind and let AB and A0 B0 be two diameters of C. Then f (A) +
that the desired condition means that G acts on Fnp as a f (B) = f (A0 ) + f (B0 ).
cyclic permutation. Assume by way of contradiction
that (p, n) has the desired property but does not appear Proof. By continuity, it suffices to check the case where α =
in our list. In particular, we have n ≥ 2. arcsin d22 is an irrational multiple of 2π. Let β be the ra-
Let I be the n × n identity matrix over F p . Decom- dian measure of the counterclockwise arc from A to A0 . By
pose Fnp as a direct sum of two subspaces V,W such that Lemma 1, the claim holds when β = α. By induction, the
M − I is nilpotent on V and invertible on W . Suppose claim also holds when β ≡ nα (mod 2π) for any positive in-
that W 6= 0. Split v as v1 +v2 with v1 ∈ V , v2 ∈ W . Since teger n. Since α is an irrational multiple of 2π, the positive
M − I is invertible on W , there exists a unique w ∈ W multiples of α fill out a dense subset of the real numbers mod-
such that (M − I)w = −v2 . Then G(k) (w) − w ∈ V for ulo 2π, so by continuity the claim holds for all β .
all nonnegative integers k. Let k be the least positive in-
teger such that G(k) (w) = w; then k is at most the cardi- Lemma 3. Let R be a rectangular region of arbitrary (posi-
nality of V , which is strictly less than pn because W 6= 0. tive) area with corners A, B,C, D labeled in counterclockwise
This gives a contradiction and thus forces W = 0. order. Then f (A) + f (C) = f (B) + f (D).
In other words, the matrix N = M − I is nilpotent; con- Proof. Let EF be a segment such that AEFD and√BEFC
sequently, N n = 0. For any positive integer k, we have are rectangles whose diagonals have length at least 2. By
Lemma 2,
G(k) (0) = v + Mv + · · · + M k−1 v
k−1 n−1  
j i f (A) + f (F) = f (D) + f (E)
=∑∑ Nv
f (C) + f (E) = f (B) + f (F),
j=0 i=0 i
n−1  
k yielding the claim.
=∑ N i v.
i=0 i + 1
Lemma 4. The restriction of f to any straight line is constant.
If n ≥ 2 and (p, n) 6= (2, 2), then pn−1
> n and so
Proof. We may choose coordinates so that the line in question
Gk (0) = 0 for k = pn−1 (because all of the binomial
is the x-axis. Define the function g(y) by
coefficients are divisible by p). This contradiction com-
pletes the proof. g(y) = f (0, y) − f (0, 0).
A6 First solution. Yes, f (x, y) must be identically 0. We
By Lemma 3, for all x ∈ R,
proceed using a series of lemmas.
Lemma 1. Let R be a rectangular region of area 1 with f (x, y) = f (x, 0) + g(y).
corners A, B,C, D labeled in counterclockwise order. Then
For any c > 0, by the original hypothesis we have
f (A) + f (C) = f (B) + f (D).
Z x+c Z y+1/c
Proof. We may choose coordinates so that for some c > 0, 0= f (u, v) du dv
x y
Z x+c Z y+1/c
A = (0, 0), B = (c, 0),C = (c, 1/c), D = (0, 1/c).
= ( f (u, 0) + g(v)) du dv
x y
Define the functions x+c Z y+1/c
1
Z
Z x+c = f (u, 0) du + c g(v) dv.
g(x, y) = f (t, y) dt c x y

Zx y
In particular, the function F(x) = xx+c f (u, 0) du is constant.
R
h(x, y) = g(x, u) du. By the fundamental theorem of calculus, we may differentiate
0
to conclude that f (x + c, 0) = f (x, 0) for all x ∈ R. Since c
For any x, y ∈ R, was also arbitrary, we deduce the claim.
Z x+c Z y+1/c
h(x, y + 1/c) − h(x, y) = f (t, u) dt du = 0
x y
 3

To complete the proof, note that since any two points in P, as otherwise the vanishing of the zero over any rect-
R2 are joined by a straight line, Lemma 4 implies that f angle would force f to vanish identically. By continuity,
is constant. This constant equals the integral of f over there must exist an open disc U such that f (P) > 0 for
any rectangular region of area 1, and hence must be 0 all P ∈ U. Choose a rectangle R of area 1 with sides
as desired. parallel to the coordinate axes with one horizontal edge
Second solution (by Eric Larson, communicated by contained in U. Since the integral of f over R is zero,
Noam Elkies). In this solution, we fix coordinates and there must exist a point Q ∈ R such that f (Q) < 0. Take
assume only that the double integral vanishes on each P to be the vertical projection of Q onto the edge of R
rectangular region of area 1 with sides parallel to the contained in U.
coordinate axes, and still conclude that f must be iden- By translating coordinates, we may assume that P =
tically 0. (0, 0) and Q = (0, a) for some a > 0. For s sufficiently
small, f is positive on the square of side length 2s cen-
Lemma. Let R be a rectangular region of area 1 with sides tered at P, which we call S, and negative on the square
parallel to the coordinate axes. Then the averages of f over of side length 2s centered at Q, which we call S0 . Since
any two adjacent sides of R are equal. the ratio 2s/(1−4s2 ) tends to 0 as s does, we can choose
s so that 2s/(1 − 4s2 ) = a/n for some positive integer n.
Proof. Without loss of generality, we may take R to have cor-
ners (0, 0), (c, 0), (c, 1/c), (0, 1/c) and consider the two sides For i ∈ Z, let Ai be the rectangle
adjacent to (c, 1/c). Differentiate the equality
1 − 4s2

Z x+c Z y+1/c (x, y) : s ≤ x ≤ s + ,
2s
0= f (u, v) du dv 
x y 2s 2s
−s + i ≤ y ≤ s+i
1 − 4s2 1 − 4s2
with respect to c to obtain
Z y+1/c Z x+c and let Bi be the rectangle
1
0= f (x + c, v) dv − 2 f (u, y + 1/c) du.
y c x

1 − 4s2
(x, y) : s ≤ x ≤ s + ,
Rearranging yields 2s

2s 2s
Z y+1/c
1
Z x+c s+i ≤ y ≤ −s + (i + 1) .
c f (x + c, v) dv = f (u, y + 1/c) du, 1 − 4s2 1 − 4s2
y c x
Then for all i ∈ Z,
which asserts the desired result.
S ∪ A0 , An ∪ S0 , Ai ∪ Bi , Bi ∪ Ai+1
Returning to the original problem, given any c > 0, we
can tile the plane with rectangles of area 1 whose ver- are all rectangles of area 1 with sides parallel to the co-
tices lie in the lattice {(mc, n/c) : m, n ∈ Z}. By re- ordinate axes, so the integral over f over each of these
peated application of the lemma, we deduce that for any rectangles is zero. Since the integral over S is positive,
positive integer n, the integral over A0 must be negative; by induction, for
all i ∈ Z the integral over Ai is negative and the integral
over Bi is positive. But this forces the integral over S0
Z c Z (n+1)c
f (u, 0) du = f (u, 0) du. to be positive whereas f is negative everywhere on S0 , a
0 nc
contradiction.
Replacing c with c/n, we obtain
B1 Each of the following functions belongs to S for the rea-
Z c/n Z c+1/n
sons indicated.
f (u, 0) du = f (u, 0) du.
0 c
f (x), g(x) given
Fixing c and taking the limit as n → ∞ yields f (0, 0) = ln(x + 1) (i)
f (c, 0). By similar reasoning, f is constant on any hor- ln( f (x) + 1), ln(g(x) + 1) (ii) plus two previous lines
izontal line and on any vertical line, and as in the first ln( f (x) + 1) + ln(g(x) + 1) (ii)
solution the constant value is forced to equal 0. ex − 1 (i)
( f (x) + 1)(g(x) + 1) − 1 (ii) plus two previous lines
Third solution. (by Sergei Artamoshin) We retain the
f (x)g(x) + f (x) + g(x) previous line
weaker hypothesis of the second solution. Assume by
f (x) + g(x) (ii) plus first line
way of contradiction that f is not identically zero.
f (x)g(x) (iii) plus two previous lines
We first exhibit a vertical segment PQ with f (P) > 0
and f (Q) < 0. It cannot be the case that f (P) ≤ 0 for all B2 Fix a face F of the polyhedron with area A. Suppose F
is completely covered by balls of radii r1 , . . . , rn whose
 4

volumes sum to V . Then on one hand, Second solution. Put bn = ean , so that bn+1 = bn e1/bn .
In terms of the bn , the problem is to prove that bn /n has
n
4 a limit as n → ∞; we will show that the limit is in fact
∑ 3 πri3 = V. equal to 1.
i=1
Expanding e1/bn as a Taylor series in 1/bn , we have
On the other hand, the intersection of a ball of radius r
with the plane containing F is a disc of radius at most r, bn+1 = bn + 1 + Rn
which covers a piece of F of area at most πr2 ; therefore
n
where 0 ≤ Rn ≤ c/bn for some absolute constant c > 0.
πri2 ≥ A. By writing
∑
i=1
n−1
By writing n as ∑ni=1 1 and applying Hölder’s inequality, bn = n + e + ∑ Ri ,
i=0
we obtain
2/3 !3 we see first that bn ≥ n + e. We then see that
n 
4 16 3
nV 2 ≥ ∑ πri3 ≥ A . bn
i=1 3 9π 0≤ −1
n
Consequently, any value of c(P) less than 16 3
9π A works. e n−1 Ri
≤ +∑
n i=0 n
B3 The answer is yes. We first note that for any collection
of m days with 1 ≤ m ≤ 2n − 1, there are at least m e n−1 c
distinct teams that won a game on at least one of those ≤ +∑
n i=0 nbi
days. If not, then any of the teams that lost games on
all of those days must in particular have lost to m other e n−1 c
≤ +∑
teams, a contradiction. n i=0 n(i + e)
If we now construct a bipartite graph whose vertices are e c log n
≤ + .
the 2n teams and the 2n − 1 days, with an edge linking n n
a day to a team if that team won their game on that day,
then any collection of m days is connected to a total of at It follows that bn /n → 1 as n → ∞.
least m teams. It follows from Hall’s Marriage Theorem Remark. This problem is an example of the general
that one can match the 2n − 1 days with 2n − 1 distinct principle that one can often predict the asymptotic be-
teams that won on their respective days, as desired. havior of a recursive sequence by studying solutions of
a sufficiently similar-looking differential equation. In
B4 First solution. We will show that the answer is yes. this case, we start with the equation an+1 − an = e−an ,
First note that for all x > −1, ex ≥ 1 + x and thus then replace an with a function y(x) and replace the dif-
ference an+1 − an with the derivative y0 (x) to obtain the
x ≥ log(1 + x). (2)
differential equation y0 = e−y , which indeed has the so-
We next claim that an > log(n + 1) (and in particular lution y = log x.
that an − log n > 0) for all n, by induction on n. For B5 Define the function
n = 0 this follows from a0 = 1. Now suppose that
an > log(n + 1), and define f (x) = x + e−x , which is f (x) = sup{x log g1 (s) + log g2 (s)}.
an increasing function in x > 0; then s∈R

an+1 = f (an ) > f (log(n + 1)) As a function of x, f is the supremum of a collection of

affine functions, so it is convex. The function e f (x) is
= log(n + 1) + 1/(n + 1) ≥ log(n + 2),
then also convex, as may be checked directly from the
where the last inequality is (2) with x = 1/(n + 1). This definition: for x1 , x2 ∈ R and t ∈ [0, 1], by the weighted
completes the induction step. AM-GM inequality
It follows that an − log n is a decreasing function in n: te f (x1 ) + (1 − t)e f (x2 ) ≥ et f (x1 )+(1−t) f (x2 )
we have
≥ e f (tx1 +(1−t)x2 ) .
(an+1 − log(n + 1)) − (an − log n)
For each t ∈ R, draw a supporting line to the graph of
= e−an + log(n/(n + 1)) e f (x) at x = t; it has the form y = xh1 (t) + h2 (t) for some
< 1/(n + 1) + log(n/(n + 1)) ≤ 0, h1 (t), h2 (t) ∈ R. For all x, we then have
where the final inequality is (2) with x = −1/(n + 1). sup{g1 (s)x g2 (s)} ≥ xh1 (t) + h2 (t)
Thus {an − log n}∞n=0 is a decreasing sequence of posi-
s∈R
tive numbers, and so it has a limit as n → ∞.
 5

with equality for x = t. This proves the desired equality p. Since p ≡ 2 (mod 3), by the law of quadratic reci-
(including the fact that the maximum on the right side procity we have χ(−3) = +1, so χ(c/3) = χ(−c). It
is achieved). p−1 (p+χ(−c))/2
thus remains to evaluate the product ∏c=1 c
Remark. This problem demonstrates an example of du- modulo p.
ality for convex functions. If p ≡ 3 (mod 4), this is easy: each factor is a quadratic
residue (this is clear if c is a residue, and otherwise
B6 First solution. Since fixed points do not affect the sig-
χ(−c) = +1 so p + χ(−c) is divisible by 4) and −1
nature of a permutation, we may ignore the residue class
is not, so we must get +1 modulo p.
of 0 and consider π as a permutation on the nonzero
residue classes modulo p. These form a cyclic group of If p ≡ 1 (mod 4), we must do more work: we choose a
order p − 1, so the signature of π is also the signature primitive root g modulo p and rewrite the product as
of multiplication by 3 as a permutation σ of the residue
p−2
classes modulo p − 1. If we identify these classes with i

the integers 0, . . . , p − 2, then the signature equals the ∏ gi(p+(−1) )/2 .

i=0
parity of the number of inversions: these are the pairs
(i, j) with 0 ≤ i < j ≤ p − 2 for which σ (i) > σ ( j). We The sum of the exponents, split into sums over i odd
may write and i even, gives
 
3i (p−3)/2 
(2 j + 1)(p − 1)

σ (i) = 3i − (p − 1) j(p + 1) +
p−1 ∑ 2
j=0
from which we see that (i, j) cannot be an inversion un-
3j 3i which simplifies to
less b p−1 c > b p−1 c. In particular, we only obtain in-
versions when i < 2(p − 1)/3. (p − 3)(p − 1)(p + 1) (p − 1)3 p2 − 1
 
p−1
+ = −p .
If i < (p − 1)/3, the elements j of {0, . . . , p − 2} for 8 8 2 2
which (i, j) is an inversion correspond to the elements
of {0, . . . , 3i} which are not multiples of 3, which are Hence the product we are trying to evaluate is congruent
2i in number. This contributes a total of 0 + 2 + · · · + to g(p−1)/2 ≡ −1 modulo p.
2(p − 2)/3 = (p − 2)(p + 1)/9 inversions. Third solution (by Mark van Hoeij). We compute the
If (p − 1)/3 < i < 2(p − 1)/3, the elements j of parity of π as the parity of the number of cycles of even
{0, . . . , p−2} for which (i, j) is an inversion correspond length in the cycle decomposition of π. For x a nonzero
to the elements of {0, . . . , 3i − p + 1} congruent to 1 residue class modulo p of multiplicative order d, the
modulo 3, which are (3i − p + 2)/3 = i − (p − 2)/3 in elements of the orbit of x under π also have order d
number. This contributes a total of 1+· · ·+(p−2)/3 = (because d divides p − 1 and hence is coprime to 3).
(p − 2)(p + 1)/18 inversions. Since the group of nonzero residue classes modulo p
is cyclic of order p − 1, the elements of order d fall
Summing up, the total number of inversions is (p − into ϕ(d)/ f (d) orbits under π, where ϕ is the Euler
2)(p + 1)/6, which is even if and only if p ≡ 3 phi function and f (d) is the multiplicative order of 3
(mod 4). This proves the claim. modulo d. The parity of π is then the parity of the sum
Second solution (by Noam Elkies). Recall that the sign of ϕ(d)/ f (d) over all divisors d of p−1 for which f (d)
of π (which is +1 if π is even and −1 if π is odd) can is even.
be computed as If d is odd, then ϕ(d)/ f (d) = ϕ(2d)/ f (2d), so the
π(x) − π(y) summands corresponding to d and 2d coincide. It thus
∏ suffices to consider those d divisible by 4. If p ≡ 3
0≤x<y<p x−y (mod 4), then there are no such summands, so the sum
is trivially even.
(because composing π with a transposition changes the
sign of the product). Reducing modulo p, we get a con- If p ≡ 1 (mod 4), then d = 4 contributes a summand
gruence with of ϕ(4)/ f (4) = 2/2 = 1. For each d which is a larger
multiple of 4, the group (Z/dZ)∗ is isomorphic to the
x3 − y3 product of Z/2Z with another group of even order,
∏ = ∏ (x2 + xy + y2 ). so the maximal power of 2 dividing f (d) is strictly
0≤x<y<p x − y 0≤x<y<p
smaller than the maximal power of 2 dividing d. Hence
ϕ(d)/ f (d) is even, and so the overall sum is odd.
It thus suffices to count the number of times each pos-
sible value of x2 + xy + y2 occurs. Each nonzero value Remark. Note that the second proof uses quadratic
c modulo p occurs p + 1 times as x2 + xy + y2 with 0 ≤ reciprocity, whereas the first and third proofs are sim-
x, y < p and hence (p + χ(c/3))/2 times with 0 ≤ x < ilar to several classical proofs of quadratic reciprocity.
y < p, where χ denotes the quadratic character modulo Abhinav Kumar notes that the problem itself is a spe-
cial case of the Duke-Hopkins quadratic reciprocity
 6

law for abelian groups (Quadratic reciprocity in a finite pdf).

group, Amer. Math. Monthly 112 (2005), 251–256; see
also http://math.uga.edu/~pete/morequadrec.
 The 75th William Lowell Putnam Mathematical Competition
Saturday, December 6, 2014

A1 Prove that every nonzero coefficient of the Taylor series B2 Suppose that f is a function on the interval [1, 3] such
of that −1 ≤ f (x) ≤ 1 for all x and 13 f (x) dx = 0. How
R

large can 13 f (x)

R
(1 − x + x2 )ex x dx be?
B3 Let A be an m × n matrix with rational entries. Sup-
about x = 0 is a rational number whose numerator (in pose that there are at least m + n distinct prime numbers
lowest terms) is either 1 or a prime number. among the absolute values of the entries of A. Show that
the rank of A is at least 2.
A2 Let A be the n × n matrix whose entry in the i-th row
and j-th column is B4 Show that for each positive integer n, all the roots of the
polynomial
1
min(i, j) n
∑ 2k(n−k) xk
for 1 ≤ i, j ≤ n. Compute det(A). k=0

A3 Let a0 = 5/2 and ak = a2k−1 − 2 for k ≥ 1. Compute are real numbers.

∞   B5 In the 75th annual Putnam Games, participants compete

1 at mathematical games. Patniss and Keeta play a game
∏ 1 −
k=0 ak in which they take turns choosing an element from the
group of invertible n × n matrices with entries in the
in closed form. field Z/pZ of integers modulo p, where n is a fixed
A4 Suppose X is a random variable that takeson only non- positive integer and p is a fixed prime number. The
2 rules of the game are:
 3  integer values, with E [X] = 1, E X = 2, and
negative
E X = 5. (Here E [y] denotes the expectation of the (1) A player cannot choose an element that has been
random variable Y .) Determine the smallest possible chosen by either player on any previous turn.
value of the probability of the event X = 0.
(2) A player can only choose an element that com-
A5 Let mutes with all previously chosen elements.
(3) A player who cannot choose an element on his/her
Pn (x) = 1 + 2x + 3x2 + · · · + nxn−1 . turn loses the game.
Prove that the polynomials Pj (x) and Pk (x) are relatively Patniss takes the first turn. Which player has a winning
prime for all positive integers j and k with j 6= k. strategy? (Your answer may depend on n and p.)
A6 Let n be a positive integer. What is the largest k B6 Let f : [0, 1] → R be a function for which there exists a
for which there exist n × n matrices M1 , . . . , Mk and constant K > 0 such that | f (x) − f (y)| ≤ K |x − y| for all
N1 , . . . , Nk with real entries such that for all i and j, the x, y ∈ [0, 1]. Suppose also that for each rational number
matrix product Mi N j has a zero entry somewhere on its r ∈ [0, 1], there exist integers a and b such that f (r) =
diagonal if and only if i 6= j? a + br. Prove that there exist finitely many intervals
I1 , . . . , InSsuch that f is a linear function on each Ii and
B1 A base 10 over-expansion of a positive integer N is an [0, 1] = ni=1 Ii .
expression of the form

N = dk 10k + dk−1 10k−1 + · · · + d0 100

with dk 6= 0 and di ∈ {0, 1, 2, . . . , 10} for all i. For

instance, the integer N = 10 has two base 10 over-
expansions: 10 = 10 · 100 and the usual base 10 expan-
sion 10 = 1 · 101 + 0 · 100 . Which positive integers have
a unique base 10 over-expansion?
 Solutions to the 75th William Lowell Putnam Mathematical Competition
Saturday, December 6, 2014
Kiran Kedlaya and Lenny Ng

A1 The coefficient of xn in the Taylor series of (1 − x + For example, for n = 5,

x2 )ex for n = 0, 1, 2 is 1, 0, 12 , respectively. For n ≥ 3,
the coefficient of xn is −1 2 0 0 0
 
2 −8 6 0 0 
1 1 1 1 − n + n(n − 1) B= 0

6 −18 12 0 .

− + =
n! (n − 1)! (n − 2)! n! 0 0 12 −32 20 
n−1 0 0 0 20 −20
= .
n(n − 2)!
Let C denote the matrix obtained from B by replacing
If n − 1 is prime, then the lowest-terms numerator is the bottom-right entry with −2n2 (for consistency with
clearly either 1 or the prime n − 1 (and in fact the latter, the rest of the diagonal). Expanding in minors along
since n−1 is relatively prime to n and to (n−2)!). If n− the bottom row produces a second-order recursion for
1 is composite, either it can be written as ab for some det(C) solving to det(C) = (−1)n (n!)2 ; a similar ex-
a 6= b, in which case both a and b appear separately in pansion then yields det(B) = (−1)n−1 n!(n − 1)!.
(n − 2)! and so the numerator is 1, or n − 1 = p2 for Remark: This problem and solution are due to one of
some prime p, in which case p appears in (n − 2)! and us (Kedlaya). The statement appears in the comments
so the numerator is either 1 or p. (In the latter case, on sequence A010790 (i.e., the sequence (n − 1)!n!) in
the numerator is actually 1 unless p = 2, as in all other the On-Line Encyclopedia of Integer Sequences (http:
cases both p and 2p appear in (n − 2)!.) //oeis.org), attributed to Benoit Cloitre in 2002.
A2 Let v1 , . . . , vn denote the rows of A. The determinant is A3 First solution: Using the identity
unchanged if we replace vn by vn − vn−1 , and then vn−1
by vn−1 − vn−2 , and so forth, eventually replacing vk by (x + x−1 )2 − 2 = x2 + x−2 ,
vk − vk−1 for k ≥ 2. Since vk−1 and vk agree in their first
k k
k − 1 entries, and the k-th entry of vk − vk−1 is 1k − k−1 1
, we may check by induction on k that ak = 22 + 2−2 ; in
the result of these row operations is an upper triangular particular, the product is absolutely convergent. Using
matrix with diagonal entries 1, 21 −1, 31 − 12 , . . . , n1 − n−1
1
. the identities
The determinant is then
x2 + 1 + x−2
n 
1 1
 n 
−1
 = x − 1 + x−1 ,
− = x + 1 + x−1
∏ k−1 ∏
k=2 k k=2 k(k − 1) x2 − x−2
= x + x−1 ,
(−1)n−1 x − x−1
= .
(n − 1)!n! we may telescope the product to obtain
Note that a similar calculation can be made whenever k k
22 − 1 + 2−2
∞   ∞
A has the form Ai j = min{ai , a j } for any monotone se- 1
∏ 1 − = ∏ 2k −2k
quence a1 , . . . , an . Note also that the standard Gaussian k=0 ak k=0 2 + 2
elimination algorithm leads to the same upper triangu- k+1 k+1 k k
∞
22 + 1 + 2−2 22 − 2−2
lar matrix, but the nonstandard order of operations used =∏ · −k−1
2k −2k k+1
22 − 22
here makes the computations somewhat easier. k=0 2 + 1 + 2
0 0
Remark: The inverse of A can be identified explicitly: 22 − 2−2 3
for n ≥ 2, it is the matrix B given by = 0 0 = .
2
2 +1+2 −2 7


 −1 i= j=1 Second solution: (by Catalin Zara) In this solution, we
2
−2i


 1<i= j<n do not use the explicit formula for ak . We instead note
Bi j = −(n − 1)n i = j = n first that the ak form an increasing sequence which can-



 ij |i − j| = 1 not approach a finite limit (since the equation L = L2 −2
 has no real solution L > 2), and is thus unbounded. Us-
0 otherwise.

ing the identity

ak+1 + 1 = (ak − 1)(ak + 1),

 2

one checks by induction on n that note crucially that these equations also hold for n ∈
{0, 1}. Therefore, the function f : [0, +∞) → R given
n  
1 2 an+1 + 1 by
∏ 1 − ak = 7 a0 a1 · · · an .
k=0
f (t) = |w|2t − t 2 |w|2 + 2t(t − 1)Re(w) − (t − 1)2
Using the identity
satisfies f (t) = 0 for t ∈ {0, 1, i + 1, j + 1}. On the other
a2n+2 − 4 = a4n+1 − 4a2n+1 , hand, for all t ≥ 0 we have

one also checks by induction on n that f 000 (t) = (2 log |w|)3 |w|2t > 0,
2 so by Rolle’s theorem, the equation f (3−k) (t) = 0 has at
q
a0 a1 · · · an = a2n+1 − 4.
3 most k distinct solutions for k = 0, 1, 2, 3. This yields
the desired contradiction.
Hence
Remark: By similar reasoning, an equation of the form
n
 
1 3 an+1 + 1 ex = P(x) in which P is a real polynomial of degree
∏ 1 − = q
k=0 ak 7 a2 − 4 d has at most d + 1 real solutions. This turns out to
n+1
be closely related to a concept in mathematical logic
3 known as o-minimality, which in turn has deep conse-
tends to 7 as an+1 tends to infinity, hence as n tends to quences for the solution of Diophantine equations.
infinity.
Second solution: (by Noam Elkies) We recall a result
A4 The answer is 13 . commonly known as the Eneström-Kakeya theorem.
First solution: Let an = P(X = n); we want the min- Lemma 1. Let
imum value for a0 . If we write Sk = ∑∞ k
n=1 n an , then
the given expectation values imply that S1 = 1, S2 = 2, f (x) = a0 + a1 x + · · · + an xn
S3 = 5. Now define f (n) = 11n − 6n2 + n3 , and note
that f (0) = 0, f (1) = f (2) = f (3) = 6, and f (n) > 6 be a polynomial with real coefficients such that 0 < a0 ≤ a1 ≤
for n ≥ 4; thus 4 = 11S1 − 6S2 + S3 = ∑∞ n=1 f (n)an ≥ · · · ≤ an . Then every root z ∈ C of f satisfies |z| ≤ 1.
6 ∑n=1 an . Since ∑n=0 an = 1, it follows that a0 ≥ 31 .
∞ ∞

Equality is achieved when a0 = 31 , a1 = 12 , a3 = 16 , and Proof. If f (z) = 0, then we may rearrange the equality 0 =
an = 0 for all other n, and so the answer is 31 . f (z)(z − 1) to obtain
Second solution: (by Tony Qiao) Define the probabil- an zn+1 = (an − an−1 )zn + · · · + (a1 − a0 )z + a0 .
ity generating function of P as the power series
∞ But if |z| > 1, then
n
G(z) = ∑ P(x = n)z .
n=0 |an zn+1 | ≤ (|an − an−1 | + · · · + |a1 − a0 |)|z|n ≤ |an zn |,

We compute that G(1) = G0 (1) = G00 (1) = G000 (1) = 1. contradiction.

By Taylor’s theorem with remainder, for any x ∈ [0, 1],
there exists c ∈ [x, 1] such that Corollary 2. Let

(x − 1)2 (x − 1)3 G0000 (c) f (x) = a0 + a1 x + · · · + an xn

G(x) = 1 + (x − 1) + + + (x − 1)4 .
2! 3! 4! be a polynomial with positive real coefficients. Then every
In particular, G(0) = + 1 1 0000
for some c ∈ [0, 1]. root z ∈ C of f satisfies r ≤ |z| ≤ R for
3 24 G (c)
However, since G has nonnegative coefficients and c ≥
r = min{a0 /a1 , . . . , an−1 /an }
0, we must have G0000 (c) ≥ 0, and so G(0) ≥ 31 . As in the
first solution, we see that this bound is best possible. R = max{a0 /a1 , . . . , an−1 /an }.

A5 First solution: Suppose to the contrary that there exist Proof. The bound |z| ≤ R follows by applying the lemma to
positive integers i 6= j and a complex number z such that the polynomial f (x/R). The bound |z| ≥ r follows by applying
Pi (z) = Pj (z) = 0. Note that z cannot be a nonnegative the lemma to the reverse of the polynomial f (x/r).
real number or else Pi (z), Pj (z) > 0; we may put w =
z−1 6= 0, 1. For n ∈ {i + 1, j + 1} we compute that Suppose now that Pi (z) = Pj (z) = 0 for some z ∈ C and
some integers i < j. We clearly cannot have j = i+1, as
wn = nw − n + 1, wn = nw − n + 1; then Pi (0) 6= 0 and so Pj (z) − Pi (z) = (i + 1)zi 6= 0; we
thus have j − i ≥ 2. By applying Corollary 2 to Pi (x),
we see that |z| ≤ 1 − 1i . On the other hand, by applying
 3

Corollary 2 to (Pj (x) − Pi (x))/xi−1 , we see that |z| ≥ Remark: The reader may notice a strong similarity be-
1 tween this solution and the first solution. The primary
1 − i+2 , contradiction.
difference is we compute that Ez0 (0) ≥ 0 instead of dis-
Remark: Elkies also reports that this problem is his
covering that Ez (−1) = 0.
submission, dating back to 2005 and arising from work
of Joe Harris. It dates back further to Example 3.7 in: Remark: It is also possible to solve this prob-
Hajime Kaji, On the tangentially degenerate curves, J. lem using a p-adic valuation on the field of alge-
London Math. Soc. (2) 33 (1986), 430–440, in which braic numbers in place of the complex absolute value;
the second solution is given. however, this leads to a substantially more compli-
cated solution. In lieu of including such a solution
Remark: Elkies points out a mild generalization which
here, we refer to the approach described by Victor
may be treated using the first solution but not the sec-
Wang here: http://www.artofproblemsolving.
ond: for integers a < b < c < d and z ∈ C which is
com/Forum/viewtopic.php?f=80&t=616731.
neither zero nor a root of unity, the matrix
  A6 The largest such k is nn . We first show that this
1 1 1 1 value can be achieved by an explicit construction. Let
a b c d
e1 , . . . , en be the standard basis of Rn . For i1 , . . . , in ∈
za zb zc zd {1, . . . , n}, let Mi1 ,...,in be the matrix with row vectors
ei1 , . . . , ein , and let Ni1 ,...,in be the transpose of Mi1 ,...,in .
has rank 3 (the problem at hand being the case a = Then Mi1 ,...,in N j1 ,..., jn has k-th diagonal entry eik · e jk ,
0, b = 1, c = i + 1, d = j + 1). proving the claim.
Remark: It seems likely that the individual polynomi- We next show that for any families of matrices Mi , N j as
als Pk (x) are all irreducible, but this appears difficult to described, we must have k ≤ nn . Let V be the n-fold ten-
prove. sor product of Rn , i.e., the vector space with orthonor-
Third solution: (by David Feldman) Note that mal basis ei1 ⊗ · · · ⊗ ein for i1 , . . . , in ∈ {1, . . . , n}. Let mi
be the tensor product of the rows of Mi ; that is,
Pn (x)(1 − x) = 1 + x + · · · + xn−1 − nxn .
n
If |z| ≥ 1, then mi = ∑ (Mi )1,i1 · · · (Mi )n,in ei1 ⊗ · · · ⊗ ein .
i1 ,...,in =1

n|z|n ≥ |z|n−1 + · · · + 1 ≥ |zn−1 + · · · + 1|,

Similarly, let n j be the tensor product of the columns of
with the first equality occurring only if |z| = 1 and the N j . One computes easily that mi · n j equals the product
second equality occurring only if z is a positive real of the diagonal entries of Mi N j , and so vanishes if and
number. Hence the equation Pn (z)(1 − z) = 0 has no so- only if i 6= j. For any ci ∈ R such that ∑i ci mi = 0, for
lutions with |z| ≥ 1 other than the trivial solution z = 1. each j we have
Since !
0= ∑ ci mi · n j = ∑ ci (mi · n j ) = c j .
Pn (x)(1 − x)2 = 1 − (n + 1)xn + nxn+1 , i i

it now suffices to check that the curves Therefore the vectors m1 , . . . , mk in V are linearly inde-
pendent, implying k ≤ nn as desired.
Cn = {z ∈ C : 0 < |z| < 1, |z|n |n + 1 − zn| = 1}
Remark: Noam Elkies points out that similar argument
are pairwise disjoint as n varies over positive integers. may be made in the case that the Mi are m × n matrices
and the N j are n × m matrices.
Write z = u + iv; we may assume without loss of gener-
ality that v ≥ 0. Define the function B1 These are the integers with no 0’s in their usual base
10 expansion. If the usual base 10 expansion of N is
Ez (n) = n log |z| + log |n + 1 − zn|. dk 10k + · · · + d0 100 and one of the digits is 0, then there
exists an i ≤ k − 1 such that di = 0 and di+1 > 0; then
One computes that for n ∈ R, Ez00 (n) < 0 if and only if we can replace di+1 10i+1 + (0)10i by (di+1 − 1)10i+1 +
(10)10i to obtain a second base 10 over-expansion.
u−v−1 u+v−1
<n< . We claim conversely that if N has no 0’s in its usual
(1 − u)2 + v2 (1 − u)2 + v2
base 10 expansion, then this standard form is the unique
In addition, Ez (0) = 0 and base 10 over-expansion for N. This holds by induc-
tion on the number of digits of N: if 1 ≤ N ≤ 9,
1 then the result is clear. Otherwise, any base 10 over-
Ez0 (0) = log(u2 + v2 ) + (1 − u) ≥ log(u) + 1 − u ≥ 0
2 expansion N = dk 10k + · · · + d1 10 + d0 100 must have
d0 ≡ N (mod 10), which uniquely determines d0 since
since log(u) is concave. From this, it follows that the
equation Ez (n) = 0 can have at most one solution with
n > 0.
 4

N is not a multiple of 10; then (N − d0 )/10 inherits the 1 ≤ x ≤ 3; then F(1) = F(3) = 0 and F(x) ≤ min{x −
base 10 over-expansion dk 10k−1 + · · · + d1 100 , which 1, 3 − x}. Using integration by parts, we obtain
must be unique by the induction hypothesis. Z 3 Z 3
f (x) F(x)
Remark: Karl Mahlburg suggests an alternate proof of dx = dx
uniqueness (due to Shawn Williams): write the usual 1 x 1 x2
Z 2 Z 3
expansion N = dk 10k + · · · + d0 100 and suppose di 6= 0 x−1 3−x
≤ dx + dx
for all i. Let M = cl 10l + · · · + c0 100 be an over- 1 x2 2 x2
expansion with at least one 10. To have M = N, we 4
must have l ≤ k; we may pad the expansion of M with = log .
3
zeroes to force l = k. Now define ei = ci − di ; since
1 ≤ di ≤ 9 and 0 ≤ ci ≤ 10, we have 0 ≤ |ei | ≤ 9. More- (Some minor adjustment is needed to make this com-
over, there exists at least one index i with ei 6= 0, since pletely rigorous, e.g., approximating f uniformly by
any index for which ci = 10 has this property. But if i is continuous functions.)
the largest such index, we have
B3 First solution: Assume by way of contradiction that A
i−1 has rank at most 1; in this case, we can find rational
10i ≤ ei 10i = − ∑ ei 10i numbers a1 , . . . , am , b1 , . . . , bn such that Ai j = ai b j for
j=0 all i, j. By deleting rows or columns, we may reduce to
i−1 the case where the ai ’s and b j ’s are all nonzero.
≤ ∑ ei |10i ≤ 9 · 10i−1 + · · · + 9 · 100 , Recall that any nonzero rational number q has a unique
j=0
prime factorization
a contradiction. q = ±2c1 3c2 5c3 · · ·
B2 In all solutions, we assume that the function f is inte-
grable. with exponents in Z. Set

First solution: Let g(x) be 1 for 1 ≤ x ≤ 2 and −1 c(q) = (c1 , c2 , c3 , . . . ).

for 2 < x ≤ 3, and define h(x) = g(x) − f (x). Then
R3
1 h(x) dx = 0 and h(x) ≥ 0 for 1 ≤ x ≤ 2, h(x) ≤ 0
Note that |ai b j | is prime if and only if c(ai ) + c(b j ) has
for 2 < x ≤ 3. Now one entry equal to 1 and all others equal to 0. The con-
Z 3 Z 2 Z 3
dition that m + n distinct primes appear in the matrix
h(x) |h(x)| |h(x)| implies that the vector space
dx = dx − dx
1 x 1 x 2 x ( )
Z 2 Z 3
|h(x)| |h(x)|
≥
1 2
dx −
2 2
dx = 0, ∑ xi c(ai ) + ∑ yi c(b j ) : xi , y j ∈ R, ∑ xi = ∑ y j
i j i j
R 3 f (x) R 3 g(x)
and thus 1 x dx ≤ 1 x dx = 2 log 2−log 3 = log 43 . contains a linearly independent set of size m + n. But
Since g(x) achieves the upper bound, the answer is that space evidently has dimension at most m + n − 1,
log 34 . contradiction.
Reformulation: (by Karl Mahlburg and Karthik Second solution: In this solution, we use standard
Adimurthi) Since f is integrable, it can be expressed terminology of graph theory, considering only sim-
in terms of the Hadamard basis ple undirected graphs (with no self-loops or multiple
 edges). We first recall the quick induction proof that
1
 x ∈ [1, 2) that a graph on k vertices with no cycles contains at
H0 (x) = −1 x ∈ [2, 3] most k − 1 edges: for k = 1, the claim is trivially true

0 x∈/ [1, 3] because there can be no edges. For k > 1, choose any
vertex v and let d be its degree. Removing the vertex v
Hn+1 (x) = Hn (2(x − 1) + 1) + Hn (2(x − 2) + 1). and the edges incident to it leaves a disjoint union of d
different graphs, each having no cycles. If the numbers
More precisely, we have f (x) = ∑n cn Hn (x) for some cn
of vertices in these graphs are k1 , . . . , kd , by induction
with |c0 | + |c1 | + | · · · | ≤ 1. Let gn = 13 (Hn (x)/x) dx; it
R
the total number of edges in the original graph is at most
is easy to show that the gn are strictly decreasing in n. (k1 − 1) + · · · + (kd − 1) + d = k − 1.
Thus
Returning to the original problem, suppose that A has
Z 3
4 rank at most 1. Draw a bipartite graph whose vertices
( f (x)/x)dx = c0 g0 + c1 g1 + · · · ≤ 1 · g0 = log . correspond to the rows and columns of A, with an edge
1 3
joining a particular row and column if the entry where
Second solution: (Art of Problem Solving, user they intersect has prime absolute value. By the previ-
libra_gold) Define the function F(x) = 1x f (t) dt for ous paragraph, this graph must contain a cycle. Since
R
 5

the graph is bipartite, this cycle must be of length 2k for so fn (−2−n+2 j ) has overall sign (−1) j , proving the
some integer k ≥ 2 (we cannot have k = 1 because the claimed alternation.
graph has no repeated edges). Without loss of gener- Remark: Karl Mahlburg suggests an alternate interpre-
ality, we may assume that the cycle consists of row 1, 2
tation of the preceding algebra: write 2− j fn (2−n+2 j ) as
column 1, row 2, column 2, and so on. There must then
exist distinct prime numbers p1 , . . . , p2k such that 2 2
2− j − 2−( j−1) + · · · + (−1) j−1 2−1 + (−1) j 2−1
|A11 | = p1 , |A21 | = p2 , . . . , |Akk | = p2k−1 , |A1k | = p2k . +(−1) j 2−1 + (−1) j+1 2−1 + (−1) j+2 2−2 + · · · ,
However, since A has rank 1, the 2 × 2 minor A11 Ai j − where the two central terms (−1) j 2−1 arise from split-
Ai1 A1 j must vanish for all i, j. If we put ri = |Ai1 | and ting the term arising from x j . Then each row is an alter-
c j = Ai j /A11 , we have nating series whose sum carries the sign of (−1) j unless
it has only two terms. Since n ≥ 3, one of the two sums
p1 · · · p2k = (r1 c1 )(r2 c1 ) · · · (rk ck )(r1 ck ) is forced to be nonzero.
= (r1 c1 · · · rk ck )2 , Remark: One of us (Kedlaya) received this problem
and solution from David Speyer in 2009 and submitted
which contradicts the existence of unique prime factor- it to the problem committee.
izations for positive rational numbers: the prime p1 oc-
curs with exponent 1 on the left, but with some even B5 We show that Patniss wins if p = 2 and Keeta wins if
exponent on the right. This contradiction completes the p > 2 (for all n). We first analyze the analogous game
proof. played using an arbitrary finite group G. Recall that for
any subset S of G, the set of elements g ∈ G which com-
B4 Define the polynomial fn (x) = ∑nk=0 2k(n−k) xk . Since mute with all elements of S forms a subgroup Z(S) of
G, called the centralizer (or commutant) of S. At any
f1 (x) = 1 + x, f2 (x) = 1 + 2x + x2 = (1 + x)2 , given point in the game, the set S of previously cho-
sen elements is contained in Z(S). Initially S = 0/ and
the claim holds for for n = 1, 2. For n ≥ 3, we show that
Z(S) = G; after each turn, S is increased by one ele-
the quantities
ment and Z(S) is replaced by a subgroup. In particular,
fn (−2−n ), fn (−2−n+2 ), . . . , fn (−2n ) if the order of Z(S) is odd at some point, it remains
odd thereafter; conversely, if S contains an element of
alternate in sign; by the intermediate value theorem, this even order, then the order of Z(S) remains even there-
will imply that fn has a root in each of the n intervals after. Therefore, any element g ∈ G for which Z({g})
(−2−n , −2−n+2 ), . . . , (−2n−2 , −2n ), forcing fn to have has odd order is a winning first move for Patniss, while
as many distinct real roots as its degree. any other first move by Patniss loses if Keeta responds
with some h ∈ Z({g}) of even order (e.g., an element of
For j ∈ {0, . . . , n}, group the terms of fn (x) as a 2-Sylow subgroup of Z({g})). In both cases, the win
is guaranteed no matter what moves follow.
···
Now let G be the group of invertible n × n matrices with
+ 2( j−5)(n− j+5) x j−5 + 2( j−4)(n− j+4) x j−4 entries in Z/pZ. If p > 2, then Z(S) will always contain
+ 2( j−3)(n− j+3) x j−3 + 2( j−2)(n− j+2) x j−2 the scalar matrix −1 of order 2, so the win for Keeta is
guaranteed. (An explicit winning strategy is to answer
+ 2( j−1)(n− j+1) x j−1 + 2 j(n− j) x j + 2( j+1)(n− j−1) x j+1 any move g with the move −g.)
+ 2( j+2)(n− j−2) x j+2 + 2( j+3)(n− j−3) x j+3 If p = 2, we establish the existence of g ∈ G such that
+ 2( j+4)(n− j−4) x j+4 + 2( j+5)(n− j−5) x j+5 Z({g}) has odd order using the existence of an irre-
ducible polynomial P(x) of degree n over Z/pZ (see
··· .
remark). We construct an n × n matrix over Z/pZ with
characteristic polynomial P(x) by taking the companion
Depending on the parity of j and of n − j, there may matrix of P(x): write P(x) = xn + Pn−1 xn−1 + · · · + P0
be a single monomial left on each end. When evaluat- and set
ing at x = −2−n+2 j , the trinomial evaluates to 0. In the
binomials preceding the trinomial, the right-hand term 0 0 · · · 0 −P0
 
dominates, so each of these binomials contributes with 1 0 · · · 0 −P1 
the sign of x j−2k , which is (−1) j . In the binomials fol-
0 1 · · · 0 −P 
g= 2 .
. . .
lowing the trinomial, the left-hand term dominates, so  .. .. . . ... .. 
. 
again the contribution has sign (−1) j .
0 0 · · · 1 −Pn−1
Any monomials which are left over on the ends also
contribute with sign (−1) j . Since n ≥ 3, there ex- In particular, det(g) = (−1)n P0 6= 0, so g ∈ G. Over
ists at least one contribution other than the trinomial, an algebraic closure of Z/pZ, g becomes diagonaliz-
 6

able with distinct eigenvalues, so any matrix commut- contradiction

ing with g must also be diagonalizable, and hence of
odd order. In particular, Z({g}) is of odd order, so Pat- n = (ms − nr)s0 + (r0 n − ms0 ) ≥ s + s0 > n;
niss has a winning strategy. 0
hence rs , rs0 remain consecutive in Fn .
Remark: It can be shown that in the case p = 2, the
only elements g ∈ G for which Z({g}) has odd order Let fn : [0, 1] → R be the piecewise linear function
are those for which g has distinct eigenvalues: in any which agrees with f at each element of Fn and is linear
other case, Z({g}) contains a subgroup isomorphic to between any two consecutive elements of Fn . Between
0
the group of k × k invertible matrices over Z/2Z for any two consecutive elements rs , rs0 of Fn , fn coincides
some k > 1, and this group has order (2k − 1)(2k − 0
with some linear function a + bx. Since s f ( rs ), s0 f ( rs0 ) ∈
2) · · · (2k − 2k−1 ). Z, we deduce first that
Remark: We sketch two ways to verify the existence
of an irreducible polynomial of degree n over Z/pZ for r0 r
b = ss0 ( f ( 0 ) − f ( ))
any positive integer n and any prime number p. One s s
is to use Möbius inversion to count the number of ir-
is an integer of absolute value at most K, and second
reducible polynomials of degree n over Z/pZ and then 0
give a positive lower bound for this count. The other that both as = s f ( rs ) − br and as0 = s0 f ( rs0 ) − br0 are
is to first establish the existence of a finite field F of integral. It follows that fn is integral.
cardinality pn , e.g., as the set of roots of the polynomial We now check that if n > 2K, then fn = fn−1 . For this,
n
x p −1 inside a splitting field, and then take the minimal it suffices to check that for any consecutive elements
polynomial of a nonzero element of F over Z/pZ which r m r0
s , n , s0 in Fn , the linear function a0 +b0 x matching f n−1
is a primitive (pn − 1)-st root of unity in F (which exist 0
from rs to rs0 has the property that f ( mn ) = a0 + b0 mn .
because the multiplicative group of F contains at most Define the integer t = n f ( mn ) − a0 n − b0 m. We then
one cyclic subgroup of any given order). One might be compute that the slope of fn from rs to mn is b0 + st,
tempted to apply the primitive element theorem for F 0
over Z/pZ, but in fact one of the preceding techniques while the slope of fn from mn to rs0 is at most b0 − s0t.
0
In order to have |b0 + st| , |b0 − s t| ≤ K, we must have
is needed in order to verify this result for finite fields,
as the standard argument that “most” elements of the (s + s0 ) |t| ≤ 2K; since s + s0 = n > 2K, this is only pos-
upper field are primitive breaks down for finite fields. sible if t = 0. Hence fn = fn−1 , as claimed.

One may also describe the preceding analysis in terms It follows that for any n > 2K, we must have fn =
of an identification of F as a Z/pZ-vector space with fn+1 = · · · . Since the condition on f and K implies that
the space of column vectors of length n. Under such f is continuous, we must also have fn = f , completing
an identification, if we take g to be an element of F − the proof.
{0} generating this group, then any element of Z({g}) Remark: The condition on f and K is called Lipschitz
commutes with all of F − {0} and hence must define an continuity.
F-linear endomorphism of F. Any such endomorphism Remark: An alternate approach is to prove that for
is itself multiplication by an element of F, so Z({g}) each x ∈ [0, 1), there exists ε ∈ (0, 1 − x) such that the
is identified with the multiplicative group of F, whose restriction of f to [x, x + ε) is linear; one may then de-
order is the odd number 2n − 1. duce the claim using the compactness of [0, 1]. In this
B6 Let us say that a linear function g on an interval is inte- approach, the role of the Farey sequence may also be
gral if it has the form g(x) = a + bx for some a, b ∈ Z, played by the convergents of the continued fraction of x
and that a piecewise linear function is integral if on ev- (at least in the case where x is irrational).
ery interval where it is linear, it is also integral. Remark: This problem and solution are due to one
For each positive integer n, define the n-th Farey se- of us (Kedlaya). Some related results can be proved
quence Fn as the sequence of rational numbers in [0, 1] with the Lipschitz continuity condition replaced by suit-
with denominators at most n. It is easily shown by in- able convexity conditions. See for example: Kiran S.
0 Kedlaya and Philip Tynan, Detecting integral polyhe-
duction on n that any two consecutive elements rs , rs0
dral functions, Confluentes Mathematici 1 (2009), 87–
of Fn , written in lowest terms, satisfy gcd(s, s0 ) = 1,
109. Such results arise in the theory of p-adic differen-
s + s0 > n, and r0 s − rs0 = 1. Namely, this is obvious
tial equations; see for example: Kiran S. Kedlaya and
for n = 1 because F1 = 01 , 11 . To deduce the claim for
0 Liang Xiao, Differential modules on p-adic polyannuli,
Fn from the claim for Fn−1 , let rs , rs0 be consecutive el- J. Inst. Math. Jusssieu 9 (2010), 155–201 (errata, ibid.,
ements of Fn−1 . If s + s = n, then for m = r + r0 we
0
669–671).
0 0
have rs < mn < rs0 and the pairs rs , mn and mn , rs0 satisfy the
0
desired conditions. Conversely, if s + s > n, then we
0
cannot have rs < mn < rs0 for a ∈ Z, as this yields the
 The 76th William Lowell Putnam Mathematical Competition
Saturday, December 5, 2015

A1 Let A and B be points on the same branch of the hyper- four numbers off the list. Repeat with the three smallest
bola xy = 1. Suppose that P is a point lying between remaining numbers 4, 5, 7 and their sum 16. Continue
A and B on this hyperbola, such that the area of the tri- in this way, crossing off the three smallest remaining
angle APB is as large as possible. Show that the re- numbers and their sum, and consider the sequence of
gion bounded by the hyperbola and the chord AP has sums produced: 6, 16, 27, 36, . . . . Prove or disprove that
the same area as the region bounded by the hyperbola there is some number in the sequence whose base 10
and the chord PB. representation ends with 2015.
A2 Let a0 = 1, a1 = 2, and an = 4an−1 − an−2 for n ≥ 2. B3 Let S be the set of all 2 × 2 real matrices
Find an odd prime factor of a2015 . 
a b

M=
A3 Compute c d

2015 2015
! whose entries a, b, c, d (in that order) form an arithmetic
2πiab/2015 progression. Find all matrices M in S for which there is
log2 ∏ ∏ (1 + e )
a=1 b=1 some integer k > 1 such that M k is also in S.

Here i is the imaginary unit (that is, i2 = −1). B4 Let T be the set of all triples (a, b, c) of positive integers
for which there exist triangles with side lengths a, b, c.
A4 For each real number x, let Express
1 2a
f (x) = ∑ n
, ∑
n∈Sx 2 3b 5c
(a,b,c)∈T

where Sx is the set of positive integers n for which bnxc as a rational number in lowest terms.
is even. What is the largest real number L such that
f (x) ≥ L for all x ∈ [0, 1)? (As usual, bzc denotes the B5 Let Pn be the number of permutations π of {1, 2, . . . , n}
greatest integer less than or equal to z.) such that
A5 Let q be an odd positive integer, and let Nq denote |i − j| = 1 implies |π(i) − π( j)| ≤ 2
the number of integers a such that 0 < a < q/4 and
gcd(a, q) = 1. Show that Nq is odd if and only if q is for all i, j in {1, 2, . . . , n}. Show that for n ≥ 2, the quan-
of the form pk with k a positive integer and p a prime tity
congruent to 5 or 7 modulo 8.
Pn+5 − Pn+4 − Pn+3 + Pn
A6 Let n be a positive integer. Suppose that A, B, and M are
n×n matrices with real entries such that AM = MB, and does not depend on n, and find its value.
such that A and B have the same characteristic polyno-
mial. Prove that det(A − MX) = det(B − XM) for every B6 For each positive integer k, let A(k)√ be the number of
n × n matrix X with real entries. odd divisors of k in the interval [1, 2k). Evaluate
∞
B1 Let f be a three times differentiable function (defined A(k)
on R and real-valued) such that f has at least five dis- ∑ (−1)k−1 k
.
k=1
tinct real zeros. Prove that f + 6 f 0 + 12 f 00 + 8 f 000 has at
least two distinct real zeros.
B2 Given a list of the positive integers 1, 2, 3, 4, . . . , take the
first three numbers 1, 2, 3 and their sum 6 and cross all
 Solutions to the 76th William Lowell Putnam Mathematical Competition
Saturday, December 5, 2015
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A1 First solution: Without loss of generality, assume that any integer m and any prime p dividing am , p also di-
A and B lie in the first quadrant with A = (t1 , 1/t1 ), B = vides a−m ; on the other hand, p cannot divide a−m+1 ,
(t2 , 1/t2 ), and t1 < t2 . If P = (t, 1/t) with t1 ≤ t ≤ t2 , as otherwise p would also divide a−m+2 , . . . , a0 = 1, a
then the area of triangle APB is contradiction. We can thus find an integer k such that
am+1 ≡ ka−m+1 (mod p); by induction on n, we see
1 1 1 1 t2 − t1 that an ≡ kan−2m (mod p) for all n. In particular, if k is
t1 t t2 = (t1 + t2 − t − t1t2 /t). odd, then p also divides akm ; we thus conclude (again)
2 1/t 1/t 1/t 2t1t2
1 2 that a2015 is divisible by a5 = 362 and thus by 181.
Remark: Although it was not needed in the solution,
When t1 ,t2 are fixed, this is maximized when t + t1t2 /t
we note in passing that if an ≡ 0 (mod p), then a2n+k ≡
is minimized,
√ which by AM-GM exactly holds when
−ak (mod p) for all k.
t = t1t2 .
Remark: One can find other odd prime factors of a2015
The line AP is given by y = t1 +t−x tt1 , and so the area of in the same manner. For example, a2015 is divisible by
the region bounded by the hyperbola and AP is each of the following quantities. (The prime factoriza-
Z t    tions were computed using the Magma computer algebra
t1 + t − x 1 t t1 t
− dx = − − log , system.)
t1 tt 1 x 2t 1 2t t1
√ a13 = 2 × 6811741
2 −t1
which at t = t1t2 is equal to 2t√
p
t1 t2 − log( t2 /t1 ). a31 = 2 × 373 × 360250962984637
Similarly, the area of the region bounded by the√hyper-
bola and PB is 2t t2
− 2tt 2 − log tt2 , which at t = t1t2 is a5·13 = 2 × 181 × 6811741
2 −t1 × 3045046274679316654761356161
also 2t√
p
t t − log( t2 /t1 ), as desired.
12
a5·31 = 1215497709121 × 28572709494917432101
Second solution: For any λ > 0, the map (x, y) 7→
× 13277360555506179816997827126375881581
(λ x, λ −1 y) preserves both areas and the hyperbola xy =
1. We may thus rescale the picture so that A, B are sym- a13·31 = 2 × 373 × 193441 × 6811741 × 360250962984637
metric across the line y = x, with A above the line. As × 16866100753000669
P moves from A to B, the area of APB increases until P × 79988387992470656916594531961 × p156
passes through the point (1, 1), then decreases. Conse-
quently, P = (1, 1) achieves the maximum area, and the where p156 is a prime of 156 decimal digits. Dividing
desired equality is obvious by symmetry. Alternatively, a2015 by the product of the primes appearing in this list
since the hyperbola is convex, the maximum is uniquely yields a number N of 824 decimal digits which is defi-
achieved at the point where the tangent line is parallel nitely not prime, because 2N 6≡ 2 (mod N), but whose
to AB, and by symmetry that point is P. prime factorization we have been unable to establish.
Note that N is larger than a 2048-bit RSA modulus, so
A2 First solution: One possible√answer is 181. √ nBy in- the difficulty of factoring it is not surprising.
duction, we have an = ((2 + 3)n + (2 − √ 3) )/2 =
(α n √
+ β n )/2 for all n, where α = 2 + 3 and β = One thing we can show is that each prime factor of N is
congruent to 1 modulo 6 × 2015 = 12090, thanks to the
2 − 3. Now note that if k is an odd positive integer
kn +β kn following lemma.
and an 6= 0, then aaknn = αα n +β n =α
(k−1)n −α (k−2)n β n +

· · · − α n β (k−2)n + β (k−1)n . This expression is both ra- Lemma. Let n be an odd integer. Then any odd prime factor
tional√(because an and akn are integers) and of the form p of an which does not divide am for any divisor m of n is
a + b 3 for some integers a, b by the expressions for congruent to 1 modulo lcm(6, n). (By either solution of the
α, β ; it follows that it must be an integer, and so akn is original problem, p also does not divide am for any positive
divisible by an . Applying this to n = 5 and k = 403, we integer m < n.)
find that a2015 is divisible by a5 = 362 and thus by 181. √
Proof. We first check that p ≡ 1 (mod 3). In Fq = F p ( 3)
Second solution: By rewriting the formula for an as we have (α/β )n ≡ −1. If p ≡ 2 (mod 3), then q = p2 and
an−2 = 4an−1 − an , we may extend the sequence back- n
wards to define an for all integers n. Since a−1 = 2,
α and β are conjugate in p; consequently,
n n
√ n
√ equality α =
the
−β in Fq2 means that α = c 3, β = −c 3 for some c ∈
we may see by induction that a−n = an for all n. For
F p . But then −3c2 = α n β n = 1 in Fq and hence in F p , which
contradicts p ≡ 2 (mod 3) by quadratic reciprocity.
 2

By the previous paragraph, α and β may be identified with and note (as above) that ω 2 , ω 4 , . . . , ω 2(n1 −1) is a per-
elements of F p , and we have (α/β )n ≡ −1, but the same does mutation of ω, . . . , ω n1 −1 , so the two products in the
not hold with n replaced by any smaller value. Since F× p is fraction are equal.
a cyclic group of order p − 1, this forces p ≡ 1 (mod n) as Remark: The function f (n) = ∑d|n d · φ (n/d) is multi-
claimed. plicative: for any two coprime positive integers m, n, we
have f (mn) = f (m) f (n). This follows from the fact that
A3 The answer is 13725. We first claim that if n is odd, f (n) is the convolution of the two multiplicative func-
then ∏nb=1 (1 + e2πiab/n ) = 2gcd(a,n) . To see this, write tions n 7→ n and n 7→ φ (n); it can also be seen directly
d = gcd(a, n) and a = da1 , n = dn1 with gcd(a1 , n1 ) = using the Chinese remainder theorem.
1. Then a1 , 2a1 , . . . , n1 a1 modulo n1 is a permutation
of 1, 2, . . . , n1 modulo n1 , and so ω a1 , ω 2a1 , . . . , ω n1 a1 A4 The answer is L = 4/7. For S ⊂ N, let F(S) =
is a permutation of ω, ω 2 , . . . , ω n1 ; it follows that for ∑n∈S 1/2n , so that f (x) = F(Sx ). Note that for T =
ω = e2πi/n1 , {1, 4, 7, 10, . . .}, we have F(T ) = 4/7.
n1 n1 n1 We first show by contradiction that for any x ∈ [0, 1),
∏ (1 + e2πiab/n ) = ∏ (1 + e2πia1 b/n1 ) = ∏ (1 + ω b ). f (x) ≥ 4/7. Since each term in the geometric series
b=1 b=1 b=1 ∑n 1/2n is equal to the sum of all subsequent terms, if
S, S0 are different subsets of N and the smallest positive
Now since the roots of zn1 − 1 are ω, ω 2 , . . . , ω n1 , it fol- integer in one of S, S0 but not in the other is in S, then
lows that zn1 − 1 = ∏nb=11
(z − ω b ). Setting z = −1 and F(S) ≥ F(S0 ). Assume f (x) < 4/7; then the smallest
using the fact that n1 is odd gives ∏nb=1
1
(1 + ω b ) = 2. integer in one of Sx , T but not in the other is in T . Now
Finally, ∏nb=1 (1 + e2πiab/n ) = (∏nb=1
1
(1 + e2πiab/n ))d = 1 ∈ Sx for any x ∈ [0, 1), and we conclude that there
d are three consecutive integers n, n + 1, n + 2 that are not
2 , and we have proven the claim.
in Sx : that is, bnxc, b(n + 1)xc, b(n + 2)xc are all odd.
From the claim, we find that Since the difference between consecutive terms in nx,
! (n + 1)x, (n + 2)x is x < 1, we conclude that bnxc =
2015 2015
2πiab/2015 b(n + 1)xc = b(n + 2)xc and so x < 1/2. But then 2 ∈ Sx
log2 ∏ ∏ (1 + e )
and so f (x) ≥ 3/4, contradicting our assumption.
a=1 b=1
2015 2015
! It remains to show that 4/7 is the greatest lower bound
2πiab/2015 for f (x), x ∈ [0, 1). For any n, choose x = 2/3 − ε with
= ∑ log2 ∏ (1 + e )
a=1 b=1 0 < ε < 1/(9n); then for 1 ≤ k ≤ n, we have 0 < mε <
2015 1/3 for m ≤ 3n, and so
= ∑ gcd(a, 2015).
a=1 b(3k − 2)xc = b(2k − 2) + 2/3 − (3k − 2)εc = 2k − 2
b(3k − 1)xc = b(2k − 1) + 1/3 − (3k − 1)εc = 2k − 1
Now for each divisor d of 2015, there are φ (2015/d)
integers between 1 and 2015 inclusive whose gcd with b(3k)xc = b(2k − 1) + 1 − 3kεc = 2k − 1.
2015 is d. Thus
It follows that Sx is a subset of S = {1, 4, 7, . . . , 3n −
2015 2, 3n + 1, 3n + 2, 3n + 3, . . .}, and so f (x) = F(Sx ) ≤
∑ gcd(a, 2015) = ∑ d · φ (2015/d). f (S) = (1/2 + 1/24 + · · · + 1/23n+1 ) + 1/23n+1 . This
a=1 d|2015 last expression tends to 4/7 as n → ∞, and so no num-
ber greater than 4/7 can be a lower bound for f (x) for
We factor 2015 = pqr with p = 5, q = 13, and r = 31,
all x ∈ [0, 1).
and calculate
A5 First solution: By inclusion-exclusion, we have
∑ d · φ (pqr/d)
 
d|pqr bq/4c
Nq = ∑ µ(d)
= 1 · (p − 1)(q − 1)(r − 1) + p · (q − 1)(r − 1) d|q
d
+ q · (p − 1)(r − 1) + r · (p − 1)(q − 1) + pq · (r − 1) 
q/d

+ pr · (q − 1) + qr · (p − 1) + pqr · 1 = ∑ µ(d)
d|q
4
= (2p − 1)(2q − 1)(2r − 1).  
q/d
≡ ∑ (mod 2),
When (p, q, r) = (5, 13, 31), this is equal to 13725. d|q squarefree
4
Remark: Noam Elkies suggests the following similar
but shorter derivation of the equality ∏nb=1
1
(1+ω b ) = 2: where µ is the Möbius function. Now
write   (
q/d 0 (mod 2) if q/d ≡ 1, 3 (mod 8)
n1 −1 n1 −1
∏b=1 (1 − ω 2b ) ≡
b 4 1 (mod 2) if q/d ≡ 5, 7 (mod 8).
∏ (1 + ω )= n1 −1
(1 − ω b )
b=1 ∏b=1
 3

So Nq is odd if and only if q has an odd number of If q = 1, then Nq is even. If q has more than one
squarefree factors q/d congruent to 5 or 7 (mod 8). prime factor, then the group (Z/qZ)× has exponent di-
If q has a prime factor p congruent to 1 or 3 (mod 8), viding φ (q)/2, so (−1)Nq ≡ (−2)φ (q)/2 ≡ 1 (mod q),
then the squarefree factors d of q occur in pairs c, pc, and thus Nq must be even in this case as well. Fi-
which are either both 1 or 3 (mod 8) or both 5 or 7 nally, suppose that q is a prime power pk with p odd
(mod 8). Hence q must have an even number of factors and k positive. Since (Z/qZ)× is a cyclic group of
that are congruent to 5 or 7 (mod 8), and so Nq is even order φ (q) = pk−1 (p − 1), in which the only square
in this case. roots of unity are ±1, it follows that (−2)φ (q)/2 ≡ ±1
If q has two prime factors p1 and p2 , each congruent to (mod q) in accordance with whether (−2)(p−1)/2 ≡ ±1
either 5 or 7 (mod 8), then the squarefree factors d of q (mod p), i.e., whether −2 is a quadratic residue or non-
occur in quadruples d, p1 d, q1 d, p1 q1 d, which are then residue. But recall that −2 is a quadratic residue mod-
congruent respectively to some permutation of 1,3,5,7 ulo p if and only if p ≡ 1, 3 (mod 8). Thus Nq is odd
(mod 8) (if p1 and p2 are distinct mod 8) or are congru- in this case if and only if p ≡ 5 or 7 (mod 8).
ent respectively to d, p1 d, p1 d, d (mod 8). Either way, We conclude that for any odd integer q ≥ 1, the quantity
we see that exactly two of the four residues are congru- Nq is odd if and only if q = pk with k positive and p a
ent to 5 or 7 (mod 8). Thus again q must have an even prime that is 5 or 7 (mod 8).
number of factors that are 5 or 7 (mod 8), and so Nq is Remark: The combination of the two solutions recov-
even in this case as well. ers Gauss’s criterion for when −2 is a quadratic residue
If q = 1, then Nq = 0 is even. The only case that remains modulo p, with essentially the original proof.
is that q = pk is a positive power of a prime p congruent
to 5 or 7 (mod 8). In this case, q has two squarefree A6 First solution: (by Noam Elkies) Using row and col-
factors, 1 and p, of which exactly one is congruent to 5 umn operations, we may construct invertible matrices
or 7 (mod 8). We conclude that Nq is odd in this case, U,V such that U −1 MV is a block diagonal matrix of
as desired. the form
 
Second solution: I 0
.
Consider the set S of all integers in {1, . . . , q − 1} that 0 0
are even and relatively prime to q. Then the product of
all elements in S is Put A0 = U −1 AU, M 0 = U −1 MV, B0 = V −1 BV , X 0 =
V −1 XU, so that A0 M 0 = M 0 B0 , det(A − MX) =
2φ (q)/2 ∏ a. det(U −1 (A − MX)U) = det(A0 − M 0 X 0 ), and det(B −
1≤a≤(q−1)/2 XM) = det(V −1 (B − XM)V ) = det(B0 − X 0 M 0 ). Form
(a,q)=1
the corresponding block decompositions
On the other hand, we can rewrite the set of ele-      
A11 A12 B11 B12 X11 X12
ments in S (mod q) as a set T of residues in the in- A0 = , B0 = ,X0 = .
A21 A22 B21 B22 X21 X22
terval [−(q − 1)/2, (q − 1)/2]. Then for each 1 ≤ a ≤
(q − 1)/2 with (a, q) = 1, T contains exactly one el- We then have
ement from {a, −a}: if −2r ≡ 2s (mod q) for some
r, s ∈ {1, . . . , (q − 1)/2}, then r ≡ −s (mod q), which is
   
A11 0 B11 B12
impossible given the ranges of r and s. Thus the product A0 M 0 = , M 0 B0 = ,
A21 0 0 0
of all elements in T is
so we must have A11 = B11 and A21 = B12 = 0; in partic-
(−1)n ∏ a, ular, the characteristic polynomial of A is the product of
1≤a≤(q−1)/2
(a,q)=1
the characteristic polynomials of A11 and A22 , and the
characteristic polynomial of B is the product of the char-
where n denotes the number of elements of S greater acteristic polynomials of B11 and B22 . Since A11 = B11 ,
than (q − 1)/2. We conclude that (−1)n ≡ 2φ (q)/2 it follows that A22 and B22 have the same characteristic
(mod q). polynomial. Since
However, note that the number of elements of S less    
X11 0 X11 X12
than (q − 1)/2 is equal to Nq , since dividing these num- X 0M0 = , M0X 0 = ,
X21 0 0 0
bers by 2 gives exactly the numbers counted by Nq .
Hence the total cardinality of S is Nq + n; however,
this cardinality also equals φ (q)/2 because the num-
bers in {1, . . . , q − 1} relatively prime to q come in pairs
{a, q − a} in each of which exactly one member is even.
We thus obtain

(−1)Nq = (−1)φ (q)/2+n

≡ (−1)φ (q)/2 2φ (q)/2 = (−2)φ (q)/2 (mod q).
 4

we conclude that To establish this equality, first apply the remark follow-
ing the previous solution to write
det(A − MX) = det(A0 − M 0 X 0 )

A11 − X11 A12 − X12
 Trace(Ai1 MXAi2 MX · · · Aim−1 MXAim )
= det
0 A22 = Trace(Aim +i1 MXAi2 MX · · · Aim−1 MX).
= det(A11 − X11 ) det(A22 )
Then apply the relation AM = MB repeatedly to com-
= det(B11 − X11 ) det(B22 ) mute M past A, to obtain
 
B − X11 0
= det 11 Trace(MBim +i1 XMBi2 XM · · · XMBim−1 X).
B21 − X21 B22
= det(B0 − X 0 M 0 ) Finally, apply the remark again to shift MBim from the
= det(B − XM), left end to the right end.
Remark: The conclusion holds with R replaced by an
as desired. (By similar arguments, A − MX and B − XM arbitrary field. In the second solution, one must reduce
have the same characteristic polynomial.) to the case of an infinite field, e.g., by replacing the orig-
Second solution: We prove directly that A − MX and inal field with an algebraic closure. The third solution
B − XM have the same characteristic polynomial, i.e., only applies to fields of characteristic 0 or positive char-
for any t ∈ R, writing At = A − tI, Bt = B − tI, we have acteristic greater than n.
Remark: It is tempting to try to reduce to the case
det(At − MX) = det(Bt − XM).
where M is invertible, as in this case A − MX and
For fixed A, B, M, the stated result is a polynomial iden- B − XM are in fact similar. However, it is not clear how
tity in t and the entries of X. It thus suffices to check to make such an argument work.
it assuming that At , Bt , X are all invertible. Since AM = B1 Let g(x) = ex/2 f (x). Then g has at least 5 distinct real
MB, we also have At M = MBt , so At MBt−1 = M. Since zeroes, and by repeated applications of Rolle’s theorem,
det(At ) = det(Bt ) by hypothesis, g0 , g00 , g000 have at least 4, 3, 2 distinct real zeroes, respec-
tively. But
det(At − MX) = det(At − At MBt−1 X)
= det(At ) det(1 − MBt−1 X) 1
g000 (x) = ex/2 ( f (x) + 6 f 0 (x) + 12 f 00 (x) + 8 f 000 (x))
8
= det(At ) det(X) det(Bt )−1 det(X −1 Bt − M)
= det(X) det(X −1 Bt − M) and ex/2 is never zero, so we obtain the desired result.
= det(Bt − XM). B2 We will prove that 42015 is such a number in the se-
quence. Label the sequence of sums s0 , s1 , . . . , and let
Remark: One can also assert directly that det(1 − an , bn , cn be the summands of sn in ascending order. We
MBt−1 X) = det(1 − XMBt−1 ) using the fact that for any prove the following two statements for each nonnega-
square matrices U and V , UV and VU have the same tive integer n:
characteristic polynomial; the latter is again proved by
reducing to the case where one of the two matrices is (a)n The sequence
invertible, in which case the two matrices are similar.
a3n , b3n , c3n , a3n+1 , b3n+1 , c3n+1 , a3n+2 , b3n+2 , c3n+2
Third solution: (by Lev Borisov) We will check that
for each positive integer k, is obtained from the sequence 10n + 1, . . . , 10n +
10 by removing one of 10n + 5, 10n + 6, 10n + 7.
Trace((A − MX)k ) = Trace((B − XM)k ).
(b)n We have
This will imply that A − MX and B − XM have the same
s3n = 30n + 6,
characteristic polynomial, yielding the desired result.
s3n+1 ∈ {30n + 15, 30n + 16, 30n + 17},
We establish the claim by expanding both sides
and comparing individual terms. By hypothesis, Ak s3n+2 = 30n + 27.
and Bk have the same characteristic polynomial, so
Trace(Ak ) = Trace(Bk ). To compare the other terms, These statements follow by induction from the follow-
it suffices to check that for any sequence i1 , i2 , . . . , im of ing simple observations:
nonnegative integers, – by computing the table of values
Trace(Ai1 MXAi2 MX · · · Aim−1 MXAim ) n an bn cn sn
= Trace(Bi1 XMBi2 XM · · · Bim−1 XMBim ). 0 1 2 3 6
1 4 5 7 16
2 8 9 10 27

we see that (a)0 holds;

 5

– (a)n implies (b)n ; Second solution: If a, b, c, d are in arithmetic progres-

– (a)n and (b)1 , . . . , (b)n together imply (a)n+1 . sion, then we may write

To produce a value of n for which sn ≡ 2015 a = r − 3s, b = r − s, c = r + s, d = r + 3s

(mod 10000), we take n = 3m + 1 for some nonnega-
tive integer m for which s3m+1 = 30m + 15. We must for some r, s. If s = 0, then clearly all powers of M are
also have 30m ≡ 2000 (mod 10000), or equivalently in xS. Also, if r = 0, then one easily checks that M 3 is
m ≡ 400 (mod 1000). By taking m = 1400, we ensure in S.
that m ≡ 2 (mod 3), so sm = 10m + 7; this ensures that We now assume rs 6= 0, and show that in that case M
sn does indeed equal 30m + 15 = 42015, as desired. cannot be in S. First, note that the characteristic polyno-
Remark: With a bit more work, we can give a complete mial of M is x2 − 2rx − 8s2 , and since M is nonsingular
description of sn , and in particular find the first term in (as s 6= 0), this is also the minimal polynomial of M by
the sequence whose decimal expansion ends in 2015. the Cayley-Hamilton theorem. By repeatedly using the
Define the function on nonnegative integers relation M 2 = 2rM + 8s2 I, we see that for each positive
integer, we have M k = tk M + uk I for unique real con-
f (n) = s3n+1 − (30n + 16) stants tk , uk (uniqueness follows from the independence
of M and I). Since M is in S, we see that M k lies in S
which takes values in {−1, 0, 1}; we then have only if uk = 0.
 On the other hand, we claim that if k > 1, then rtk > 0
0
 n ≡ 0 (mod 3) and uk > 0 if k is even, and tk > 0 and ruk > 0 if k is
f (n) = − f ((n − 1)/3) n ≡ 1 (mod 3) odd (in particular, uk can never be zero). The claim is


−1 n ≡ 2 (mod 3). true for k = 2 by the relation M 2 = 2rM + 8s2 I. Assum-
ing the claim for k, and multiplying both sides of the
Consequently, if we write n in base 3, then f (n) = 0 relation M k = tk M + uk I by M, yields
unless the expansion ends with 2 followed by a string
M k+1 = tk (2rM + 8s2 I) + uk M = (2rtk + uk )M + 8s2tk I,
of 1s of length k ≥ 0, in which case f (n) = (−1)k+1 .
In this notation, we have sn ≡ 2015 (mod 10000) if and implying the claim for k + 1.
only if n = 3m + 1 for some nonnegative integer m for Remark: (from artofproblemsolving.com, user
which m ≡ 400 (mod 1000) and f (m) = −1. Since hoeij) Once one has uk = 0, one can also finish using
400 = 112211(3) , the first such term in the sequence is the relation M · M k = M k · M.
in fact s1201 = 12015.
B4 First solution: The answer is 17/21. For fixed b, c,
B3 First solution: Any element be
written as M =

of S can−1 there is a triangle of side lengths a, b, c if and only if
αA + β B, where A = 11 11 , B = −3 1 3
, and α, β ∈ R. |b − c| < a < b + c. It follows that the desired sum is
Note that A2 = 44 44 and B3 = −24 −8


8 24 are both in S,
and so any matrix of the form αA or β B, α, β ∈ R,
!
1 b+c−1
2b+c − 2|b−c|+1
satisfies the given condition. S=∑ b c ∑ 2a = ∑ .
b,c 3 5 a=|b−c|+1 b,c 3b 5c
We claim that these are also the only matrices in S
satisfying the given condition. Indeed,  suppose
√ 
M= We write this as S = S1 +S2 where S1 sums over positive
1 1/ 2
αA + β B where α, β 6= 0. Let C = √ with in- integers b, c with b ≤ c and S2 sums over b > c. Then
  −1 1/ 2
1/2 −1/2
verse C−1 = 1/√2 1/√2 . If we define D = C−1 MC, ∞ ∞
2b+c − 2c−b+1
√ S1 =
 
then D = 2α 0γ 1γ where γ = − β α 2 . Now suppose
∑∑ 3b 5c
b=1 c=b
!  !
that M k is in S with k ≥ 2. Since ( 1 −1 ) A −1 1
2 b 2 2 c


= ∞   ∞
1 =∑ − b ∑
= 0, we have ( 1 −1 ) M k −1 1



( 1 −1 ) B −1 = 0, and b=1 3 6 c=b 5
so the upper left entry of C−1 M kC = Dk is 0. On the b
!  
2 5 2 b
∞  
other hand, from the expression for 2
 2D, an easy induc- =∑ − b
k
tion on k shows that D = (2α) k γ pk−1 γ pk 3 6 3 5
γ pk pk+1 , where
b=1
  !
pk is defined inductively by p0 = 0, p1 = 1, pk+2 = 5 4 b 10 1 b
∞  
γ 2 pk + pk+1 . In particular, it follows from the inductive =∑ −
b=1 3 15 3 15
definition that pk > 0 when k ≥ 1, whence the upper left
entry of Dk is nonzero when k ≥ 2, a contradiction. 85
= .
Remark: A variant of this solution can be obtained by 231
diagonalizing the matrix M.
 6

Similarly, be the number which has n − 1, n consecutively in that

order, but not at the beginning or end. It is clear that
∞
2b+c − 2b−c+1
∞
every permutation π counted by Pn either lies in exactly
S2 = ∑ ∑
c=1 b=c+1 3b 5c one of the sets counted by Un ,Vn ,Wn , Tn , Sn , or is the
 c  ∞  b ! reverse of such a permutation. Therefore
∞
2 2 2
=∑ − c ∑ 3
5 10 Pn = 2(Un +Vn +Wn + Tn + Sn ).
c=1 b=c+1
∞  c   c+1
2 2 2 By examining how each of the elements in the sets
=∑ − c 3
5 10 3 counted by Un+1 ,Vn+1 ,Wn+1 , Tn+1 , Sn+1 can be ob-
c=1
c tained from a (unique) element in one of the sets
1 c
∞      
4 counted by Un ,Vn ,Wn , Tn , Sn by suitably inserting the el-
=∑ 2 −4
c=1 15 15 ement n + 1, we obtain the recurrence relations
34
= . Un+1 = Un +Wn + Tn ,
77
Vn+1 = Un ,
17
We conclude that S = S1 + S2 = 21 . Wn+1 = Wn ,
Second solution: Recall that the real numbers a, b, c Tn+1 = Vn ,
form the side lengths of a triangle if and only if
Sn+1 = Sn +Vn .
a+b+c
s − a, s − b, s − c > 0 s= , Also, it is clear that Wn = 1 for all n.
2
So far we have assumed n ≥ 3, but it is straightforward
and that if we put x = 2(s−a), y = 2(s−b), z = 2(s−c), to extrapolate the sequences Pn ,Un ,Vn ,Wn , Tn , Sn back
to n = 2 to preserve the preceding identities. Hence for
y+z z+x x+y
a= ,b = ,c = . all n ≥ 2,
2 2 2
To generate all integer triples (a, b, c) which form the Pn+5 = 2(Un+5 +Vn+5 +Wn+5 + Tn+5 + Sn+5 )
side lengths of a triangle, we must also assume that = 2((Un+4 +Wn+4 + Tn+4 ) +Un+4
x, y, z are either all even or all odd. We may therefore +Wn+4 +Vn+4 + (Sn+4 +Vn+4 ))
write the original sum as = Pn+4 + 2(Un+4 +Wn+4 +Vn+4 )
2(y+z)/2 2(y+z)/2 = Pn+4 + 2((Un+3 +Wn+3 + Tn+3 ) +Wn+3 +Un+3 )
∑ + ∑ . = Pn+4 + Pn+3 + 2(Un+3 −Vn+3 +Wn+3 − Sn+3 )
x,y,z>0 odd
3(z+x)/2 5(x+y)/2 x,y,z>0 even 3(z+x)/2 5(x+y)/2
= Pn+4 + Pn+3 + 2((Un+2 +Wn+2 + Tn+2 ) −Un+2
To unify the two sums, we substitute in the first case +Wn+2 − (Sn+2 −Vn+2 ))
x = 2u + 1, y = 2v + 1, z = 2w + 1 and in the second = Pn+4 + Pn+3 + 2(2Wn+2 + Tn+2 − Sn+2 −Vn+2 )
case x = 2u + 2, y = 2v + 2, z = 2w + 2 to obtain
= Pn+4 + Pn+3 + 2(2Wn+1 +Vn+1
2a 2v+w 2−1
∞  
− (Sn+1 +Vn+1 ) −Un+1 )
=
∑ 3b 5c ∑ 3w+u 5u+v 1 +
(a,b,c)∈T u,v,w=1 3−1 5−1 = Pn+4 + Pn+3 + 2(2Wn +Un − (Sn +Vn ) −Un
17 ∞
 u ∞  v ∞  w
1 2 2 − (Un +Wn + Tn ))
= ∑ 15 ∑ 5 ∑ 3 = Pn+4 + Pn+3 − Pn + 4,
2 u=1 v=1 w=1
17 1/15 2/5 2/3 as desired.
=
2 1 − 1/15 1 − 2/5 1 − 2/3
Remark: There are many possible variants of the
17 above solution obtained by dividing the permutations
= .
21 up according to different features. For example, Karl
Mahlburg suggests writing
B5 The answer is 4.
Assume n ≥ 3 for the moment. We write Pn = 2Pn0 , Pn0 = Q0n + R0n
the permutations π counted by Pn as sequences
π(1), π(2), . . . , π(n). Let Un be the number of permuta- where Pn0 counts those permutations counted by Pn for
tions counted by Pn that end with n − 1, n; let Vn be the which 1 occurs before 2, and Q0n counts those permuta-
number ending in n, n − 1; let Wn be the number starting tions counted by Pn0 for which π(1) = 1. One then has
with n − 1 and ending in n − 2, n; let Tn be the number the recursion
ending in n − 2, n but not starting with n − 1; and let Sn
Q0n = Q0n−1 + Q0n−3 + 1
 7
√ √ √
corresponding to the cases where π(1), π(2) = 1, 2; of summands is at most n (since n(2√ n − 1) ≥ n),
where π(1), π(2), π(3) = 1, 3, 2; and the unique case and so the total is bounded above by 1/ n. Hence the
1, 3, 5, . . . , 6, 4, 2. Meanwhile, one has difference converges to zero as n → ∞; that is, S1 con-
verges and equals S2 .
R0n = R0n−1 + Q0n−2
We may thus focus hereafter on computing S2 . We be-
corresponding to the cases containing 3, 1, 2, 4 (where gin by writing
removing 1 and reversing gives a permutation counted ∞ ∞ Z 1
1
by R0n−1 ); and where 4 occurs before 3, 1, 2 (where re- S2 = ∑ 2` − 1 ∑ (−1)m−1 t m−1 dt.
moving 1, 2 and reversing gives a permutation counted `=1 m=` 0
by Q0n−2 ).
Our next step will be to interchange the inner sum and
Remark: The permutations counted by Pn are known as the integral, but again this requires some justification.
key permutations, and have been studied by E.S. Page,
Systematic generation of ordered sequences using re- Lemma 1. Let f0 , f1 , . . . be a sequence of continuous func-
currence relations, The Computer Journal 14 (1971), tions on [0, 1] such that for each x ∈ [0, 1], we have
no. 2, 150–153. We have used the same notation for
consistency with the literature. The sequence of the Pn f0 (x) ≥ f1 (x) ≥ · · · ≥ 0.
also appears as entry A003274 in the On-line Encyclo-
pedia of Integer Sequences (http://oeis.org). Then
Z 1 Z 1
!
∞ ∞
B6 (from artofproblemsolving.com) We will prove n n
that the sum converges to π 2 /16. Note first that the ∑ (−1) 0
fn (t) dt =
0
∑ (−1) fn (t) dt
n=0 n=0
sum does not converge absolutely, so we are not free to
rearrange it arbitrarily. For that matter, the standard al- provided that both sums converge.
ternating sum test does not apply because the absolute
values of the terms does not decrease to 0, so even the Proof. Put gn (t) = f2n (t) − f2n+1 (t) ≥ 0; we may then rewrite
convergence of the sum must be established by hand. the desired equality as
Setting these issues aside momentarily, note that the el- !
∞ Z 1 Z 1 ∞
ements of the set counted by A(k) are those odd posi-
∑ gn (t) dt = ∑ gn (t) dt,
tive integers
√ d for which m = k/d is also an integer and n=0 0 0 n=0
d < 2dm; if we write d = 2` − 1, then the condition
on m reduces to m ≥ `. In other words, the original sum which is a case of the Lebesgue monotone convergence theo-
equals rem.
∞
(−1)m−1 By Lemma 1, we have
S1 := ∑ ∑ ,
k=1 `≥1,m≥` m(2` − 1) !
∞ Z 1 ∞
k=m(2`−1) 1 m−1 m−1
S2 = ∑ ∑ (−1) t dt
and we would like to rearrange this to `=1 2` − 1 0 m=`
(−t)`−1
∞ Z 1
1
∞
1 ∞
(−1)m−1 = ∑ 2` − 1 dt.
S2 := ∑ ∑ , `=1 0 1+t
`=1 2` − 1 m=` m
Since the outer sum is absolutely convergent, we may
in which both sums converge by the alternating sum freely interchange it with the integral:
test. In fact a bit more is true: we have
!
1 (−t)`−1
Z 1 ∞
∞
(−1)m−1 1 S2 = ∑ dt
< ,
∑ m ` 0 `=1 2` − 1 1 + t
m=` !
(−1)`−1t `−1/2
Z 1 ∞
1
so the outer sum converges absolutely. In particular, S2 = √ ∑ 2` − 1 dt
0 t(1 + t) `=1
is the limit of the truncated sums
Z 1
1 √
1 ∞
(−1)m−1 = √ arctan( t) dt
S2,n = ∑ ∑ . 0 t(1 + t)
`(2`−1)≤n
2` − 1 m=` m Z 1 √
2
= 2
arctan(u) du (u = t)
0 1+u
To see that S1 converges to the same value as S2 , write
π2
n = arctan(1)2 − arctan(0)2 = .
A(k) 1 ∞
(−1)m−1 16
S2,n − ∑ (−1)k−1 = ∑ ∑ .
k=1 k `(2`−1)≤n
2` − 1 m=b n +1c
m
2`−1
The expression on the right is bounded above in abso-
lute value by the sum ∑`(2`−1)≤n n1 , in which the number
 The 77th William Lowell Putnam Mathematical Competition
Saturday, December 3, 2016

A1 Find the smallest positive integer j such that for every A6 Find the smallest constant C such that for every real
polynomial p(x) with integer coefficients and for every polynomial P(x) of degree 3 that has a root in the in-
integer k, the integer terval [0, 1],

dj
Z 1
p( j) (k) = p(x) |P(x)| dx ≤ C max |P(x)| .
dx j x=k 0 x∈[0,1]

(the j-th derivative of p(x) at k) is divisible by 2016. B1 Let x0 , x1 , x2 , . . . be the sequence such that x0 = 1 and
A2 Given a positive integer n, let M(n) be the largest inte- for n ≥ 0,
ger m such that
xn+1 = ln(exn − xn )
   
m m−1
> . (as usual, the function ln is the natural logarithm). Show
n−1 n that the infinite series
Evaluate x0 + x1 + x2 + · · ·
M(n)
lim . converges and find its sum.
n→∞ n
B2 Define a positive integer n to be squarish if either n is
A3 Suppose that f is a function from R to R such that itself a perfect square or the distance from n to the near-
  est perfect square is a perfect square. For example, 2016
1
f (x) + f 1 − = arctan x is squarish, because the nearest perfect square to 2016
x is 452 = 2025 and 2025 − 2016 = 9 is a perfect square.
(Of the positive integers between 1 and 10, only 6 and
for all real x 6= 0. (As usual, y = arctan x means −π/2 < 7 are not squarish.)
y < π/2 and tan y = x.) Find
For a positive integer N, let S(N) be the number of
Z 1
squarish integers between 1 and N, inclusive. Find pos-
f (x) dx. itive constants α and β such that
0

S(N)
A4 Consider a (2m − 1) × (2n − 1) rectangular region, lim = β,
N→∞ N α
where m and n are integers such that m, n ≥ 4. This
region is to be tiled using tiles of the two types shown: or show that no such constants exist.
B3 Suppose that S is a finite set of points in the plane such
that the area of triangle 4ABC is at most 1 whenever A,
B, and C are in S. Show that there exists a triangle of
area 4 that (together with its interior) covers the set S.
B4 Let A be a 2n × 2n matrix, with entries chosen indepen-
(The dotted lines divide the tiles into 1 × 1 squares.) dently at random. Every entry is chosen to be 0 or 1,
The tiles may be rotated and reflected, as long as their each with probability 1/2. Find the expected value of
sides are parallel to the sides of the rectangular region. det(A − At ) (as a function of n), where At is the trans-
They must all fit within the region, and they must cover pose of A.
it completely without overlapping. B5 Find all functions f from the interval (1, ∞) to (1, ∞)
What is the minimum number of tiles required to tile with the following property: if x, y ∈ (1, ∞) and x2 ≤
the region? y ≤ x3 , then ( f (x))2 ≤ f (y) ≤ ( f (x))3 .

A5 Suppose that G is a finite group generated by the two B6 Evaluate

elements g and h, where the order of g is odd. Show
that every element of G can be written in the form
∞
(−1)k−1 ∞ 1
∑ k ∑ k2n + 1 .
k=1 n=0
gm1 hn1 gm2 hn2 · · · gmr hnr

with 1 ≤ r ≤ |G| and m1 , n1 , m2 , n2 , . . . , mr , nr ∈

{−1, 1}. (Here |G| is the number of elements of G.)
 The 78th William Lowell Putnam Mathematical Competition
Saturday, December 2, 2017

A1 Let S be the smallest set of positive integers such that are there to paint each edge red, white, or blue such that
each of the 20 triangular faces of the icosahedron has
(a) 2 is in S, two edges of the same color and a third edge of a dif-
(b) n is in S whenever n2 is in S, and ferent color? [Note: the top matter on each exam paper
(c) (n + 5)2 is in S whenever n is in S. included the logo of the Mathematical Association of
America, which is itself an icosahedron.]
Which positive integers are not in S?
B1 Let L1 and L2 be distinct lines in the plane. Prove that
(The set S is “smallest” in the sense that S is contained L1 and L2 intersect if and only if, for every real number
in any other such set.) λ 6= 0 and every point P not on L1 or L2 , there exist
−→ −→
A2 Let Q0 (x) = 1, Q1 (x) = x, and points A1 on L1 and A2 on L2 such that PA2 = λ PA1 .
B2 Suppose that a positive integer N can be expressed as
(Qn−1 (x))2 − 1
Qn (x) = the sum of k consecutive positive integers
Qn−2 (x)
N = a + (a + 1) + (a + 2) + · · · + (a + k − 1)
for all n ≥ 2. Show that, whenever n is a positive integer,
Qn (x) is equal to a polynomial with integer coefficients. for k = 2017 but for no other values of k > 1. Consid-
A3 Let a and b be real numbers with a < b, and let f and ering all positive integers N with this property, what is
g be continuous functions from [a, b] to (0, ∞) such that the smallest positive integer a that occurs in any of these
Rb Rb expressions?
a f (x) dx = a g(x) dx but f 6= g. For every positive
integer n, define B3 Suppose that f (x) = ∑∞ i
i=0 ci x is a power series for
which each coefficient ci is 0 or 1. Show that if
( f (x))n+1
Z b
In = dx. f (2/3) = 3/2, then f (1/2) must be irrational.
a (g(x))n
B4 Evaluate the sum
Show that I1 , I2 , I3 , . . . is an increasing sequence with ∞  
limn→∞ In = ∞. ln(4k + 2) ln(4k + 3) ln(4k + 4) ln(4k + 5)
∑ 3 · 4k + 2 − 4k + 3 − 4k + 4 − 4k + 5
k=0
A4 A class with 2N students took a quiz, on which the pos-
ln 2 ln 3 ln 4 ln 5 ln 6 ln 7
sible scores were 0, 1, . . . , 10. Each of these scores oc- = 3· − − − +3· −
curred at least once, and the average score was exactly 2 3 4 5 6 7
7.4. Show that the class can be divided into two groups ln 8 ln 9 ln 10
− − +3· −··· .
of N students in such a way that the average score for 8 9 10
each group was exactly 7.4. (As usual, ln x denotes the natural logarithm of x.)
A5 Each of the integers from 1 to n is written on a separate B5 A line in the plane of a triangle T is called an equal-
card, and then the cards are combined into a deck and izer if it divides T into two regions having equal area
shuffled. Three players, A, B, and C, take turns in the and equal perimeter. Find positive integers a > b > c,
order A, B,C, A, . . . choosing one card at random from with a as small as possible, such that there exists a trian-
the deck. (Each card in the deck is equally likely to gle with side lengths a, b, c that has exactly two distinct
be chosen.) After a card is chosen, that card and all equalizers.
higher-numbered cards are removed from the deck, and
the remaining cards are reshuffled before the next turn. B6 Find the number of ordered 64-tuples (x0 , x1 , . . . , x63 )
Play continues until one of the three players wins the such that x0 , x1 , . . . , x63 are distinct elements of
game by drawing the card numbered 1. {1, 2, . . . , 2017} and
Show that for each of the three players, there are arbi-
trarily large values of n for which that player has the x0 + x1 + 2x2 + 3x3 + · · · + 63x63
highest probability among the three players of winning
is divisible by 2017.
the game.
A6 The 30 edges of a regular icosahedron are distinguished
by labeling them 1, 2, . . . , 30. How many different ways
 Solutions to the 78th William Lowell Putnam Mathematical Competition
Saturday, December 2, 2017
Kiran Kedlaya and Lenny Ng

A1 We claim that the positive integers not in S are 1 and all Second solution. We establish directly that Qn (x) =
multiples of 5. If S consists of all other natural numbers, xQn−1 (x) − Qn−2 (x), which again suffices. From the
then S satisfies the given conditions: note that the only equation
perfect squares not in S are 1 and numbers of the form
(5k)2 for some positive integer k, and it readily follows 1 = Qn−1 (x)2 − Qn (x)Qn−2 (x) = Qn (x)2 − Qn+1 (x)Qn−1 (x)
that both (b) and (c) hold.
we deduce that
Now suppose that T is another set of positive integers
satisfying (a), (b), and (c). Note from (b) and (c) that if Qn−1 (x)(Qn−1 (x) + Qn+1 (x)) = Qn (x)(Qn (x) + Qn−2 (x)).
n ∈ T then n + 5 ∈ T , and so T satisfies the following
property: Since deg(Qn (x)) = n by an obvious induction, the
polynomials Qn (x) are all nonzero. We may thus
(d) if n ∈ T , then n + 5k ∈ T for all k ≥ 0. rewrite the previous equation as
The following must then be in T , with implications la-
Qn+1 (x) + Qn−1 (x) Qn (x) + Qn−2 (x)
beled by conditions (b) through (d): = ,
Qn (x) Qn−1 (x)
c c d b d b
2 ⇒ 49 ⇒ 542 ⇒ 562 ⇒ 56 ⇒ 121 ⇒ 11 Q (x)+Q (x)
meaning that the rational functions n Q (x) n−2
are all
d b d b n−1
11 ⇒ 16 ⇒ 4 ⇒ 9 ⇒ 3 equal to a constant value. By taking n = 2 and comput-
d b ing from the definition that Q2 (x) = x2 − 1, we find the
16 ⇒ 36 ⇒ 6
constant value to be x; this yields the desired recurrence.
Since 2, 3, 4, 6 ∈ T , by (d) S ⊆ T , and so S is smallest. Remark: By induction, one may also obtain the ex-
plicit formula
A2 First solution. Define Pn (x) for P0 (x) = 1, P1 (x) = x,
bn/2c
and Pn (x) = xPn−1 (x) − Pn−2 (x). We claim that Pn (x) =
 
n − k n−2k
k
Qn (x) for all n ≥ 0; since Pn (x) clearly is a polynomial Qn (x) = ∑ (−1) x .
k=0 k
with integer coefficients for all n, this will imply the
desired result.
Remark: In light of the explicit formula for Qn (x),
Since {Pn } and {Qn } are uniquely determined by their Karl Mahlburg suggests the following bijective inter-
respective recurrence relations and the initial conditions pretation of the identity Qn−1 (x)2 − Qn (x)Qn−2 (x) = 1.
P0 , P1 or Q0 , Q1 , it suffices to check that {Pn } satis- Consider the set Cn of integer compositions of n with
fies the same recurrence as Q: that is, (Pn−1 (x))2 − all parts 1 or 2; these are ordered tuples (c1 , . . . , ck ) such
Pn (x)Pn−2 (x) = 1 for all n ≥ 2. Here is one proof of that c1 + · · · + ck = n and ci ∈ {1, 2} for all i. For a given
this: for n ≥ 1, define the 2 × 2 matrices composition c, let o(c) and d(c) denote the number of
    1’s and 2’s, respectively. Define the generating function
P (x) Pn (x) x −1
Mn = n−1 , T=
Pn−2 (x) Pn−1 (x) 1 0 Rn (x) = ∑ xo(c) ;
c∈Cn
with P−1 (x) = 0 (this value being consistent with the
recurrence). Then det(T ) = 1 and T Mn = Mn+1 , so by then Rn (x) = ∑ j n−j j xn−2 j , so that Qn (x) =


induction on n we have
i−n/2 Rn (ix). (The polynomials Rn (x) are some-
(Pn−1 (x))2 − Pn (x)Pn−2 (x) = det(Mn ) = det(M1 ) = 1. times called Fibonacci polynomials; they satisfy
Rn (1) = Fn . This interpretation of Fn as the cardinality
of Cn first arose in the study of Sanskrit prosody,
Remark: A similar argument shows that any second-
specifically the analysis of a line of verse as a sequence
order linear recurrent sequence also satisfies a quadratic
of long and short syllables, at least 500 years prior to
second-order recurrence relation. A familiar example
the work of Fibonacci.)
is the identity Fn−1 Fn+1 − Fn2 = (−1)n for Fn the n-
th Fibonacci number. More examples come from var- The original identity is equivalent to the identity
ious classes of orthogonal polynomials, including the
Chebyshev polynomials mentioned below. Rn+1 (x)Rn−1 (x) − Rn (x)2 = (−1)n−1 .
 2
Rb
This follows because if we identify the composition c MSTang) Since a ( f (x) − g(x)) dx = 0, we have
with a tiling of a 1 × n rectangle by 1 × 1 squares and
( f (x))n+2 ( f (x))n+1
Z b 
1 × 2 dominoes, it is almost a bijection to place two
In+1 − In = − dx
tilings of length n on top of each other, offset by one a (g(x))n+1 (g(x))n
square, and hinge at the first possible point (which is
( f (x))n+1
Z b
the first square in either). This only fails when both = ( f (x) − g(x)) dx
tilings are all dominoes, which gives the term (−1)n−1 . (g(x))n+1
a
( f (x))n+1
Z b 
Remark: This problem appeared on the = − 1 ( f (x) − g(x)) dx
2012 India National Math Olympiad; see a (g(x))n+1
( f (x) − g(x))2 (( f (x))n + · · · + g(x)n )
https://artofproblemsolving.com/community/ Z b
c6h1219629. Another problem based on the same idea = dx.
a (g(x))n+1
is problem A2 from the 1993 Putnam.
The integrand is continuous, nonnegative, and not iden-
A3 First solution. Extend the definition of In to n = 0, so tically zero; hence In+1 − In > 0.
that I0 = ab f (x) dx > 0. Since ab ( f (x) − g(x)) dx = 0,
R R

we have To prove that limn→∞ In = ∞, note that we cannot

have f (x) ≤ g(x) identically, as then the equality
Z b Rb Rb
f (x)
I1 − I0 = ( f (x) − g(x)) dx a f (x) dx = a g(x) dx would imply f (x) = g(x) iden-
a g(x) tically. That is, there exists some t ∈ [a, b] such that
Z b
( f (x) − g(x))2 f (t) > g(t). By continuity, there exist a quantity c > 1
= dx > 0, and an interval J = [t0 ,t1 ] in [a, b] such that f (x) ≥ cg(x)
a g(x)
for all x ∈ J. We then have
where the inequality follows from the fact that the in-
( f (x))n+1
Z t1 Z t1
tegrand is a nonnegative continuous function on [a, b] n
In ≥ dx ≥ c f (x) dx;
that is not identically 0. Now for n ≥ 0, the Cauchy– t0 (g(x))n t0
Schwarz inequality gives
since f (x) > 0 everywhere, we have tt01 f (x) dx > 0 and
R
Z b
( f (x))n+1
 Z b
( f (x))n+3
 hence In is bounded below by a quantity which tends to
In In+2 = dx dx ∞.
a (g(x))n a (g(x))
n+2
2 Remark: One can also give a variation of the second
( f (x))n+2
Z b 
≥ dx 2
= In+1 . half of the solution which shows directly that In+1 −
n+1
a (g(x)) In ≥ cn d for some c > 1, d > 0, thus proving both asser-
tions at once.
It follows that the sequence {In+1 /In }∞n=0 is nondecreas-
ing. Since I1 /I0 > 1, this implies that In+1 > In for Third solution. (from David Savitt, via Art of Problem
all n; also, In /I0 = ∏n−1 n Solving) Extend the definition of In to all real n, and
k=0 (Ik+1 /Ik ) ≥ (I1 /I0 ) , and so
limn→∞ In = ∞ since I1 /I0 > 1 and I0 > 0. note that
Z b Z b
Remark: Noam Elkies suggests the following vari-
I−1 = g(x) dx = f (x) dx = I0 .
ant of the previous solution, which eliminates the need a a
to separately check that I1 > I0 . First, the proof that
2
In In+2 ≥ In+1 applies also for n = −1 under the con- By writing
vention that I−1 = ab g(x) dx (as in the fourth solution
R
Z b
below). Second, this equality must be strict for each In = exp((n + 1) log f (x) − n log g(x)) dx,
a
n ≥ −1: otherwise, the equality condition in Cauchy–
Schwarz would imply that g(x) = c f (x) identically for we see that the integrand is a strictly convex function of
some c > 0, and the equality ab f (x) dx = ab g(x) dx
R R
n, as then is In . It follows that In is strictly increasing
would then force c = 1, contrary to assumption. Con- and unbounded for n ≥ 1.
sequently, the sequence In+1 /In is strictly increasing; Fourth solution. (by David Rusin) Again, extend the
since I0 /I−1 = 1, it follows that for n ≥ 0, we again definition of In to n = −1. Now note that for n ≥ 0 and
have In+1 /In ≥ I1 /I0 > 1 and so on. x ∈ [a, b], we have
Second solution. (from Art of Problem Solving, user
 !
f (x) n+1 f (x) n
  
( f (x) − g(x)) − ≥0
g(x) g(x)

because both factors have the same sign (depending

on the comparison between f (x) and g(x)); moreover,
equality only occurs when f (x) = g(x). Since f and g
are not identically equal, we deduce that

In+1 − In > In − In−1

 3

and so in particular whence S1 ≤ 7.4N − 27.5 ≤ SN−4 . It follows that there

is some k such that Sk ∈ [7.4N − 40, 7.4N − 15], since
In+1 − In ≥ I1 − I0 > I0 − I−1 = 0. this interval has length 25 and 7.4N − 27.5 lies inside it.

This proves both claims. For this value of k, note that both Sk and 7.4N are inte-
gers (the latter since the sum of all scores in the class is
Remark: This problem appeared in 2005 on the integer (7.4)(2N) and so N must be divisible by 5).
an undergraduate math olympiad in Brazil. See Thus there is an integer m with 15 ≤ m ≤ 40 for which
https://artofproblemsolving.com/community/ Sk = 7.4N − m. By our first claim, we can choose five
c7h57686p354392 for discussion. scores from a1 , . . . , a10 whose sum is m. When we add
these to the sum of the N − 5 scores ak+10 , . . . , ak+N+4 ,
A4 First solution. Let a1 , . . . , a2N be the scores in non-
we get precisely 7.4N. We have now found N scores
decreasing order, and define the sums si = ∑i+N j=i+1 a j whose sum is 7.4N and thus whose average is 7.4.
for i = 0, . . . , N. Then s0 ≤ · · · ≤ sN and s0 + sN =
∑2N
j=1 a j = 7.4(2N), so s0 ≤ 7.4N ≤ sN . Let i be the
Third solution. It will suffices to show that given any
largest index for which si ≤ 7.4N; note that we can- partition of the students into two groups of N, if the
not have i = N, as otherwise s0 = sN = 7.4N and hence sums are not equal we can bring them closer together by
a1 = · · · = a2N = 7.4, contradiction. Then 7.4N − si < swapping one pair of students between the two groups.
si+1 − si = ai+N+1 − ai and so To state this symbolically, let S be the set of students
and, for any subset T of S, let ΣT denote the sum of
ai < si + ai+N+1 − 7.4N ≤ ai+N+1 ; the scores of the students in T ; we then show that if
S = A ∪ B is a partition into two N-element sets with
since all possible scores occur, this means that we can ΣA > ΣB, then there exist students a ∈ A, B ∈ B such
find N scores with sum 7.4N by taking ai+1 , . . . , ai+N+1 that the sets
and omitting one occurrence of the value si + ai+N+1 −
7.4N. A0 = A \ {a} ∪ {b}, B0 = A \ {b} ∪ {a}
Remark: David Savitt (via Art of Problem Solving)
satisfy
points out that a similar argument applies provided that
there are an even number of students, the total score is 0 ≤ ΣA0 − ΣB0 < ΣA − ΣB.
even, and the achieved scores form a block of consecu-
tive integers. In fact, this argument will apply at the same level of
Second solution. We first claim that for any integer m generality as in the remark following the first solution.
with 15 ≤ m ≤ 40, we can find five distinct elements To prove the claim, let a1 , . . . , an be the scores in A and
of the set {1, 2, . . . , 10} whose sum is m. Indeed, for let b1 , . . . , bn be the scores in B (in any order). Since
0 ≤ k ≤ 4 and 1 ≤ ` ≤ 6, we have ΣA − ΣB ≡ ΣS (mod 2) and the latter is even, we must
k
!
10
! have ΣA − ΣB ≥ 2. In particular, there must exist in-
+ (k + `) + = 34 − 5k + `, dices i, j ∈ {1, . . . , n} such that ai > b j . Consequently,
∑j ∑ j
if we sort the sequence a1 , . . . , an , b1 , . . . , bn into nonde-
j=1 j=k+7
creasing order, it must be the case that some term b j is
and for fixed k this takes all values from 35 − 5k to 40 − followed by some term ai . Moreover, since the achieved
5k inclusive; then as k ranges from 0 to 4, this takes all scores form a range of consecutive integers, we must in
values from 15 to 40 inclusive. fact have ai = b j + 1. Consequently, if we take a = ai ,
Now suppose that the scores are a1 , . . . , a2N , where we b = b j , we then have ΣA0 − Σ0 B = ΣA − ΣB − 2, which
order the scores so that ak = k for k ≤ 10 and the subse- proves the claim.
quence a11 , a12 , . . . , a2N is nondecreasing. For 1 ≤ k ≤ A5 First solution. Let an , bn , cn be the probabilities that
N − 4, define Sk = ∑k+N+4 j=k+10 a j . Note that for each k, players A, B, C, respectively, will win the game. We
Sk+1 − Sk = ak+N+5 − ak+10 and so 0 ≤ Sk+1 − Sk ≤ 10. compute these by induction on n, starting with the val-
Thus S1 , . . . , SN−4 is a nondecreasing sequence of inte- ues
gers where each term is at most 10 more than the previ-
ous one. On the other hand, we have a1 = 1, b1 = 0, c1 = 0.
2N
If player A draws card k, then the resulting game state
S1 + SN−4 = ∑ aj
j=11
is that of a deck of k − 1 cards with the players taking
turns in the order B,C, A, B, . . . . In this state, the proba-
10
= (7.4)(2N) − ∑ a j bilities that players A, B,C will win are ck−1 , ak−1 , bk−1
j=1
provided that we adopt the convention that
= (7.4)(2N) − 55, a0 = 0, b0 = 0, c0 = 1.
 4

We thus have by the residue of m mod 3). We use the convention that
p0 (0) = 1, p0 (m) = 0 for m > 0. Define the generating
1 n 1 n 1 n function Pn (X) = ∑nm=0 pn (m)xm . We will establish that
an = ∑ ck−1 , bn = ∑ ak−1 , cn = ∑ bk−1 .
n k=1 n k=1 n k=1
X(X + 1) · · · (X + n − 1)
Put Pn (X) =
n!
xn = an − bn , yn = bn − cn , zn = cn − an ; (which may be guessed by computing pn (m) for small
n by hand). There are several ways to do this; for in-
we then have stance, this follows from the recursion
n 1 1 (n − 1)
xn+1 = xn + zn , Pn (X) = XPn−1 (X) + Pn−1 (X).
n+1 n+1 n n
n 1
yn+1 = yn + xn , (In this recursion, the first term corresponds to condi-
n+1 n+1
n 1 tional probabilities given that the first card drawn is
zn+1 = zn + yn . n, and the second term corresponds to the remaining
n+1 n+1
cases.)
Note that if an+1 = bn+1 = cn+1 = 0, then Let ω be a primitive cube root of 1. With notation as in
the first solution, we have
xn = −nzn = n2 yn = −n3 xn = n4 zn
Pn (ω) = an + bn ω + cn ω;
and so xn = zn = 0, or in other words an = bn = cn . By
induction on n, we deduce that an , bn , cn cannot all be combining this with the explicit formula for Pn (X) and
equal. That is, the quantities xn , yn , zn add up to zero the observation that
and at most one
p of them vanishes; consequently, the √
quantity rn = xn2 + y2n + z2n is always positive and the 3
arg(w + n) = arctan
quantities 2n − 1
xn yn zn recovers the geometric description of an , bn , cn given in
xn0 = , y0n = , z0n =
rn rn rn the first solution (as well as the remark following the
first solution).
form the coordinates of a point Pn on a fixed circle C in
R3 . Third solution. For this argument, we use the auxiliary
quantities
Let Pn0 be the point (zn , xn , yn ) obtained from Pn by a
clockwise rotation of angle 2π 3 . The point Pn+1 then lies 1 1 1
on the ray through the origin passing through the point a0n = an − , b0n = bn − , c0n = cn − ;
3 3 3
dividing the chord from Pn to Pn0 in the ratio 1 : n. The
(clockwise) arc from Pn to Pn+1 therefore has a measure these satisfy the relations
of
1 n 0 1 n 0 1 n 0
√ √ a0n = ∑ ck−1 , b0n = ∑ ak−1 , c0n = ∑ bk−1
3 3 n k=1 n k=1 n k=1
arctan = + O(n−3 );
2n − 1 2n − 1
as well as
these measures form a null sequence whose sum di-
verges. It follows that any arc of C contains infinitely 1
a0n+1 = a0n + (c0 − a0n )
many of√
the Pn√
; taking a suitably short arc around the n+1 n
point ( 2 , 0, − 22 ), we deduce that for infinitely many
2 1
b0n+1 = b0n + (a0 − b0n )
n, A has the highest winning probability, and similarly n+1 n
for B and C. 1
c0n+1 = c0n + (b0 − c0n ).
Remark: From the previous analysis, we also deduce n+1 n
that We now show that ∑∞ 0
n=1 an cannot diverge to +∞ (and
√ 0
likewise for ∑n=1 bn and ∑∞
∞ 0
rn+1 n2 − n + 1 3 n=1 cn by similar reasoning).
= = 1− + O(n−2 ), Suppose the contrary; then there exists some ε > 0
rn n+1 2(n + 1)
and some n0 > 0 such that ∑nk=1 a0k ≥ ε for all n ≥ n0 .
For n > n0 , we have b0n ≥ ε; this in turn implies that
from which it follows that rn ∼ cn−3/2 for some c > 0. 0
∑∞ n=1 bn diverges to +∞. Continuing around the cir-
Second solution. (by Noam Elkies) In this approach, cle, we deduce that for n sufficiently large, all three of
we instead compute the probability pn (m) that the game a0n , b0n , c0n are positive; but this contradicts the identity
ends after exactly m turns (the winner being determined
 5

a0n + b0n + c0n = 0. We thus conclude that ∑∞ 0

n=1 an does Consequently, for n0 the smallest value for which xn0 6=
0
not diverge to +∞; in particular, lim infn→∞ an ≤ 0. xn , we must have xn0 = 2. By this and two similar argu-
By the same token, we may see that ∑∞ 0 ments, we deduce that
n=1 an cannot con-
verge to a positive limit L (and likewise for ∑∞ 0
n=1 bn and 1 ⇒ 5, 2 ⇒ 6, 3 ⇒ 4.
∞
c0 by similar reasoning). Namely, this would im-
∑n=1 n
ply that b0n ≥ L/2 for n sufficiently large, contradicting Suppose that xn = 4. By the earlier discussion, we must
the previous argument. have a0n0 < 0 for some n0 > n, and so we cannot have
0
By similar reasoning, ∑∞ n=1 an cannot diverge to −∞ or xn0 = 4 for all n0 > n. On the other hand, as long as
converge to a negative limit L (and likewise for ∑∞ 0
n=1 bn xn = 4, we have
∞ 0
and ∑n=1 cn by similar reasoning).
1
We next establish that there are infinitely many n for b0n+1 − a0n+1 = b0n − a0n + (2a0n − b0n − c0n )
n+1
which a0n > 0 (and likewise for b0n and c0n by similar
n−1 0 1
reasoning). Suppose to the contrary that for n suffi- = (b − a0n ) + (b0 − c0n ) > 0
ciently large, we have a0n ≤ 0. By the previous argu- n+1 n n+1 n
ments, the sum ∑∞ 0 1
n=1 an cannot diverge to ∞ or con- c0n+1 − a0n+1 = c0n − a0n + (a0 + b0n − 2c0n )
verge to a nonzero limit; it must therefore converge n+1 n
to 0. In particular, for n sufficiently large, we have n−1 0 1
= (c − a0n ) + (b0 − a0n ) > 0.
b0n = ∑nk=1 a0k−1 ≥ 0. Iterating the construction, we see n+1 n n+1 n
that for n sufficiently large, we must have c0n ≤ 0, a0n ≥ 0,
b0n ≤ 0, and c0n ≥ 0. As a result, for n sufficiently large Consequently, for n0 the smallest value for which xn0 6=
we must have a0n = b0n = c0n = 0; but we may rule this xn , we must have xn0 = 1. By this and two similar argu-
out as in the original solution. ments, we deduce that
By similar reasoning, we may deduce that there are in- 4 ⇒ 1, 5 ⇒ 2, 6 ⇒ 3.
finitely many n for which a0n < 0 (and likewise for b0n
and c0n by similar reasoning). We now continue us- Combining, we obtain
ing a suggestion of Jon Atkins. Define the values of
the sequence xn according to the relative comparison of 1⇒5⇒2⇒6⇒3⇒4⇒1
a0n , b0n , c0n (using the fact that these cannot all be equal):
and hence the desired result.
xn = 1 : a0n ≤ b0n < c0n
A6 The number of such colorings is 220 310 =
xn = 2 : b0n ≤ c0n < a0n 61917364224.
xn = 3 : c0n ≤ a0n < b0n First solution: Identify the three colors red, white, and
xn = 4 : a0n < c0n ≤ b0n blue with (in some order) the elements of the field F3 of
xn = 5 : b0n < a0n ≤ c0n three elements (i.e., the ring of integers mod 3). The set
of colorings may then be identified with the F3 -vector
xn = 6 : c0n < b0n ≤ a0n . space FE3 generated by the set E of edges. Let F be
the set of faces, and let FF3 be the F3 -vector space on
We consider these values as states and say that there is
the basis F; we may then define a linear transformation
a transition from state i to state j, and write i ⇒ j, if for
T : FE3 → FF3 taking a coloring to the vector whose com-
every n ≥ 2 with xn = i there exists n0 > n with xn0 = j.
ponent corresponding to a given face equals the sum of
(In all cases when we use this notation, it will in fact be
the three edges of that face. The colorings we wish to
the case that the first value of n0 > n for which xn0 6= i
count are the ones whose images under T consist of vec-
satisfes xn0 = j, but this is not logically necessary for
tors with no zero components.
our final conclusion.)
We now show that T is surjective. (There are many pos-
Suppose that xn = 1. By the earlier discussion, we must
sible approaches to this step; for instance, see the fol-
have a0n0 > 0 for some n0 > n, and so we cannot have
lowing remark.) Let Γ be the dual graph of the icosa-
xn0 = 1 for all n0 > n. On the other hand, as long as
hedron, that is, Γ has vertex set F and two elements of
xn = 1, we have
F are adjacent in Γ if they share an edge in the icosa-
1 hedron. The graph Γ admits a hamiltonian path, that is,
c0n+1 − b0n+1 = c0n − b0n + (2b0n − a0n − c0n ) there exists an ordering f1 , . . . , f20 of the faces such that
n+1
n−1 0 1 any two consecutive faces are adjacent in Γ. For exam-
= (cn − b0n ) + (c0 − a0n ) > 0 ple, such an ordering can be constructed with f1 , . . . , f5
n+1 n+1 n being the five faces sharing a vertex of the icosahedron
1
c0n+1 − a0n+1 = c0n − a0n + (a0 + b0n − 2c0n ) and f16 , . . . , f20 being the five faces sharing the antipo-
n+1 n dal vertex.
n−1 0 1
= (c − a0n ) + (b0 − a0n ) > 0. For i = 1, . . . , 19, let ei be the common edge of fi and
n+1 n n+1 n fi+1 ; these are obviously all distinct. By prescribing
 6

components for e1 , . . . , e19 in turn and setting the oth- the plane by a factor of λ with center P carries L1 to
ers to zero, we can construct an element of FE3 whose another line parallel to L1 and hence not parallel to L2 .
image under T matches any given vector of FF3 in the Let A2 be the unique intersection of L2 with the image
components of f1 , . . . , f19 . The vectors in FF3 obtained of L1 , and let A1 be the point on L1 whose image under
in this way thus form a 19-dimensional subspace; this −→ −→
the dilation is A2 ; then PA2 = λ PA1 .
subspace may also be described as the vectors for which
In the other direction, suppose that L1 and L2 are par-
the components of f1 , . . . , f19 have the same sum as the
allel. Let P be any point in the region between L1 and
components of f2 , . . . , f20 .
L2 and take λ = 1. Then for any point A1 on L1 and
By performing a mirror reflection, we can construct a −→ −→
any point A2 on L2 , the vectors PA1 and PA2 have com-
second hamiltonian path g1 , . . . , g20 with the property ponents perpendicular to L1 pointing in opposite direc-
that tions; in particular, the two vectors cannot be equal.
g1 = f1 , g2 = f5 , g3 = f4 , g4 = f3 , g5 = f2 . Reinterpretation: (by Karl Mahlburg) In terms of vec-
tors, we may find vectors ~v1 ,~v2 and scalars c1 , c2 such
Repeating the previous construction, we obtain a differ- that Li = {~x ∈ R2 : ~vi ·~x = ci }. The condition in the
ent 19-dimensional subspace of FF3 which is contained problem amounts to finding a vector ~w and a scalar t
in the image of T . This implies that T is surjective, as such that P + ~w ∈ L1 , P + λ w ∈ L2 ; this comes down to
asserted earlier. solving the linear system
Since T is a surjective homomorphism from a 30-
~v1 · (P + ~w) = c1
dimensional vector space to a 20-dimensional vector
space, it has a 10-dimensional kernel. Each of the 220 ~v2 · (P + λ ~w) = c2
elements of FF3 with no zero components is then the im-
which is nondegenerate and solvable for all λ if and
age of exactly 310 colorings of the desired form, yield-
only if ~v1 ,~v2 are linearly independent.
ing the result.
Remark: There are many ways to check that T is sur- B2 We prove that the smallest value of a is 16.
jective. One of the simplest is the following (from Art Note that the expression for N can be rewritten as
of Problem Solving, user Ravi12346): form a vector in k(2a + k − 1)/2, so that 2N = k(2a + k − 1). In this
FE with components 2, 1, 2, 1, 2 at the five edges around expression, k > 1 by requirement; k < 2a + k − 1 be-
some vertex and all other components 0. This maps to cause a > 1; and obviously k and 2a + k − 1 have oppo-
a vector in FF with only a single nonzero component; site parity. Conversely, for any factorization 2N = mn
by symmetry, every standard basis vector of FF arises with 1 < m < n and m, n of opposite parity, we obtain
in this way. an expression of N in the desired form by taking k = m,
Second solution: (from Bill Huang, via Art of Problem a = (n + 1 − m)/2.
Solving user superpi83) Let v and w be two antipodal We now note that 2017 is prime. (On the exam, solvers
vertices of the icosahedron. Let Sv (resp. Sw ) be the set would have had to verify this by hand. Since 2017 <
of five edges incident to v (resp. w). Let Tv (resp. Tw ) be 452 , this can be done by trial division by the primes up
the set of five edges of the pentagon formed by the op- to 43.) For 2N = 2017(2a + 2016) not to have another
posite endpoints of the five edges in Sv (resp. Sw ). Let U expression of the specified form, it must be the case that
be the set of the ten remaining edges of the icosahedron. 2a + 2016 has no odd divisor greater than 1; that is,
Consider any one of the 310 possible colorings of U. 2a + 2016 must be a power of 2. This first occurs for
The edges of Tv ∪ U form the boundaries of five faces 2a + 2016 = 2048, yielding the claimed result.
with no edges in common; thus each edge of Tv can be Reinterpretation: (by Karl Mahlburg) To avoid N hav-
colored in one of two ways consistent with the given ing another representation, for k = 2, . . . , 2016, we must
condition, and similarly for Tw . That is, there are 310 210 have
possible colorings of Tv ∪ Tw ∪ U consistent with the (
given condition. k/2 k ≡ 0 (mod 2)
N 6≡
To complete the count, it suffices to check that there are 0 k ≡ 1 (mod 2).
exactly 25 ways to color Sv consistent with any given
coloring of Tv . Using the linear-algebraic interpretation Consequently, N 6≡ 0 (mod p) for any odd prime p <
from the first solution, this follows by observing that 2017 and N ≡ 0 (mod 1024). Since N must be divisible
(by the previous remark) the map from FS3v to the F3 - by 2017, this again yields the claimed value of a.
vector space on the faces incident to v is surjective, and
hence an isomorphism for dimensional reasons. A di- B3 Suppose by way of contradiction that f (1/2) is rational.
Then ∑∞ −i is the binary expansion of a rational
rect combinatorial proof is also possible. i=0 ci 2
number, and hence must be eventually periodic; that is,
B1 Recall that L1 and L2 intersect if and only if they are not there exist some integers m, n such that ci = cm+i for all
parallel. In one direction, suppose that L1 and L2 inter-
sect. Then for any P and λ , the dilation (homothety) of
 7

i ≥ n. We may then write Finally, we have

n−1 ∞ ∞
xn m−1
f (x) = ∑ ci xi + ∑ cn+i xi .
1 − xm i=0
S= ∑ a2k + ∑ a2k+1 + (log 2)2
i=0 k=1 k=0
∞
Evaluating at x = 2/3, we may equate f (2/3) = 3/2 = ∑ ak + (log 2)2 = (log 2)2 .
with k=1

n−1 m−1
1 2n 3m Second solution. We start with the following observa-
ci 2i 3n−i−1 + n+m−1 m cn+i 2i 3m−1−i ;
3n−1 ∑ i=0 3 (3 − 2m ) ∑ i=0
tion: for any positive integer n,

d −s
since all terms on the right-hand side have odd denomi- n = −(log n)n−s .
nator, the same must be true of the sum, a contradiction. ds s=1
Remark: Greg Marks asks whether the assumption that (Throughout, we view s as a real parameter, but see the
f (2/3) = 3/2 further ensures that f (1/2) is transcen- remark below.) For s > 0, consider the absolutely con-
dental. We do not know of any existing results that vergent series
would imply this. However, the following result fol-
lows from a theorem of T. Tanaka (Algebraic indepen- ∞
dence of the values of power series generated by linear L(s) = ∑ (3(4k +2)−s −(4k +3)−s −(4k +4)−s −(4k +5)−s );
k=0
recurrences, Acta Arith. 74 (1996), 177–190), building
upon work of Mahler. Let {an }∞ n=0 be a linear recurrent in the same range we have
sequence of positive integers with characteristic poly-
nomial P. Suppose that P(0), P(1), P(−1) 6= 0 and that ∞ 
log(4k + 2) log(4k + 3)
no two distinct roots of P have ratio which is a root L0 (s) = ∑ 3 −
an k=0 (4k + 2)s (4k + 3)s
of unity. Then for f (x) = ∑∞ n=0 x , the values f (1/2) 
and f (2/3) are algebraically independent over Q. (Note log(4k + 4) log(4k + 5)
+ − ,
that for f as in the original problem, the condition on (4k + 4)s (4k + 5)s
ratios of roots of P fails.)
so we may interchange the summation with taking the
B4 We prove that the sum equals (log 2)2 ; as usual, we limit at s = 1 to equate the original sum with −L0 (1).
write log x for the natural logarithm of x instead of ln x. To make further progress, we introduce the Riemann
Note that of the two given expressions of the original zeta function ζ (s) = ∑∞ −s
n=1 n , which converges abso-
sum, the first is absolutely convergent (the summands lutely for s > 1. In that region, we may freely rearrange
decay as log(x)/x2 ) but the second one is not; we must sums to write
thus be slightly careful when rearranging terms.
First solution. Define ak = logk k − log(k+1) L(s) + ζ (s) = 1 + 4(2−s + 6−s + 10−s + · · · )
k+1 . The infinite
n
sum ∑∞ a
k=1 k converges to 0 since ∑ k=1 ak telescopes to = 1 + 22−s (1 + 3−s + 5−s + · · · )
log(n+1)
− n+1 and this converges to 0 as n → ∞. Note that = 1 + 22−s (ζ (s) − 2−s − 4−s − · · · )
ak > 0 for k ≥ 3 since logx x is a decreasing function of x = 1 + 22−s ζ (s) − 22−2s ζ (s).
for x > e, and so the convergence of ∑∞ k=1 ak is absolute.
Write S for the desired sum. Then since 3a4k+2 + In other words, for s > 1, we have
2a4k+3 + a4k+4 = (a4k+2 + a4k+4 ) + 2(a4k+2 + a4k+3 ),
we have L(s) = 1 + ζ (s)(−1 + 22−s − 22−2s ).
∞ s
Now recall that ζ (s) − s−1 extends to a C∞ function for
S= ∑ (3a4k+2 + 2a4k+3 + a4k+4 ) s > 0, e.g., by applying Abel summation to obtain
k=0
∞ ∞ s s
= ∑ a2k + ∑ 2(a4k+2 + a4k+3 ), ζ (s) − = ∑ n(n−s − (n + 1)−s ) −
k=1 k=0
s − 1 n=1 s−1
∞ Z n+1
s
where we are allowed to rearrange the terms in the =s∑n x−s−1 dx −
infinite sum since ∑ ak converges absolutely. Now n=1 n s−1
Z ∞
2(a4k+2 + a4k+3 ) = log(4k+2)
2k+1 − log(4k+4)
2k+2 = a2k+1 + = −s (x − bxc)x−s−1 dx.
1 1 1
(log 2)( 2k+1 − 2k+2 ), and summing over k gives
Also by writing 22−s = 2 exp((1 − s) log 2 and 22−2s =
∞ ∞
(−1)k+1 ∞
exp(2(1 − s) log 2), we may use the exponential series
∑ 2(a4k+2 + a4k+3 ) = ∑ a2k+1 + (log 2) ∑ k to compute the Taylor expansion of
k=0 k=0 k=1
∞
= ∑ a2k+1 + (log 2)2 . f (s) =
−1 + 22−s − 22−2s
k=0 s−1
 8

at s = 1; we get equalizers intersecting T on the sides AB and AC, we

want to solve g(x) = p/2 where g(x) = x + bc/(2x)
f (s) = −(log 2)2 (s − 1)2 + O((s − 1)3 ). and x ∈ [c/2, c]; since g is convex and g(c/2) < p/2,
g(c) < p/2, there are no such solutions.
Consequently, if we rewrite the previous expression for
L(s) as It follows that if T has exactly two equalizers, then it
must have exactly one equalizer intersecting T on the
−1 + 22−s − 22−2s sides AC and BC. Here we want to solve h(x) = p/2
L(s) = 1 + (s − 1)ζ (s) · , where h(x) = x + ab/(2x) and x ∈ [a/2, a]. Now h is
s−1
convex and h(a/2) > p/2, h(a) > p/2; thus h(x) = p/2
then we have an equality of C∞ functions for s > 1, and has exactly one solution x ∈ [a/2, a] if and only if there
hence (by continuity) an equality of Taylor series about is x0 ∈ [a/2, a] with h0 (x0 ) =p0 and h(x0 ) = p/2. The
s = 1. That is, first condition implies x0 = ab/2, and then p the sec-
ond condition gives 8ab = p2 . Note that ab/2 is in
L(s) = 1 − (log 2)2 (s − 1) + O((s − 1)2 ), [a/2, a] since a > b and a < b + c < 2b.
which yields the desired result. We conclude that T has two equalizers if and only if
8ab = (a + b + c)2 . Note that (a, b, c) = (9, 8, 7) works.
Remark: We claim that this is the only possibility when a > b > c
The use of series ∑∞ −s as functions of a real pa-
n=1 cn n are integers and a ≤ 9. Indeed, the only integers (a, b)
rameter s dates back to Euler, who observed that the di- such that 2 ≤ b < a ≤ 9 and 8ab is a perfect square
vergence of ζ (s) as s → 1 gives a proof of the infinitude are (a, b) = (4, 2), (6, 3), (8, 4), (9, 2), and (9, 8), and
of primes distinct from Euclid’s approach, and Dirich- the first four possibilities do not produce triangles since
let, who upgraded this idea to prove his theorem on the they do not satisfy a < 2b. This gives the claimed result.
distribution of primes across arithmetic progressions. It 2016!
was Riemann who introduced the idea of viewing these B6 First solution. The desired count is 1953! − 63! · 2016,
series as functions of a complex parameter, thus mak- which we compute using the principle of inclusion-
ing it possible to use the tools of complex analysis (e.g., exclusion. As in A2, we use the fact that 2017 is prime;
the residue theorem) and leading to the original proof this means that we can do linear algebra over the field
of the prime number theorem by Hadamard and de la F2017 . In particular, every nonzero homogeneous linear
Vallée Poussin. equation in n variables over F2017 has exactly 2017n−1
solutions.
In the language of complex analysis, one may handle
the convergence issues in the second solution in a dif- For π a partition of {0, . . . , 63}, let |π| denote the num-
ferent way: use the preceding calculation to establish ber of distinct parts of π, Let π0 denote the partition of
the equality {0, . . . , 63} into 64 singleton parts. Let π1 denote the
partition of {0, . . . , 63} into one 64-element part. For
L(s) = 1 + ζ (s)(−1 + 22−s − 22−2s ) π, σ two partitions of {0, . . . , 63}, write π|σ if π is a re-
finement of σ (that is, every part in σ is a union of parts
for Real(s) > 1, then observe that both sides are holo- in π). By induction on |π|, we may construct a collec-
morphic for Real(s) > 0 and so the equality extends to tion of integers µπ , one for each π, with the properties
that larger domain. that
(
B5 The desired integers are (a, b, c) = (9, 8, 7). 1 σ = π0
Suppose we have a triangle T = 4ABC with BC = a, ∑ µπ = 0 σ 6= π0 .
π|σ
CA = b, AB = c and a > b > c. Say that a line is
an area equalizer if it divides T into two regions of Define the sequence c0 , . . . , c63 by setting c0 = 1 and
equal area. A line intersecting T must intersect two of ci = i for i > 1. Let Nπ be the number of ordered 64-
the three sides of T . First consider a line intersecting tuples (x0 , . . . , x63 ) of elements of F2017 such that xi = x j
the segments AB at X and BC at Y , and let BX = x, whenever i and j belong to the same part and ∑63 i=0 ci xi is
BY = y. This line is an area equalizer if and only if
divisible by 2017. Then Nπ equals 2017|π|−1 unless for
xy sin B = 2 area(4XBY ) = area(4ABC) = 21 ac sin B,
each part S of π, the sum ∑i∈S ci vanishes; in that case,
that is, 2xy = ac. Since x ≤ c and y ≤ a, the area equal-
izers correspond to values of x, y with xy = ac/2 and Nπ instead equals 2017|π| . Since c0 , . . . , c63 are positive
x ∈ [c/2, c]. Such an area equalizer is also an equal- integers which sum to 1 + 63·64 2 = 2017, the second out-
izer if and only if p/2 = x + y, where p = a + b + c come only occurs for π = π1 . By inclusion-exclusion,
is the perimeter of T . If we write f (x) = x + ac/(2x), the desired count may be written as
then we want to solve f (x) = p/2 for x ∈ [c/2, c]. Now
note that f is convex, f (c/2) = a + c/2 > p/2, and ∑ µπ Nπ = 2016 · µπ1 + ∑ µπ 2017|π|−1 .
π π
f (c) = a/2 + c < p/2; it follows that there is exactly
one solution to f (x) = p/2 in [c/2, c]. Similarly, for
 9

Similarly, the number of ordered 64-tuples with no re- some i. Since then x j 6= 0 for all j 6= i, we may apply the in-
peated elements may be written as duction hypothesis to see that there are f (k − 1, p) solutions
  that arise this way for a given i (and these do not overlap).
2017 This proves the claim.
64! = ∑ µπ 2017|π| .
64 π
To compute f (k, p) explicitly, it is convenient to work
2016! with the auxiliary function
The desired quantity may thus be written as 1953! +
2016µπ1 .
p f (k, p)
It remains to compute µπ1 . We adopt an approach sug- g(k, p) = ;
gested by David Savitt: apply inclusion-exclusion to k!
count distinct 64-tuples in an arbitrary set A. As above, by the lemma, this satisfies g(1, p) = 0 and
this yields  
p
|A|(|A| − 1) · · · (|A| − 63) = ∑ µπ |A||π| . g(k, p) = − g(k − 1, p)
k
π    
p−1 p−1
= + − g(k − 1, p) (k > 1).
Viewing both sides as polynomials in |A| and comparing k k−1
coefficients in degree 1 yields µπ = −63! and thus the By induction on k, we deduce that
claimed answer.
    
Second solution. (from Art of Problem Solving, user p−1 k−1 p−1
g(k, p) − = (−1) g(1, p) −
ABCDE) We first prove an auxiliary result. k 1
Lemma. Fix a prime p and define the function f (k) on posi- = (−1)k (p − 1)
tive integers by the conditions p−1

and hence g(k, p) = k + (−1)k (p − 1).
f (1, p) = 0 We now set p = 2017 and count the tuples in ques-
(p − 1)! tion. Define c0 , . . . , c63 as in the first solution. Since
f (k, p) = − k f (k − 1, p) (k > 1). c0 + · · · + c63 = p, the translation action of F p preserves
(p − k)!
the set of tuples; we may thus assume without loss of
Then for any positive integers a1 , . . . , ak with a1 +· · ·+ak < p, generality that x0 = 0 and multiply the count by p at the
there are exactly f (p) solutions to the equation a1 x1 + · · · + end. That is, the desired answer is
ak xk = 0 with x1 , . . . , xk ∈ F p nonzero and pairwise distinct.
2017 f (63, 2017) = 63!g(63, 2017)
Proof. We check the claim by induction, with the base case
  
2016
k = 1 being obvious. For the induction step, assume the claim = 63! − 2016
63
for k − 1. Let S be the set of k-tuples of distinct elements of
p!
F p ; it consists of (p−k)! elements. This set is stable under the as claimed.
action of i ∈ F p by translation:

(x1 , . . . , xk ) 7→ (x1 + i, . . . , xk + i).

Since 0 < a1 · · · + ak < p, exactly one element of each orbit

gives a solution of a1 x1 + · · · + ak xk = 0. Each of these solu-
tions contributes to f (k) except for those in which xi = 0 for
 The 79th William Lowell Putnam Mathematical Competition
Saturday, December 1, 2018

A1 Find all ordered pairs (a, b) of positive integers for is a rational number.
which
B1 Let P be the set of vectors defined by
1 1 3
+ = .   
a b 2018 a
P= 0 ≤ a ≤ 2, 0 ≤ b ≤ 100, and a, b ∈ Z .
b
A2 Let S1 , S2 , . . . , S2n −1 be the nonempty subsets of
{1, 2, . . . , n} in some order, and let M be the (2n − 1) × Find all v ∈ P such that the set P \ {v} obtained by
(2n − 1) matrix whose (i, j) entry is omitting vector v from P can be partitioned into two
( sets of equal size and equal sum.
0 if Si ∩ S j = 0;
/ B2 Let n be a positive integer, and let fn (z) = n+(n−1)z+
mi j = (n − 2)z2 + · · · + zn−1 . Prove that fn has no roots in the
1 otherwise.
closed unit disk {z ∈ C : |z| ≤ 1}.
Calculate the determinant of M. B3 Find all positive integers n < 10100 for which simulta-
neously n divides 2n , n − 1 divides 2n − 1, and n − 2
A3 Determine the greatest possible value of ∑10 i=1 cos(3xi ) divides 2n − 2.
for real numbers x1 , x2 , . . . , x10 satisfying ∑10
i=1 cos(xi ) =
0. B4 Given a real number a, we define a sequence by x0 = 1,
x1 = x2 = a, and xn+1 = 2xn xn−1 −xn−2 for n ≥ 2. Prove
A4 Let m and n be positive integers with gcd(m, n) = 1, and
that if xn = 0 for some n, then the sequence is periodic.
let
    B5 Let f = ( f1 , f2 ) be a function from R2 to R2 with con-
mk m(k − 1)
ak = − tinuous partial derivatives ∂∂ xfij that are positive every-
n n
where. Suppose that
for k = 1, 2, . . . , n. Suppose that g and h are elements in  2
a group G and that ∂ f1 ∂ f2 1 ∂ f1 ∂ f2
− + >0
a1 a2 an
∂ x1 ∂ x2 4 ∂ x2 ∂ x1
gh gh · · · gh = e,
everywhere. Prove that f is one-to-one.
where e is the identity element. Show that gh = hg. (As
usual, bxc denotes the greatest integer less than or equal B6 Let S be the set of sequences of length 2018 whose
to x.) terms are in the set {1, 2, 3, 4, 5, 6, 10} and sum to 3860.
Prove that the cardinality of S is at most
A5 Let f : R → R be an infinitely differentiable function
2018
satisfying f (0) = 0, f (1) = 1, and f (x) ≥ 0 for all x ∈

3860 2018
R. Show that there exist a positive integer n and a real 2 · .
2048
number x such that f (n) (x) < 0.
A6 Suppose that A, B,C, and D are distinct points, no three
of which lie on a line, in the Euclidean plane. Show that
if the squares of the lengths of the line segments AB,
AC, AD, BC, BD, and CD are rational numbers, then the
quotient

area(4ABC)
area(4ABD)
 Solutions to the 79th William Lowell Putnam Mathematical Competition
Saturday, December 1, 2018
Kiran Kedlaya and Lenny Ng

While preparing these solutions, we learned of the Novem- In Mn , perform the following operations, which do not
ber 27 death of Kent Merryfield, who moderated the Put- change the determinant: subtract the final row from
nam discussions on the Art of Problem Solving Forums and rows 2n−1 through 2n − 2, and then subtract the final
thereby contributed directly and indirectly to these solutions column from columns 2n−1 through 2n − 2. The result
over many years. His presence will be dearly missed. is the matrix
 
A1 By clearing denominators and regrouping, we see that 0
the given equation is equivalent to  .. 
 Mn−1 Mn−1 . 
 
(3a − 2018)(3b − 2018) = 20182 .
 

 0 
0 ··· 0 0  .

Each of the factors is congruent to 1 (mod 3). There 
. . . .
.. . . .. .. 
 
are 6 positive factors of 20182 = 22 · 10092 that are  M
 n−1 
congruent to 1 (mod 3): 1, 22 , 1009, 22 · 1009, 
 0 ··· 0 0 

10092 , 22 · 10092 . These lead to the 6 possible pairs:
0 ··· 0 0 ··· 0 1
(a, b) = (673, 1358114), (674, 340033), (1009, 2018),
(2018, 1009), (340033, 674), and (1358114, 673).
We can remove the final row and column with-
As for negative factors, the ones that are congruent to 1 out changing the determinant. Now swap the
(mod 3) are −2, −2 · 1009, −2 · 10092 . However, all of first 2n−1 − 1 rows with the final 2n−1 − 1 rows:
these lead to pairs where a ≤ 0 or b ≤ 0. this changes the determinant by an overall factor
n−1 2
A2 The answer is 1 if n = 1 and −1 if n > 1. Write Mn of (−1)(2 −1) = −1. The ! result is the block-
for a (2n − 1) × (2n − 1) matrix of the given form, and Mn−1 0
note that det Mn does not depend on the ordering of the diagonal matrix , whose determinant is
Mn−1 Mn−1
subsets: transposing two subsets has the effect of trans-
(det Mn−1 )2 . Thus det Mn = −(det Mn−1 )2 as desired.
posing two rows and then transposing two columns in
Mn , and this does not change the determinant. A3 The maximum value is 480/49. Since cos(3xi ) =
Clearly det M1 = 1. We claim that for n > 1, det Mn = 4 cos(xi )3 − 3 cos(xi ), it is equivalent to maximize
−(det Mn−1 )2 , and the desired answer will follow by in- 4 ∑10 3 10
i=1 yi for y1 , . . . , y10 ∈ [−1, 1] with ∑i=1 yi = 0; note
duction. Let S10 , . . . , S20 n−1 −1 denote the nonempty sub- that this domain is compact, so the maximum value
sets of {1, . . . , n − 1} in any order, with resulting matrix is guaranteed to exist. For convenience, we establish
Mn−1 . Let m0i j denote the (i, j) entry of Mn−1 . Now or- something slightly stronger: we maximize 4 ∑ni=1 y3i for
der the nonempty subsets S1 , . . . , S2n −1 of {1, . . . , n} as y1 , . . . , yn ∈ [−1, 1] with ∑ni=1 yi = 0, where n may be
follows: any even nonnegative integer up to 10, and show that
 the maximum is achieved when n = 10.
0 i ≤ 2n−1 − 1
Si
 We first study the effect of varying yi and y j while
0
Si = Si−2n−1 +1 ∪ {n} 2n−1 ≤ i ≤ 2n − 2 fixing their sum. If that sum is s, then the function

{n} i = 2n − 1. y 7→ y3 + (s − y)3 has constant second derivative 6s,
so it is either everywhere convex or everywhere con-
(For example, if S10 , . . . , S20 n−1 −1 are ordered in lex- cave. Consequently, if (y1 , . . . , yn ) achieves the maxi-
icographic order as binary strings, then so are mum, then for any two indices i < j, at least one of the
S1 , . . . , S2n −1 .) Let Mn be the resulting matrix. Then following must be true:
we have: – one of yi , y j is extremal (i.e., equal to 1 or −1);
 
0 – yi = y j < 0 (in which case s < 0 and the local max-
 ..  imum is achieved above);
 Mn−1 Mn−1 . 

 0
 – yi = −y j (in which case s = 0 above).
 
Mn = 
 1 ··· 1 1  . In the third case, we may discard yi and y j and achieve a
 .. . . .. ..  case with smaller n; we may thus assume that this does
 M
n−1 . . . . 
not occur. In this case, all of the non-extremal values
 
1 ··· 1 1 
 
 are equal to some common value y < 0, and moreover
0 ··· 0 1 ··· 1 1
 2

we cannot have both 1 and -1. We cannot omit 1, as To clarify this last equality, consider a lattice walk start-
otherwise the condition ∑ni=1 yi = 0 cannot be achieved; ing from (0, 0), ending at (n, m), staying below the line
we must thus have only the terms 1 and y, occurring y = mx/n, and keeping as close to this line as possible.
with some positive multiplicities a and b adding up to If one follows this walk and records the element g for
n. Since a + b = n and a + by = 0, we can solve for y to each horizontal step and h for each vertical step, one ob-
obtain y = −a/b; we then have tains the word gha1 · · · ghan . Now take this walk, reflect
across the line y = x, rotate by a half-turn, then trans-
n
a2
 
3 3 late to put the endpoints at (0, 0) and (m, n); this is the
4 ∑ yi = a + by = 4a 1 − 2 .
i=1 b analogous walk for the pair (n, m).
Remark. By tracing more carefully through the argu-
Since y > −1, we must have a < b. For fixed a, the ment, one sees in addition that there exists an element k
target function increases as b increases, so the optimal of G for which g = km , h = k−n .
case must occur when a + b = 10. The possible pairs
(a, b) at this point are Second solution. (by Greg Martin) Since gcd(m, n) =
1, there exist integers x, y such that mx+ny = 1; we may
(1, 9), (2, 8), (3, 7), (4, 6); further assume that x ∈ {1, . . . , n}. We first establish the
identity
computing the target function for these values yields re- 
spectively ak − 1 if k ≡ 0 (mod n)

ak−x = ak + 1 if k ≡ 1 (mod n)
32 15 480 80 
, , , , 
ak otherwise.
9 2 49 9
yielding 480/49 as the maximum value. Namely, by writing −mx = ny − 1, we see that
Remark. Using Lagrange multipliers yields a similar 
m(k − x)
 
m(k − x − 1)

derivation, but with a slight detour required to separate ak−x = −
n n
local minima and maxima. For general n, the above    
argument shows that the target function is maximized mk + ny − 1 m(k − 1) + ny − 1
= −
when a + b = n. n n
   
mk − 1 m(k − 1) − 1
A4 First solution. We prove the claim by induction on = −
m + n. For the base case, suppose that n = 1; we then n n
have m = 1 and the given equation becomes gh = e.
and so
The claim then reduces to the fact that a one-sided in-
verse in G is also a two-sided inverse. (Because G is
   
mk − 1 mk
a group, g has an inverse g−1 ; since gh = e, we have ak−x − ak = −
n n
h = g−1 (gh) = g−1 e = g−1 , so hg = e = gh.)    
m(k − 1) − 1 m(k − 1)
Suppose now that n > 1. In case m > n, set g̃ = gh, − − .
n n
h̃ = h, and
    The first parenthesized expression equals 1 if n divides
(m − n)k (m − n)(k − 1) mk, or equivalently n divides k, and 0 otherwise. Sim-
bk = − (k = 1, . . . , n).
n n ilarly, the second parenthesized expression equals 1 if
n divides k − 1 and 0 otherwise. This proves the stated
then
identity.
g̃h̃b1 · · · g̃h̃bn = gha1 · · · ghan = e, We now use the given relation gha1 · · · ghan = e to write

so the induction hypothesis implies that g̃ and h̃ com- ghg−1 h−1 = gh(ha1 gha2 · · · ghan−1 ghan )h−1
mute; this implies that g and h commute. = gha1 +1 gha2 · · · ghan−1 ghan −1
In case m < n, note that ak ∈ {0, 1} for all k. Set g̃ = = gha1−x · · · ghan−x
h−1 , h̃ = g−1 , and
= (ghan+1−x · · · ghan )(gha1 · · · ghan−x ).
   
n` n(` − 1)
bl = − (` = 1, . . . , m); The two parenthesized expressions multiply in the
m m
opposite order to gha1 · · · ghan = e, so they must be
we claim that (two-sided) inverses of each other. We deduce that
ghg−1 h−1 = e, meaning that g and h commute.
g̃h̃b1 · · · g̃h̃bm = (gha1 · · · ghan )−1 = e, Third solution. (by Sucharit Sarkar) Let T denote the
torus R2 /Z2 . The line segments from (0, 0) to (1, 0)
so the induction hypothesis implies that h̃ and g̃ com-
mute; this implies that g and h commute.
 3

and from (0, 0) to (0, 1) are closed loops in T , and we then integrate both sides from 0 to x to conclude.
denote them by g and h respectively. Now let p be the (n)
(image of the) point (ε, −ε) in T for some 0 < ε  1. Now for f ∈ S, we have 0 ≤ f (1) ≤ f n!(1) for all n ≥ 0.
The punctured torus T \ {p} deformation retracts onto On the other hand, by Taylor’s theorem with remainder,
the union of the loops g and h, and so π1 (T \ {p}), the n
fundamental group of T \ {p} based at (0, 0), is the free f (k) (1)
f (x) ≥ ∑ (x − 1)k (x ≥ 1).
group on two generators, hg, hi. k=0 k!
Let γ and γ̃ denote the following loops based at (0, 0)
f (k) (1)
in T : γ is the image of the line segment from (0, 0) to Applying this with x = 2, we obtain f (2) ≥ ∑nk=0 k!
(n, m) under the projection R2 → T , and γ̃ is the image for all n; this implies that limn→∞ f (n) (1)
= 0. Since
n!
of the lattice walk from (0, 0) to (n, m), staying just be- f (n) (1)
low the line y = mx/n, that was described in the first f (1) ≤ n! , we must have f (1) = 0.
solution. There is a straight-line homotopy with fixed For f ∈ S, we proved earlier that f (x) = 0 for all x ≤ 0,
endpoints between the two paths in R2 from (0, 0) to as well as for x = 1. Since the function g(x) = f (cx) is
(n, m), the line segment and the lattice walk, and this also ultraconvex for c > 0, we also have f (x) = 0 for all
homotopy does not pass through any point of the form x > 0; hence f is identically zero.
(a + ε, b − ε) for a, b ∈ Z by the construction of the lat-
To sum up, if f : R → R is infinitely differentiable,
tice walk. It follows that γ and γ̃ are homotopic loops
f (0) = 0, and f (1) = 1, then f cannot be ultraconvex.
in T \ {p}. Since the class of γ̃ in π1 (T \ {p}) is evi-
This implies the desired result.
dently gha1 gha2 · · · ghan , it follows that the class of γ in
π1 (T \ {p}) is the same. Variant. (by Yakov Berchenko-Kogan) Another way to
show that any f ∈ S is identically zero is to show that
Now since gcd(m, n) = 1, there is an element φ ∈
for f ∈ S and k a positive integer,
GL2 (Z) sending (n, m) to (1, 0), which then sends the
line segment from (0, 0) to (n, m) to the segment from x 0
(0, 0) to (1, 0). Then φ induces a homeomorphism of T f (x) ≤ f (x) (x ≥ 0).
k
sending γ to g, which in turn induces an isomorphism
φ∗ : π1 (T \ {p}) → π1 (T \ {φ −1 (p)}). Both fundamen- We prove this by induction on k. For the base case k = 1,
tal groups are equal to hg, hi, and we conclude that φ∗ note that f 00 (x) ≥ 0 implies that f 0 is nondecreasing. For
sends gha1 gha2 · · · ghan to g. It follows that φ∗ induces x ≥ 0, we thus have
an isomorphism Z x Z x
f (x) = f 0 (t) dt ≤ f 0 (x) dt = x f 0 (x).
hg, h | gha1 gha2 · · · ghan i → hg, h | gi ∼
= hhi ∼
= Z. 0 0

To pass from k to k + 1, apply the induction hypothesis

Since Z is abelian, g and h must commute in to f 0 and integrate by parts to obtain
hg, h | gha1 gha2 · · · ghan i, whence they must also com- Z x
mute in G. k f (x) = k f 0 (t) dt
0
A5 First solution. Call a function f : R → R ultraconvex Z x
if f is infinitely differentiable and f (n) (x) ≥ 0 for all ≤ t f 00 (t) dt
0
n ≥ 0 and all x ∈ R, where f (0) (x) = f (x); note that if Z x
f is ultraconvex, then so is f 0 . Define the set = x f 0 (x) − f 0 (t) dt = x f 0 (x) − f (x).
0

S = { f : R → R | f ultraconvex and f (0) = 0}.

Remark. Noam Elkies points out that one can refine
For f ∈ S, we must have f (x) = 0 for all x < 0: if the argument to show that if f is ultraconvex, then it is
f (x0 ) > 0 for some x0 < 0, then by the mean value theo- analytic (i.e., it is represented by an entire Taylor series
about any point, as opposed to a function like f (x) =
rem there exists x ∈ (0, x0 ) for which f 0 (x) = f (x 0)
x0 < 0. 2
e−1/x whose Taylor series at 0 is identically zero); he
In particular, f 0 (0) = 0, so f 0 ∈ S also.
attributes the following argument to Peter Shalen. Let
We show by induction that for all n ≥ 0, gn (x) = ∑nk=0 k!1 f (k )(0)xk be the n-th order Taylor poly-
nomial of f . By Taylor’s theorem with remainder (a/k/a
f (n) (1) n Lagrange’s theorem), f (x) − gn (x) is everywhere non-
f (x) ≤ x ( f ∈ S, x ∈ [0, 1]).
n! negative; consequently, for all x ≥ 0, the Taylor series
1 (n)
∑∞ n=0 n! f (0)xn converges and is bounded above by f .
We induct with base case n = 0, which holds because
any f ∈ S is nondecreasing. Given the claim for n = m, But since f (n+1) (x) is nondecreasing, Lagrange’s the-
1
we apply the induction hypothesis to f 0 ∈ S to see that orem also implies that f (x) − gn (x) ≤ (n+1)! f (n+1) (x);
for fixed x ≥ 0, the right side tends to 0 as n → ∞. Hence
f (n+1) (1) n f is represented by its Taylor series for x ≥ 0, and so is
f 0 (t) ≤ t (t ∈ [0, 1]),
n!
 4

analytic for x > 0; by replacing f (x) with f (x − c), we and so f (c)/ f (1) ≤ cn+1 . (Here for convenience, we
may conclude that f is everywhere analytic. extend f continuously to [0, 1].) That is, f (c)/cn+1 ≤
Remark. We record some properties of the class of f (1) for all c ∈ (0, 1). For any b ∈ (0, 1), we may apply
ultraconvex functions. the same logic to the function f (bx) to deduce that if
f 0 (c) > 0, then f (bc)/cn+1 ≤ f (b), or equivalently
– Any nonnegative constant function is ultraconvex.
The exponential function is ultraconvex. f (bc) f (b)
≤ n+1 .
– If f is ultraconvex, then f0
is ultracon- (bc)n+1 b
vex. Conversely, if f 0 is ultraconvex and
This yields the claim unless f 0 is identically 0 on (0, 1),
lim infx→−∞ f (x) ≥ 0, then f is ultraconvex.
but in that case the claim is obvious anyway.
– The class of ultraconvex functions is closed under
We now apply the claim to show that for f as in the
addition, multiplication, and composition.
problem statement, it cannot be the case that f (n) (x) is
Second solution. (by Zachary Chase) In this solution, nonnegative on (0, 1) for all n. Suppose the contrary;
we use Bernstein’s theorem on monotone functions. To then for any fixed x ∈ (0, 1), we may apply the previous
state this result, we say that a function f : [0, ∞) → R is claim with arbitrarily large n to deduce that f (x) = 0.
totally monotone if f is continuous, f is infinitely differ- By continuity, we also then have f (1) = 0, a contradic-
entiable on (0, ∞), and (−1)n f (n) (x) is nonnegative for tion.
all positive integers n and all x > 0. For such a function, Fourth solution. (by Alexander Karabegov) As in the
Bernstein’s theorem asserts that there is a nonnegative first solution, we may see that f (n) (0) = 0 for all n. Con-
finite Borel measure µ on [0, ∞) such that sequently, for all n we have
Z ∞ Z x
e−tx dµ(t) (x ≥ 0). 1
f (x) = f (x) = (x − t)n−1 f (n) (t) dt (x ∈ R)
0 (n − 1)! 0

For f as in the problem statement, for any M > 0, the and hence
restriction of f (M − x) to [0, ∞) is totally monotone, Z 1 Z 1
so Bernstein’s theorem
R provides a Borel measure µ for
1
f (x) dx = (1 − t)n f (n) (t) dt.
which f (M − x) = R0∞ e−tx dµ(t) for all x ≥ 0. Taking 0 n! 0
x = M, we see that 0∞ e−Mt dµ(t) = f (0) = 0; since µ
is a nonnegative measure, it must be identically zero. Suppose now that f is infinitely differentiable, f (1) =
Hence f (x) is identically zero for x ≤ M; varying over 1, and f (n) (x) ≥ 0 for all n and all x ∈ [0, 1]. Then
all M, we deduce the desired result. Z 1
1 1 1 Z
Third solution. (from Art of Problem Solving user f (x) dx = · (1 − t)n f (n) (t) dt
chronondecay) In this solution, we only consider the 0 n (n − 1)! 0
Z 1
behavior of f on [0, 1]. We first establish the following 1 1
≤ · (1 − t)n−1 f (n) (t) dt
result. Let f : (0, 1) → R be a function such that for n (n − 1)! 0
each positive integer n, f (n) (x) is nonnegative on (0, 1), 1 1
tends to 0 as x → 0+ , and tends to some limit as x → 1− . = f (1) = .
n n
Then for each nonnegative integer n, f (x)x−n is nonde-
Since this holds for all n, we have 01 f (x) dx = 0, and
R
creasing on (0, 1).
To prove the claimed result, we proceed by induction on so f (x) = 0 for x ∈ [0, 1]; this yields the desired contra-
n, the case n = 0 being a consequence of the assump- diction.
tion that f 0 (x) is nonnegative on (0, 1). Given the claim A6 First solution. Choose a Cartesian coordinate sys-
for some n ≥ 0, note that since f 0 also satisfies the hy- tem with origin at the midpoint of AB and positive x-
potheses of the problem, f 0 (x)x−n is also nondecreasing axis containing A. By
on (0, 1). Choose c ∈ (0, 1) and consider the function √ √ the condition on AB, we have
A = ( a, 0), B = (− a, 0) for some positive rational
f 0 (c) n number a. Let (x1 , y1 ) and (x2 , y2 ) be the respective co-
g(x) = x (x ∈ [0, 1)). ordinates of C and D; by computing the lengths of the
cn segments AC, BC, AD, BD,CD, we see that the quanti-
For x ∈ (0, c), f 0 (x)x−n ≤ f 0 (c)c−n , so f 0 (x) ≤ g(x); ties
similarly, for x ∈ (c, 1), f 0 (x) ≥ g(x). It follows that √ √
if f 0 (c) > 0, then (x1 − a)2 + y21 , (x1 + a)2 + y21 ,
√ √
R1 0 R1 Rc 0 Rc (x2 − a)2 + y22 , (x2 + a)2 + y22 ,
f (x) dx c g(x) dx 0 f (x) dx g(x) dx
(x1 − x2 )2 + (y1 − y2 )2
Rcc
0
≥ c
R ⇒ R1 ≤ R01
0 f (x) dx 0 g(x) dx 0 f 0 (x) dx 0 g(x) dx
 5

are all rational numbers. By adding and subtracting the In particular, the determinant vanishes if and only if
first two quantities, and similarly for the next two, we A, B,C, D are coplanar. From the identity
see that the quantities
√ √ 64(4Area(4ABC)2 Area(4ABD)2 − 9AB2V 2 )
x12 + y21 , x1 a, x22 + y22 , x2 a
= (AB4 − AB2 (AC2 + AD2 + BC2 + BD2 − 2CD2 )
are rational numbers. Since a is a rational number, so +(AC2 − BC2 )(AD2 − BD2 ))2
then are
√ we see that Area(4ABC)Area(4ABD) is rational;
2 (x1 a)2 since each of the areas has rational square, we deduce
x1 =
a the claim.
√
2 (x2 a)2 Fourth solution. (by Greg Martin) Define the signed
x2 =
a
√ √ angles α = ∠BAC, β = ∠BAD, γ = ∠CAD, so that α +
(x1 a)(x2 a) γ = β . By the Law of Cosines,
x1 x2 =
a
2AB · AC cos α = AB2 + AC2 − BC2 ∈ Q
y21 = (x12 + y21 ) − x12
2AB · AD cos β = AB2 + AD2 − BD2 ∈ Q
y22 = (x22 + y22 ) − x22 .
2AC · AD cos γ = AC2 + AD2 −CD2 ∈ Q.
Now note that the quantity
In particular, (2AB · AC cos α)2 ∈ Q, and so cos2 α ∈ Q
(x1 − x2 )2 + (y1 − y2 )2 = x12 − 2x1 x2 + x22 + y21 − 2y1 y2 + y22 and sin2 α = 1 − cos2 α ∈ Q, and similarly for the other
two angles.
is known to be rational, as is every summand on the
right except −2y1 y2 ; thus y1 y2 is also rational. Since y21 Applying the addition formula to cos β , we deduce that
is also rational, so then is y1 /y2 = (y1 y2 )/(y21 ); since
2AB · AD cos α cos γ − 2AB · AD sin α sin γ ∈ Q.
√ √
area(4ABC) = ay1 , area(4ABD) = ay2 ,
The first of these terms equals
this yields the desired result. (2AB · AC cos α)(2AB · AC cos α)
Second solution. (by Manjul Bhargava) Let b, c, d be ∈ Q,
AC2
the vectors AB, AC, AD viewed as column vectors. The
desired ratio is given by so the second term must also be rational. But now
Area(4ABC) AB · AC sin α
det(b, c) det(b, c)T det(b, c) =
= Area(4ACD) AC · AD sin γ
det(b, d) det(b, c)T det(b, d)
!−1 2AB · AD sin α sin γ
= ∈Q
!
b·b b·c b·b b·d 2AD2 sin2 γ
= det det .
c·b c·c c·b c·d
as desired.
The square of the length of AB is b · b, so this quantity is Remark. Derek Smith observes that this result is
rational. The square of the lengths of AC and BC are c·c Proposition 1 of: M. Knopf, J. Milzman, D. Smith, D.
and (c − b) · (c − b) = b · b + c · c − 2b · c, so b · c = c · b Zhu and D. Zirlin, Lattice embeddings of planar point
is rational. Similarly, using AD and BD, we deduce that sets, Discrete and Computational Geometry 56 (2016),
d · d and b · d is rational; then using CD, we deduce that 693–710.
c · d is rational. Remark. It is worth pointing out that it is indeed possi-
Third solution. (by David Rusin) Recall that Heron’s ble to choose points A, B,C, D satisfying the conditions
formula (for the area of a triangle in terms of its side of the problem; one can even ensure that the lengths
length) admits the following three-dimensional ana- of all four segments are themselves rational. For ex-
logue due to Piero della Francesca: if V denotes the ample, it was originally observed by Euler that one can
volume of a tetrahedron with vertices A, B,C, D ∈ R3 , find an infinite set of points on the unit circle whose
then pairwise distances are all rational numbers. One way to
see this is to apply the linear fractional transformation
0 AB2 AC2 AD2 1 f (z) = z+i
 
z−i to the Riemann sphere to carry the real axis
 AB2 0 BC2 BD2 1
  (plus ∞) to the unit circle, then compute that
288V 2 = det  AC2 BC2 0 CD2 1
 
2|z1 − z2 ||
 2
AD BD2 CD2 0 1
 | f (z1 ) − f (z2 )| = .
|(z1 − i)(z2 − i)|
1 1 1 1 0
 6

Let S be the set of rational numbers z for which 2(z2 +1) have norm 1 and the same argument, which only hap-
is a perfect square; the set f (S) has the desired property pens for z = 1. Thus there can be no root of fn with
provided that it is infinite. That can be checked in var- |z| ≤ 1.
ious ways; for instance, the equation 2(x2 + 1) = (2y)2 Second solution. (by Karl Mahlburg) Define the poly-
equates to x2 − 2y2 = −1 (a modified Brahmagupta- nomial
Pell equation), which has infinitely many solutions even
over the integers: gn (z) = nzn−1 + · · · + 2z + 1
√ √
x + y 2 = (1 + 2)2n+1 . and note that zn−1 gn (z−1 ) = fn (z). Since fn (0) 6= 0, to
prove the claim it is equivalent to show that gn has no
B1 The answer is the collection of vectors (1, b) where 0 ≤ roots in the region |z| ≥ 1.
b ≤ 100 and b is even. (For ease of typography, we Now note that gn (z) = h0n (z) for
write tuples instead of column vectors.)
First we show that if P \ {v} can be partitioned into hn (z) = zn + · · · + z + 1,
subsets S1 and S2 of equal size and equal sum, then v
must be of the form (1, b) where b is even. For a finite a polynomial with roots e2πi j/(n+1) for j = 0, . . . , n. By
nonempty set S of vectors in Z2 , let Σ(S) denote the the Gauss-Lucas theorem, the roots of gn lie in the con-
sum of the vectors in S. Since the average x- and y- vex hull of the roots of hn , and moreover cannot be
coordinates in P are 1 and 50, respectively, and there vertices of the convex hull because hn has no repeated
are 3 · 101 elements in P, we have roots. This implies the claim.
Remark. Yet another approach is to use the Eneström-
Σ(P) = 303 · (1, 50) = (303, 15150). Kakeya theorem: if Pn (z) = a0 + · · · + an zn is a polyno-
mial with real coefficients satisfying |an | ≥ · · · ≥ |a0 | >
On the other hand, 0, then the roots of Pn (z) all satisfy |z| ≤ 1. Namely, ap-
plying this to the polynomial gn (z/c) for c = n/(n − 1)
Σ(P) = v + Σ(S1 ) + Σ(S2 ) = v + 2Σ(S1 ).
shows that the roots of gn all satisfy |z| ≤ 1/c.
By parity considerations, the entries of v must be odd Remark. For a related problem, see problem A5 from
and even, respectively, and thus v is of the claimed the 2014 Putnam competition.
form. `
B3 The values of n with this property are 22 for ` =
Next suppose v = (1, b) where b is even. Note that
1, 2, 4, 8. First, note that n divides 2n if and only if n
P \ {(1, 50)} can be partitioned into 151 pairs of (dis-
is itself a power of 2; we may thus write n = 2m and
tinct) vectors (x, y) and (2 − x, 100 − y), each sum-
note that if n < 10100 , then
ming to (2, 100). If b 6= 50 then three of these
pairs are {(1, b), (1, 100−b)},{(2, b), (0, 100−b)}, and 2m = n < 10100 < (103 )34 < (210 )34 = 2340 .
{(2, 25 + b/2), (0, 75 − b/2)}. Of the remaining 148
pairs, assign half of them to S1 and half to S2 , and Moreover, the case m = 0 does not lead to a solution
then complete the partition of P \ {v} by assign- because for n = 1, n − 1 = 0 does not divide 2n − 1 = 1;
ing (0, 100 − b), (2, 25 + b/2), and (1, 50) to S1 and we may thus assume 1 ≤ m ≤ 340.
(1, 100 − b), (2, b), and (0, 75 − b/2) to S2 . (Note that
Next, note that modulo n − 1 = 2m − 1, the powers of
the three vectors assigned to each of S1 and S2 have the
2 cycle with period m (the terms 20 , . . . , 2m−1 remain
same sum (3, 175 − b/2).) By construction, S1 and S2
the same upon reduction, and then the next term repeats
have the same number of elements, and Σ(S1 ) = Σ(S2 ).
the initial 1); consequently, n − 1 divides 2n − 1 if and
For b = 50, this construction does not work be- only if m divides n, which happens if and only if m is a
cause (1, b) = (100 − b), but a slight variation can power of 2. Write m = 2` and note that 2` < 340 < 512,
be made. In this case, three of the pairs in P \ so ` < 9. The case ` = 0 does not lead to a solution
{(1, 50)} are {(2, 50), (0, 50)}, {(1, 51), (1, 49)}, and because for n = 2, n − 2 = 0 does not divide 2n − 2 = 2;
{(0, 49), (2, 51)}. Assign half of the other 148 pairs we may thus assume 1 ≤ ` ≤ 8.
to S1 and half to S2 , and complete the partition of
Finally, note that n − 2 = 2m − 2 divides 2n − 2 if and
P \ {(1, 50)} by assigning (2, 50), (1, 51), and (0, 49)
only if 2m−1 − 1 divides 2n−1 − 1. By the same logic
to S1 and (0, 50), (1, 49), and (2, 51) to S2 .
as the previous paragraph, this happens if and only if
B2 First solution. Note first that fn (1) > 0, so 1 is not a m − 1 divides n − 1, that is, if 2` − 1 divides 2m − 1.
root of fn . Next, note that This in turn happens if and only if ` divides m = 2` ,
which happens if and only if ` is a power of 2. The
(z − 1) fn (z) = zn + · · · + z − n; values allowed by the bound ` < 9 are ` = 1, 2, 4, 8; for
these values, m ≤ 28 = 256 and
however, for |z| ≤ 1, we have |zn + · · · + z| ≤ n by the
triangle inequality; equality can only occur if z, . . . , zn n = 2m ≤ 2256 ≤ (23 )86 < 1086 < 10100 ,

so the solutions listed do satisfy the original inequality.

 7

B4 We first rule out the case |a| > 1. In this case, we prove B5 Let (a1 , a2 ) and (a01 , a02 ) be distinct points in R2 ;
that |xn+1 | ≥ |xn | for all n, meaning that we cannot have we want to show that f (a1 , a2 ) 6= f (a01 , a02 ). Write
xn = 0. We proceed by induction; the claim is true for (v1 , v2 ) = (a01 , a02 ) − (a1 , a2 ), and let γ(t) = (a1 , a2 ) +
n = 0, 1 by hypothesis. To prove the claim for n ≥ 2, t(v1 , v2 ), t ∈ [0, 1], be the path between (a1 , a2 ) and
write (a01 , a02 ). Define a real-valued function g by g(t) =
(v1 , v2 ) · f (γ(t)). By the Chain Rule,
|xn+1 | = |2xn xn−1 − xn−2 | ! !
≥ 2|xn ||xn−1 | − |xn−2 | 0 ∂ f1 /∂ x1 ∂ f1 /∂ x2 v1
f (γ(t)) = .
≥ |xn |(2|xn−1 | − 1) ≥ |xn |, ∂ f2 /∂ x1 ∂ f2 /∂ x2 v2

where the last step follows from |xn−1 | ≥ |xn−2 | ≥ · · · ≥ Abbreviate ∂ fi /∂ x j by fi j ; then
|x0 | = 1. ! !
We may thus assume hereafter that |a| ≤ 1. We can f f v
 
g0 (t) = v1 v2 11 12 1
then write a = cos b for some b ∈ [0, π]. Let {Fn } be the f21 f22 v2
Fibonacci sequence, defined as usual by F1 = F2 = 1
and Fn+1 = Fn + Fn−1 . We show by induction that = f11 v21 + ( f12 + f21 )v1 v2 + f22 v22
2
4 f11 f22 − ( f12 + f21 )2 2

f12 + f21
xn = cos(Fn b) (n ≥ 0). = f11 v1 + v2 + v2
2 f11 4 f11
Indeed, this is true for n = 0, 1, 2; given that it is true for ≥0
n ≤ m, then
since f11 and f11 f22 − ( f12 + f21 )2 /4 are positive by as-
2xm xm−1 = 2 cos(Fm b) cos(Fm−1 b) sumption. Since the only way that equality could hold
= cos((Fm − Fm−1 )b) + cos((Fm + Fm−1 )b) is if v1 and v2 are both 0, we in fact have g0 (t) > 0 for
= cos(Fm−2 b) + cos(Fm+1 b) all t. But if f (a1 , a2 ) = f (a01 , a02 ), then g(0) = g(1), a
contradiction.
and so xm+1 = 2xm xm−1 − xm−2 = cos(Fm+1 b). This Remark. A similar argument shows more generally
completes the induction. that f : Rn → Rn is injective if at all points in Rn , the
Since xn = cos(Fn b), if xn = 0 for some n then Fn b = Jacobian matrix D f satisfies the following property: the
k quadratic form associated to the bilinear form with ma-
2 π for some odd integer k. In particular, we can write trix D f (or the symmetrized bilinear form with matrix
b = dc (2π) where c = k and d = 4Fn are integers.
(D f + (D f )T )/2) is positive definite. In the setting of
Let xn denote the pair (Fn , Fn+1 ), where each entry in the problem, the symmetrized matrix is
this pair is viewed as an element of Z/dZ. Since there
are only finitely many possibilities for xn , there must
!
f11 ( f12 + f21 )/2
be some n2 > n1 such that xn1 = xn2 . Now xn uniquely ,
determines both xn+1 and xn−1 , and it follows that the ( f12 + f21 )/2 f22
sequence {xn } is periodic: for ` = n2 − n1 , xn+` = xn
for all n ≥ 0. In particular, Fn+` ≡ Fn (mod d) for all n. and this is positive definite if and only if f11 and the
F c Fn c determinant of the matrix are both positive (Sylvester’s
But then n+` d − d is an integer, and so criterion). Note that the assumptions that f12 , f21 > 0
  are unnecessary for the argument; it is also easy to
Fn+` c
xn+` = cos (2π) see that the hypotheses f11 , f12 > 0 are also superflu-
d ous. (The assumption f11 f22 − ( f12 + f21 )2 > 0 implies
 
Fn c f11 f22 > 0, so both are nonzero and of the same sign;
= cos (2π) = xn
d by continuity, this common sign must be constant over
all of R2 . If it is negative, then apply the same logic to
for all n. Thus the sequence {xn } is periodic, as desired. (− f1 , − f2 ).)
Remark. Karl Mahlburg points out that one can moti- B6 (by Manjul Bhargava) Let a(k, n) denote the num-
vate the previous solution by computing the terms ber of sequences of length k taken from the set
{1, 2, 3, 4, 5, 6, 10} and having sum n. We prove that
x2 = 2a2 − 1, x3 = 4a3 − 3a, x4 = 16a5 − 20a3 + 5a
 k
and recognizing these as the Chebyshev polynomials n 2018
a(k, n) < 2
T2 , T3 , T5 . (Note that T3 was used in the solution of prob- 2048
lem A3.)
by double induction on n + k and n − k. The claim is
Remark. It is not necessary to handle the case |a| > 1 clearly true when n − k ≤ 0 and in particular when n =
separately; the cosine function extends to a surjective k = 1, the smallest case for n + k.
analytic function on C and continues to satisfy the addi-
tion formula, so one can write a = cos b for some b ∈ C
and then proceed as above.
 8

We categorize the sequences counted by a(k, n) by In order for this to hold for x = 1/2, k = 2018, one must
whether they end in 1, 2, 3, 4, 5, 6, 10; removing the last take n = 3860.
term of such a sequence yields a sequence counted Remark. For purposes of comparison, the stated bound
by a(k − 1, n − 1), a(k − 1, n − 2), a(k − 1, n − 3), a(k − is about 101149 , while the trivial upper bound given by
1, n − 4), a(k − 1, n − 5), a(k − 1, n − 6), a(k − 1, n − 10), counting all sequences of length 2018 of positive inte-
respectively. Therefore, gers that sum to 3860 is
a(k, n) = a(k − 1, n − 1) + · · · 
3859

+ a(k − 1, n − 6) + a(k − 1, n − 10) ∼ 101158 .
2017
2018 k−1
 
< (2n−1 + · · · + 2n−6 + 2n−10 ) The latter can be easily derived by a “stars and bars”
2048 argument: visualize each sequence of this form by rep-
2018 k−1 resenting the value n by n stars and inserting a bar be-
  
n 1 1 1
=2 +···+ + tween adjacent terms of the sequence. The resulting
2 64 1024 2048
  k−1 string of symbols consists of one star at the beginning,
1009 2018 2017 bar-star combinations, and 3860-2018 more stars.
= 2n
1024 2048 Using a computer, it is practical to compute the exact

2018
k cardinality of S by finding the coefficient of x3860 in (x+
= 2n x2 +· · ·+x6 +x10 )2018 . For example, this can be done in
2048 Sage in a couple of seconds as follows. (The truncation
where we used directly the induction hypothesis to ob- is truncated modulo x4000 for efficiency.)
tain the inequality on the second line. The case k =
2018, n = 3860 yields the desired result. sage: P.<x> = PowerSeriesRing(ZZ, 4000)
sage: f = (x + x^2 + x^3 + x^4 + \
Remark. K. Soundararajan suggests the following rein-
....: x^5 + x^6 + x^10)^2018
terpretation of this argument. The quantity a(k, n) can
sage: m = list(f)[3860]
be interpreted as the coefficient of xn in (x + x2 + · · · +
sage: N(m)
x6 + x10 )k . Since this polynomial has nonnegative coef-
8.04809122940636e1146
ficients, for any x, we have

a(k, n)xn < (x + x2 + · · · + x6 + x10 )k . This computation shows that the upper bound of the
problem differs from the true value by a factor of about
1
Substituting x = 2 yields the bound stated above. 150.
On a related note, Alexander Givental suggests that the
value n = 3860 (which is otherwise irrelevant to the
problem) may have been chosen for the following rea-
son: as a function of x, the upper bound x−n (x + x2 +
· · · + x6 + x10 )k is minimized when

x(1 + 2x + · · · + 6x5 + x9 ) n
= .
x + x2 + · · · + x6 + x10 k
 The 80th William Lowell Putnam Mathematical Competition
Saturday, December 7, 2019

A1 Determine all possible values of the expression B1 Denote by Z2 the set of all points (x, y) in the plane with
integer coordinates. For each integer n ≥ 0, let Pn be the
A3 + B3 +C3 − 3ABC subset of Z2 consisting of the point (0, 0) together with
all points (x, y) such that x2 + y2 = 2k for some integer
where A, B, and C are nonnegative integers. k ≤ n. Determine, as a function of n, the number of
A2 In the triangle 4ABC, let G be the centroid, and let I four-point subsets of Pn whose elements are the vertices
be the center of the inscribed circle. Let α and β be of a square.
the angles at the vertices A and B, respectively. Sup- B2 For all n ≥ 1, let
pose that the segment IG is parallel to AB and that
β = 2 tan−1 (1/3). Find α.
 
(2k−1)π
n−1 sin 2n
an = ∑
.
A3 Given real numbers b0 , b1 , . . . , b2019 with b2019 6= 0, let
 
(k−1)π kπ
k=1 cos2 2n cos2 2n
z1 , z2 , . . . , z2019 be the roots in the complex plane of the
polynomial
Determine
2019 an
P(z) = ∑ bk zk . lim
n→∞ n3
.
k=0

Let µ = (|z1 | + · · · + |z2019 |)/2019 be the average of the B3 Let Q be an n-by-n real orthogonal matrix, and let
distances from z1 , z2 , . . . , z2019 to the origin. Determine u ∈ Rn be a unit column vector (that is, uT u = 1). Let
the largest constant M such that µ ≥ M for all choices P = I − 2uuT , where I is the n-by-n identity matrix.
of b0 , b1 , . . . , b2019 that satisfy Show that if 1 is not an eigenvalue of Q, then 1 is an
eigenvalue of PQ.
1 ≤ b0 < b1 < b2 < · · · < b2019 ≤ 2019.
B4 Let F be the set of functions f (x, y) that are twice con-
tinuously differentiable for x ≥ 1, y ≥ 1 and that satisfy
A4 Let f be a continuous real-valued function on R3 .
Sup- the following two equations (where subscripts denote
pose that for every sphere S of radius 1, the integral of partial derivatives):
f (x, y, z) over the surface of S equals 0. Must f (x, y, z)
be identically 0? x fx + y fy = xy ln(xy),
A5 Let p be an odd prime number, and let F p denote the x2 fxx + y2 fyy = xy.
field of integers modulo p. Let F p [x] be the ring of poly-
nomials over F p , and let q(x) ∈ F p [x] be given by For each f ∈ F , let

p−1 m( f ) = min ( f (s + 1, s + 1) − f (s + 1, s) − f (s, s + 1) + f (s, s)) .

s≥1
q(x) = ∑ ak xk ,
k=1
Determine m( f ), and show that it is independent of the
where choice of f .

ak = k(p−1)/2 mod p. B5 Let Fm be the mth Fibonacci number, defined by F1 =

F2 = 1 and Fm = Fm−1 + Fm−2 for all m ≥ 3. Let p(x) be
Find the greatest nonnegative integer n such that (x − the polynomial of degree 1008 such that p(2n + 1) =
1)n divides q(x) in F p [x]. F2n+1 for n = 0, 1, 2, . . . , 1008. Find integers j and k
such that p(2019) = Fj − Fk .
A6 Let g be a real-valued function that is continuous on
the closed interval [0, 1] and twice differentiable on the B6 Let Zn be the integer lattice in Rn . Two points in Zn
open interval (0, 1). Suppose that for some real number are called neighbors if they differ by exactly 1 in one
r > 1, coordinate and are equal in all other coordinates. For
which integers n ≥ 1 does there exist a set of points
g(x) S ⊂ Zn satisfying the following two conditions?
lim = 0.
x→0+ xr
(1) If p is in S, then none of the neighbors of p is in S.
Prove that either (2) If p ∈ Zn is not in S, then exactly one of the neigh-
bors of p is in S.
lim g0 (x) = 0 or lim sup xr |g00 (x)| = ∞.
x→0+ x→0+
 Solutions to the 80th William Lowell Putnam Mathematical Competition
Saturday, December 7, 2019
Kiran Kedlaya and Lenny Ng

A1 The answer is all nonnegative integers not congruent to and let r be the inradius of 4ABC. Since C, G, M are
3 or 6 (mod 9). Let X denote the given expression; collinear with CM = 3GM, the distance from C to line
we first show that we can make X equal to each of the AB is 3 times the distance from G to AB, and the latter is
claimed values. Write B = A + b and C = A + c, so that r since IG k AB; hence the altitude CD has length 3r. By
3
the double angle formula for tangent, CDDB = tan β = 4 ,
X = (b2 − bc + c2 )(3A + b + c). and so DB = 4r. Let E be the point where the incircle
meets AB; then EB = r/ tan( β2 ) = 3r. It follows that
By taking (b, c) = (0, 1) or (b, c) = (1, 1), we obtain re- ED = r, whence the incircle is tangent to the altitude
spectively X = 3A + 1 and X = 3A + 2; consequently, CD. This implies that D = A, ABC is a right triangle,
as A varies, we achieve every nonnegative integer not and α = π2 .
divisible by 3. By taking (b, c) = (1, 2), we obtain
X = 9A + 9; consequently, as A varies, we achieve ev- Remark. One can obtain a similar solution by fixing a
ery positive integer divisible by 9. We may also achieve coordinate system with B at the origin and A on the pos-
X = 0 by taking (b, c) = (0, 0). itive x-axis. Since tan β2 = 13 , we may assume without
In the other direction, X is always nonnegative: either loss of generality that I = (3, 1). Then C lies on the in-
apply the arithmetic mean-geometric mean inequality, tersection of the line y = 3 (because CD = 3r as above)
or write b2 − bc + c2 = (b − c/2)2 + 3c2 /4 to see that it with the line y = 34 x (because tan β = 34 as above), forc-
is nonnegative. It thus only remains to show that if X ing C = (4, 3) and so forth.
is a multiple of 3, then it is a multiple of 9. Note that Solution 2. Let a, b, c be the lengths of BC,CA, AB,
3A + b + c ≡ b + c (mod 3) and b2 − bc + c2 ≡ (b + c)2 respectively. Let r, s, and K denote the inradius,
(mod 3); consequently, if X is divisible by 3, then b + c semiperimeter, and area of 4ABC. By Heron’s For-
must be divisible by 3, so each factor in X = (b2 − bc + mula,
c2 )(3A + b + c) is divisible by 3. This proves the claim.
r2 s2 = K 2 = s(s − a)(s − b)(s − c).
Remark. The factorization of X used above can be
written more symmetrically as
If IG is parallel to AB, then
2 2 2
X = (A + B +C)(A + B +C − AB − BC −CA).
1 1 1
rc = area(4ABI) = area(4ABG) = K = rs
One interpretation of the factorization is that X is the 2 3 3
determinant of the circulant matrix 3(a+b)
  and so c = a+b2 . Since s = 4 and s − c = a+b 4 , we
A B C have 3r2 = (s − a)(s − b). Let E be the point at which
C A B
the incircle meets AB; then s − b = EB = r/ tan( β2 ) and
B C A
s − a = EA = r/ tan( α2 ). It follows that tan( α2 ) tan( β2 ) =
1 α π
which has the vector (1, 1, 1) as an eigenvector (on ei- 3 and so tan( 2 ) = 1. This implies that α = 2 .
ther side) with eigenvalue A + B + C. The other eigen- Remark. The equality c = a+b 2 can also be derived
values are A + ζ B + ζ 2C where ζ is a primitive cube from the vector representations
root of unity; in fact, X is the norm form for the ring
Z[T ]/(T 3 − 1), from which it follows directly that the A + B +C aA + bB + cC
image of X is closed under multiplication. (This is G= , I= .
3 a+b+c
similar to the fact that the image of A2 + B2 , which is
the norm form for the ring Z[i] of Gaussian integers, is Solution 3. (by Catalin Zara) It is straightforward to
closed under multiplication.) check that a right triangle with AC = 3, AB = 4, BC = 5
One can also the unique factorization property of the works. For example, in a coordinate system with A =
ring Z[ζ ] of Eisenstein integers as follows. The three (0, 0), B = (4, 0),C = (0, 3), we have
factors of X over Z[ζ3 ] are pairwise congruent modulo  
1 − ζ3 ; consequently, if X is divisible by 3, then it is di- 4
G= ,1 , I = (1, 1)
visible by (1 − ζ3 )3 = −3ζ3 (1 − ζ3 ) and hence (because 3
it is a rational integer) by 32 .
and for D = (1, 0),
A2 Solution 1. Let M and D denote the midpoint of AB
and the foot of the altitude from C to AB, respectively, β ID 1
tan = = .
2 BD 3
 2

It thus suffices to suggest that this example is unique up Noam Elkies points out that a similar result holds in Rn
to similarity. for any n. Also, there exist nonzero continuous func-
Let C0 be the foot of the angle bisector at C. Then tions on Rn whose integral over any unit ball vanishes;
this implies certain negative results about image recon-
CI CA +CB struction.
=
IC0 AB A5 The answer is p−1 d
2 . Define the operator D = x dx ,
and so IG is parallel to AB if and only if CA + CB = d
where dx indicates formal differentiation of polyno-
2AB. We may assume without loss of generality that A mials. For n as in the problem statement, we have
and B are fixed, in which case this condition restricts C q(x) = (x − 1)n r(x) for some polynomial r(x) in F p not
to an ellipse with foci at A and B. Since the angle β divisible by x − 1. For m = 0, . . . , n, by the product rule
is also fixed, up to symmetry C is further restricted to we have
a half-line starting at B; this intersects the ellipse in a
unique point. (Dm q)(x) ≡ nm xm (x − 1)n−m r(x) (mod (x − 1)n−m+1 ).
Remark. Given that CA +CB = 2AB, one can also re-
Since r(1) 6= 0 and n 6≡ 0 (mod p) (because n ≤
cover the ratio of side lengths using the law of cosines.
deg(q) = p − 1), we may identify n as the smallest non-
A3 The answer is M = 2019−1/2019 . For any choices of negative integer for which (Dn q)(1) 6= 0.
b0 , . . . , b2019 as specified, AM-GM gives Now note that q = D(p−1)/2 s for

µ ≥ |z1 · · · z2019 |1/2019 = |b0 /b2019 |1/2019 ≥ 2019−1/2019 . xp − 1

s(x) = 1 + x + · · · + x p−1 = = (x − 1) p−1
x−1
To see that this is best possible, consider b0 , . . . , b2019
given by bk = 2019k/2019 for all k. Then since (x − 1) p = x p − 1 in F p [x]. By the same logic
as above, (Dn s)(1) = 0 for n = 0, . . . , p − 2 but not for
2019
z2020 − 1 n = p − 1. This implies the claimed result.
P(z/20191/2019 ) = ∑ zk =
k=0 z−1 Remark. One may also finish by checking directly that
for any positive integer m,
has all of its roots on the unit circle. It follows that all
of the roots of P(z) have modulus 2019−1/2019 , and so
(
p−1
m −1 (mod p) if (p − 1)|m
µ = 2019−1/2019 in this case. ∑ k ≡ 0 (mod p) otherwise.
k=1
A4 The answer is no. Let g : R → R be any continuous
function with g(t + 2) = g(t) for all t and 02 g(t) dt = 0
R If (p − 1)|m, then km ≡ 1 (mod p) by the little Fermat
(for instance, g(t) = sin(πt)). Define f (x,RR
y, z) = g(z). theorem, and so the sum is congruent to p − 1 ≡ −1
We claim that for any sphere S of radius 1, S f dS = 0. (mod p). Otherwise, for any primitive root ` mod p,
multiplying the sum by `m permutes the terms modulo
Indeed, let S be the unit sphere centered at (x0 , y0 , z0 ). p and hence does not change the sum modulo p; since
We can parametrize S by S(φ , θ ) = (x0 , y0 , z0 ) + `n 6≡ 1 (mod p), this is only possible if the sum is zero
(sin φ cos θ , sin φ sin θ , cos φ ) for φ ∈ [0, π] and θ ∈ modulo p.
[0, 2π]. Then we have
A6 Solution 1. (by Harm Derksen) We assume
that lim supx→0+ xr |g00 (x)| < ∞ and deduce that
limx→0+ g0 (x) = 0. Note that
Z π Z 2π
∂S ∂S
ZZ
f (x, y, z) dS = f (S(φ , θ )) × dθ dφ
S 0 0 ∂φ ∂θ
Z π Z 2π lim sup xr sup{|g00 (ξ )| : ξ ∈ [x/2, x]} < ∞.
x→0+
= g(z0 + cos φ ) sin φ dθ dφ
0 0
Z 1 Suppose for the moment that there exists a function h
= 2π g(z0 + t) dt, on (0, 1) which is positive, nondecreasing, and satisfies
−1
g(x) h(x)
where we have used the substitution t = cos φ ; but this lim = lim = 0.
x→0+ h(x) x→0+ xr
last integral is 0 for any z0 by construction.
Remark. The solution recovers the famous observation For some c > 0, h(x) < xr < x for x ∈ (0, c). By Taylor’s
of Archimedes that the surface area of a spherical cap theorem with remainder, we can find a function ξ on
is linear in the height of the cap. In place (0, c) such that ξ (x) ∈ [x − h(x), x] and
RR of spheri-
cal coordinates, one may also compute S f (x, y, z) dS
1
by computing the integral over a ball of radius r, then g(x − h(x)) = g(x) − g0 (x)h(x) + g00 (ξ (x))h(x)2 .
computing the derivative with respect to r and evaluat- 2
ing at r = 1.
 3

We can thus express g0 (x) as B1 The answer is 5n + 1.

We first determine the set Pn . Let Qn be the set of points
g(x) 1 r 00 h(x) g(x − h(x)) h(x − h(x))
+ x g (ξ (x)) r − . in Z2 of the form (0, ±2k ) or (±2k , 0) for some k ≤ n.
h(x) 2 x h(x − h(x)) h(x) Let Rn be the set of points in Z2 of the form (±2k , ±2k )
for some k ≤ n (the two signs being chosen indepen-
As x → 0+ , g(x)/h(x), g(x − h(x))/h(x − h(x)), and
dently). We prove by induction on n that
h(x)/xr tend to 0, while xr g00 (ξ (x)) remains bounded
(because ξ (x) ≥ x − h(x) ≥ x − xr ≥ x/2 for x small) Pn = {(0, 0)} ∪ Qbn/2c ∪ Rb(n−1)/2c .
and h(x − h(x))/h(x) is bounded in (0, 1]. Hence
limx→0+ g0 (x) = 0 as desired. We take as base cases the straightforward computations
It thus only remains to produce a function h with the
desired properties; this amounts to “inserting” a func- P0 = {(0, 0), (±1, 0), (0, ±1)}
tion between g(x) and xr while taking care to ensure the P1 = P0 ∪ {(±1, ±1)}.
positive and nondecreasing properties. One of many op-
For n ≥ 2, it is clear that {(0, 0)} ∪ Qbn/2c ∪ Rb(n−1)/2c ⊆
p
tions is h(x) = xr f (x) where
Pn , so it remains to prove the reverse inclusion. For
f (x) = sup{|z−r g(z)| : z ∈ (0, x)}, (x, y) ∈ Pn , note that x2 + y2 ≡ 0 (mod 4); since every
perfect square is congruent to either 0 or 1 modulo 4,
so that x and y must both be even. Consequently, (x/2, y/2) ∈
Pn−2 , so we may appeal to the induction hypothesis to
h(x) p g(x) p
= f (x), = f (x)x−r g(x). conclude.
xr h(x)
We next identify all of the squares with vertices in Pn .
In the following discussion, let (a, b) and (c, d) be two
Solution 2. We argue by contradiction. Assume that
opposite vertices of a square, so that the other two ver-
lim supx→0+ xr |g00 (x)| < ∞, so that there is an M such
tices are
that |g00 (x)| < Mx−r for all x; and that limx→0+ g0 (x) 6= 0,
so that there is an ε0 > 0 and a sequence xn → 0 with
 
a−b+c+d a+b−c+d
|g0 (xn )| > ε0 for all n. ,
2 2
Now let ε > 0 be arbitrary. Since limx→0+ g(x)x−r = 0,
there is a δ > 0 for which |g(x)| < εxr for all x < δ . and
r
0 xn
Choose n sufficiently large that ε2M < xn and xn < δ /2; 
a + b + c − d −a + b + c + d

ε0 xnr , .
then xn + 2M < 2xn < δ . In addition, we have |g0 (x)| > 2 2
r
0 xn
ε0 /2 for all x ∈ [xn , xn + ε2M ] since |g0 (xn )| > ε0 and
00 −r −r
|g (x)| < Mx ≤ Mxn in this range. It follows that – Suppose that (a, b) = (0, 0). Then (c, d) may be
any element of Pn not contained in P0 . The number
ε02 xnr ε0 xnr of such squares is 4n.
< |g(xn + ) − g(xn )|
2 2M 2M – Suppose that (a, b), (c, d) ∈ Qk for some k. There
ε0 xn r is one such square with vertices
≤ |g(xn + )| + |g(xn )|
2M

ε0 xnr r
 {(0, 2k ), (0, 2−k ), (2k , 0), (2−k , 0)}
< ε (xn + ) + xnr
2M for k = 0, . . . , b n2 c, for a total of b n2 c + 1. To
< ε(1 + 2r )xnr , show that there are no others, by symmetry it suf-
fices to rule out the existence of a square with
whence 4M(1 + 2r )ε > ε02 . Since ε > 0 is arbitrary and opposite vertices (a, 0) and (c, 0) where a > |c|.
M, r, ε0 are fixed, this gives the desired contradiction. The other two vertices of this square would be
Remark. Harm Derksen points out that the “or” in the ((a+c)/2, (a−c)/2) and ((a+c)/2, (−a+c)/2).
problem need not be exclusive. For example, take These cannot belong to any Qk , or be equal to
(0, 0), because |a + c|, |a − c| ≥ a − |c| > 0 by the
triangle inequality. These also cannot belong to
(
x5 sin(x−3 ) x ∈ (0, 1]
g(x) = any Rk because (a+|c|)/2 > (a−|c|)/2. (One can
0 x = 0. also phrase this argument in geometric terms.)
Then for x ∈ (0, 1), – Suppose that (a, b), (c, d) ∈ Rk for some k. There
is one such square with vertices
g0 (x) = 5x4 sin(x−3 ) − 3x cos(x−3 )
{(2k , 2k ), (2k , −2k ), (−2k , 2k ), (−2k , −2k )}
00 3 −3 −3 −3
g (x) = (20x − 9x ) sin(x ) − 18 cos(x ).
for k = 0, . . . , b n−1 n+1
2 c, for a total of b 2 c. To show
For r = 2, limx→0+ x−r g(x) = limx→0+ x3 sin(x−3 ) =
0, that there are no others, we may reduce to the pre-
limx→0+ g0 (x) = 0 and xr g00 (x) = (20x5 − vious case: rotating by an angle of π4 and then
9x−1 ) sin(x−3 ) − 18x2 cos(x−3 ) is unbounded as
x → 0+ . (Note that g0 (x) is not differentiable at x = 0.)
 4
√
rescaling by a factor of 2 would yield a square We then use the estimate
with two opposite vertices in some Qk not cen-
sin x
tered at (0, 0), which we have already ruled out. = 1 + O(x2 ) (x ∈ [0, π])
x
– It remains to show that we cannot have (a, b) ∈
Qk and (c, d) ∈ Rk for some k. By symmetry, we to rewrite the summand as
may reduce to the case where (a, b) = (0, 2k ) and  
(2k−1)π
(c, d) = (2` , ±2` ). If d > 0, then the third vertex 2n
  2 
k
(2k−1 , 2k−1 + 2` ) is impossible. If d < 0, then the 1+O 2
(k+1)π 2 kπ
2 n
 
third vertex (−2k−1 , 2k−1 − 2` ) is impossible. 2n 2n

Summing up, we obtain which simplifies to

 
jnk n+1 8(2k − 1)n3 n
4n + +1+ = 5n + 1 + O .
2 2 k2 (k + 1)2 π 3 k
squares, proving the claim. Consequently,
Remark. Given the computation of Pn , we can alter-
an n−1
  
natively show that the number of squares with vertices 8(2k − 1) 1
in Pn is 5n + 1 as follows. Since this is clearly true for = ∑ k2 (k + 1)2 π 3 + O
n3 k=1 kn2
n = 1, it suffices to show that for n ≥ 2, there are ex-
8 n−1 (2k − 1)
 
actly 5 squares with vertices in Pn , at least one of which log n
= 3 ∑ 2 + O .
is not in Pn−1 . Note that the convex hull of Pn is a square π k=1 k (k + 1)2 n2
S whose four vertices are the four points in Pn \ Pn−1 . If
v is one of these points, then a square with a vertex at Finally, note that
v can only lie in S if its two sides containing v are in
n−1 n−1  
line with the two sides of S containing v. It follows that (2k − 1) 1 1 1
there are exactly two squares with a vertex at v and all ∑ k2 (k + 1)2 = ∑ k2 − (k + 1)2 = 1 − n2
k=1 k=1
vertices in Pn : the square corresponding to S itself, and a
square whose vertex diagonally opposite to v is the ori- converges to 1, and so limn→∞ ann3 = 8
.
gin. Taking the union over the four points in Pn \ Pn−1 π3

gives a total of 5 squares, as desired. B3 Solution 1. We first note that P corresponds to the lin-
8
ear transformation on Rn given by reflection in the hy-
B2 The answer is π3
. perplane perpendicular to u: P(u) = −u, and for any v
Solution 1. By the double angle and sum-product iden- with hu, vi = 0, P(v) = v. In particular, P is an orthogo-
tities for cosine, we have nal matrix of determinant −1.
        We next claim that if Q is an n × n orthogonal matrix
2 (k − 1)π 2 kπ (k − 1)π kπ
2 cos − 2 cos = cos − cos that does not have 1 as an eigenvalue, then det Q =
2n 2n n n (−1)n . To see this, recall that the roots of the char-
   
(2k − 1)π π acteristic polynomial p(t) = det(tI − Q) all lie on the
= 2 sin sin ,
2n 2n unit circle in C, and all non-real roots occur in conju-
gate pairs (p(t) has real coefficients, and orthogonality
and it follows that the summand in an can be written as implies that p(t) = ±t n p(t −1 )). The product of each
  conjugate pair of roots is 1; thus det Q = (−1)k where
1 1 1 k is the multiplicity of −1 as a root of p(t). Since 1 is

−  +
. not a root and all other roots appear in conjugate pairs,
sin 2nπ
cos2 (k−1)π cos2 kπ
2n 2n k and n have the same parity, and so det Q = (−1)n .
Thus the sum telescopes and we find that Finally, if neither of the orthogonal matrices Q nor PQ
has 1 as an eigenvalue, then det Q = det(PQ) = (−1)n ,
contradicting the fact that det P = −1. The result fol-
 
1 1 1 1 lows.
an = π

−1 +   = − π +

3 π

.
sin 2n (n−1)π sin 2n sin 2n
cos2 2n Remark. It can be shown that any n × n orthogonal
matrix Q can be written as a product of at most n hy-
Finally, since limx→0 sinx x = 1, we have perplane reflections (Householder matrices). If equality
π
limn→∞ n sin 2n = 2 , and thus limn→∞ ann3 = π83 .

π occurs, then det(Q) = (−1)n ; if equality does not occur,
then Q has 1 as an eigenvalue. Consequently, equality
Solution 2. We first substitute n − k for k to obtain fails for one of Q and PQ, and that matrix has 1 as an
  eigenvalue.
n−1 sin (2k+1)π
2n
an = ∑  
.
2 (k+1)π
k=1 sin 2n sin2 kπ
2n
 5

Since ln(x) is increasing, ss+1 ln(x) dx is an increasing

R
Sucharit Sarkar suggests the following topological in-
terpretation: an orthogonal matrix without 1 as an function of s, and so it is minimized over s ∈ [1, ∞)
eigenvalue induces a fixed-point-free map from the when s = 1. We conclude that
(n − 1)-sphere to itself, and the degree of such a map Z 2
must be (−1)n . 1 1
m( f ) = + ln(x) dx = 2 ln 2 −
2 1 2
Solution 2. This solution uses the (reverse) Cayley
transform: if Q is an orthogonal matrix not having 1 independent of f .
as an eigenvalue, then
Remark. The phrasing of the question suggests that
A = (I − Q)(I + Q) −1 solvers were not expected to prove that F is nonempty,
even though this is necessary to make the definition of
is a skew-symmetric matrix (that is, AT = −A). m( f ) logically meaningful. Existence will be explicitly
established in the next solution.
Suppose then that Q does not have 1 as an eigenvalue.
Let V be the orthogonal complement of u in Rn . On one Solution 2. We first verify that
hand, for v ∈ V ,
1
f (x, y) = (xy ln(xy) − xy)
−1 −1 2
(I − Q) (I − QP)v = (I − Q) (I − Q)v = v.
is an element of F , by computing that
On the other hand,
1
(I − Q)−1 (I − QP)u = (I − Q)−1 (I + Q)u = Au x fx = y fy = xy ln(xy)
2
and hu, Aui = hAT u, ui = h−Au, ui, so Au ∈ V . Put x2 fxx = y2 fyy = xy.
w = (1 − A)u; then (1 − QP)w = 0, so QP has 1 as an
eigenvalue, and the same for PQ because PQ and QP (See the following remark for motivation for this guess.)
have the same characteristic polynomial. We next show that the only elements of F are f +
Remark. The Cayley transform is the following con- a ln(x/y) + b where a, b are constants. Suppose that
struction: if A is a skew-symmetric matrix, then I + A is f + g is a second element of F . As in the first solu-
invertible and tion, we deduce that gxy = 0; this implies that g(x, y) =
u(x) + v(y) for some twice continuously differentiable
Q = (I − A)(I + A)−1 functions u and v. We also have xgx + ygy = 0, which
now asserts that xgx = −ygy is equal to some constant
is an orthogonal matrix. a. This yields that g = a ln(x/y) + b as desired.
Remark. (by Steven Klee) A related argument is We next observe that
to compute det(PQ − I) using the matrix determinant
lemma: if A is an invertible n × n matrix and v, w are g(s + 1, s + 1) − g(s + 1, s) − g(s, s + 1) + g(s, s) = 0,
1 × n column vectors, then
so m( f ) = m( f + g). It thus remains to compute m( f ).
det(A + vwT ) = det(A)(1 + wT A−1 v). To do this, we verify that

This reduces to the case A = I, in which case it again f (s + 1, s + 1) − f (s + 1, s) − f (s, s + 1) + f (s, s)

comes down to the fact that the product of two square
matrices (in this case, obtained from v and w by padding is nondecreasing in s by computing its derivative to be
with zeroes) retains the same characteristic polynomial ln(s + 1) − ln(s) (either directly or using the integral
when the factors are reversed. representation from the first solution). We thus mini-
mize by taking s = 1 as in the first solution.
B4 Solution 1. We compute that m( f ) = 2 ln 2 − 21 . La- Remark. One way to make a correct guess for f is to
bel the given differential equations by (1) and (2). If notice that the given equations are both symmetric in x
we write, e.g., x ∂∂x (1) for the result of differentiat- and y and posit that f should also be symmetric. Any
ing (1) by x and multiplying the resulting equation by symmetric function of x and y can be written in terms
x, then the combination x ∂∂x (1) + y ∂∂y (1) − (1) − (2) of the variables u = x + y and v = xy, so in principle we
gives the equation 2xy fxy = xy ln(xy)+xy, whence fxy = could translate the equations into those variables and
1 solve. However, before trying this, we observe that xy
2 (ln(x) + ln(y) + 1).
appears explicitly in the equations, so it is reasonable to
Now we observe that
make a first guess of the form f (x, y) = h(xy). For such
f (s + 1, s + 1) − f (s + 1, s) − f (s, s + 1) + f (s, s) a choice, we have

x fx + y fy = 2xyh0 = xy ln(xy)
Z s+1 Z s+1
= fxy dy dx
s s
1 s+1 s+1
Z Z which forces us to set h(t) = 21 (t ln(t) − t).
= (ln(x) + ln(y) + 1) dy dx
2 s s
Z s+1
1
= + ln(x) dx.
2 s
 6

B5 Solution 1. We prove that ( j, k) = (2019, 1010) is a We thus deduce that p(x) = F2019 − F1010 as claimed.
valid solution. More generally, let p(x) be the polyno- Remark. Karl Mahlburg suggests the following variant
mial of degree N such that p(2n + 1) = F2n+1 for 0 ≤ of this. As above, use Lagrange interpolation to write
n ≤ N. We will show that p(2N + 3) = F2N+3 − FN+2 .
1008  
Define a sequence of polynomials p0 (x), . . . , pN (x) by 1009
p0 (x) = p(x) and pk (x) = pk−1 (x) − pk−1 (x + 2) for k ≥ p(2019) = ∑ Fj ;
j=0 j
1. Then by induction on k, it is the case that pk (2n +
1) = F2n+1+k for 0 ≤ n ≤ N − k, and also that pk has it will thus suffice to verify (by substiting j 7→ 1009 − j)
degree (at most) N − k for k ≥ 1. Thus pN (x) = FN+1 that
since pN (1) = FN+1 and pN is constant.
1009  
We now claim that for 0 ≤ k ≤ N, pN−k (2k + 3) = 1009
∑ j Fj+1 = F2019 .
∑kj=0 FN+1+ j . We prove this again by induction on k: j=0
for the induction step, we have
This identity has the following combinatorial interpre-
pN−k (2k + 3) = pN−k (2k + 1) + pN−k+1 (2k + 1) tation. Recall that Fn+1 counts the number of ways to
k−1 tile a 1 × n rectangle with 1 × 1 squares and 1 × 2 domi-
= FN+1+k + ∑ FN+1+ j . noes (see below). In any such tiling with n = 2018, let
j=0 j be the number of squares among the first 1009 tiles.
These can be ordered in 1009


j ways, and the remaining
Thus we have p(2N + 3) = p0 (2N + 3) = ∑Nj=0 FN+1+ j . 2018 − j − 2(1009 − j) = j squares can be tiled in Fj+1
Now one final induction shows that ∑mj=1 Fj = Fm+2 −1, ways.
and so p(2N + 3) = F2N+3 − FN+2 , as claimed. In the As an aside, this interpretation of Fn+1 is the oldest
case N = 1008, we thus have p(2019) = F2019 − F1010 . known interpretation of the Fibonacci sequence, long
Solution 2. This solution uses the Lagrange interpola- predating Fibonacci himself. In ancient Sanskrit, sylla-
tion formula: given x0 , . . . , xn and y0 , . . . , yn , the unique bles were classified as long or short, and a long syllable
polynomial P of degree at most n satisfying P(xi ) = yi was considered to be twice as long as a short syllable;
for i = 0, . . . , n is consequently, the number of syllable patterns of total
length n equals Fn+1 .
n
x−xj Remark. It is not difficult to show that the solution
∑ P(xi ) ∏ xi − x j =
i=0 j6=i ( j, k) = (2019, 2010) is unique (in positive integers).
First, note that to have Fj − Fk > 0, we must have k < j.
Write If j < 2019, then
√ √
1 1+ 5 1− 5 F2019 − F1010 = F2018 + F2017 − F1010 > Fj > Fj − Fk .
Fn = √ (α n − β −n ), α= ,β = .
5 2 2
If j > 2020, then
For γ ∈ R, let pγ (x) be the unique polynomial of degree
at most 1008 satisfying Fj − Fk ≥ Fj − Fj−1 = Fj−2 ≥ F2019 > F2019 − F1010 .

p1 (2n + 1) = γ 2n+1 , p2 (2n + 1) = γ 2n+1 (n = 0, . . . , 1008); Since j = 2019 obviously forces k = 1010, the only
other possible solution would be with j = 2020. But
then p(x) = √1 (pα (x) − pβ (x)). then
5
By Lagrange interpolation,
(Fj − Fk ) − (F2019 − F1010 ) = (F2018 − Fk ) + F1010
1008
2019 − (2 j + 1) which is negative for k = 2019 (it equals F1010 − F2017 )
pγ (2019) = ∑ γ 2n+1 ∏
n=0 0≤ j≤1008, j6=n (2n + 1) − (2 j + 1) and positive for k ≤ 2018.
1008
1009 − j B6 Such a set exists for every n. To construct an example,
= ∑ γ 2n+1 ∏ define the function f : Zn → Z/(2n + 1)Z by
n=0 0≤ j≤1008, j6=n n − j
1008  
2n+1 1008−n 1009 f (x1 , . . . , xn ) = x1 + 2x2 + · · · + nxn (mod 2n + 1),
= ∑γ (−1)
n=0 n
then let S be the preimage of 0.
= −γ((γ 2 − 1)1009 − (γ 2 )1009 ).
To check condition (1), note that if p ∈ S and q is a
neighbor of p differing only in coordinate i, then
For γ ∈ {α, β } we have γ2 = γ + 1 and so
f (q) = f (p) ± i ≡ ±i (mod 2n + 1)
pγ (2019) = γ 2019 − γ 1010 .
 7

and so q ∈
/ S. Remark. According to Art of Problem Solving (thread
To check condition (2), note that if p ∈ Znis not in S, c6h366290), this problem was a 1985 IMO submission
then there exists a unique choice of i ∈ {1, . . . , n} such from Czechoslovakia. For an application to steganog-
that f (p) is congruent to one of +i or −i modulo 2n + raphy, see: J. Fridrich and P. Lisoněk, Grid colorings in
1. The unique neighbor q of p in S is then obtained steganography, IEEE Transactions on Information The-
by either subtracting 1 from, or adding 1 to, the i-th ory 53 (2007), 1547–1549.
coordinate of p.
 The 81st William Lowell Putnam Mathematical Competition
Saturday, February 20, 2021

A1 How many positive integers N satisfy all of the follow- B1 For a positive integer n, define d(n) to be the sum of the
ing three conditions? digits of n when written in binary (for example, d(13) =
1 + 1 + 0 + 1 = 3). Let
(i) N is divisible by 2020.
2020
(ii) N has at most 2020 decimal digits.
(iii) The decimal digits of N are a string of consecutive
S= ∑ (−1)d(k) k3 .
k=1
ones followed by a string of consecutive zeros.
Determine S modulo 2020.
A2 Let k be a nonnegative integer. Evaluate
B2 Let k and n be integers with 1 ≤ k < n. Alice and Bob
k   play a game with k pegs in a line of n holes. At the
k− j k + j
∑ 2 . beginning of the game, the pegs occupy the k leftmost
j=0 j
holes. A legal move consists of moving a single peg to
any vacant hole that is further to the right. The play-
A3 Let a0 = π/2, and let an = sin(an−1 ) for n ≥ 1. Deter- ers alternate moves, with Alice playing first. The game
mine whether ends when the pegs are in the k rightmost holes, so who-
∞ ever is next to play cannot move and therefore loses.
∑ a2n For what values of n and k does Alice have a winning
n=1 strategy?
converges. B3 Let x0 = 1, and let δ be some constant satisfying 0 <
δ < 1. Iteratively, for n = 0, 1, 2, . . . , a point xn+1 is
A4 Consider a horizontal strip of N +2 squares in which the
chosen uniformly from the interval [0, xn ]. Let Z be the
first and the last square are black and the remaining N
smallest value of n for which xn < δ . Find the expected
squares are all white. Choose a white square uniformly
value of Z, as a function of δ .
at random, choose one of its two neighbors with equal
probability, and color this neighboring square black if it B4 Let n be a positive integer, and let Vn be the set of inte-
is not already black. Repeat this process until all the re- ger (2n + 1)-tuples v = (s0 , s1 , · · · , s2n−1 , s2n ) for which
maining white squares have only black neighbors. Let s0 = s2n = 0 and |s j − s j−1 | = 1 for j = 1, 2, · · · , 2n. De-
w(N) be the expected number of white squares remain- fine
ing. Find
2n−1
w(N) q(v) = 1 + ∑ 3s j ,
lim . j=1
N→∞ N
1
and let M(n) be the average of over all v ∈ Vn . Eval-
A5 Let an be the number of sets S of positive integers for q(v)
which uate M(2020).
B5 For j ∈ {1, 2, 3, 4}, let z j be a complex number with
∑ Fk = n, |z j | = 1 and z j 6= 1. Prove that
k∈S

where the Fibonacci sequence (Fk )k≥1 satisfies Fk+2 = 3 − z1 − z2 − z3 − z4 + z1 z2 z3 z4 6= 0.

Fk+1 +Fk and begins F1 = 1, F2 = 1, F3 = 2, F4 = 3. Find
the largest integer n such that an = 2020. B6 Let n be a positive integer. Prove that
A6 For a positive integer N, let fN 1 be the function defined n √
by ∑ (−1)bk( 2−1)c
≥ 0.
k=1
N
N + 1/2 − n (As usual, bxc denotes the greatest integer less than or
fN (x) = ∑ (N + 1)(2n + 1) sin((2n + 1)x).
n=0 equal to x.)

Determine the smallest constant M such that fN (x) ≤ M

for all N and all real x.

1 Corrected from FN in the source.

 Solutions to the 81st William Lowell Putnam Mathematical Competition
Saturday, February 20, 2021
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A1 The values of N that satisfy (ii) and (iii) are precisely the heads among the first 2k + 1 flips, and there are exactly
numbers of the form N = (10a − 10b )/9 for 0 ≤ b < a ≤ as many outcomes with at most k heads.
2020; this expression represents the integer with a digits Third solution. (by Pankaj Sinha) The sum in question
beginning with a string of 1’s and ending with b 0’s. A in the coefficient of xk in the formal power series
value N of this form is divisible by 2020 = 22 · 5 · 101 if
and only if 10b (10a−b −1) is divisible by each of 32 , 22 · k k
5, and 101. Divisibility by 32 is a trivial condition since ∑ 2k− j (1 + x)k+ j = 2k (1 + x)k ∑ 2− j (1 + x) j
10 ≡ 1 (mod 9). Since 10a−b − 1 is odd, divisibility by j=0 j=0
22 ·5 occurs if and only if b ≥ 2. Finally, since 102 ≡ −1 1 − (1 + x)k+1 /2k+1
(mod 101), we see that 10a−b is congruent to 10, −1, = 2k (1 + x)k
1 − (1 + x)/2
−10, or 1 (mod 101) depending on whether a − b is
congruent to 1, 2, 3, or 0 (mod 4); thus 10a−b − 1 is 2k+1 (1 + x)k − (1 + x)2k+1
=
divisible by 101 if and only if a − b is divisible by 4. 1−x
It follows that we need to count the number of (a, b) = (2 (1 + x)k − (1 + x)2k+1 )(1 + x + · · · ).
k+1

with 2 ≤ b < a ≤ 2020 with 4 | a − b. For given b, there

This evidently equals
are b 2020−b
4 c possible values of a. Thus the answer is
k   k  
504 + 504 + 504 + 503 + 503 + 503 + 503 + · · · + 1 + 1 + 1 + 1 k 2k + 1 1
2 k+1
∑ j − ∑ j = 2k+1 (2k ) − 2 22k+1
= 4(504 + 503 + · · · + 1) − 504 = 504 · 1009 = 508536. j=0 j=0

= 22k+1 − 22k = 22k = 4k .

A2 The answer is 4k .
First solution. Let Sk denote the given sum. Then, with Remark. This sum belongs to a general class that
the convention that −1n

= 0 for any n ≥ 0, we have for can be evaluated mechanically using the WZ method.
k ≥ 1, See for example the book A = B by Petvoksek–Wilf–
Zeilberger.
k    
k−1+ j k−1+ j
Sk = ∑ 2k− j j
+
j−1
A3 The series diverges. First note that since sin(x) < x for
j=0 all x > 0, the sequence {an } is positive and decreas-
k−1     k   ing, with a1 = 1. Next, we observe that for x ∈ [0, 1],
k−1− j k − 1 + j 2k − 1 k− j k − 1 + j
=2∑2 + +∑2 sin(x) ≥ x − x3 /6: this follows from Taylor’s theorem
j=0 j k j=1 j−1 with remainder, since sin(x) = x − x3 /6 + (sin c)x4 /24

2k − 1
 k−1 
k+ j
 for some c between 0 and x.
= 2Sk−1 + + ∑ 2k− j−1 √
k j We now claim that an ≥ 1/ n for all n ≥ 1; it follows
j=0
that ∑ a2n diverges since ∑ 1/n diverges. To prove the
= 2Sk−1 + Sk /2 claim, we induct on √ n, with n = 1 being trivial.√ Sup-
pose that an ≥ 1/ n. To prove √ sin(a n ) ≥ 1/ n + 1,
and so Sk = 4Sk−1 . Since S0 = 1, it follows that Sk = 4k note that since sin(an ) ≥ sin(1/ n), it suffices to prove
for all k. √
that x − x3 /6 ≥ (n + 1)−1/2 where x = 1/ n. Squar-
Second solution. Consider a sequence of fair coin flips ing both sides and clearing denominators, we find that
a1 , a2 , . . . and define the random variable X to be the this is equivalent to (n + 1)(6n − 1)2 ≥ 36n3 , or 24n2 −
index of the (k + 1)-st occurrence of heads. Then 11n + 1 ≥ 0. But this last inequality is true since
  24n2 − 11n + 1 = (3n − 1)(8n − 1), and the induction
n − 1 −n is complete.
P[X = n] = 2 ;
k
A4 The answer is 1/e. We first establish a recurrence for
writing n = k + j + 1, we may thus rewrite the given w(N). Number the squares 1 to N + 2 from left to right.
sum as There are 2(N − 1) equally likely events leading to the
first new square being colored black: either we choose
22k+1 P[X ≤ 2k + 1]. one of squares 3, . . . , N + 1 and color the square to its
left, or we choose one of squares 2, . . . , N and color the
It now suffices to observe that P[X ≤ 2k + 1] = 21 : we square to its right. Thus the probability of square i being
have X ≤ 2k + 1 if and only if there are at least k + 1
 2

1
the first new square colored black is 2(N−1) if i = 2 or numbers {1, . . . , N + 1} all originally colored white.
i = N + 1 and 1
if 3 ≤ i ≤ N. Once we have changed Choose a permutation π ∈ SN+1 uniformly at random.
N−1
the first square i from white to black, then the strip di- For i = 1, . . . , N + 1 in succession, color π(i) black in
vides into two separate systems, squares 1 through i and case π(i + 1) is currently white (regarding i + 1 mod-
squares i through N + 2, each with first and last square ulo N + 1). After this, the expected number of white
black and the rest white, and we can view the remain- squares remaining is w(N).
ing process as continuing independently for each sys- Remark. Andrew Bernoff reports that this problem
tem. Thus if square i is the first square to change color, was inspired by a similar question of Jordan Ellenberg
the expected number of white squares at the end of the (disseminated via Twitter), which in turn was inspired
process is w(i − 2) + w(N + 1 − i). It follows that by the final question of the 2017 MATHCOUNTS
competition. See http://bit-player.org/2017/
1 counting-your-chickens-before-theyre-pecked
w(N) = (w(0) + w(N − 1))+
2(N − 1) for more discussion.
!
N
1 A5 The answer is n = F4040 − 1. In both solutions, we use
N −1 ∑ (w(i − 2) + w(N + 1 − i)) freely the identity
i=3
1
+ (w(N − 1) + w(0)) F1 + F2 + · · · + Fm−2 = Fm − 1 (1)
2(N − 1)
which follows by a straightforward induction on m. We
and so
also use the directly computed values
(N − 1)w(N) = 2(w(1) + · · · + w(N − 2)) + w(N − 1).
a1 = a2 = 2, a3 = a4 = 3. (2)
If we replace N by N − 1 in this equation and subtract
from the original equation, then we obtain the recur- First solution. (by George Gilbert)
rence We extend the definition of an by setting a0 = 1.
w(N − 2) Lemma 1. For m > 0 and Fm ≤ n < Fm+1 ,
w(N) = w(N − 1) + .
N −1
an = an−Fm + aFm+1 −n−1 . (3)
(−1)k
We now claim that w(N) = (N + 1) ∑N+1
k=0 k! for N ≥
Proof. Consider a set S for which ∑k∈S Fk = n. If m ∈ S then
0. To prove this, we induct on N. The formula holds for
S \ {m} gives a representation of n − Fm , and this construction
N = 0 and N = 1 by inspection: w(0) = 0 and w(1) = 1.
k is reversible because n − Fm < Fm−1 ≤ Fm . If m ∈ / S, then
Now suppose that N ≥ 2 and w(N − 1) = N ∑Nk=0 (−1)
k! , {1, . . . , m − 1} \ S gives a representation of Fm+1 − n − 1, and
(−1) k
w(N − 2) = (N − 1) ∑N−1 this construction is also reversible. This implies the desired
k=0 k! . Then
equality.
w(N − 2)
w(N) = w(N − 1) + Lemma 2. For m ≥ 2,
N −1
N N−1
 
(−1)k (−1)k m+2
=N ∑ +∑ aFm = aFm+1 −1 = .
2
k=0 k! k=0 k!
N−1
(−1)k N(−1)N Proof. By (2), this holds for m = 2, 3, 4. We now proceed by
= (N + 1) ∑ +
k=0 k! N! induction; for m ≥ 5, given all preceding cases, we have by
N+1 Lemma 1 that
(−1)k
= (N + 1) ∑ jmk m+2
k!
k=0 aFm = a0 + aFm−1 −1 = 1 + =
2 2
and the induction is complete. aFm+1 −1 = aFm−1 −1 + a0 = aFm .
Finally, we compute that
Using Lemma 2, we see that an = 2020 for n = F4040 −
w(N) w(N) 1.
lim = lim
N→∞ N N→∞ N + 1
∞
(−1)k 1 Lemma 3. For Fm ≤ n < Fm+1 , an ≥ aFm .
=∑ = .
k=0 k! e
Proof. We again induct on m. By Lemma 2, we may assume
that
Remark. AoPS user pieater314159 suggests the fol-
lowing alternate description of w(N). Consider the 1 ≤ n − Fm ≤ (Fm+1 − 2) − Fm = Fm−1 − 2. (4)
 3

By (2), we may also assume n ≥ 6, so that m ≥ 5. We apply For a = 1, . . . , k and b = 0, . . . , b(` − 1)/2c, we can re-
Lemma 1, keeping in mind that place tk,` with the string of the same length

(n − Fm ) + (Fm+1 − n − 1) = Fm−1 − 1. (10)k−a (0)(1)2a−1 (01)b 10`−2b

If max{n − Fm , Fm+1 − n − 1} ≥ Fm−2 , then one of the sum- to obtain a new bitstring corresponding to a set S with
mands in (5) is at least aFm−2 (by the induction hypothesis) ∑k∈S Fk = n. Consequently,
and the other is at least 2 (by (4) and the induction hypothe-
sis), so r   
`i + 1
an ≥ ∏ 1 + ki . (5)
  i=1 2
m+4
an ≥ aFm−2 + 2 = .
2
For integers k, ` ≥ 1, we have
Otherwise, min{n − Fm , Fm+1 − n − 1} ≥ Fm−3 and so by the    
`+1 `+1
induction hypothesis again, 1+k ≥ k+ ≥ 2.
2 2
   
m−1 m−2 m+2
an ≥ 2aFm−3 = 2 ≥2 ≥ . Combining this with repeated use of the inequality
2 2 2
xy ≥ x + y (x, y ≥ 2),
Combining Lemma 2 and Lemma 3, we deduce that for
n > F4040 − 1, we have an ≥ aF4040 = 2021. This com- we deduce that
pletes the proof. r 
1 + ∑ri=1 (2ki + `i )
   
`i + 1
Second solution. We again start with a computation of an ≥ ∑ ki + ≥ .
some special values of an . i=1 2 2

Lemma 1. For all m ≥ 1, In particular, for any even m ≥ 2, we have an > m2 for
  all n ≥ Fm . Taking m = 4040 yields the desired result.
m+1
aFm −1 = Remark. It can be shown with a bit more work that
2 the set S0 gives the unique representation of n as a sum
of distinct Virahanka–Fibonacci numbers, no two con-
Proof. We proceed by induction on m. The result holds for secutive; this is commonly called the Zeckendorf repre-
m = 1 and m = 2 by (2). For m > 2, among the sets S counted sentation of n, but was first described by Lekkerkerker.
by aFm −1 , by (1) the only one not containing m − 1 is S = Using this property, one can show that the lower bound
{1, 2, . . . , m − 2}, and there are aFm −Fm−1 −1 others. Therefore, in (5) is sharp.
aFm −1 = aFm −Fm−1 −1 + 1 A6 The smallest constant M is π/4.
   
m−1 m+1 We start from the expression
= aFm−2 −1 + 1 = +1 = .
2 2
N  
1 2 1
fN (x) = ∑2 − sin((2n + 1)x). (6)
Given an arbitrary positive integer n, define the set S0 n=0 2n + 1 N + 1
as follows: start with the largest k1 for which Fk1 ≤ n,
then add the largest k2 for which Fk1 + Fk2 ≤ n, and so Note that if sin(x) > 0, then
on, stopping once ∑k∈S0 Fk = n. Then form the bitstring
N N
1
(
1 k ∈ S0
∑ sin((2n + 1)x) = 2i ∑ (ei(2n+1)x − e−i(2n+1)x )
sn = · · · e1 e0 , ek = n=0 n=0
0 k∈/ S0 ;
!
1 ei(2N+3)x − eix e−i(2N+3)x − e−ix
= −
note that no two 1s in this string are consecutive. We 2i e2ix − 1 e−2ix − 1
can thus divide sn into segments
!
1 ei(2N+2)x − 1 e−i(2N+2)x − 1
= −
tk1 ,`1 · · ·tkr ,`r (ki , `i ≥ 1) 2i eix − e−ix e−ix − eix
1 ei(2N+2)x + e−i(2N+2)x − 2
where the bitstring tk,` is given by =
2i eix − e−ix
tk,` = (10)k (0)` 2 cos((2N + 2)x) − 2
=
2i(2i sin(x))
(that is, k repetitions of 10 followed by ` repetitions of 1 − cos((2N + 2)x)
0). Note that `r ≥ 1 because e1 = e0 = 0. = ≥ 0.
2 sin(x)
 4

We use this to compare the expressions of fN (x) and and the limit is equal to π/4. We conclude that fN (x) ≤
fN+1 (x) given by (6). For x ∈ (0, π) with sin((2N + M holds for M = π/4 but not for any smaller M, as
3)x) ≥ 0, we may omit the summand n = N + 1 from desired.
fN+1 (x) to obtain Remark. It is also possible to replace the use of the
convergence of the Fourier series with a more direct ar-
fN+1 (x) − fN (x)
  N gument; it is sufficient to do this for x in a dense subset
1 1 1 of (0, π), such as the rational multiples of π.
≥ − ∑ sin((2n + 1)x) ≥ 0.
2 N + 1 N + 2 n=0 Another alternative (described at https:
//how-did-i-get-here.com/2020-putnam-a6/)
For x ∈ (0, π) with sin((2N + 3)x) ≤ 0, we may insert is to deduce from (7) and a second geometric series
the summand n = N + 1 into fN+1 (x) to obtain computation (omitted here) that
fN+1 (x) − fN (x) N
d

1 − cos((2N + 2)x)

  N+1 fN0 (x) = ∑ cos((2n + 1)x) −
1 1 1 n=0 dx 4(N + 1) sin(x)
≥ − ∑ sin((2n + 1)x) ≥ 0.
2 N + 1 N + 2 n=0 sin((2N + 2)x)
=
2 sin(x)
In either case, we deduce that for x ∈ (0, π), the se-
(2N + 2) sin((2N + 2)x) − cos(x)(1 − cos((N + 2)x)
quence { fN (x)}N is nondecreasing. −
4(N + 1) sin(x)2
Now rewrite (6) as
cos(x)(1 − cos((N + 2)x)
= ,
N
sin((2n + 1)x) 1 − cos((2N + 2)x) 4(N + 1) sin(x)2
fN (x) = ∑ − (7)
2n + 1 4(N + 1) sin(x)
n=0 which is nonnegative for x ∈ (0, π/2] and nonpositive
for x ∈ [π/2, π). This implies that fN (x) always has a
and note that the last term tends to 0 as N → ∞. Conse-
global maximum at x = π/2, so it suffices to check the
quently, limN→∞ fN (x) equals the sum of the series
convergence of the Fourier series for the square wave
∞
1 at that point. This reduces to the Madhava–Gregory–
∑ 2n + 1 sin((2n + 1)x), Newton series evaluation
n=0
1 1 1 π
which is the Fourier series for the “square wave” func- 1− + − + · · · = arctan(1) = .
3 5 7 4
tion defined on (−π, π] by
 B1 Note that
π
− 4 x ∈ (−π, 0)

2047
x 7→ π4 x ∈ (0, π) (1 − x)(1 − x2 )(1 − x4 ) · · · (1 − x1024 ) =
 ∑ (−1)d(k) xk
0 x = 0, π k=0

and extended periodically. Since this function is contin- and

uous on (0, π), we deduce that the Fourier series con- 2047
verges to the value of the function; that is, x2016 (1 − x)(1 − x2 ) · · · (1 − x16 ) = ∑ (−1)d(k) xk .
k=2016
π
lim fN (x) = (x ∈ (0, π)). d
N→∞ 4 Applying x dx to both sides of each of these two equa-
tions three times, and then setting x = 1, shows that
This is enough to deduce the desired result as follows.
Since 2047 2047

fN (x + 2π) = fN (x), fN (−x) = − fN (x),

∑ (−1)d(k) k3 = ∑ (−1)d(k) k3 = 0,
k=0 k=2016

it suffices to check the bound fN (x) ≤ π for x ∈ (−π, π]. and therefore
For x = 0, π we have fN (x) = 0 for all N. For x ∈
2015
(−π, 0), the previous arguments imply that
∑ (−1)d(k) k3 = 0.
k=1
0 ≥ f0 (x) ≥ f1 (x) ≥ · · ·

For x ∈ (0, π), the previous arguments imply that

π
0 ≤ f0 (x) ≤ f1 (x) ≤ · · · ≤
4
 5

Hence we may write [0, 1] and noting that this sends δ to δ /c, we see that
this latter expected value is equal to f (δ /c). That is,
2020
for c ≥ δ , g(δ , c) = 1 + f (δ /c). It follows that we have
S= ∑ (−1)d(k) k3
k=2016 Z 1
4 f (δ ) = g(δ , c) dc
= ∑ (−1)d(k) (k + 2016)3 0
Z 1 Z 1
k=0
=δ+ (1 + f (δ /c)) dc = 1 + f (δ /c) dc.
≡ (−4)3 + (−1)(−3)3 + (−1)(−2)3 + (1)(−1)3 δ δ

= −64 + 27 + 8 − 1 Now define h : [1, ∞) → R by h(x) = f (1/x); then we

≡ −30 ≡ 1990 (mod 2020). have
Z 1 Z x
1
Remark. The function d(n) appears in the OEIS as h(x) = 1 + h(cx) dc = 1 + h(c) dc.
sequence A000120. 1/x x 1

Rewriting this as xh(x) − x = 1x h(c) dc and differenti-

R
B2 We refer to this two-player game, with n holes and k
pegs, as the (n, k)-game. We will show that Alice has ating with respect to x gives h(x) + xh0 (x) − 1 = h(x),
a winning strategy for the (n, k)-game if and only if at whence h0 (x) = 1/x and so h(x) = log(x) +C for some
least one of n and k is odd; otherwise Bob has a winning constant C. Since h(1) = f (1) = 1, we conclude that
strategy. C = 1, h(x) = 1 + log(x), and finally f (δ ) = 1 − log(δ ).
This gives the claimed answer.
We reduce the first claim to the second as follows. If
n and k are both odd, then Alice can move the k-th peg 1
B4 The answer is 4040 . We will show the following more
to the last hole; this renders the last hole, and the peg general fact. Let a be any nonzero number and define
in it, totally out of play, thus reducing the (n, k)-game q(v) = 1 + ∑2n−1 sj 1
j=1 a ; then the average of q(v) over all
to the (n − 1, k − 1)-game, for which Alice now has a 1
v ∈ Vn is equal to 2n , independent of a.
winning strategy by the second claim. Similarly, if n is
odd but k is even, then Alice may move the first peg to Let Wn denote the set of (2n)-tuples w = (w1 , . . . , w2n )
the (k +1)-st hole, removing the first hole from play and such that n of the wi ’s are equal to +1 and the other
reducing the (n, k)-game to the (n − 1, k) game. Finally, n are equal to −1. Define a map φ : Wn → Wn by
if n is even but k is odd, then Alice can move the first φ (w1 , w2 , . . . , w2n ) = (w2 , . . . , w2n , w1 ); that is, φ moves
peg to the last hole, taking the first and last holes, and the first entry to the end. For w ∈ Wn , define the orbit
the peg in the last hole, out of play, and reducing the of w to be the collection of elements of Wn of the form
(n, k)-game to the (n − 2, k − 1)-game. φ k (w), k ≥ 1, where φ k denotes the k-th iterate of φ ,
and note that φ 2n (w) = w. Then Wn is a disjoint union
We now assume n and k are both even and describe a
of orbits. For a given w ∈ Wn , the orbit of w consists
winning strategy for the (n, k)-game for Bob. Subdivide
of w, φ (w), . . . , φ m−1 (w), where m is the smallest posi-
the n holes into n/2 disjoint pairs of adjacent holes. Call
tive integer with φ m (w) = w; the list φ (w), . . . , φ 2n (w)
a configuration of k pegs good if for each pair of holes,
runs through the orbit of w completely 2n/m times, with
both or neither is occupied by pegs, and note that the
each element of the orbit appearing the same number of
starting position is good. Bob can ensure that after each
times.
of his moves, he leaves Alice with a good configuration:
presented with a good configuration, Alice must move Now define the map f : Wn → Vn by f (w) = v =
j
a peg from a pair of occupied holes to a hole in an un- (s0 , . . . , s2n ) with s j = ∑i=1 wi ; this is a one-to-one cor-
occupied pair; then Bob can move the other peg from respondence between Wn and Vn , with the inverse map
the first pair to the remaining hole in the second pair, given by w j = s j − s j−1 for j = 1, . . . , 2n. We claim that
resulting in another good configuration. In particular, 1
for any w ∈ Wn , the average of q(v) , where v runs over
this ensures that Bob always has a move to make. Since vectors in the image of the orbit of w under f , is equal
the game must terminate, this is a winning strategy for to 2n1
. Since Wn is a disjoint union of orbits, Vn is a dis-
Bob. joint union of the images of these orbits under f , and it
1
B3 Let f (δ ) denote the desired expected value of Z as a then follows that the overall average of q(v) over v ∈ Vn
1
function of δ . We prove that f (δ ) = 1 − log(δ ), where is 2n .
log denotes natural logarithm. To prove the claim, we compute the average of
1
For c ∈ [0, 1], let g(δ , c) denote the expected value of q( f (φ k (w)))
over k = 1, . . . , 2n; since φ k (w) for k =
Z given that x1 = c, and note that f (δ ) = 01 g(δ , c) dc.
R
1, . . . , 2n runs over the orbit of w with each element in
Clearly g(δ , c) = 1 if c < δ . On the other hand, if c ≥ the orbit appearing equally, this is equal to the desired
δ , then g(δ , c) is 1 more than the expected value of Z average. Now if we adopt the convention that the in-
would be if we used the initial condition x0 = c rather dices in wi are considered mod 2n, so that w2n+i = wi
than x0 = 1. By rescaling the interval [0, c] linearly to for all i, then the i-th entry of φ k (w) is wi+k ; we can
 6

j
then define s j = ∑i=1 wi for all j ≥ 1, and s2n+i = si for Lemma 1. Let z1 , z2 , z3 be three distinct complex numbers
2n
all i since ∑i=1 wi = 0. We now have with |z j | = 1 and z1 + z2 + z3 ∈ [0, +∞). Then there exist an-
other three complex numbers z01 , z02 , z03 , not all distinct, with
2n j 2n 2n |z0j | = 1 and
q( f (φ k (w))) = ∑ a∑i=1 wi+k = ∑ as j+k −sk = a−sk ∑ as j .
j=1 j=1 j=1
z01 + z02 + z03 ∈ (z1 + z2 + z3 , +∞), z1 z2 z3 = z01 z02 z03 .

Thus Proof. Write z j = eiθ j for j = 1, 2, 3. We are then trying to

2n 2n maximize the target function
1 ask
∑ q( f (φ k (w))) = ∑ ∑2n sj
= 1,
k=1 k=1 j=1 a cos θ1 + cos θ2 + cos θ3
1 1 given the constraints
and the average of q( f (φ k (w)))
over k = 1, . . . , 2n is 2n ,
as desired.
0 = sin θ1 + sin θ2 + sin θ3
B5 First solution. (by Mitja Mastnak) It will suffice to ∗ = θ1 + θ2 + θ3
show that for any z1 , z2 , z3 , z4 ∈ C of modulus 1 such
that |3 − z1 − z2 − z3 − z4 | = |z1 z2 z3 z4 |, at least one of Since z1 , z2 , z3 run over a compact region without boundary,
z1 , z2 , z3 is equal to 1. the maximum must be achieved at a point where the matrix
To this end, let z1 = eαi , z2 = eβ i , z3 = eγi and  
sin θ1 sin θ2 sin θ3
cos θ1 cos θ2 cos θ3 
f (α, β , γ) = |3 − z1 − z2 − z3 |2 − |1 − z1 z2 z3 |2 .
1 1 1
A routine calculation shows that
is singular. Since the determinant of this matrix computes (up
f (α, β , γ) = 10 − 6 cos(α) − 6 cos(β ) − 6 cos(γ) to a sign and a factor of 2) the area of the triangle with ver-
tices z1 , z2 , z3 , it cannot vanish unless some two of z1 , z2 , z3 are
+ 2 cos(α + β + γ) + 2 cos(α − β )
equal. This proves the claim.
+ 2 cos(β − γ) + 2 cos(γ − α).
For n a positive integer, let Hn be the hypocycloid curve
Since the function f is continuously differentiable, and
in C given by
periodic in each variable, f has a maximum and a min-
imum and it attains these values only at points where Hn = {(n − 1)z + z−n+1 : z ∈ C, |z| = 1}.
∇ f = (0, 0, 0). A routine calculation now shows that
In geometric terms, Hn is the curve traced out by a
∂f ∂f ∂f marked point on a circle of radius 1 rolling one full cir-
+ + = 6(sin(α) + sin(β ) + sin(γ) − sin(α + β + γ))
∂α ∂β ∂γ cuit along the interior of a circle of radius 1, starting
     
α +β β +γ γ +α from the point z = 1. Note that the interior of Hn is not
= 24 sin sin sin . convex, but it is star-shaped: it is closed under multi-
2 2 2
plication by any number in [0, 1].
Hence every critical point of f must satisfy one of
z1 z2 = 1, z2 z3 = 1, or z3 z1 = 1. By symmetry, let us Lemma 2. For n a positive integer, let Sn be the set of complex
assume that z1 z2 = 1. Then numbers of the form w1 + · · · + wn for some w1 , . . . , wn ∈ C
with |w j | = 1 and w1 · · · wn = 1. Then for n ≤ 4, Sn is the
f = |3 − 2Re(z1 ) − z3 |2 − |1 − z3 |2 ; closed interior of Hn (i.e., including the boundary).

since 3 − 2Re(z1 ) ≥ 1, f is nonnegative and can be zero Proof. By considering n-tuples of the form (z, . . . , z, z−n+1 ),
only if the real part of z1 , and hence also z1 itself, is we see that Hn ⊆ Sn . It thus remains to check that Sn lies
equal to 1. in the closed interior of Hn . We ignore the easy cases n = 1
Remark. If z1 = 1, we may then apply the same logic to (where H1 = S1 = {1}) and n = 2 (where H2 = S2 = [−2, 2])
deduce that one of z2 , z3 , z4 is equal to 1. If z1 = z2 = 1, and assume hereafter that n ≥ 3.
we may factor the expression By Lemma 1, for each ray emanating from the the origin, the
extreme intersection point of Sn with this ray (which exists be-
3 − z1 − z2 − z3 − z4 + z1 z2 z3 z4 cause Sn is compact) is achieved by some tuple (w1 , . . . , wn )
with at most two distinct values. For n = 3, this immediately
as (1 − z3 )(1 − z4 ) to deduce that at least three of implies that this point lies on Hn . For n = 4, we must also
z1 , . . . , z4 are equal to 1. consider tuples consisting of two pairs of equal values; how-
Second solution. We begin with an “unsmoothing” ever, these only give rise to points in [−4, 4], which are indeed
construction. contained in H4 .
 7

Turning to the original problem, consider z1 , . . . , z4 ∈ C Thus for i > 0,

with |z j | = 1 and √ √
ci = 2 + b(i + 1)( 2 − 1)c − bi( 2 − 1)c. (8)
3 − z1 − z2 − z3 − z4 + z1 z2 z3 z4 = 0; √ √
Now note that b(i + 1)( 2 − 1)c − bi( 2 − 1)c is either
we must prove that at least one z j is equal to 1. Let z 1 or 0 depending
√ on whether or √not there is an integer j
be any fourth root of z1 z2 z3 z4 , put w j = z j /z, and put between i( 2 − 1) and (i + 1)( 2 − 1): this condition
√
s = w1 + · · · + w4 . In this notation, we have is equivalent to i < j( 2 + 1) < i + 1. That is, for i > 0,
s = z3 + 3z−1 ,
( √
3 if i = b j( 2 + 1)c for some integer j,
ci = (9)
where s ∈ S4 and z3 + 3z−1 ∈ H4 . That is, s is a bound- 2 otherwise;
ary point of S4 , so in particular it is the extremal point
of S4 on the ray emanating from the origin through s. by inspection, this also holds for i = 0.
By Lemma 1, this implies that w1 , . . . , w4 take at most Now we are asked to prove that
two distinct values. As in the proof of Lemma 2, we
n
distinguish two cases.
∑ (−1)ak ≥ 1 (10)
– If w1 = w2 = w3 , then k=0

w−3 3 −1 for all n ≥ 1. We will prove that if (10) holds for all
1 + 3w1 = z + 3z . n ≤ N, then (10) holds for all n ≤ 4N; since (10) clearly
From the geometric description of Hn , we see that holds for n = 1, this will imply the desired result.
this forces w−1
1 = z and hence z1 = 1. Suppose that (10) holds for n ≤ N. To prove that (10)
– If w1 = w2 and w3 = w4 , then s ∈ [−4, 4] and hence holds for n ≤ 4N, it suffices to show that the partial
s = ±4. This can only be achieved by taking w1 = sums
· · · = w4 = ±1; since s = z3 + 3z−1 we must also m
have z = ±1, yielding z1 = · · · = z4 = 1. ∑ (−1)i ci
i=0
Remark. With slightly more work, one can show that
Lemma 2 remains true for all positive integers n. The of the sequence {(−1)ak } are positive for all m such
missing extra step is to check that for m = 1, . . . , n − 1, that c0 + · · · + cm−1 < 4N + 3, since these partial sums
the hypocycloid curve cover all clusters through a4N . Now if c0 + · · · + cm−1 <
4N + 3, then since each ci is at least 2, we must have
{mzn−m + (n − m)z−m : z ∈ C, |z| = 1} m < 2N + 2. From (9), we see that if m is odd, then
m m
is contained in the filled interior of Hn . In fact, this
curve only touches Hn at points where they both touch ∑ (−1)i ci = ∑ (−1)i (ci − 2)
i=0 i=0
the unit circle (i.e., at d-th roots of unity for d = √
gcd(m, n)); this can be used to formulate a correspond- = ∑(−1)b j( 2+1)c
= ∑(−1)a j
ing version of the original problem, which we leave to j j
the reader.
where
√ the sum in j is over √ nonnegative integers
√ j with
B6 First {ak }∞
by ak = j( 2 + 1) < m, i.e., j < m( 2 − 1); since m( 2 − 1) <
√ solution. Define the sequence k=0
m/2 < N + 1, ∑ j (−1)a j is positive by the induction hy-
bk( 2 − 1)c. The first few terms of the sequence
{(−1)ak } are pothesis. Similarly, if m is even, then ∑m i
i=0 (−1) ci =
a
cm + ∑ j (−1) and this is again positive by the induc-
j

1, 1, 1, −1, −1, 1, 1, 1, −1, −1, 1, 1, 1, . . . . tion hypothesis. This concludes the induction step and
the proof.
Define a new sequence {ci }∞ i=0 given by 3, 2, 3, 2, 3, . . ., Remark. More generally,
whose members alternately are the lengths of the clus- √ using the same proof
√ we can
establish the result with 2−1 replaced by n2 + 1−n
ters of consecutive 1’s and the lengths of the clusters of
for any positive integer n.
consecutive −1’s in the sequence {(−1)ak }. Then for
any i, c0 + · · · + ci is Second solution. For n ≥ 0, define the function
√ the number of nonnegative inte-
gers k such that √ bk( 2 − 1)c is strictly less than i + 1, n √
i.e., such that k( 2 − 1) √ < i + 1. This last condition is f (n) = ∑ (−1)bk( 2−1)c

equivalent to k < (i + 1)( 2 + 1), and we conclude that k=1

√ with the convention that f (0) = 0.

c0 + · · · + ci = b(i + 1)( 2 + 1)c + 1
√ Define the sequence q0 , q1 , . . . by the initial conditions
= 2i + 3 + b(i + 1)( 2 − 1)c.
q0 = 0, q1 = 1
 8

and the recurrence relation If j is odd, then

n √
q j = 2q j−1 + q j−2 .
f (n) = f (n − q j ) + ∑ (−1)bk( 2−1)c
k=n−q j +1
This is OEIS sequence A000129; its first few terms are
n
0, 1, 2, 5, 12, 29, 70, . . . . = f (n − q j ) − 2 + ∑ (−1)bkq j−1 /q j c .
k=n−q j +1
Note that q j ≡ j (mod 2).
Since
We now observe that the fractions q j−1 /q j are the
√ con-
vergents of the continued fraction expansion of 2 − 1. b(k + q j )q j−1 /q j c ≡ bkq j−1 /q j c (mod 2),
This implies the following additional properties of the
sequence. we also have
qj
– For all j ≥ 0,
f (n) = f (n − q j ) + ∑ (−1)bkq j−1 /q j c .
q2 j √ q2 j+1 k=1
< 2−1 < .
q2 j+1 q2 j+2 In this sum, the summand indexed by q j contributes 1, and the
summands indexed by k and q j − k cancel each other out for
√ r/s with s < q j + q j+1 such
– There is no fraction k = 1, . . . , q j − 1. We thus have
that rs separates 2 − 1 from q j /q j−1 . In particu-
lar, for k < q j + q j+1 , f (n) = f (n − q j ) + 1
√
 
kq j−1 as claimed.
bk( 2 − 1)c =
qj
From Lemma 1, we have
except when j is even and k ∈ {q j , 2q j }, in which
case they differ by 1. f (q2 j ) = f (q2 j − 2q2 j−1 ) + 2 = f (q2 j−2 ) + 2.

We use this to deduce a “self-similarity” property of By induction on j, f (q2 j ) = 2 j for all j ≥ 0; by simi-
f (n). lar logic, we have f (n) ≤ f (q2 j ) = 2 j for all n ≤ q2 j .
We can now apply Lemma 1 once more to deduce that
Lemma 1. Let n, j be nonnegative integers with q j ≤ n < q j + f (n) ≥ 0 for all j.
q j+1 .
Remark. As a byproduct of the first solution, we con-
(a) If j is even, then firm the equality of two sequences that were entered
separately in the OEIS but conjectured to be equal:
f (n) = f (q j ) − f (n − q j ). A097509 (indexed from 0) matches the definition of
{ci }, while A276862 (indexed from 1) matches the
(b) If j is odd, then characterization of {ci−1 } given by (8).

f (n) = f (n − q j ) + 1. Remark. We can confirm an additional conjecture

from the OEIS by showing that in the notation of the
first solution, the sequence a(n) = cn+1 indexed from
Proof. If j is even, then
1 equals A082844: “Start with 3,2 and apply the rule
n √ a(a(1) + a(2) + · · · + a(n)) = a(n), fill in any unde-
f (n) = f (q j ) + ∑ (−1)bk( 2−1)c
fined terms with a(t) = 2 if a(t − 1) = 3 and a(t) = 3 if
k=q j +1 a(t − 1) = 2.” We first verify the recursion. By (10),
n
= f (q j ) + ∑ (−1)bkq j−1 /q j c + ∗ a(1) + · · · + a(n) = c0 + · · · + cn+1 − c0 − c1
k=q j +1 √
= b(n + 2)( 2 + 1)c − 4.
where ∗ equals 2 if n ≥ 2q j (accounting for the term k = 2q j )
and 0 otherwise. Continuing, From (9), we see that a(a(1) + · · · + a(n) + 3) = 3.
Consequently, exactly one of a(a(1) + · · · + a(n)) or
n−q j a(a(1) + · · · + a(n) + 1) equals 3, and it is the former
f (n) = f (q j ) + ∑ (−1)q j−1 +bkq j−1 /q j c + ∗ if and only if
1
√ √
n−q j √ b(n + 2)( 2 + 1)c − 3 = b(n + 1)( 2 + 1)c,
= f (q j ) − ∑ (−1)q j−1 +bk( 2−1)c
1 i.e., if and only if a(n) = cn+1 = 3.
= f (q j ) − f (n − q j ).
 9

We next check that the definition correctly fills in val- Lemma 3. For j > 0, the sequence c0 , . . . , c j−1 is palindromic
ues not determined by the recursion. If a(n) = 3, then if and only if
a(a(1)+· · ·+a(n)+1) = 2 because no two consecutive
values can both equal 3; by the same token, a(n+1) = 2 j = q2i+1 or j = q2i+1 + q2i+2
and so there are no further values to fill in. If a(n) = 2,
then a(a(1) + · · · + a(n) + 1) = 3 by the previous para- for some nonnegative integer i. (That is, j must belong to one
graph; this in turn implies a(a(1) + · · · + a(n) + 2) = 2, of the sequences A001653 or A001541.) In particular, j must
at which point there are no further values to fill in. be odd.
Remark. We can confirm an additional conjecture from Proof. Let j be an index for which {c0 , . . . , c j−1 } is palin-
the OEIS by showing that in the notation of the first dromic. In particular, c j−1 = c0 = 3, so from (9), we see that
solution, the sequence {ci } equals A245219. This de- √
pends on some additional lemmas. j − 1 = bk( 2 + 1)c for some k. Given this, the sequence is
palindromic if and only if
Lemma 2. Let k be a positive integer. Then √ √ √
n √ o n √ o bi( 2+1)c+b(k −i)( 2+1)c = bk( 2+1)c (i = 0, . . . , k),
i( 2 − 1) < k( 2 − 1) (i = 0, . . . , k − 1)
or equivalently
if and only if k = q2 j or k = q2 j + q2 j−1 for some j > 0. n √ o n √ o n √ o
i( 2 − 1) + (k − i)( 2 − 1) = k( 2 − 1) (i = 0, . . . , k)
Proof. For each j > 0, we have
where the braces denote fractional parts. This holds if and
q2 j−2 q2 j q2 j−1 + 2q2 j−2 √ q2 j+1 q2 j−1 only if
< = < 2−1 < < .
q2 j−1 q2 j+1 q2 j + 2q2 j−1 q2 j+2 q2 j n √ o n √ o
i( 2 − 1) < k( 2 − 1) (i = 0, . . . , k − 1),
We also have
q2 j−2 q2 j q2 j−1 + 2q2 j−2 q2 j−1 + q2 j−2 q2 j−1 so we may apply Lemma 2 to identify k and hence j.
< = < < .
q2 j−1 q2 j+1 q2 j + 2q2 j−1 q2 j + q2 j−1 q2 j Lemma 4. For j > 0, if there exists a positive integer k such
q2 j−1 +q2 j−2 √ that
Moreover, cannot be less than 2 − 1, or else it
q2 j +q2 j−1
√ (c0 , . . . , c j−2 ) = (ck , . . . , ck+ j−2 ) but c j−1 6= ck+ j−1 ,
would be a better approximation to 2 − 1 than the conver-
gent q2 j /q2 j+1 with q2 j+1 > q2 j + q2 j−1 . By the same token, then
q2 j−1 +q2 j−2 √
q2 j +q2 j−1 cannot be a better approximation to 2 − 1 than
q2 j+1 /q2 j+2 . We thus have j = q2i+1 or j = q2i+1 + q2i+2

q2 j √ q2 j+1 q2 j−1 + q2 j−2 q2 j−1 for some nonnegative integer i. In particular, j is odd and (by
< 2−1 < < < . Lemma 3) the sequence (c0 , . . . , c j−1 ) is palindromic.
q2 j+1 q2 j+2 q2 j + q2 j−1 q2 j

From this, we see that Proof. Since the sequence {ci } consists of 2s and 3s, we
√ √ √ must have {c j−1 , ck+ j−1 } = {2, 3}. Since each pair of 3s is
{q2 j ( 2−1)} < {(q2 j +q2 j−1 )( 2−1)} < {q2 j+2 ( 2−1)}. separated by either one or two 2s, we must have c j−2 = 2,
c j−3 = 3. In particular,
√ by (9) there is an integer i for which
It will now suffice to show that for q2 j < k < q2 j + q2 j−1 , j − 3 =√b(i − 1)( 2 + 1)c; there is also an integer l such that
√ √ k = bl( 2 + 1)c. By hypothesis, we have
{k( 2 − 1)} < {q2 j ( 2 − 1)}
√ √ √
b(h + l)( 2 + 1)c = bh( 2 + 1)c + bl( 2 + 1)c
while for q2 j + q2 j−1 < k < q2 j+2 ,
√ √ for h = 0, . . . , i − 1 but not for h = i. In other words,
{k( 2 − 1)} < {(q2 j + q2 j−1 )( 2 − 1)}.
n √ o n √ o n √ o
The first of these assertion is an immediate consequence of the (h + l)( 2 − 1) = h( 2 − 1) + l( 2 − 1)
“best approximation” property of the convergent q2 j−1 /q2 j . √
As for the second assertion, note that for k in this range, no for h = 0, . . . , i − 1 but not for h√= i. That is, {h( 2 − 1)}
q j
fraction with denominator k can lie strictly between q2 2j+1 and belongs to the interval (0, 1 − {l( 2 − 1)}) for h = 0, . . . , i − 1
q2 j−1 +q2 j−2 but not for h = i; in particular,
because these fractions are consecutive terms in a
q2 j +q2 j−1
n √ o n √
Farey sequence (that is, their difference has numerator 1 in
o
h( 2 − 1) < i( 2 − 1) (h = 0, . . . , i − 1),
lowest terms); in particular,
√ such a fraction cannot be a better
q +q2 j−2
upper approximation to 2 − 1 than 2q2j−1j +q2 j−1
. so we may apply Lemma 2 to identify i and hence j.
 10

The sequence A245219 is defined as the sequence of follows.√ It is enough to prove that K ≥ bi when i =
coefficients of the continued fraction of sup{bi } where b j(2 + 2)c for some integer j.
b1 = 1 and for i > 1,
– If c0 , . . . , c j−1 is palindromic, then Lemma 3 im-
( √ plies that j is odd; that is, the continued fraction
bi + 1 if i = b j 2c for some integer j;
bi+1 = [c j−1 , . . . , c0 ] has odd length. In this case, replac-
1/bi otherwise.
ing the final term c0 = c j−1 by the larger quantity
[c j−1 , c j , . . . ] increases the value of the continued
It is equivalent to take the supremum over values of i
fraction.
for which bi+1 = 1/bi ; by Beatty’s
√ theorem, this occurs
precisely when i = b j(2 + 2)c for some integer j. In – If c0 , . . . , c j−1 is not palindromic, then there is
this case, bi has continued fraction a least integer k ∈ {0, . . . , j − 1} such that ck 6=
c j−1−k . By Lemma 3, the sequence c0 , c1 , . . .
[c j−1 , . . . , c0 ]. has arbitrarily long palindromic initial segments,
so the sequence (c j−1 , . . . , c j−1−k ) also occurs as
Let K be the real number with continued fraction ch , . . . , ch+k for some h > 0. By Lemma 4, k is
[c0 , c1 , . . . ]; we must show that K = sup{bi }. In one even and ck = 3 > 2 = c j−1−k ; hence in the con-
direction, by Lemma 3, there are infinitely many values tinued fraction for bi , replacing the final segment
of i for which [c j−1 , . . . , c0 ] = [c0 , . . . , c j−1 ]; the corre- c j−1−k , . . . , c0 by ck , ck+1 , . . . increases the value.
sponding values bi accumulate at K, so K ≤ sup{bi }.
In the other direction, we show that K ≥ sup{bi } as
 The 82nd William Lowell Putnam Mathematical Competition
Saturday, December 4, 2021

A1 A grasshopper starts at the origin in the coordinate plane B2 Determine the maximum value of the sum
and makes a sequence of hops. Each hop has length 5,
∞
and after each hop the grasshopper is at a point whose n
coordinates are both integers; thus, there are 12 possible
S= ∑ 2n (a1 a2 · · · an )1/n
n=1
locations for the grasshopper after the first hop. What is over all sequences a1 , a2 , a3 , · · · of nonnegative real
the smallest number of hops needed for the grasshopper numbers satisfying
to reach the point (2021, 2021)?
∞
A2 For every positive real number x, let ∑ ak = 1.
k=1
r+1 r+1 1r
g(x) = lim ((x + 1) −x ) .
r→0
B3 Let h(x, y) be a real-valued function that is twice con-
tinuously differentiable throughout R2 , and define
Find limx→∞ g(x)
x .

A3 Determine all positive integers N for which the sphere ρ(x, y) = yhx − xhy .

x2 + y2 + z2 = N Prove or disprove: For any positive constants d and r

with d > r, there is a circle S of radius r whose cen-
has an inscribed regular tetrahedron whose vertices ter is a distance d away from the origin such that the
have integer coordinates. integral of ρ over the interior of S is zero.
A4 Let
B4 Let F0 , F1 , . . . be the sequence of Fibonacci numbers,
ZZ 
1 + 2x2 1 + y2
 with F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2.
I(R) = − dx dy. For m > 2, let Rm be the remainder when the product
x2 +y2 ≤R2 1 + x4 + 6x2 y2 + y4 2 + x4 + y4 Fm −1 k
∏k=1 k is divided by Fm . Prove that Rm is also a Fi-
Find bonacci number.

lim I(R), B5 Say that an n-by-n matrix A = (ai j )1≤i, j≤n with integer
R→∞
entries is very odd if, for every nonempty subset S of
or show that this limit does not exist. {1, 2, . . . , n}, the |S|-by-|S| submatrix (ai j )i, j∈S has odd
determinant. Prove that if A is very odd, then Ak is very
A5 Let A be the set of all integers n such that 1 ≤ n ≤ 2021 odd for every k ≥ 1.
and gcd(n, 2021) = 1. For every nonnegative integer j,
let B6 Given an ordered list of 3N real numbers, we can trim
j it to form a list of N numbers as follows: We divide
S( j) = ∑n . the list into N groups of 3 consecutive numbers, and
n∈A
within each group, discard the highest and lowest num-
Determine all values of j such that S( j) is a multiple of bers, keeping only the median.
2021. Consider generating a random number X by the follow-
A6 Let P(x) be a polynomial whose coefficients are all ei- ing procedure: Start with a list of 32021 numbers, drawn
ther 0 or 1. Suppose that P(x) can be written as a prod- independently and uniformly at random between 0 and
uct of two nonconstant polynomials with integer coeffi- 1. Then trim this list as defined above, leaving a list
cients. Does it follow that P(2) is a composite integer? of 32020 numbers. Then trim again repeatedly until just
one number remains; let X be this number. Let µ be the
B1 Suppose that the plane is tiled with an infinite checker- expected value of |X − 21 |. Show that
board of unit squares. If another unit square is dropped
 2021
on the plane at random with position and orientation in- 1 2
dependent of the checkerboard tiling, what is the prob- µ≥ .
4 3
ability that it does not cover any of the corners of the
squares of the checkerboard?
 Solutions to the 82nd William Lowell Putnam Mathematical Competition
Saturday, December 4, 2021
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A1 The answer is 578. the error term O(x−1 ) is bounded in absolute value by
Each hop corresponds to adding one of the 12 vectors (r + 1)r/x. For x ≥ 1, r ≤ 1 this quantity is bounded
(0, ±5), (±5, 0), (±3, ±4), (±4, ±3) to the position in absolute value by (r + 1)r, independently of x. This
of the grasshopper. Since (2021, 2021) = 288(3, 4) + allows us to continue by interchanging the order of the
288(4, 3) + (0, 5) + (5, 0), the grasshopper can reach limits, obtaining
(2021, 2021) in 288 + 288 + 1 + 1 = 578 hops.
lim lim (r + 1 + O(x−1 ))1/r
On the other hand, let z = x + y denote the sum of the x r→0 x→∞
and y coordinates of the grasshopper, so that it starts at = lim (r + 1)1/r
z = 0 and ends at z = 4042. Each hop changes the sum r→0
of the x and y coordinates of the grasshopper by at most = lim (1 + 1/s)s = e,
s→∞
7, and 4042 > 577 × 7; it follows immediately that the
grasshopper must take more than 577 hops to get from where in the last step we take s = 1/r.
(0, 0) to (2021, 2021).
Third solution. (by Clayton Lungstrum) We first ob-
Remark. This solution implicitly uses the distance serve that
function Z x+1 1/r
r+1 r+1 1/r r
d((x1 , y1 ), (x2 , y2 )) = |x1 − x2 | + |y1 − y2 | ((x + 1) − x ) = (r + 1)u du
x
on the plane, variously called the taxicab metric, the Z x+1
1/r
Manhattan metric, or the L1 -norm (or ℓ1 -norm). = (r + 1) 1/r r
u du .
x
A2 The limit is e.
First solution. By l’Hôpital’s Rule, we have Since limr→0 (r + 1)1/r = e, we deduce that

log((x + 1)r+1 − xr+1 )

Z x+1
1/r
lim
r→0 r g(x) = e lim ur du .
r→0 x
d
= lim log((x + 1)r+1 − xr+1 ) For r > 0, ur is increasing for x ≤ u ≤ x + 1, so
r→0 dr
(x + 1)r+1 log(x + 1) − xr+1 log x Z x+1
= lim xr ≤ ur du ≤ (x + 1)r ;
r→0 (x + 1)r+1 − xr+1
x
= (x + 1) log(x + 1) − x log x,
for r < 0, ur is decreasing for x ≤ u ≤ x + 1, so
where log denotes natural logarithm. It follows that Z x+1
x+1
g(x) = e(x+1) log(x+1)−x log x = (x+1)
xx . Thus xr ≥ ur du ≥ (x + 1)r .
    x  x
g(x) x+1 1
lim = lim · lim 1 + = 1 · e = e. In both cases, we deduce that
x→∞ x x→∞ x x→∞ x
Z x+1
1/r
Second solution. We first write x≤ ur du ≤ x + 1;
x
g(x) ((x + 1)r+1 − xr+1 )1/r
lim = lim lim
x→∞ x x→∞ r→0 x applying the squeeze theorem to the resulting inequality
e ≤ g(x) 1


((r + 1)xr + O(xr−1 ))1/r x ≤ e 1 + x yields the claimed limit.
= lim lim .
x→∞ r→0 x A3 The integers N with this property are those of the form
We would like to interchange the order of the limits, but 3m2 for some positive integer m.
this requires some justification. Using Taylor’s theorem In one direction, for N = 3m2 , the points
with remainder, for x ≥ 1, r ≤ 1 we can bound the error
term O(xr−1 ) in absolute value by (r + 1)rxr−1 . This (m, m, m), (m, −m, −m), (−m, m, −m), (−m, −m, m)
means that if we continue to rewrite the orginial limit as
form the vertices of a regular tetrahedron inscribed in
lim lim (r + 1 + O(x−1 ))1/r , the sphere x2 + y2 + z2 = N.
r→0 x→∞
 2

Conversely, suppose that Pi = (xi , yi , zi ) for i = 1, . . . , 4 may take the limit over R through the integrals to obtain
are the vertices of an inscribed regular tetrahedron. Z √2 Z 2π
Then the center of this tetrahedron must equal the cen- r2 /2
lim I(R) = r dr dθ
ter of the sphere, namely (0, 0, 0). Consequently, these R→∞ 1 0 r4 (1 − (sin2 2θ )/2)
four vertices together with Qi = (−xi , −yi , −zi ) for i = Z √2 Z 2π
dr 1
1, . . . , 4 form the vertices of an inscribed cube in the = dθ
sphere. The side length of this cube is (N/3)1/2 , so its 1 r 0 2 − sin2 2θ
√ Z 2π 1
volume is (N/3)3/2 ; on the other hand, this volume also = log 2 dθ
equals the determinant of the matrix with row vectors 0 1 + cos2 2θ
Z 2π
Q2 − Q1 , Q3 − Q1 , Q4 − Q1 , which is an integer. Hence 1 2
= log 2 dθ .
(N/3)3 is a perfect square, as then is N/3. 2 0 3 + cos 4θ
√ It thus remains to evaluate
2
A4 The limit exists and equals 2 π log 2.
Z 2π
2 2
Z π
dθ = 2 dθ .
We first note that we can interchange x and y to obtain 0 3 + cos 4θ 0 3 + cos θ
One option for this is to use the half-angle substitution
1 + 2y2 1 + x2
ZZ  
I(R) = − dx dy. t = tan(θ /2) to get
4 2 2
1 + x + 6x y + y4 2 + x4 + y4
x2 +y2 ≤R2
4 2
Z ∞ Z ∞
2 2
dt = dt
Averaging the two expressions for I(R) yields −∞ 3(1 + t ) + (1 − t ) 2 + t2
−∞
√ x ∞
 
ZZ = 2 arctan √
I(R) = ( f (x, y) − g(x, y)) dx dy 2 −∞
x2 +y2 ≤R2
√
= 2π.
where Putting this together yields the claimed result.

1 + x2 + y2 A5 The values of j in question are those not divisible by

f (x, y) = either 42 or 46.
1 + x4 + 6x2 y2 + y4
We first check that for p prime,
1 + x2 /2 + y2 /2
g(x, y) = . p−1
2 + x4 + y4
∑ nj ≡ 0 (mod p) ⇔ j ̸≡ 0 (mod p − 1).
n=1
Now note that
If j ≡ 0 (mod p − 1), then n j ≡ 1 (mod p) for each
f (x, y) = 2g(x + y, x − y). p−1 j
n, so ∑n=1 n ≡ p − 1 (mod p). If j ̸≡ 0 (mod p − 1),
we can pick a primitive root m modulo p, observe that
We can thus write m j ̸≡ 1 (mod p), and then note that
ZZ p−1 p−1 p−1
I(R) =
R2 ≤x2 +y2 ≤2R2
g(x, y) dx dy. ∑ n j ≡ ∑ (mn) j = m j ∑ n j (mod p),
n=1 n=1 n=1
p−1 j
To compute this integral, we switch to polar coordi- which is only possible if ∑n=1 n ≡ 0 (mod p).
nates: We now note that the prime factorization of 2021 is 43×
Z R√2 Z 2π 47, so it suffices to determine when S( j) is divisible by
I(R) = g(r cos θ , r sin θ )r dr dθ each of 43 and 47. We have
R 0 42
Z R√2 Z 2π
1 + r2 /2 S( j) ≡ 46 ∑ n j (mod 43)
= r dr dθ . n=1
R 0 2 + r4 (1 − (sin2 2θ )/2)
46
S( j) ≡ 42 ∑ n j (mod 47).
We rescale r to remove the factor of R from the limits n=1
of integration:
Since 46 and 42 are coprime to 43 and 47, respectively,
Z √2 Z 2π we have
1 + R2 r2 /2
I(R) = R2 r dr dθ .
1 0 2 + R4 r4 (1 − (sin2 2θ )/2) S( j) ≡ 0 (mod 43) ⇔ j ̸≡ 0 (mod 42)
S( j) ≡ 0 (mod 47) ⇔ j ̸≡ 0 (mod 46).
Since the integrand is uniformly bounded for R ≫ 0, we This yields the claimed result.
 3

A6 Yes, it follows that P(2) is a composite integer. (Note: B1 The probability is 2 − π6 .

1 is neither prime nor composite.)
Set coordinates so that the original tiling includes the
Write P(x) = a0 + a1 x + · · · + an xn with ai ∈ {0, 1} and (filled) square S = {(x, y) : 0 ≤ x, y ≤ 1}. It is then
an = 1. Let α be an arbitrary root of P. Since P(α) = 0, equivalent to choose the second square by first choosing
α cannot be a positive real number. In addition, if α ̸= 0 a point uniformly at random in S to be the center of the
then square, then choosing an angle of rotation uniformly at
|1 + an−1 α −1 | = |an−2 α −2 + · · · + a0 α −n | random from the interval [0, π/2].
≤ |α|−2 + · · · + |α|−n . For each θ ∈ [0, π/2], circumscribe a square Sθ around
S with angle of rotation θ relative to S; this square has
If α ̸= 0 and Re(α) ≥ 0, then Re(1 + an−1 α −1 ) ≥ 1 and side length sin θ + cos θ . Inside Sθ , draw the smaller
|α|−2 square Sθ′ consisting of points at distance greater than
1 ≤ |α|−2 + · · · + |α|−n < ; 1/2 from each side of Sθ ; this square has side length
1 − |α|−1
√ sin θ + cos θ − 1.
this yields |α| < (1 + 5)/2.
We now verify that a unit square with angle of rotation
By the same token, if α ̸= 0 then θ fails to cover any corners of S if and only if its center
|1 + an−1 α −1 + an−2 α −2 | ≤ |α|−3 + · · · + |α|−n . lies in the interior of Sθ′ . In one direction, if one of the
corners of S is covered, then that corner lies on a side of
We deduce from this that Re(α) ≤ 3/2 as follows. Sθ which meets the dropped square, so the center of the
– There is nothing to check if Re(α) ≤ 0. dropped square is at distance less than 1/2 from that
side of Sθ . To check the converse, note that there are
– If the argument of α belongs to [−π/4, π/4], then
two ways to dissect the square Sθ into the square Sθ′ plus
Re(α −1 ), Re(α −2 ) ≥ 0, so
four sin θ × cos θ rectangles. If θ ̸= 0, π/4, then one of
|α|−3 these dissections has the property that each corner P of
1 ≤ |α|−3 + · · · + |α|−n < . S appears as an interior point of a side (not a corner) of
1 − |α|−1
one of the rectangles R. It will suffice to check that if the
Hence |α|−1 is greater than the unique positive center of the dropped square is in R, then the dropped
root of x3 + x − 1, which is greater than 2/3. square covers P; this follows from the fact that sin θ and
– Otherwise, α has argument in (−π/2, √ π/4) ∪ cos θ are both at most 1.
(π/4, π/2), so the bound √ |α| <√ (1 + 5)/2 im-
It follows that the conditional probability, given that the
plies that Re(α) < (1 + 5)/(2 2) < 3/2. angle of rotation is chosen to be θ , that the dropped
By hypothesis, there exists a factorization P(x) = square does not cover any corners of S is (sin θ +
Q(x)R(x) into two nonconstant integer polynomials, cos θ − 1)2 . We then compute the original probability
which we may assume are monic. Q(x + 3/2) is a prod- as the integral
uct of polynomials, each of the form x − α where α is a
2
Z π/2
real root of P or of the form
   (sin θ + cos θ − 1)2 dθ
3 3 π 0
x+ −α x+ −α
2
Z π/2
2 2
= (2 + sin 2θ − 2 sin θ − 2 cos θ ) dθ
  2 π 0
2 3 3
= x + 2Re −α x+ −α 2

1
π/2
2 2 = 2θ − cos 2θ + 2 cos θ − 2 sin θ
π 2 0
where α is a nonreal root of P. It follows that Q(x +
3/2) has positive coefficients; comparing its values at 2 6
= (π + 1 − 2 − 2) = 2 − .
x = 1/2 and x = −1/2 yields Q(2) > Q(1). We can- π π
not have Q(1) ≤ 0, as otherwise the intermediate value
theorem would imply that Q has a real root in [1, ∞); Remark: Noam Elkies has some pictures illustrating
hence Q(1) ≥ 1 and so Q(2) ≥ 2. Similarly R(2) ≥ 2, this problem: image 1, image 2.
so P(2) = Q(2)R(2) is composite.
Remark. A theorem of Brillhart, Filaseta, and Odlyzko B2 The answer is 2/3.
from 1981 states that if a prime p is written as ∑i ai bi
in any base b ≥ 2, the polynomial ∑i ai xi is irreducible. By AM-GM, we have
(The case b = 10 is an older result of Cohn.) The solu-
tion given above is taken from: Ram Murty, Prime num- 2n+1 (a1 · · · an )1/n = (4a1 )(42 a2 ) · · · (4n an )

1/n
bers and irreducible polynomials, Amer. Math. Monthly
109 (2002), 452–458). The final step is due to Pólya and ∑nk=1 (4k ak )
≤ .
Szegő. n
 4

Thus where the final congruence follows from Wilson’s The-

orem. Now observe that when m is odd, p = Fm must be
∞
∑nk=1 (4k ak ) congruent to either 1 or 2 (mod 4): this follows from
2S ≤ ∑
n=1 4n inspection of the Virahanka–Fibonacci sequence mod
∞ n ∞ ∞ 4, which has period 6: 1, 1, 2, 3, 1, 0, 1, 1, . . .. It follows
= ∑ ∑ (4k−n ak ) = ∑ ∑ (4k−n ak ) that A2 ≡ (−1) p(p+1)/2 = −1.
n=1 k=1 k=1 n=k
∞
On the other hand, by the Kepler–Cassini identity
4ak 4
= ∑ = Fn2 = (−1)n+1 + Fn−1 Fn+1
k=1 3 3
with n = m − 1, we have Fm−1 2 ≡ (−1)m = −1. Thus
and S ≤ 2/3. Equality is achieved when ak = 43k for all 2 2
we have 0 ≡ A − Fm−1 ≡ (A − Fm−1 )(A − Fm−2 ). Since
k, since in this case 4a1 = 42 a2 = · · · = 4n an for all n. p is prime, it must be the case that either A = Fm−1 or
A = Fm−2 , and we are done.
B3 We prove the given statement.
Remark. The Kepler–Cassini identity first appears in a
For any circle S of radius r whose center is at distance letter of Kepler from 1608. Noam Elkies has scanned
d from the origin, express the integral in polar coordi- the relevant page of Kepler’s collected works (slightly
nates s, θ : NSFW if your boss can read Latin).
ZZ Z s2 Z θ2 (s) B5 For convenience, throughout we work with matrices
ρ= (yhx − xhy )(s sin θ , s cos θ )s dθ ds. over the field of 2 elements. In this language, if there
S s1 θ1 (s)
exists a permutation matrix P such that P−1 AP is unipo-
For fixed s, the integral over θ is a line integral of grad h, tent (i.e., has 1s on the main diagonal and 0s below
which evaluates to h(P2 ) − h(P1 ) where P1 , P2 are the it), then A is very odd: any principal submatrix of A is
endpoints of the endpoints of the arc of the circle of ra- conjugate to a principal submatrix of P−1 AP, which is
RR lying within S . If we now
dius s centered at the origin again unipotent and in particular nonsingular. We will
fix r and d and integrate S ρ over all choices of S solve the problem by showing that conversely, for any
(this amounts to a single integral over an angle in the very odd matrix A, there exists a permutation matrix P
range [0, 2π]), we may interchange the order of integra- such that P−1 AP is unipotent. Since the latter condi-
tion to first integrate over θ , then over the choice of S , tion is preserved by taking powers, this will prove the
and at this point desired result.
RR we get 0 for every s. We conclude that
the integral of S over all choices of S vanishes; since To begin, we may take S = {i} to see that aii = 1. We
the given integral varies continuously in S , by the in- next form a (loopless) directed graph on the vertex set
termediate value theorem there must be some S where {1, . . . , n} with an edge from i to j whenever ai j = 1,
the given integral is 0. and claim that this graph has no cycles. To see this,
suppose the contrary, choose a cycle of minimal length
B4 We can check directly that R3 = R4 = 1 are Virahanka– m ≥ 2, and let i1 , . . . , im be the vertices in order. The
Fibonacci numbers; henceforth we will assume m ≥ 5. minimality of the cycle implies that
m −1 k
Denote the product ∏Fk=1 k by A. Note that if Fm is (
composite, say Fm = ab for a, b > 1 integers, then A is 1 if k − j ≡ 0 or 1 (mod m)
ai j ik =
divisible by aa bb and thus by Fm = ab; it follows that 0 otherwise.
Rm = 0 = F0 when Fm is composite.
The submatrix corresponding to S = {i1 , . . . , im } has
Now suppose that Fm is prime. Since F2n = Fn (Fn+1 + row sum 0 and hence is singular, a contradiction.
Fn−1 ) for all n, Fm is composite if m > 4 is even; thus
we must have that m is odd. Write p = Fm , and use ≡ We now proceed by induction on n. Since the directed
to denote congruence (mod p). Then we have graph has no cycles, there must be some vertex which
is not the starting point of any edge (e.g., the endpoint
p−1 p−1 p−1 of any path of maximal length). We may conjugate by
A= ∏ (p − k) p−k ≡ ∏ (−k) p−k = (−1) p(p−1)/2 ∏ k p−k a permutation matrix so that this vertex becomes 1. We
k=1 k=1 k=1 now apply the induction hypothesis to the submatrix
corresponding to S = {2, . . . , n} to conclude.
and consequently
Remark. A directed graph without cycles, as in our
p−1 solution, is commonly called a DAG (directed acyclic
A2 ≡ (−1) p(p−1)/2 ∏ (kk k p−k ) graph). It is a standard fact that a directed graph is a
k=1 TAG if and only if there is a linear ordering of its ver-
= (−1) p(p−1)/2
((p − 1)!) p tices consistent with all edge directions. See for exam-
ple https://en.wikipedia.org/wiki/Directed_
≡ (−1) p(p+1)/2 , acyclic_graph.
 5

Remark. An n × n matrix A = (ai j ) for which the as desired.

value of ai j depends only on i − j (mod n) is called Second solution. Retain notation as in the first solution.
a circulant matrix. The circulant matrix with first row Again Fk ( 12 ) = 12 , so (2) implies
(1, 1, 0, . . . , 0) is an example of an n × n matrix whose
determinant is even, but whose other principal minors  
1
 
1 1 1
are all odd. fk = 6 fk−1 × × .
2 2 2 2
B6 First solution. (based on a suggestion of Noam Elkies)
Let fk (x) be the probability distribution of Xk , the last By induction on k, we deduce that fk ( 21 ) = ( 32 )k and
number remaining when one repeatedly trims a list of fk (x) is nondecreasing on [0, 21 ]. (More precisely, be-
3k random variables chosen with respect to the uniform sides (2), the second assertion uses that Fk−1 (x) in-
R [0, 1]; note that f0 (x) = 1 for x ∈ [0, 1].
distribution on creases from 0 to 1/2 and y 7→ y − y2 is nondecreasing
Let Fk (x) = 0x fk (t) dt be the cumulative distribution on [0, 1/2].)
function; by symmetry, Fk ( 21 ) = 21 . Let µk be the ex- The expected value of |Xk − 12 | equals
pected value of Xk − 12 ; then µ0 = 14 , so it will suffice to
Z 1/2 
prove that µk ≥ 23 µk−1 for k > 0.

1
µk = 2 − x fk (x) dx
By integration by parts and symmetry, we have 0 2
Z 1/2  
1
Z 1/2  Z 1/2
=2 − x dx.

1 x fk
µk = 2 − x fk (x) dx = 2 Fk (x) dx; 0 2
0 2 0
Define the function
that is, µk computes twice the area under the curve
y = Fk (x) for 0 ≤ x ≤ 12 . Since Fk is a monotone func- (
k h i
3 2 k
x ∈ 0, 12


2 3
tion from [0, 12 ] with Fk (0) = 0 and Fk ( 21 ) = 12 , we may gk (x) =
transpose the axes to obtain 0 otherwise.

Note that for x ∈ [0, 1/2] we have

Z 1/2  
1 −1
µk = 2 − Fk (y) dy. (1)
0 2 Z x
(gk (t) − fk (1/2 − t)) dt ≥ 0
Since fk (x) is the probability distribution of the median 0
of three random variables chosen with respect to the dis- with equality at x = 0 or x = 1/2. (On the in-
tribution fk−1 (x), terval [0, (1/2)(2/3)k ] the integrand is nonnegative,
fk (x) = 6 fk−1 (x)Fk−1 (x)(1 − Fk−1 (x)) (2) so the function increases from 0; on the interval
[(1/2)(2/3)k , 1/2] the integrand is nonpositive, so the
or equivalently function decreases to 0.) Hence by integration by parts,

Fk (x) = 3Fk−1 (x)2 − 2Fk−1 (x)3 .

Z 1/2
(3)
µk − 2 xgk (x) dx
0
By induction, Fk is the k-th iterate of F1 (x) = 3x2 − 2x3 , Z 1/2  
so 1
= 2x( fk − x − gk (x)) dx
0 2
Fk (x) = Fk−1 (F1 (x)). (4) Z 1/2 Z x Z x   
2 1
3
= x gk (t) − fk − t dt dt dx ≥ 0.
Since f1 (t) = 6t(1 − t) ≤ for t ∈ [0, 12 ],
2 0 0 0 2
Z 1/2 (This can also be interpreted as an instance of the rear-
 
1 3 1
− F1 (x) = 6t(1 − t) dt ≤ −x ; rangement inequality.)
2 x 2 2
We now see that
for y ∈ [0, 12 ], we may take x = Fk−1 (y) to obtain Z 1/2
1

2 1
 µk ≥ 2 xgk (x) dx
− Fk−1 (y) ≥ −1
− Fk−1 (y) . (5) 0
2 3 2  k Z (1/2)(2/3)k
3
≥2 x dx
Using (4) and (5), we obtain 2 0
k
1 2 (1/2)(2/3)
 k
Z 1/2  3

1
µk = 2 − Fk−1 (y) dy =2
2
x
2 0
0 2
 k  2k
1 2 k
Z 1/2    
4 1 −1 2 3 1 2
≥ − Fk−1 (y) dy = µk−1 =2 =
3 0 2 3 2 8 3 4 3
 6

as desired. 1/2 and the median is

Z 1/2  
Remark. For comparison, if we instead take the me- 2n + 1
2 (1/2 − x)P2n+1 (x)dx = 2−2n−2 .
dian of a list of n numbers, the probability distribution 0 n
is given by
For n = 32021 , using Stirling’s approximation this can be
estimated as 1.13(0.577)2021 < 0.25(0.667)2021 . This
(2n + 1)! n
P2n+1 (x) = x (1 − x)n . shows that the trimming procedure produces a quantity
n!n! that is on average further away from 1/2 than the me-
dian.
The expected value of the absolute difference between
 The 83rd William Lowell Putnam Mathematical Competition
Saturday, December 3, 2022

A1 Determine all ordered pairs of real numbers (a, b) such B2 Let × represent the cross product in R3 . For what posi-
that the line y = ax + b intersects the curve y = ln(1 + tive integers n does there exist a set S ⊂ R3 with exactly
x2 ) in exactly one point. n elements such that

A2 Let n be an integer with n ≥ 2. Over all real polynomials S = {v × w : v, w ∈ S}?

p(x) of degree n, what is the largest possible number of
negative coefficients of p(x)2 ? B3 Assign to each positive real number a color, either red
or blue. Let D be the set of all distances d > 0 such
A3 Let p be a prime number greater than 5. Let f (p) denote that there are two points of the same color at distance
the number of infinite sequences a1 , a2 , a3 , . . . such that d apart. Recolor the positive reals so that the numbers
an ∈ {1, 2, . . . , p − 1} and an an+2 ≡ 1 + an+1 (mod p) in D are red and the numbers not in D are blue. If we
for all n ≥ 1. Prove that f (p) is congruent to 0 or 2 iterate this recoloring process, will we always end up
(mod 5). with all the numbers red after a finite number of steps?
B4 Find all integers n with n ≥ 4 for which there exists a
A4 Suppose that X1 , X2 , . . . are real numbers between 0 and sequence of distinct real numbers x1 , . . . , xn such that
1 that are chosen independently and uniformly at ran- each of the sets
dom. Let S = ∑ki=1 Xi /2i , where k is the least positive
integer such that Xk < Xk+1 , or k = ∞ if there is no such {x1 , x2 , x3 }, {x2 , x3 , x4 }, . . . ,
integer. Find the expected value of S. {xn−2 , xn−1 , xn }, {xn−1 , xn , x1 }, and {xn , x1 , x2 }
A5 Alice and Bob play a game on a board consisting of forms a 3-term arithmetic progression when arranged in
one row of 2022 consecutive squares. They take turns increasing order.
placing tiles that cover two adjacent squares, with Alice
going first. By rule, a tile must not cover a square that B5 For 0 ≤ p ≤ 1/2, let X1 , X2 , . . . be independent random
is already covered by another tile. The game ends when variables such that
no tile can be placed according to this rule. Alice’s goal 
is to maximize the number of uncovered squares when 1
 with probability p,
the game ends; Bob’s goal is to minimize it. What is Xi = −1 with probability p,
the greatest number of uncovered squares that Alice can

0 with probability 1 − 2p,
ensure at the end of the game, no matter how Bob plays?
for all i ≥ 1. Given a positive integer n and integers
A6 Let n be a positive integer. Determine, in terms of n, b, a1 , . . . , an , let P(b, a1 , . . . , an ) denote the probability
the largest integer m with the following property: There that a1 X1 + · · · + an Xn = b. For which values of p is it
exist real numbers x1 , . . . , x2n with −1 < x1 < x2 < · · · < the case that
x2n < 1 such that the sum of the lengths of the n intervals
P(0, a1 , . . . , an ) ≥ P(b, a1 , . . . , an )
[x12k−1 , x22k−1 ], [x32k−1 , x42k−1 ], . . . , [x2n−1
2k−1 2k−1
, x2n ]
for all positive integers n and all integers b, a1 , . . . , an ?
is equal to 1 for all integers k with 1 ≤ k ≤ m. B6 Find all continuous functions f : R+ → R+ such that

B1 Suppose that P(x) = a1 x + a2 x2 + · · · + an xn is a poly- f (x f (y)) + f (y f (x)) = 1 + f (x + y)

nomial with integer coefficients, with a1 odd. Suppose
that eP(x) = b0 + b1 x + b2 x2 + · · · for all x. Prove that bk for all x, y > 0.
is nonzero for all k ≥ 0.
 Solutions to the 83rd William Lowell Putnam Mathematical Competition
Saturday, December 3, 2022
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A1 Write f (x) = ln(1 + x2 ). We show that y = ax + b inter- A2 The answer is 2n−2. Write p(x) = an xn +· · ·+a1 x+a0
sects y = f (x) in exactly one point if and only if (a, b) and p(x)2 = b2n x2n + · · · + b1 x + b0 . Note that b0 = a20
lies in one of the following groups: and b2n = a2n . We claim that not all of the remain-
ing 2n − 1 coefficients b1 , . . . , b2n−1 can be negative,
– a=b=0
whence the largest possible number of negative coef-
– |a| ≥ 1, arbitrary b ficients is ≤ 2n − 2. Indeed, suppose bi < 0 for 1 ≤ i ≤
– 0 < |a| < 1, and b < ln(1 − r− )2 − |a|r− or b > 2n − 1. Since b1 = 2a0 a1 , we have a0 ̸= 0. Assume
ln(1 − r+ )2 − |a|r+ , where a0 > 0 (or else replace p(x) by −p(x)). We claim by
√ induction on i that ai < 0 for 1 ≤ i ≤ n. For i = 1, this
1 ± 1 − a2 follows from 2a0 a1 = b1 < 0. If ai < 0 for 1 ≤ i ≤ k − 1,
r± = .
a then
Since the graph of y = f (x) is symmetric under re- k−1
flection in the y-axis, it suffices to consider the case 2a0 ak = bk − ∑ ai ak−i < bk < 0
i=1
a ≥ 0: y = ax + b and y = −ax + b intersect y = f (x) the
same number of times. For a = 0, by the symmetry of and thus ak < 0, completing the induction step. But now
y = f (x) and the fact that f (x) > 0 for all x ̸= 0 implies b2n−1 = 2an−1 an > 0, contradiction.
that the only line y = b that intersects y = f (x) exactly
It remains to show that there is a polynomial p(x) such
once is the line y = 0.
that p(x)2 has 2n−2 negative coefficients. For example,
We next observe that on [0, ∞), f ′ (x) = 1+x2x
2 increases we may take
′ ′
on [0, 1] from f (0) = 0 to a maximum at f (1) = 1, and
then decreases on [1, ∞) with limx→∞ f ′ (x) = 0. In par- p(x) = n(xn + 1) − 2(xn−1 + · · · + x),
ticular, f ′ (x) ≤ 1 for all x (including x < 0 since then
f ′ (x) < 0) and f ′ (x) achieves each value in (0, 1) ex- so that
actly twice on [0, ∞).
p(x)2 = n2 (x2n + xn + 1) − 2n(xn + 1)(xn−1 + · · · + x)
For a ≥ 1, we claim that any line y = ax + b inter-
sects y = f (x) exactly once. They must intersect at + (xn−1 + · · · + x)2 .
least once by the intermediate value theorem: for x ≪ 0,
For i ∈ {1, . . . , n − 1, n + 1, . . . , n − 1}, the coefficient of
ax + b < 0 < f (x), while for x ≫ 0, ax + b > f (x) since
ln(1+x2 )
xi in p(x)2 is at most −2n (coming from the cross term)
limx→∞ x = 0. On the other hand, they cannot plus −2n + 2 (from expanding (xn−1 + · · · + x)2 ), and
intersect more than once: for a > 1, this follows from hence negative.
the mean value theorem, since f ′ (x) < a for all x. For
a = 1, suppose that they intersect at two points (x0 , y0 ) A3 First solution. We view the sequence a1 , a2 , . . . as
and (x1 , y1 ). Then lying in F×p ⊂ F p . Then the sequence is determined
R x1 ′ by the values of a1 and a2 , via the recurrence an+2 =
y1 − y0 x0 f (x) dx (1 + an+1 )/an . Using this recurrence, we compute
1= = <1
x1 − x0 x1 − x0
1 + a2 1 + a1 + a2
a3 = , a4 = ,
since f ′ (x) is continuous and f ′ (x) ≤ 1 with equality a1 a1 a2
only at one point. 1 + a1
a5 = , a6 = a1 , a7 = a2
Finally we consider 0 < a < 1. The equation f ′ (x) = a2
a has exactly two solutions, at x = r− and x = r+ for
r± as defined above. If we define g(x) = f (x) − ax, and thus the sequence is periodic with period 5. The
then g′ (r± ) = 0; g′ is strictly decreasing on (−∞, r− ), values for a1 and a2 may thus be any values in F× p pro-
strictly increasing on (r− , r+ ), and strictly decreasing vided that a1 ̸= p − 1, a2 ̸= p − 1, and a1 + a2 ̸= p − 1.
on (r+ , ∞); and limx→−∞ g(x) = ∞ while limx→∞ g(x) = The number of choices for a1 , a2 ∈ {1, . . . , p − 2} such
−∞. It follows that g(x) = b has exactly one solution that a1 + a2 ̸= p − 1 is thus (p − 2)2 − (p − 2) = (p −
for b < g(r− ) or b > g(r+ ), exactly three solutions for 2)(p − 3).
g(r− ) < b < g(r+ ), and exactly two solutions for b = Since p is not a multiple of 5, (p − 2)(p − 3) is a prod-
g(r± ). That is, y = ax + b intersects y = f (x) in exactly uct of two consecutive integers a, a + 1, where a ̸≡ 2
one point if and only if b < g(r− ) or b > g(r+ ). (mod 5). Now 0 · 1 ≡ 0, 1 · 2 ≡ 2, 3 · 4 ≡ 2, and
 2

4 · 0 ≡ 0 (mod 5). Thus the number of possible se- Since 2022 ≡ 6 (mod 7), this will yield a(2022) = 2 +
quences a1 , a2 , . . . is 0 or 2 (mod 5), as desired. ⌊ 2022
7 ⌋ = 290.
Second solution. Say that a sequence is admissible if We proceed by induction, starting with the base cases
it satisfies the given conditions. As in the first solution, n ≤ 6. Since the number of odd intervals never de-
any admissible sequence is 5-periodic. creases, we have a(n), b(n) ≥ n − 2⌊ 2n ⌋; by looking at
Now consider the collection S of possible 5-tuples of the possible final positions, we see that equality holds
numbers mod p given by (a1 , a2 , a3 , a4 , a5 ) for admis- for n = 0, 1, 2, 3, 5. For n = 4, 6, Alice moving first can
sible sequences {an }. Each of these 5-tuples in S split the original interval into two odd intervals, guar-
comes from a unique admissible sequence, and there anteeing at least two odd intervals in the final position;
is a 5-periodic action on S given by cyclic permutation: whereas Bob can move to leave behind one or two in-
(a, b, c, d, e) → (b, c, d, e, a). This action divides S into tervals of length 2, guaranteeing no odd intervals in the
finitely many orbits, and each orbit either consists of 5 final position.
distinct tuples (if a, b, c, d, e are not all the same) or 1 We now proceed to the induction step. Suppose that
tuple (a, a, a, a, a). It follows that the number of admis- n ≥ 7 and the claim is known for all m < n. In particular,
sible sequences is a multiple of 5 plus the number of this means that a(m) ≥ b(m); consequently, it does not
constant admissible sequences. change the analysis to allow a player to pass their turn
Constant admissible sequences correspond to nonzero after the first move, as both players will still have an
numbers a (mod p) such that a2 ≡ 1 + a (mod p). optimal strategy which involves never passing.
Since the quadratic x2 − x − 1 has discriminant 5, for It will suffice to check that
p > 5 it has either 2 roots (if the discriminant is a
quadratic residue mod p) or 0 roots mod p. a(n) = a(n − 7) + 1, b(n) = b(n − 7) + 1.
A4 The expected value is 2e1/2 − 3. Moving first, Alice can leave behind two intervals of
Extend S to an infinite sum by including zero sum- length 1 and n − 3. This shows that
mands for i > k. We may then compute the expected
value as the sum of the expected value of the i-th sum- a(n) ≥ 1 + b(n − 3) = a(n − 7) + 1.
mand over all i. This summand occurs if and only
if X1 , . . . , Xi−1 ∈ [Xi , 1] and X1 , . . . , Xi−1 occur in non- On the other hand, if Alice leaves behind intervals of
increasing order. These two events are independent length i and n − 2 − i, Bob can choose to play in ei-
and occur with respective probabilities (1 − Xi )i−1 and ther one of these intervals and then follow Alice’s lead
1 thereafter (exercising the pass option if Alice makes the
(i−1)! ; the expectation of this summand is therefore
last legal move in one of the intervals). This shows that
Z 1
1
i
t(1 − t)i−1 dt a(n) ≤ max{min{a(i) + b(n − 2 − i),
2 (i − 1)! 0
1
Z 1 b(i) + a(n − 2 − i)} : i = 0, 1, . . . , n − 2}
= ((1 − t)i−1 − (1 − t)i ) dt = a(n − 7) + 1.
2i (i − 1)! 0
 
1 1 1 1
= i − = i . Moving first, Bob can leave behind two intervals of
2 (i − 1)! i i + 1 2 (i + 1)!
lengths 2 and n − 4. This shows that
Summing over i, we obtain
∞ ∞   b(n) ≤ a(n − 4) = b(n − 7) + 1.
1 1 1/2 1
∑ i = 2 ∑ i = 2 e − 1 − .
i=1 2 (i + 1)! i=2 2 i! 2 On the other hand, if Bob leaves behind intervals of
length i and n − 2 − i, Alice can choose to play in ei-
A5 We show that the number in question equals 290. More ther one of these intervals and then follow Bob’s lead
generally, let a(n) (resp. b(n)) be the optimal final score thereafter (again passing as needed). This shows that
for Alice (resp. Bob) moving first in a position with n
consecutive squares. We show that b(n) ≥ min{max{a(i) + b(n − 2 − i),
jnk  j n k b(i) + a(n − 2 − i)} : i = 0, 1, . . . , n − 2}
a(n) = +a n−7 ,
= b(n − 7) + 1.
j n7 k  j n7 k
b(n) = +b n−7 ,
7 7 This completes the induction.
and that the values for n ≤ 6 are as follows: A6 First solution. The largest such m is n. To show that
n 0 1 2 3 4 5 6 m ≥ n, we take
a(n) 0 1 0 1 2 1 2 (2n + 1 − j)π
b(n) 0 1 0 1 0 1 0 x j = cos ( j = 1, . . . , 2n).
2n + 1
 3

It is apparent that −1 < x1 < · · · < x2n < 1. The sum of Now suppose by way of contradiction that we have an
the lengths of the intervals can be interpreted as example showing that m ≥ n + 1. We then have
2n n n
− ∑ ((−1)2n+1− j x j )2k−1 12k−1 + ∑ x2i−1
2k−1 2k−1
= ∑ x2i (k = 1, . . . , n + 1).
j=1 i=1 i=1

2n 2k−1
By the lemma, this means that the multisets
 
π
= − ∑ cos(2n + 1 − j) π + {1, x1 , x3 , . . . , x2n−1 } and {x2 , x4 , . . . , x2n } become equal
j=1 2n + 1
after removing pairs of inverses until this becomes im-
2n 
2π(n + 1) j 2k−1

possible. However, of the resulting two multisets, the
= − ∑ cos . first contains 1 and the second does not, yielding the
j=1 2n + 1
desired contradiction.
For ζ = e2πi(n+1)/(2n+1) , this becomes Remark. One can also prove the lemma using the in-
vertibility of the Vandermonde matrix
2k−1
ζ j +ζ−j
2n 
=−∑ (xij )i=0,...,n; j=0,...,n
j=1 2
for x0 , . . . , xn pairwise distinct (this matrix has determi-
1 2n 2k−1 2k − 1 j(2k−1−2l)
 
= − 2k−1 ∑ ∑ ζ nant ∏0≤i< j≤n (xi − x j ) ̸= 0). For a similar argument,
2 j=1 l=0 l see Proposition 22 of: M. Bhargava, Galois groups
1 2k−1 2k − 1 2n j(2k−1−2l)
  of random integer polynomials and van der Waerden’s
= − 2k−1 ∑ ∑ζ conjecture, arXiv:2111.06507.
2 l=0 l j=1
Remark. The solution for m = n given above is not
1 2k−1 2k − 1
 
unique (see below). However, it does become unique
= − 2k−1 ∑ (−1) = 1, if we add the assumption that xi = −x2n+1−i for i =
2 l=0 l
1, . . . , 2n (i.e., the set of intervals is symmetric around
using the fact that ζ 2k−1−2l is a nontrivial root of unity 0).
of order dividing 2n + 1. Second solution. (by Evan Dummit) Define the poly-
To show that m ≤ n, we use the following lemma. We nomial
say that a multiset {x1 , . . . , xm } of complex numbers is p(x) = (x − x1 )(x + x2 ) · · · (x − x2n−1 )(x + x2n )(x + 1);
inverse-free if there are no two indices 1 ≤ i ≤ j ≤ m
such that xi + x j = 0; this implies in particular that 0 by hypothesis, p(x) has 2n + 1 distinct real roots in the
does not occur. interval [−1, 1). Let sk denote the k-th power sum of
p(x); then for any given m, the desired condition is that
Lemma. Let {x1 , . . . , xm }, {y1 , . . . , yn } be two inverse-free
s2k−1 = 0 for k = 1, . . . , m. Let ek denote the k-th ele-
multisets of complex numbers such that
mentary symmetric function of the roots of p(x); that
m n is,
∑ xi2k−1 = ∑ y2k−1
i (k = 1, . . . , max{m, n}).
2n+1
i=1 i=1
p(x) = x2n+1 + ∑ (−1)k ek x2n+1−k .
Then these two multisets are equal. i=k

By the Girard–Newton identities,

Proof. We may assume without loss of generality that m ≤ n.
Form the rational functions (2k − 1)e2k−1 = s1 e2k−2 − s2 e2k−2 + · · · − s2k e1 ;
m n
xi z yi z hence the desired condition implies that e2k−1 = 0 for
f (z) = ∑ , g(z) = ∑ ;
2 2 2 2 k = 1, . . . , m.
i=1 1 − xi z i=1 1 − yi z
If we had a solution with m = n + 1, then the vanishing
both f (z) and g(z) have total pole order at most 2n. Mean- of e1 , . . . , e2k+1 would imply that p(x) is an odd poly-
while, by expanding in power series around z = 0, we see that nomial (that is, p(x) = −p(x) for all x), which in turn
f (z) − g(z) is divisible by z2n+1 . Consequently, the two series would imply that x = 1 is also a root of p. Since we have
are equal. already identified 2n + 1 other roots of p, this yields a
However, we can uniquely recover the multiset {x1 , . . . , xm } contradiction.
from f (z): f has poles at {1/x12 , . . . , 1/xm
2 } and the residue of
By the same token, a solution with m = n corresponds to
2
the pole at z = 1/xi uniquely determines both xi (i.e., its sign) a polynomial p(x) of the form xq(x2 )+a for some poly-
and its multiplicity. Similarly, we may recover {y1 , . . . , yn } nomial q(x) of degree n and some real number a (neces-
from g(z), so the two multisets must coincide. sarily equal to q(1)). It will thus suffice to choose q(x)
 4

so that the resulting polynomial p(x) has roots consist- Second solution. Let G be the set of power series of
xn
ing of −1 plus 2n distinct values in (−1, 1). To do this, the form ∑∞n=0 cn n! with c0 = 1, cn ∈ Z; then G forms a
start with any polynomial r(x) of degree n with n dis- group under formal series multiplication because
tinct positive roots (e.g., r(x) = (x − 1) · · · (x − n)). The ! !
polynomial xr(x2 ) then has 2n + 1 distinct real roots; ∞
xn ∞
xn ∞
xn
consequently, for ε > 0 sufficiently small, xr(x2 ) + ε ∑ cn n! ∑ dn n! = ∑ en n!
n=0 n=0 n=0
also has 2n + 1 distinct real roots. Let −α be the
smallest of√these roots (so that α > 0); we then take with
q(x) = r(x α) to achieve the desired result.
n  
Remark. Brian Lawrence points out that one can also n
en = ∑ m cm dn−m .
produce solutions for m = n by starting with the degen- m=0
erate solution
By the same calculation, the subset H of series with
−an−1 , . . . , −a1 , 0, a1 , . . . , an−1 , 1 cn ∈ 2Z for all n ≥ 1 is a subgroup of G.
n
We have e2x ∈ H because 2n! ∈ 2Z for all n ≥ 1: the
(where 0 < a1 < · · · < an−1 < 1 but no other conditions exponent of 2 in the prime factorization of n! is
are imposed) and deforming it using the implicit func-
tion theorem. More precisely, there exists a differen- ∞ jnk ∞
n
tiable parametric solution x1 (t), . . . , x2n (t) with xi (t) = ∑ <∑ = n.
i=1 2i i=1 2
i
x2n−i (t) for i = 1, . . . , n − 1 specializing to the previous
solution at t = 0, such that xi′ (0) ̸= 0 for i = n, . . . , 2n; k
this is because the Jacobian matrix For any integer k ≥ 2, we have ex ∈ H because (nk)! n! ∈
2Z for all n ≥ 1: this is clear if k = 2, n = 1, and in all
J = ((2k − 1)xi (0)2k−2 )i=n,...,2n;k=1,...,n other cases the ratio is divisible by (n + 1)(n + 2).
We deduce that eP(x)−x ∈ H. By writing eP(x) as ex =
(interpreting 00 as 1) has the property that every maxi- ∞ xn
∑n=0 n! times an element of H, we deduce that k!bk is
mal minor is nonzero (these being scaled Vandermonde odd for all k ≥ 0.
matrices). In particular we may normalize so that
′ (0) < 0, and then evaluating at a small positive value Third solution. (by David Feldman) We interpret eP(x)
x2n
using the exponential formula for generating functions.
of t gives the desired example.
For each j, choose a set S j consisting of |a j | col-
In the proof that m = n + 1 cannot occur, one can simi- ors. Then bk is a weighted count over set partitions of
larly use the implicit function theorem (with some care) {1, . . . , k}, with each part of size j assigned a color in
to reduce to the case where {|x1 |, . . . , |x2n |} has cardi- S j , and the weight being (−1)i where i is the number of
nality n + 1. This can be extended to a complete solu- parts of any size j for which a j < 0.
tion, but the details are rather involved.
Since we are only looking for the parity of bk , we may
B1 We prove that bk k! is an odd integer for all k ≥ 0. dispense with the signs; that is, we may assume a j ≥ 0
(P(x)) n for all j and forget about the weights.
First solution. Since eP(x) = ∑∞ n=0 n! , the number
k Choose an involution on each S j with at most one fixed
k! bk is the coefficient of x in
point; this induces an involution on the partitions, so to
k−1 find the parity of bk we may instead count fixed points
k!
(P(x))k + ∑ (P(x))n . of the involution. That is, we may assume that a j ∈
n=0 n! {0, 1}.
In particular, b0 = 1 and b1 = a1 are both odd. Let Tk be the set of set partitions in question with the all-
singletons partition removed; it now suffices to exhibit
Now suppose k ≥ 2; we want to show that bk is odd. a fixed-point-free involution of Tk . To wit, for each par-
The coefficient of xk in (P(x))k is ak1 . It suffices to show tition in Tk , there is a smallest index i ∈ {1, . . . , k − 1}
k!
that the coefficient of xk in n! (P(x))n is an even integer for which i and i + 1 are not both singletons; we define
for any n < k. For k even or n ≤ k − 2, this follows an involution by swapping the positions of i and i + 1.
k!
immediately from the fact that n! is an even integer. For
k odd and n = k − 1, we have B2 The possible values of n are 1 and 7.

k! Clearly the set S = {0} works. Suppose that S ̸= {0}

(P(x))k−1 = k(a1 x + a2 x2 + · · · )k−1 is a finite set satisfying the given condition; in par-
(k − 1)! ticular, S does not consist of a collection of collinear
= k(ak−1
1 x
k−1
+ (k − 1)ak−2 k
1 a2 x + · · · ) vectors, since otherwise {v × w : v, w ∈ S} = {0}. We
claim that S cannot contain any nonzero vector v with
and the coefficient of xk is k(k − 1)ak−2
1 a2 , which is ∥v∥ ̸= 1. Suppose otherwise, and let w ∈ S be a vector
again an even integer. not collinear with v. Then S must contain the nonzero
 5

vector u1 = v × w, as well as the sequence of vec- We then observe that for any m ≥ 2, we obtain a se-
tors un defined inductively by un = v × un−1 . Since quence of the desired form of length 3m + 3 = (2m −
each un is orthogonal to v by construction, we have 1) + 1 + (m + 1) + 2 by concatenating the arithmetic
∥un ∥ = ∥v∥∥un−1 ∥ and so ∥un ∥ = ∥v∥n−1 ∥u1 ∥. The se- progressions
quence ∥un ∥ consists of all distinct numbers and thus S
is infinite, a contradiction. This proves the claim, and (1, 3, . . . , 4m − 3, 4m − 1),
so every nonzero vector in S is a unit vector. 4m − 2, (4m, 4m − 4, . . . , 4, 0), 2.
Next note that any pair of vectors v, w ∈ S must either be We see that no terms are repeated by noting that the
collinear or orthogonal: by the claim, v, w are both unit first parenthesized sequence consists of odd numbers;
vectors, and if v, w are not collinear then v × w ∈ S must the second sequence consists of multiples of 4; and the
be a unit vector, whence v ⊥ w. Now choose any pair of remaining numbers 2 and 4m − 2 are distinct (because
non-collinear vectors v1 , v2 ∈ S, and write v3 = v1 × v2 . m ≥ 2) but both congruent to 2 mod 4.
Then {v1 , v2 , v3 } is an orthonormal basis of R3 , and it
follows that all of these vectors are in S: 0, v1 , v2 , v3 , It remains to show that no such sequence occurs with
−v1 = v3 ×v2 , −v2 = v1 ×v3 , and −v3 = v2 ×v1 . On the n = 6. We may assume without loss of generality that
other hand, S cannot contain any vector besides these the smallest common difference among the arithmetic
seven, since any other vector w in S would have to be progressions is 1 and occurs for {x1 , x2 , x3 }; by rescal-
simultaneously orthogonal to all of v1 , v2 , v3 . ing, shifting, and reversing the sequence as needed, we
may assume that x1 = 0 and (x2 , x3 ) ∈ {(1, 2), (2, 1)}.
Thus any set S ̸= {0} satisfying the given condi- We then have x4 = 3 and
tion must be of the form {0, ±v1 , ±v2 , ±v3 } where
{v1 , v2 , v3 } is an orthonormal basis of R3 . It is clear that (x5 , x6 ) ∈ {(4, 5), (−1, −5), (−1, 7), (5, 4), (5, 7)}.
any set of this form does satisfy the given condition. We
conclude that the answer is n = 1 or n = 7. In none of these cases does {x5 , x6 , 0} form an arith-
metic progression.
B3 The answer is yes. Let R0 , B0 ⊂ R+ be the set of red and Remark. If one interprets “distinct” in the problem
blue numbers at the start of the process, and let Rn , Bn statement to mean “not all equal”, then the problem be-
be the set of red and blue numbers after n steps. We comes simpler: the same argument as above shows that
claim that R2 = R+ . n must be a multiple of 3, in which case a suitable rep-
etition of the sequence −1, 0, 1 works.
We first note that if y ∈ B1 , then y/2 ∈ R1 . Namely,
the numbers y and 2y must be of opposite colors in the B5 First solution. The answer is p ≤ 1/4. We first
original coloring, and then 3y/2 must be of the same show that p > 1/4 does not satisfy the desired con-
color as one of y or 2y. dition. For p > 1/3, P(0, 1) = 1 − 2p < p = P(1, 1).
Now suppose by way of contradiction that x ∈ B2 . Then For p = 1/3, it is easily calculated (or follows from
of the four numbers x, 2x, 3x, 4x, every other number the next calculation) that P(0, 1, 2) = 1/9 < 2/9 =
must be in R1 and the other two must be in B1 . By P(1, 1, 2). Now suppose 1/4 < p < 1/3, and consider
the previous observation, 2x and 4x cannot both be in (b, a1 , a2 , a3 , . . . , an ) = (1, 1, 2, 4, . . . , 2n−1 ). The only
B1 ; it follows that 2x, 4x ∈ R1 and x, 3x ∈ B1 . By the solution to
previous observation again, x/2 and 3x/2 must both be X1 + 2X2 + · · · + 2n−1 Xn = 0
in R1 , but then x = 3x/2 − x/2 is in R2 , contradiction.
We conclude that R2 = R+ , as desired. with X j ∈ {0, ±1} is X1 = · · · = Xn = 0; thus
P(0, 1, 2, . . . , 22n−1 ) = (1 − 2p)n . On the other hand, the
B4 The values of n in question are the multiples of 3 start- solutions to
ing with 9. Note that we interpret “distinct” in the prob-
lem statement to mean “pairwise distinct” (i.e., no two X1 + 2X2 + · · · + 2n−1 Xn = 1
equal). See the remark below.
with X j ∈ {0, ±1} are
We first show that such a sequence can only oc-
cur when n is divisible by 3. If d1 and d2 are (X1 , X2 , . . . , Xn ) = (1, 0, . . . , 0), (−1, 1, 0, . . . , 0),
the common differences of the arithmetic progressions (−1, −1, 1, 0, . . . , 0), . . . , (−1, −1, . . . , −1, 1),
{xm , xm+1 , xm+2 } and {xm+1 , xm+2 , xm+3 } for some m,
then d2 ∈ {d1 , 2d1 , d1 /2}. By scaling we may assume and so
that the smallest common difference that occurs is 1;
in this case, all of the common differences are integers. P(1, 1, 2, . . . , 2n−1 )
By shifting, we may assume that the xi are themselves = p(1 − 2p)n−1 + p2 (1 − 2p)n−2 + · · · + pn
all integers. We now observe that any three consecutive
terms in the sequence have pairwise distinct residues (1 − 2p)n − pn
=p .
modulo 3, forcing n to be divisible by 3. 1 − 3p
 6

It follows that the inequality P(0, 1, 2, . . . , 2n−1 ) ≥ For p ≤ 1/4, we have

P(1, 1, 2, . . . , 2n−1 ) is equivalent to
ϕXk (θ ) = (1 − 2p) + 2p cos(θ ) ≥ 0.
pn+1 ≥ (4p − 1)(1 − 2p)n ,
Hence for a1 , . . . , an ∈ Z, the random variable S =
but this is false for sufficiently large n since 4p − 1 > 0 a1 X1 + · · · + an Xn satisfies
and p < 1 − 2p.
n n
Now suppose p ≤ 1/4; we want to show that ϕS (θ ) = ∏ ϕak Xk (θ ) = ∏ ϕXk (ak θ ) ≥ 0.
for arbitrary a1 , . . . , an and b ̸= 0, P(0, a1 , . . . , an ) ≥ k=1 k=1
P(b, a1 , . . . , an ). Define the polynomial
We may thus conclude that P(S = b) ≤ P(S = 0) for any
f (x) = px + px −1
+ 1 − 2p, b ∈ Z, as desired.
1
and observe that P(b, a1 , . . . , an ) is the coefficient of xb B6 The only such functions are the functions f (x) = 1+cx
in f (xa1 ) f (xa2 ) · · · f (xan ). We can write for some c ≥ 0 (the case c = 0 giving the constant func-
tion f (x) = 1). Note that we interpret R+ in the prob-
f (xa1 ) f (xa2 ) · · · f (xan ) = g(x)g(x−1 ) lem statement to mean the set of positive real numbers,
excluding 0.
for some
√
real polynomial g: indeed, if we define α = For convenience, we reproduce here the given equation:
1−2p+ 1−4p
2p > 0, then f (x) = αp (x + α)(x−1 + α), and
so we can use f (x f (y)) + f (y f (x)) = 1 + f (x + y) (1)
 p n/2
g(x) = (xa1 + α) · · · (xan + α). We first prove that
α
lim f (x) = 1. (2)
It now suffices to show that in g(x)g(x−1 ), the co- x→0+
efficient of x0 is at least as large as the coefficient
of xb for any b ̸= 0. Since g(x)g(x−1 ) is symmetric Set
upon inverting x, we may assume that b > 0. If we L− = lim inf f (x), L+ = lim sup f (x).
write g(x) = c0 x0 + · · · + cm xm , then the coefficients x→0+ x→0+
of x0 and xb in g(x)g(x−1 ) are c20 + c21 + · · · + c2m and
c0 cb + c1 cb+1 + · · · + cm−b cm , respectively. But For any fixed y, we have by (1)

2(c0 cb + c1 cb+1 + · · · + cm−b cm ) L+ = lim sup f (x f (y))

x→0+
≤ (c20 + c2b ) + (c21 + c2b+1 ) + · · · + (c2m−b + c2m ) ≤ lim sup(1 + f (x + y)) = 1 + f (y) < ∞.
≤ 2(c20 + · · · + c2m ), x→0+

and the result follows. Consequently, x f (x) → 0 as x → 0+ . By (2) with y = x,

Second solution. (by Yuval Peres) We check that p ≤ 2L+ = lim sup 2 f (x f (x))
1/4 is necessary as in the first solution. To check that it x→0+
is sufficient, we introduce the following concept: for X = lim sup(1 + f (2x)) = 1 + L+
a random variable taking finitely many integer values, x→0+
define the characteristic function 2L− = lim inf 2 f (x f (x))
x→0+
iℓθ
ϕX (θ ) = ∑ P(X = ℓ)e = lim inf(1 + f (2x)) = 1 + L−
ℓ∈Z x→0+

(i.e., the expected value of eiXθ , or the Fourier transform and so L− = L+ = 1, confirming (2).
of the probability measure corresponding to X). We use We next confirm that
two evident properties of these functions:
f (x) ≥ 1 for all x > 0 =⇒ f (x) = 1 for all x > 0. (3)
– If X and Y are independent, then ϕX+Y (θ ) =
ϕX (θ ) + ϕY (θ ). Suppose that f (x) ≥ 1 for all x > 0. For 0 < c ≤ ∞, put
– For any b ∈ Z, Sc = sup{ f (x) : 0 < x ≤ c}; for c < ∞, (2) implies that
Z 2π Sc < ∞. If there exists y > 0 with f (y) > 1, then from
1 (1) we have f (x + y) − f (x f (y)) = f (y f (x)) − 1 ≥ 0;
P(X = b) = e−ibθ ϕX (θ ) dθ .
2 0 hence
 
In particular, if ϕX (θ ) ≥ 0 for all θ , then P(X = y f (y)
Sc = S(c−y) f (y) c ≥ c0 =
b) ≤ P(X = 0). f (y) − 1
 7

and (since (c − y) f (y) − c0 = f (y)(c − c0 )) iterating this By substituting y 7→ xy in (1),

construction shows that S∞ = Sc for any c > c0 . In any
case, we deduce that f (x f (xy)) + f (xy f (x)) = 1 + f (x + xy).

f (x) ≥ 1 for all x > 0 =⇒ S∞ < ∞. (4) Taking the limit as x → ∞ and applying (6) yields

f (1/y) + f (y) = 1. (8)

Still assuming that f (x) ≥ 1 for all x > 0, note that from
(1) with x = y, Combining (1) with (8) yields
1 
1

f (x f (x)) = (1 + f (2x)). f (x f (y)) = f (x + y) + f .
2 y f (x)
Multiply both sides by x f (y), then take the limit as x →
Since x f (x) → 0 as x → 0+ by (2) and x f (x) → ∞ as ∞ to obtain
x → ∞, x f (x) takes all positive real values by the inter-
mediate value theorem. We deduce that 2S∞ ≤ 1 + S∞
 
1
and hence S∞ = 1; this proves (3). 1 = lim x f (y) f (x + y) + lim x f (y) f
x→∞ x→∞ y f (x)
We may thus assume hereafter that f (x) < 1 for some = f (y) + lim x f (y)y f (x)
x→∞
x > 0. We next check that
= f (y) + y f (y)
lim f (x) = 0. (5) 1
x→∞ and solving for f (y) now yields f (y) = 1+y , as desired.

Put I = inf{ f (x) : x > 0} < 1, choose ε ∈ (0, (1 − I)/2), Remark. Some variants of the above approach are pos-
and choose y > 0 such that f (y) < I + ε. We then must sible. For example, once we have (5), we can establish
have x f (x) ̸= y for all x, or else that f is monotone decreasing as follows. We first check
that
1 + I ≤ 1 + f (2x) = 2 f (y) < 2I + 2ε,
f (x) < 1 for all x > 0. (9)
contradiction. Since x f (x) → 0 as x → 0+
by (2), we Suppose by way of contradiction that f (x) = 1 for some
have sup{x f (x) : x > 0} < ∞ by the intermediate value x. By (1),
theorem, yielding (5).
By (2) plus (5), f −1 (1/2) is nonempty and compact. f (2x) + 1 = 2 f (x f (x)) = 2 f (x) = 2
We can now simplify by noting that if f (x) satisfies the
original equation, then so does f (cx) for any c > 0; we and so f (2x) = 1. It follows that f −1 (1) is infinite, con-
may thus assume that the least element of f −1 (1/2) is tradicting (5).
1 We next check that
1, in which case we must show that f (x) = 1+x .
We next show that x < y =⇒ f (x) > f (y). (10)
lim x f (x) = 1. (6) For x < y, by substituting x 7→ y − x in (1) we obtain
x→∞

For all x > 0, by (1) with y = x, 1 + f (y) = f (x f (y − x)) + f ((y − x) f (x))

< 1 + f ((y − x) f (x)),
1 1
f (x f (x)) = (1 + f (2x)) > = f (1), (7) whence f ((y − x) f (x)) > f (y). Because (y − x) f (x) →
2 2
0 as x → y− and (y − x) f (x) → y as x → 0+ , (y − x) f (x)
so in particular x f (x) ̸= 1. As in the proof of (5), this takes all values in (0, y) as x varies over (0, y); this
implies that x f (x) < 1 for all x > 0. However, by (5) proves (10).
and (7) we have f (x f (x)) → 12 as x → ∞, yielding (6).
 The 84th William Lowell Putnam Mathematical Competition
Saturday, December 2, 2023

A1 For a positive integer n, let fn (x) = unoccupied, then a legal move is to slide the coin from
cos(x) cos(2x) cos(3x) · · · cos(nx). Find the small- (i, j) to (i + 1, j + 1). How many distinct configurations
est n such that | fn′′ (0)| > 2023. of coins can be reached starting from the initial config-
uration by a (possibly empty) sequence of legal moves?
A2 Let n be an even positive integer. Let p be a monic, real B2 For each positive integer n, let k(n) be the number of
polynomial of degree 2n; that is to say, p(x) = x2n + ones in the binary representation of 2023 · n. What is
a2n−1 x2n−1 + · · · + a1 x + a0 for some real coefficients the minimum value of k(n)?
a0 , . . . , a2n−1 . Suppose that p(1/k) = k2 for all integers
k such that 1 ≤ |k| ≤ n. Find all other real numbers x B3 A sequence y1 , y2 , . . . , yk of real numbers is called
for which p(1/x) = x2 . zigzag if k = 1, or if y2 − y1 , y3 − y2 , . . . , yk − yk−1 are
nonzero and alternate in sign. Let X1 , X2 , . . . , Xn be
A3 Determine the smallest positive real number r such
chosen independently from the uniform distribution on
that there exist differentiable functions f : R → R and
[0, 1]. Let a(X1 , X2 , . . . , Xn ) be the largest value of k for
g : R → R satisfying
which there exists an increasing sequence of integers
(a) f (0) > 0, i1 , i2 , . . . , ik such that Xi1 , Xi2 , . . . , Xik is zigzag. Find the
expected value of a(X1 , X2 , . . . , Xn ) for n ≥ 2.
(b) g(0) = 0,
(c) | f ′ (x)| ≤ |g(x)| for all x, B4 For a nonnegative integer n and a strictly increasing se-
(d) |g′ (x)| ≤ | f (x)| for all x, and quence of real numbers t0 ,t1 , . . . ,tn , let f (t) be the cor-
responding real-valued function defined for t ≥ t0 by the
(e) f (r) = 0. following properties:
A4 Let v1 , . . . , v12 be unit vectors in R3 from the origin to
(a) f (t) is continuous for t ≥ t0 , and is twice differen-
the vertices of a regular icosahedron. Show that for ev-
tiable for all t > t0 other than t1 , . . . ,tn ;
ery vector v ∈ R3 and every ε > 0, there exist integers
a1 , . . . , a12 such that ∥a1 v1 + · · · + a12 v12 − v∥ < ε. (b) f (t0 ) = 1/2;
A5 For a nonnegative integer k, let f (k) be the number of (c) limt→t + f ′ (t) = 0 for 0 ≤ k ≤ n;
k
ones in the base 3 representation of k. Find all complex
(d) For 0 ≤ k ≤ n − 1, we have f ′′ (t) = k + 1 when
numbers z such that
tk < t < tk+1 , and f ′′ (t) = n + 1 when t > tn .
31010 −1
∑ (−2) f (k) (z + k)2023 = 0. Considering all choices of n and t0 ,t1 , . . . ,tn such that
k=0 tk ≥ tk−1 + 1 for 1 ≤ k ≤ n, what is the least possible
value of T for which f (t0 + T ) = 2023?
A6 Alice and Bob play a game in which they take turns
choosing integers from 1 to n. Before any integers are B5 Determine which positive integers n have the follow-
chosen, Bob selects a goal of “odd” or “even”. On the ing property: For all integers m that are relatively
first turn, Alice chooses one of the n integers. On the prime to n, there exists a permutation π : {1, 2, . . . , n} →
second turn, Bob chooses one of the remaining inte- {1, 2, . . . , n} such that π(π(k)) ≡ mk (mod n) for all
gers. They continue alternately choosing one of the in- k ∈ {1, 2, . . . , n}.
tegers that has not yet been chosen, until the nth turn,
which is forced and ends the game. Bob wins if the B6 Let n be a positive integer. For i and j in {1, 2, . . . , n},
parity of {k : the number k was chosen on the kth turn} let s(i, j) be the number of pairs (a, b) of nonnegative
matches his goal. For which values of n does Bob have integers satisfying ai + b j = n. Let S be the n-by-n
a winning strategy? matrix whose (i, j) entry is s(i, j). For example, when
6 3 2 2 2
 
B1 Consider an m-by-n grid of unit squares, indexed by 3 0 1 0 1
(i, j) with 1 ≤ i ≤ m and 1 ≤ j ≤ n. There are (m − n = 5, we have S = 2 1 0 0 1. Compute the de-
 
1)(n − 1) coins, which are initially placed in the squares 2 0 0 0 1
(i, j) with 1 ≤ i ≤ m − 1 and 1 ≤ j ≤ n − 1. If a coin oc- 2 1 1 1 2
cupies the square (i, j) with i ≤ m − 1 and j ≤ n − 1 terminant of S.
and the squares (i + 1, j), (i, j + 1), and (i + 1, j + 1) are
 Solutions to the 84th William Lowell Putnam Mathematical Competition
Saturday, December 2, 2023
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A1 If we use the product rule to calculate fn′′ (x), the result Now define the function h : [0, r) → (−π/2, π/2) by
is a sum of terms of two types: terms where two distinct h(x) = tan−1 (g(x)/ f (x)). We compute that
factors cos(m1 x) and cos(m2 x) have each been differen-
tiated once, and terms where a single factor cos(mx) has f (x)g′ (x) − g(x) f ′ (x)
h′ (x) =
been differentiated twice. When we evaluate at x = 0, f (x)2 + g(x)2
all terms of the first type vanish since sin(0) = 0, while
the term of the second type involving (cos(mx))′′ be- and thus
comes −m2 . Thus
| f (x)||g′ (x)| + |g(x)|| f ′ (x)| | f (x)|2 + |g(x)|2
|h′ (x)| ≤ ≤ = 1.
n
n(n + 1)(2n + 1) f (x)2 + g(x)2 f (x)2 + g(x)2
| fn′′ (0)| = − ∑ m2 = .
m=1 6 Since h(0) = 0, we have |h(x)| ≤ x < r for all x ∈ [0, r).
Since r < π/2 and tan−1 is increasing on (−r, r), we
The function g(n) = n(n+1)(2n+1)
6 is increasing for n ∈ N conclude that |g(x)/ f (x)| is uniformly bounded above
and satisfies g(17) = 1785 and g(18) = 2109. It follows by tan r for all x ∈ [0, r). But this contradicts the fact
that the answer is n = 18. that f (r) = 0 and g(r) ̸= 0, since limx→r− g(x)/ f (x) =
∞. This contradiction shows that r < π/2 cannot be
A2 The only other real numbers with this property are achieved.
±1/n!. (Note that these are indeed other values than Second solution. (by Victor Lie) As in the first solu-
±1, . . . , ±n because n > 1.) tion, we may assume f (x) > 0 for x ∈ [0, r). Combining
Define the polynomial q(x) = x2n+2 − x2n p(1/x) = our hypothesis with the fundamental theorem of calcu-
x2n+2 − (a0 x2n + · · · + a2n−1 x + 1). The statement that lus, for x > 0 we obtain
p(1/x) = x2 is equivalent (for x ̸= 0) to the state- Z x
ment that x is a root of q(x). Thus we know that | f ′ (x)| ≤ |g(x)| ≤ g′ (t) dt
±1, ±2, . . . , ±n are roots of q(x), and we can write 0
Z x Z x
2 2 2 2 2 ≤ |g′ (t)| dt ≤ | f (t)| dt.
q(x) = (x + ax + b)(x − 1)(x − 4) · · · (x − n ) 0 0
Rx
for some monic quadratic polynomial x2 + ax + b. Define F(x) = 0 f (t) dt; we then have
Equating the coefficients of x2n+1 and x0 on both sides
gives 0 = a and −1 = (−1)n (n!)2 b, respectively. Since f ′ (x) + F(x) ≥ 0 (x ∈ [0, r]).
n is even, we have x2 + ax + b = x2 − (n!)−2 . We con-
clude that there are precisely two other real numbers x Now suppose by way of contradiction that r < π2 . Then
such that p(1/x) = x2 , and they are ±1/n!. cos x > 0 for x ∈ [0, r], so

A3 The answer is r = π2 , which manifestly is achieved by f ′ (x) cos x + F(x) cos x ≥ 0 (x ∈ [0, r]).
setting f (x) = cos x and g(x) = sin x.
The left-hand side is the derivative of f (x) cos x +
First solution. Suppose by way of contradiction that F(x) sin x. Integrating from x = y to x = r, we obtain
there exist some f , g satisfying the stated conditions for
some 0 < r < π2 . We first note that we can assume that F(r) sin r ≥ f (y) cos y + F(y) sin y (y ∈ [0, r]).
f (x) ̸= 0 for x ∈ [0, r). Indeed, by continuity, {x | x ≥
0 and f (x) = 0} is a closed subset of [0, ∞) and thus has We may rearrange to obtain
a minimum element r′ with 0 < r′ ≤ r. After replacing
r by r′ , we now have f (x) ̸= 0 for x ∈ [0, r). F(r) sin r sec2 y ≥ f (y) sec y + F(y) sin y sec2 y (y ∈ [0, r]).
Next we note that f (r) = 0 implies g(r) ̸= 0. In- The two sides are the derivatives of F(r) sin r tan y and
deed, define the function k : R → R by k(x) = f (x)2 + F(y) sec y, respectively. Integrating from y = 0 to y = r
g(x)2 . Then |k′ (x)| = 2| f (x) f ′ (x) + g(x)g′ (x))| ≤ and multiplying by cos2 r, we obtain
4| f (x)g(x)| ≤ 2k(x), where the last inequality fol-
lows from the AM-GM inequality. It follows that F(r) sin2 r ≥ F(r)
d
dx (log k(x)) ≤ 2 for x ∈ [0, r); since k(x) is continu-
ous at x = r, we conclude that k(r) ̸= 0. which is impossible because F(r) > 0 and 0 < sin r < 1.
 2

A4 The assumption that all vertices of the icosahedron cor- Proving (1) is a straightforward induction on i: the
n−i−1 n−i−1
respond to vectors of the same length forces the center induction step applies S3 − 2I + S−3 to the
of the icosahedron to lie at the origin, since the icosa- right hand side of (1), using the general formula for
hedron is inscribed in a unique sphere. Since scaling (Sℓ − 2I + S−ℓ )zm .
the icosahedron does not change whether or not the
Now setting i = n in (1), we find that for some Cn ,
stated conclusion is true, we may choose coordinates so
that the vertices are the cyclic permutations
√
of the vec- n−1 
9n − 1

3j −3 j 2n+3 3
1 1 1+ 5
tors (± 2 , ± 2 φ , 0) where φ = 2 is the golden ratio. ∏ (S − 2I + S )z = Cn z + z .
j=0 16
The subgroup of R3 generated by these vectors contains
G × G × G where G is the subgroup of R generated by √
n
1 and φ . Since φ is irrational, it generates a dense sub- The roots of this polynomial are 0 and ± 94−1 i, and it
group of R/Z; hence G is dense in R, and so G × G × G follows that the roots of pn (z) are these three numbers
n
is dense in R3 , proving the claim. minus 3 2−1 . In particular, when n = 1010, we find that
the roots of p1010 (z) are as indicated above.
A5 The complex numbers z with this property are
√ A6 (Communicated by Kai Wang) For all n, Bob has a win-
31010 − 1 31010 − 1 91010 − 1 ning strategy. Note that we can interpret the game play
− and − ± i.
2 2 4 as building a permutation of {1, . . . , n}, and the number
We begin by noting that for n ≥ 1, we have the follow- of times an integer k is chosen on the k-th turn is exactly
ing equality of polynomials in a parameter x: the number of fixed points of this permutation.
3n −1 n−1
For n even, Bob selects the goal “even”. Divide
f (k) k 2·3 j 3j {1, . . . , n} into the pairs {1, 2}, {3, 4}, . . . ; each time Al-
∑ (−2) x = ∏ (x − 2x + 1).
k=0 j=0 ice chooses an integer, Bob follows suit with the other
integer in the same pair. For each pair {2k − 1, 2k}, we
This is readily shown by induction on n, using the fact
see that 2k − 1 is a fixed point if and only if 2k is, so the
that for 0 ≤ k ≤ 3n−1 − 1, f (3n−1 + k) = f (k) + 1 and
number of fixed points is even.
f (2 · 3n−1 + k) = f (k).
For n odd, Bob selects the goal “odd”. On the first turn,
Now define a “shift” operator S on polynomials in z by
if Alice chooses 1 or 2, then Bob chooses the other one
S(p(z)) = p(z + 1); then we can define Sm for all m ∈ Z
to transpose into the strategy for n − 2 (with no moves
by Sm (p(z)), and in particular S0 = I is the identity map.
made). We may thus assume hereafter that Alice’s first
Write
move is some k > 2, which Bob counters with 2; at this
3n −1
point there is exactly one fixed point.
pn (z) := ∑ (−2) f (k) (z + k)2n+3
k=0 Thereafter, as long as Alice chooses j on the j-th turn
for n ≥ 1; it follows that (for j ≥ 3 odd), either j + 1 < k, in which case Bob can
choose j + 1 to keep the number of fixed points odd; or
n−1
j j j + 1 = k, in which case k is even and Bob can choose 1
pn (z) = ∏ (S2·3 − 2S3 + I)z2n+3
j=0
to transpose into the strategy for n − k (with no moves
made).
n−1
n −1)/2 j j
= S(3 ∏ (S3 − 2I + S−3 )z2n+3 . Otherwise, at some odd turn j, Alice does not choose j.
j=0 At this point, the number of fixed points is odd, and
on each subsequent turn Bob can ensure that neither
Next observe that for any ℓ, the operator Sℓ − 2I + S−ℓ
his own move nor Alice’s next move does not create
acts on polynomials in z in a way that decreases degree
a fixed point: on any turn j for Bob, if j + 1 is available
by 2. More precisely, for m ≥ 0, we have
Bob chooses it; otherwise, Bob has at least two choices
(Sℓ − 2I + S−ℓ )zm = (z + ℓ)m − 2zm + (z − ℓ)m available, so he can choose a value other than j.
   
m 2 m−2 m 4 m−4 B1 The number of such configurations is m+n−2


=2 ℓ z +2 ℓ z + O(zm−6 ). m−1 .
2 4
Initially the unoccupied squares form a path from (1, n)
We use this general calculation to establish the follow- to (m, 1) consisting of m − 1 horizontal steps and n − 1
ing: for any 1 ≤ i ≤ n, there is a nonzero constant Ci vertical steps, and every move preserves this property.
(depending on n and i but not z) such that This yields an injective map from the set of reachable
i
n− j n− j
configurations to the set of paths of this form.
∏ (S3 − 2I + S−3 )z2n+3
Since the number of such paths is evidently m+n−2


j=1 m−1 (as
  one can arrange the horizontal and vertical steps in any
= Ci z2n+3−2i + (2n+3−2i)(n+1−i)
6 (∑i
j=1 9 n− j )z2n+1−2i order), it will suffice to show that the map we just wrote
down is also surjective; that is, that one can reach any
+O(z2n−1−2i ). (1) path of this form by a sequence of moves.
 3

This is easiest to see by working backwards. Ending B3 The expected value is 2n+2 3 .
at a given path, if this path is not the initial path, then Divide the sequence X1 , . . . , Xn into alternating increas-
it contains at least one sequence of squares of the form ing and decreasing segments, with N segments in all.
(i, j) → (i, j − 1) → (i + 1, j − 1). In this case the square Note that removing one term cannot increase N: if
(i + 1, j) must be occupied, so we can undo a move by the removed term is interior to some segment then the
replacing this sequence with (i, j) → (i + 1, j) → (i + number remains unchanged, whereas if it separates two
1, j − 1). segments then one of those decreases in length by 1
(and possibly disappears). From this it follows that
B2 The minimum is 3. a(X1 , . . . , Xn ) = N + 1: in one direction, the endpoints
First solution. of the segments form a zigzag of length N + 1; in the
other, for any zigzag Xi1 , . . . , Xim , we can view it as a
We record the factorization 2023 = 7 · 172 . We first rule
sequence obtained from X1 , . . . , Xn by removing terms,
out k(n) = 1 and k(n) = 2. If k(n) = 1, then 2023n = 2a
so its number of segments (which is manifestly m − 1)
for some a, which clearly cannot happen. If k(n) = 2,
cannot exceed N.
then 2023n = 2a + 2b = 2b (1 + 2a−b ) for some a > b.
Then 1 + 2a−b ≡ 0 (mod 7); but −1 is not a power of 2 For n ≥ 3, a(X1 , . . . , Xn ) − a(X2 , . . . , Xn ) is 0 if X1 , X2 , X3
mod 7 since every power of 2 is congruent to either 1, form a monotone sequence and 1 otherwise. Since the
2, or 4 (mod 7). six possible orderings of X1 , X2 , X3 are equally likely,
We now show that there is an n such that k(n) = 3. 2
E(a(X1 , . . . , Xn ) − a(X1 , . . . , Xn−1 )) = .
It suffices to find a > b > 0 such that 2023 divides 3
2a + 2b + 1. First note that 22 + 21 + 1 = 7 and 23 ≡ 1 Moreover, we always have a(X1 , X2 ) = 2 because any
(mod 7); thus if a ≡ 2 (mod 3) and b ≡ 1 (mod 3) sequence of two distinct elements is a zigzag. By lin-
then 7 divides 2a + 2b + 1. Next, 28 + 25 + 1 = 172 and earity of expectation plus induction on n, we obtain
216·17 ≡ 1 (mod 172 ) by Euler’s Theorem; thus if a ≡ 8 E(a(X1 , . . . , Xn )) = 2n+2
(mod 16 · 17) and b ≡ 5 (mod 16 · 17) then 172 divides 3 as claimed.
2a + 2b + 1. B4 The minimum value of T is 29.
We have reduced the problem to finding a, b such that Write tn+1 = t0 + T and define sk = tk −tk−1 for 1 ≤ k ≤
a ≡ 2 (mod 3), a ≡ 8 (mod 16 · 17), b ≡ 1 (mod 3), n + 1. On [tk−1 ,tk ], we have f ′ (t) = k(t − tk−1 ) and so
b ≡ 5 (mod 16 · 17). But by the Chinese Remainder f (tk ) − f (tk−1 ) = 2k s2k . Thus if we define
Theorem, integers a and b solving these equations exist
n+1
and are unique mod 3 · 16 · 17. Thus we can find a, b sat-
isfying these congruences; by adding appropriate mul-
g(s1 , . . . , sn+1 ) = ∑ ks2k ,
k=1
tiples of 3 · 16 · 17, we can also ensure that a > b > 1.
then we want to minimize ∑n+1 k=1 sk = T (for all pos-
Second solution. We rule out k(n) ≤ 2 as in the first
sible values of n) subject to the constraints that
solution. To force k(n) = 3, we first note that 24 ≡ −1
g(s1 , . . . , sn+1 ) = 4045 and sk ≥ 1 for k ≤ n.
(mod 17) and deduce that 268 ≡ −1 (mod 172 ). (By
writing 268 = ((24 + 1) − 1)17 and expanding the bino- We first note that a minimum value for T is in-
mial, we obtain −1 plus some terms each of which is deed achieved. To see this, note that the constraints
divisible by 17.) Since (28 − 1)2 is divisible by 172 , g(s1 , . . . , sn+1 ) = 4045 and sk ≥ 1 place an upper bound
on n. For fixed n, the constraint g(s1 , . . . , sn+1 ) = 4045
0 ≡ 216 − 2 · 28 + 1 ≡ 216 + 2 · 268 · 28 + 1 places an upper bound on each sk , whence the set of
(s1 , . . . , sn+1 ) on which we want to minimize ∑ sk is a
= 277 + 216 + 1 (mod 172 ). compact subset of Rn+1 .
Now say that T0 is the minimum value of ∑n+1 k=1 sk (over
On the other hand, since 23 ≡ −1 (mod 7),
all n and s1 , . . . , sn+1 ), achieved by (s1 , . . . , sn+1 ) =
(s01 , . . . , s0n+1 ). Observe that there cannot be another
277 + 216 + 1 ≡ 22 + 21 + 1 ≡ 0 (mod 7). ′ +1
(s1 , . . . , sn′ +1 ) with the same sum, ∑nk=1 sk = T0 , satis-
Hence n = (277 + 216 + 1)/2023 is an integer with fying g(s1 , . . . , sn′ +1 ) > 4045; otherwise, the function f
k(n) = 3. for (s1 , . . . , sn′ +1 ) would satisfy f (t0 + T0 ) > 4045 and
there would be some T < T0 such that f (t0 + T ) = 4045
Remark. A short computer calculation shows that the by the intermediate value theorem.
value of n with k(n) = 3 found in the second solution is
We claim that s0n+1 ≥ 1 and s0k = 1 for 1 ≤ k ≤ n. If
the smallest possible. For example, in SageMath, this
reduces to a single command: s0n+1 < 1 then

g(s01 , . . . , s0n−1 , s0n + s0n+1 ) − g(s01 , . . . , s0n−1 , s0n , s0n+1 )

assert all((2^a+2^b+1) % 2023 != 0
for a in range(1,77) for b in range(1,a)) = s0n+1 (2ns0n − s0n+1 ) > 0,
 4

contradicting our observation from the previous para- Proof. We first check the “only if” direction. Suppose that
graph. Thus s0n+1 ≥ 1. If s0k > 1 for some 1 ≤ k ≤ n then σ = τ 2 . Then every cycle of τ of length m remains a cycle in
replacing (s0k , s0n+1 ) by (1, s0n+1 + s0k − 1) increases g: σ if m is odd, and splits into two cycles of length m/2 if m is
even.
g(s01 , . . . , 1, . . . , s0n+1 + s0k − 1) − g(s01 , . . . , s0k , . . . , s0n+1 ) We next check the “if” direction. We may partition the cycles
= (s0k − 1)((n + 1 − k)(s0k + 1) + 2(n + 1)(s0n+1 − 1)) > 0, of σ into individual cycles of odd length and pairs of cycles
of the same even length; then we may argue as above to write
again contradicting the observation. This establishes each partition as the square of another permutation.
the claim.
Given that s0k = 1 for 1 ≤ k ≤ n, we have T = s0n+1 + n Suppose now that n > 1 is odd. Write n = pe k where p
and is an odd prime, k is a positive integer, and gcd(p, k) =
1. By the Chinese remainder theorem, we have a ring
n(n + 1) isomorphism
g(s01 , . . . , s0n+1 ) = + (n + 1)(T − n)2 .
2
Z/nZ ∼
= Z/pe Z × Z/kZ.
Setting this equal to 4045 and solving for T yields
r Recall that the group (Z/pe Z)× is cyclic; choose m ∈ Z
4045 n
T = n+ − . reducing to a generator of (Z/pe Z)× and to the identity
n+1 2 in (Z/kZ)× . Then σn,m consists of k cycles (an odd
For n = 9 this yields T = 29; it thus suffices to show number) of length pe−1 (p − 1) (an even number) plus
that for all n, some shorter cycles. By Lemma 1, σn,m does not have
r a square root.
4045 n Suppose next that n ≡ 2 (mod 4). Write n = 2k with k
n+ − ≥ 29.
n+1 2 odd, so that
This is evident for n ≥ 30. For n ≤ 29, rewrite the claim Z/nZ ∼
= Z/2Z × Z/kZ.
as
r
4045 n Then σn,m acts on {0} × Z/kZ and {1} × Z/kZ with the
− ≥ 29 − n; same cycle structure, so every cycle length occurs an
n+1 2 even number of times. By Lemma 1, σn,m has a square
we then obtain an equivalent inequality by squaring root.
both sides: Finally, suppose that n is divisible by 4. For m = −1,
4045 n σn,m consists of two fixed points (0 and n/2) together
− ≥ n2 − 58n + 841. with n/2 − 1 cycles (an odd number) of length 2 (an
n+1 2 even number). By Lemma 1, σn,m does not have a
Clearing denominators, gathering all terms to one side, square root.
and factoring puts this in the form
B6 The determinant equals (−1)⌈n/2⌉−1 2⌈ n2 ⌉.
95 To begin with, we read off the following features of S.
(9 − n)(n2 − n + 356) ≥ 0.
2
– S is symmetric: Si j = S ji for all i, j, corresponding
The quadratic factor Q(n) has a minimum at 95 4 = 23.75 to (a, b) 7→ (b, a)).
and satisfies Q(8) = 40, Q(10) = −19; it is thus positive
for n ≤ 8 and negative for 10 ≤ n ≤ 29. – S11 = n + 1, corresponding to (a, b) =
(0, n), (1, n − 1), . . . , (n, 0).
B5 The desired property holds if and only if n = 1 or n ≡ 2 – If n = 2m is even, then Sm j = 3 for j = 1, m, cor-
(mod 4). responding to (a, b) = (2, 0), (1, 2nj ), (0, nj ).
Let σn,m be the permutation of Z/nZ induced by mul-
tiplication by m; the original problem asks for which n – For n2 < i ≤ n, Si j = #(Z ∩ { n−i n
j , j }), correspond-
does σn,m always have a square root. For n = 1, σn,m is ing to (a, b) = (1, n−i n
j ), (0, j ).
the identity permutation and hence has a square root.
We next identify when a general permutation admits a Let T be the matrix obtained from S by performing row
square root. and column operations as follows: for d = 2, . . . , n − 2,
subtract Snd times row n − 1 from row d and subtract
Lemma 1. A permutation σ in Sn can be written as the square Snd times column n − 1 from column d; then subtract
of another permutation if and only if for every even positive row n − 1 from row n and column n − 1 from column n.
integer m, the number of cycles of length m in σ is even. Evidently T is again symmetric and det(T ) = det(S).
 5

n
Let us examine row i of T for 2 < i < n − 1: In the odd case, we can strike the last two rows and
columns (creating another negation) and then conclude
Ti1 = Si1 − Sin S(n−1)1 = 2 − 1 · 2 = 0 at once. In the even case, the rows and columns are
Ti j = Si j − Sin S(n−1) j − Sn j Si(n−1) labeled 1, 2n , n − 1, n; by adding row/column n − 1 to
( row/column 2n , we produce
1 if j divides n − i
= (1 < j < n − 1)
0 otherwise. 
n+1 1 2 0

Ti(n−1) = Si(n−1) − Sin S(n−1)(n−1) = 0 − 1 · 0 = 0 ⌊n/2⌋  1 1 1 0
(−1) det 
2 1 0 1
Tin = Sin − Sin S(n−1)n − Si(n−1) = 1 − 1 · 1 − 0 = 0.
0 0 1 0
Now recall (e.g., from the expansion of a determinant in
minors) if a matrix contains an entry equal to 1 which is and we can again strike the last two rows and columns
the unique nonzero entry in either its row or its column, (creating another negation) and then read off the result.
then we may strike out this entry (meaning striking out
the row and column containing it) at the expense of mul- Remark. One can use a similar approach to compute
tiplying the determinant by a sign. To simplify notation, some related determinants. For example, let J be the
we do not renumber rows and columns after performing matrix with Ji j = 1 for all i, j. In terms of an indetermi-
this operation. nate q, define the matrix T by
We next verify that for the matrix T , for i = 2, . . . , ⌊ 2n ⌋
in turn, it is valid to strike out (i, n − i) and (n − Ti j = qSi j .
i, i) at the cost of multiplying the determinant by -1.
Namely, when we reach the entry (n − i, i), the only We then have
other nonzero entries in this row have the form (n − i, j)
where j > 1 divides n − i, and those entries are in pre-
viously struck columns. det(T − tJ) = (−1)⌈n/2⌉−1 q2(τ(n)−1) (q − 1)n−1 fn (q,t)
where τ(n) denotes the number of divisors of n and
We thus compute det(S) = det(T ) as:
  (
n + 1 −1 0 qn−1t + q2 − 2t for n odd,
(−1)⌊n/2⌋−1 det  −1 0 1 for n odd, fn (q,t) =
qn−1t + q2 − qt − t for n even.
0 1 0
n + 1 −1 2 0
 
Taking t = 1 and then dividing by (q − 1)n , this yields a
 −1 −1 1 −1
(−1)⌊n/2⌋−1 det  for n even. q-deformation of the original matrix S.
2 1 0 1
0 −1 1 0
 The 85th William Lowell Putnam Mathematical Competition
Saturday, December 7, 2024

A1 Determine all positive integers n for which there exist n squares from the grid, no two in the same row or col-
positive integers a, b, and c satisfying umn, such that the numbers contained in the selected
squares are exactly 1, 2, . . . , n?
2an + 3bn = 4cn .
B2 Two convex quadrilaterals are called partners if they
A2 For which real polynomials p is there a real polynomial have three vertices in common and they can be labeled
q such that ABCD and ABCE so that E is the reflection of D across
the perpendicular bisector of the diagonal AC. Is there
p(p(x)) − x = (p(x) − x)2 q(x) an infinite sequence of convex quadrilaterals such that
each quadrilateral is a partner of its successor and no
for all real x? two elements of the sequence are congruent? [A dia-
gram has been omitted.]
A3 Let S be the set of bijections
B3 Let rn be the nth smallest positive solution to tan x = x,
T : {1, 2, 3} × {1, 2, . . . , 2024} → {1, 2, . . . , 6072} where the argument of tangent is in radians. Prove that
such that T (1, j) < T (2, j) < T (3, j) for all j ∈ 1
{1, 2, . . . , 2024} and T (i, j) < T (i, j + 1) for all i ∈ 0 < rn+1 − rn − π <
(n2 + n)π
{1, 2, 3} and j ∈ {1, 2, . . . , 2023}. Do there exist a and c
in {1, 2, 3} and b and d in {1, 2, . . . , 2024} such that the for n ≥ 1.
fraction of elements T in S for which T (a, b) < T (c, d)
is at least 1/3 and at most 2/3? B4 Let n be a positive integer. Set an,0 = 1. For k ≥ 0,
choose an integer mn,k uniformly at random from the
A4 Find all primes p > 5 for which there exists an integer set {1, . . . , n}, and let
a and an integer r satisfying 1 ≤ r ≤ p − 1 with the fol-
lowing property: the sequence 1, a, a2 , . . . , a p−5 can be

an,k + 1, if mn,k > an,k ;

rearranged to form a sequence b0 , b1 , b2 , . . . , b p−5 such
an,k+1 = an,k , if mn,k = an,k ;
that bn − bn−1 − r is divisible by p for 1 ≤ n ≤ p − 5. 
a − 1, if m < a .
n,k n,k n,k
A5 Consider a circle Ω with radius 9 and center at the ori-
gin (0, 0), and a disc ∆ with radius 1 and center at (r, 0), Let E(n) be the expected value of an,n . Determine
where 0 ≤ r ≤ 8. Two points P and Q are chosen in- limn→∞ E(n)/n.
dependently and uniformly at random on Ω. Which
value(s) of r minimize the probability that the chord PQ B5 Let k and m be positive integers. For a positive in-
intersects ∆? teger n, let f (n) be the number of integer sequences
x1 , . . . , xk , y1 , . . . , ym , z satisfying 1 ≤ x1 ≤ · · · ≤ xk ≤ z ≤
A6 Let c0 , c1 , c2 , . . . be the sequence defined so that n and 1 ≤ y1 ≤ · · · ≤ ym ≤ z ≤ n. Show that f (n) can be
√ expressed as a polynomial in n with nonnegative coeffi-
1 − 3x − 1 − 14x + 9x2 ∞
cients.
= ∑ ck xk
4 k=0 2
B6 For a real number a, let Fa (x) = ∑n≥1 na e2n xn for 0 ≤
for sufficiently small x. For a positive integer n, let A x < 1. Find a real number c such that
be the n-by-n matrix with i, j-entry ci+ j−1 for i and j in
{1, . . . , n}. Find the determinant of A. lim Fa (x)e−1/(1−x) = 0 for all a < c, and
x→1−

B1 Let n and k be positive integers. The square in the ith lim Fa (x)e−1/(1−x) = ∞ for all a > c.
x→1−
row and jth column of an n-by-n grid contains the num-
ber i + j − k. For which n and k is it possible to select
 Solutions to the 85th William Lowell Putnam Mathematical Competition
Saturday, December 7, 2024
Manjul Bhargava, Kiran Kedlaya, and Lenny Ng

A1 The answer is n = 1. When n = 1, (a, b, c) = (1, 2, 2) is by r(x)2 . Then

a solution to the given equation. We claim that there are
no solutions when n ≥ 2. d
0≡ (p(p(x)) − x)
For n = 2, suppose that we have a solution to 2a2 + dx
3b2 = 4c2 with a, b, c ∈ N. By dividing each of a, b, c = p′ (x)p′ (p(x)) − 1
by gcd(a, b, c), we obtain another solution; thus we ≡ p′ (x)2 − 1
can assume that gcd(a, b, c) = 1. Note that we have
= (p′ (x) + 1)(p′ (x) − 1)
a2 + c2 ≡ 0 (mod 3), and that only 0 and 1 are per-
fect squares mod 3; thus we must have a2 ≡ c2 ≡ 0 = r′ (x)(r′ (x) + 2) (mod r(x)).
(mod 3). But then a, c are both multiples of 3; it follows
from b2 = 12(c/3)2 −6(a/3)2 that b is a multiple of 3 as If α is a root of r(x) of some multiplicity m, then the
well, contradicting our assumption that gcd(a, b, c) = 1. multiplicity of α as a root of r′ (x) is m − 1. Conse-
quently, every root of r(x) must be a root of r′ (x) + 2;
For n ≥ 3, suppose that 2an + 3bn = 4cn . As in the in particular such a root cannot also be a root of r′ (x),
previous case, we can assume gcd(a, b, c) = 1. Since so every root of r(x) is simple. Putting this together, we
3bn = 4cn − 2an , b must be even. We can then write deduce that r(x) divides r′ (x) + 2. If r(x) is constant,
an + 2n−1 · 3(b/2)n = 2cn , and so a must be even. then p(x) = x + c for some c. Otherwise, deg(r′ (x) +
Then 2n−1 (a/2)n + 2n−2 · 3(b/2)n = cn , and c must be 2) < deg(r(x)) and so r′ (x) + 2 must be zero; then
even as well. This contradicts our assumption that r(x) = −2x + c for some c, whence p(x) = −x + c.
gcd(a, b, c) = 1.
A3 Yes, such a, b, c, d exist: we take
A2 The answer is p(x) = ±x + c for any c ∈ R. Note that
any such polynomial works: if p(x) = x + c then p(x) − (a, b) = (2, 1), (c, d) = (1, 2).
x = c, while if p(x) = −x + c then p(p(x)) − x = 0. We
will show that in fact these are the only polynomials We will represent T as an 3 × n array (3 rows, n
p(x) such that p(p(x)) − x is divisible by r(x)2 , where columns) of integers in which each of 1, . . . , 3n occurs
r(x) = p(x) − x. exactly once and the rows and columns are strictly in-
First solution. Suppose that p(p(x)) − x is divisible by creasing; we will specialize to n = 2024 at the end.
r(x)2 . Then We first note that T (1, 1) = 1 and 2 ∈ {T (1, 2), T (2, 1)}.
From this, it follows that T (2, 1) < T (1, 2) if and only
x ≡ p(p(x)) if T (2, 1) = 2.
= p(x + r(x)) We next recall a restricted form of the hook length for-
≡ p(x) + p′ (x)r(x) (mod r(x)2 ). mula (see the first remark for a short proof of this re-
stricted version and the second remark for the state-
In other words, r(x)(1 + p′ (x)) is divisible by r(x)2 . ment of the general formula). Consider more gener-
From this, it follows that either r(x) = 0 or 1 + p′ (x) ally an array consisting of (up to) three rows of lengths
is divisible by r(x). In the first case, we have p(x) = x. n1 ≥ n2 ≥ n3 ≥ 0, aligned at the left. Let f (n1 , n2 , n3 ) be
In the second case, if 1 + p′ (x) = 0 then p(x) = −x + c the number of ways to fill this array with a permutation
for some constant c; otherwise, we have of the numbers 1, . . . , n1 + n2 + n3 in such a way that
each row increases from left to right and each column
deg(p) − 1 = deg(1 + p′ (x)) ≥ deg(r) increases from top to bottom. The hook length formula
then shows that f (n1 , n2 , n3 ) equals
and this is only possible if p(x) = x + c for some con-
stant c. (n1 − n2 + 1)(n1 − n3 + 2)(n2 − n3 + 1)(n1 + n2 + n3 )!
.
Second solution. Suppose that p(p(x)) − x is divisible (n1 + 2)!(n2 + 1)!n3 !

We then note that if T (2, 1) = 2, we obtain a array with

row lengths n, n − 1, n − 1 by removing 1 and 2, relabel-
ing each remaining i as 3n + 1 − i, and reflecting in both
 2

axes. The probability that T (2, 1) < T (1, 2) is thus diagram such that either i = i′ , j ≤ j′ or i ≤ i′ , j = j′ (in-
cluding s itself). Then the number of standard Young
f (n, n − 1, n − 1) (2)(3)(n + 1)n tableaux for this diagram equals
=
f (n, n, n) (1)(2)(3n)(3n − 1)
n+1 1 4 N!
= = + ; .
3n − 1 3 9n − 3 ∏s hs
For a proof along the lines of the argument given in the
this is always greater than 13 , and for n = 2024 it is vis-
previous remark, see: Kenneth Glass and Chi-Keung
ibly less than 32 . Ng, A simple proof of the hook length formula, Ameri-
Remark. We prove the claimed formula for can Mathematical Monthly 111 (2004), 700–704.
f (n1 , n2 , n3 ) by induction on n1 + n2 + n3 . To begin
with, if n2 = n3 = 0, then the desired count is indeed A4 The prime p = 7 works: choose a = 5 and r = 3, and
f (n1 , 0, 0) = 1. Next, suppose n2 > 0, n3 = 0. The entry note that 1, a, a2 can be rearranged to form b0 = 5, b1 =
n1 + n2 must go at the end of either the first or second 1, b2 = 25 satisfying the stated property.
row; counting ways to complete the diagram from these We claim that no prime p > 7 works. Suppose other-
starting points yields wise: there exist p, a, r with p > 7 and r ∤ p such that
1, a, . . . , a p−5 can be rearranged to form b0 , . . . , b p−5
f (n1 , n2 , 0) = f (n1 − 1, n2 , 0) + f (n1 , n2 − 1, 0). with bn ≡ b0 + nr (mod p) for all 0 ≤ n ≤ p − 5. Since
r ∤ p, {b0 , b0 + r, . . . , b0 + (p − 5)r} represents a collec-
(This works even if n1 = n2 , in which case the first row tion of p − 4 distinct elements of Z/pZ. It follows that
is not an option but correspondingly f (n2 − 1, n2 , 0) = all of 1, a, . . . , a p−5 are distinct mod p. In particular,
0.) The induction step then follows from the identity p ∤ a; also, since p − 5 ≥ p−1 k
2 , we conclude that a ̸≡ 1
p−1
(n1 − n2 )(n1 + 1) + (n1 − n2 + 2)n2 (mod p) for any 1 ≤ k ≤ 2 . It follows that a is a
= 1. primitive root mod p.
(n1 − n2 + 1)(n1 + n2 )
Since a is a primitive root, a−3 , a−2 , a−1 , a0 , . . . , a p−5
(As an aside, the case n1 = n2 , n3 = 0 recovers a stan- runs through all nonzero elements of Z/p exactly
dard interpretation of the Catalan numbers.) once. On the other hand, b0 − 4r, b0 − 3r, b0 − 2r, b0 −
Finally, suppose n3 > 0. We then have r, b0 , . . . , b0 + (p − 5)r runs through all elements of
Z/pZ exactly once. The given condition now implies
f (n1 , n2 , n3 ) that
= f (n1 − 1, n2 , n3 ) + f (n1 , n2 − 1, n3 ) + f (n1 , n2 , n3 − 1),
{b0 − 4r, b0 − 3r, b0 − 2r, b0 − r} = {0, c, c2 , c3 }
and the induction step now reduces to the algebraic
identity where c = a−1 ; that is, 0, c, c2 , c3 can be rearranged to
give an arithmetic sequence x1 , x2 , x3 , x4 in Z/pZ.
(n1 − n2 )(n1 − n3 + 1)(n1 + 2)
If 0, c, c2 , c3 can be arranged into a four-term arithmetic
(n1 − n2 + 1)(n1 − n3 + 2)(n1 + n2 + n3 ) progression, then by dividing the progression by c, we
(n1 − n2 + 2)(n2 − n3 )(n2 + 1) see that 0, 1, c, c2 can also be arranged into a four-term
+
(n1 − n2 + 1)(n2 − n3 + 1)(n1 + n2 + n3 ) arithmetic progression. Now no two of 1, c, c2 can both
(n1 − n3 + 3)(n2 − n3 + 2)n3 be adjacent to 0 in this arithmetic progression, or oth-
+ = 1. erwise they would be negative of each other; but this
(n1 − n3 + 2)(n2 − n3 + 1)(n1 + n2 + n3 )
is impossible because the order of c is greater than 4.
We conclude that 0 must be either the first or the last
Remark. We formulate the general hook length for- term of the progression, and by reversing the sequence
mula in standard terminology. Let N be a positive inte- if necessary, we can assume that 0 is the first term of
ger, and consider a semi-infinite checkerboard with top the progression. Now the last three terms of this pro-
and left edges. A Ferrers diagram is a finite subset of gression cannot be 1, c, c2 or c2 , c, 1 in that order, as
the squares of the board which is closed under taking a c − 1 ̸= c2 − c because c ̸= 1. Thus the only possibil-
unit step towards either edge. Given a Ferrers diagram ities for the arithmetic progression that remain are
with N squares, a standard Young tableau for this dia-
gram is a bijection of the squares of the diagram with 0, 1, c2 , c; 0, c2 , 1, c;
the integers 1, . . . , N such that the numbers always in-
crease under taking a unit step away from either edge. 0, c, 1, c2 ; 0, c, c2 , 1.
For each square s = (i, j) in the diagram, the hook As twice the second term must be the third term, and
length hs of s is the number of squares (i′ , j′ ) in the thrice the second term must be the fourth term, we im-
mediately eliminate each of the above possibilities: the
first sequence is not possible because we must have
 3

c2 = 2, c = 3, which is a valid solution only when To see this, note that the length of a minor arc of a cir-
p = 7; for the second sequence, we must have 1 = 2c2 cle is a strictly increasing function of the length of the
and 1 = 3c, which is again a valid solution only when chord connecting the two endpoints of the arc. In this
p = 7; for the third sequence, we must have 1 = 2c and case, the chord connects points on the two given paral-
c2 = 3c, implying c = 1/2 = 3, which is possible only lel lines, so the distance between these points is min-
when p = 5; and for the fourth sequence, we must have imized by having them be the endpoints of a segment
c2 = 2c and 1 = 3c, implying c = 2 = 1/3, which is perpendicular to the two lines; this achieves the situa-
again possible only when p = 5. tion described above.
A5 We will show that r = 0 (and no other value of r) mini- A6 The determinant equals 10n(n−1)/2 . We compute the
mizes the stated probability. Note that P and Q coincide corresponding determinant for the coefficients of the
with probability 0; thus we can assume that P ̸= Q. generic power series
First solution. First restrict P, Q to points on Ω such ∞
that the segment PQ makes an angle of θ with the y f (x) := ∑ cn x n , c1 = 1,
axis, where θ is a fixed number with −π/2 < θ ≤ π/2. n=1
By rotating the diagram by −θ around the origin, we
move PQ to be a vertical line and move ∆ to be cen- with associated continued fraction
tered at (r cos θ , −r sin θ ). In this rotated picture, P a0
a1 , a0 = 1.
and Q are at (9 cos φ , ±9 sin φ ) where φ is chosen uni- x−1 + b0 + x−1 +b +··· 1
formly at random in (0, π). Now the vertical tangent
lines to the boundary of ∆, x = r cos θ ± 1, intersect If we truncate by replacing an+1 = 0, we get a ra-
the y > 0 semicircle −1 )

of Ω at (9 cos φ , 9 sin φ ) where tional function which can be written as ABn (x −1 where
φ = cos−1 r cos9θ ±1 . Thus the probability that PQ in- n (x )
An (x), Bn (x) are polynomials determined by the initial
tersects ∆ for a specific value of θ is π1 f (r, θ ), where conditions
we define

r cos θ − 1
 
r cos θ + 1
 A−1 (x) = 1, A0 (x) = 0, B−1 (x) = 0, B0 (x) = 1
f (r, θ ) = cos−1 − cos−1 .
9 9 and the recurrences

If we now allow θ to vary (uniformly) in (−π/2, π/2], An+1 (x) = (x + bn )An (x) + an An−1 (x) (n > 0)
we find that the overall probability that PQ intersects ∆ Bn+1 (x) = (x + bn )Bn (x) + an Bn−1 (x) (n > 0).
is
Since each additional truncation accounts for two more
1
Z π/2
P(r) = f (r, θ ) dθ . coefficients of the power series, we have
π2 −π/2
An (x−1 )
The function P(r) is differentiable with = f (x) + O(x2n+1 ),
Bn (x−1 )
1 ∂ f (r, θ )
Z π/2
P′ (r) = dθ . or equivalently (since Bn (x) is monic of degree n)
π2 −π/2 ∂r
f (x)Bn (x−1 ) − An (x−1 ) = O(xn+1 ). (1)
Now
∂ f (r, θ )  We now reinterpret in the language of orthogonal poly-
= (cost) (80 − 2r cost − r2 cos2 t)−1/2 nomials. For a polynomial P(x) = ∑i Pi xi , define
∂r 
−(80 + 2r cost − r2 cos2 t)−1/2 , Z
P(x) = ∑ Pi ci+1 ;
µ i
which, for t ∈ (−π/2, π/2), is zero for r = 0 and strictly
positive for r > 0. It follows that P′ (0) = 0 and P′ (r) < then the vanishing of the coefficient of xi+1 in (1) (with
0 for r ∈ (0, 8], whence P(r) is minimized when r = 0. n := i) implies that
Second solution. (based on ideas from Elliott Liu, Z
Bjorn Poonen, Linus Tang, and Allen Wang) We inter- xi B j (x) = 0 ( j < i).
pret the first paragraph of the first solution as reducing µ
the original problem to the following assertion: given
By expanding 0 = µ xi−1 Bi+1 (x) using the recurrence,
R
two parallel lines at distance 2, both of which intersect
we deduce that µ xi Bi (x) + ai µ xi−1 Bi−1 (x) = 0, and
R R
a circle of radius 9, the length of either of the two con-
gruent arcs of the circle lying between the two lines is so
minimized when the line halfway between the two par- Z
allel lines passes through the center of the circle. xi Bi (x) = (−1)i a1 · · · ai .
µ
 4

We deduce that In particular, mi1 = 0 for i ≥ 2. Starting from M, for j =

( 2, . . . , n − 1 in turn, for k = j + 1, . . . , n subtract ak− j+1 times
0 i ̸= j
Z
column j from column i. The resulting matrix has first col-
Bi (x)B j (x) = (2)
µ (−1)i a1 · · · ai i = j. umn (1, 0, . . . , 0) and removing its first row and column leaves
Hn−1 (β ), yielding the claimed equality.
In other words, for U the n×n matrix such that Ui j is the
coefficient of x j in Bi (x), the matrix UAU t is a diagonal Remark. A matrix A whose i, j-entry depends only
matrix D with diagonal entries Di,i = (−1)i−1 a1 · · · ai−1 on i + j is called a Hankel matrix. The above compu-
for i = 1, . . . , n. Since U is a unipotent matrix, its deter- tation of the determinant of a Hankel matrix in terms
minant is 1; we conclude that of continued fractions is adapted from H.S. Wall, An-
alytic Theory of Continued Fractions, Theorems 50.1
det(A) = det(D) = (−1)n(n−1)/2 an−1
1 · · · an−1 . and 51.1.
The same analysis shows that if we define the sequence
We now return to the sequence {cn } given in the prob- {cn }n=1 by c1 = 1 and
lem statement, for which
√ n−1
1 − 3x − 1 − 14x + 9x2 cn = acn−1 + b ∑ ci cn−i (n > 1),
f (x) = . i=1
4
For then an = −ab − b2 , bn = −a − 2b for all n > 0 and so
√
1 − 7x − 1 − 14x + 9x2 det(A) = (ab + b2 )n(n−1)/2 ;
g(x) := ,
2x
the problem statement is the case a = 3, b = 2. The case
we have a = 0, b = 1 yields the sequence of Catalan numbers; the
case a = 1, b = 1 yields the Schröder numbers (OEIS
1 10
f (x) = , g(x) = . sequence A006318).
x−1 − 5 − g(x) x−1 − 7 − g(x)
There are a number of additional cases of Hankel deter-
This means that the continued fraction is eventually pe- minants of interest in combinatorics. For a survey, see:
riodic; in particular, a1 = a2 = · · · = −10. Plugging into A. Junod, Hankel determinants and orthogonal polyno-
the general formula for det(A) yields the desired result. mials, Expositiones Mathematicae 21 (2003), 63–74.
This yields the desired result.
B1 This is possible if and only if n is odd and k = (n+1)/2.
Reinterpretation. (suggested by Bjorn Poonen) Given
a formal Laurent series α = ∑i ai xi , define the matrices We first check that these conditions are necessary. If
Hn (α) = (ai+ j−1 )ni, j=1 and the determinants hn (α) = the pairs (a1 , b1 ), . . . , (an , bn ) index squares of the grid
with no two in the same row or column, then each of
det Hn (α). One can then recover the evaluation of the
the two sequences a1 , . . . , an and b1 , . . . , bn is a per-
determinants from the following lemma.
mutation of {1, . . . , n}, and so in particular has sum
Lemma. Suppose α = ∑∞ i
i=1 ai x is a formal power series with 1 + · · · + n = n(n+1)2 . In particular, if the selected num-
ai = 1. Define the power series β by α −1 = x−1 − β . Then for bers are 1, 2, . . . , n in some order, then
all n ≥ 1, hn (α) = hn−1 (β ).
n
n(n + 1)
= ∑ (ai + bi − k)
Proof. For m ≥ 2, by equating the coefficients of xm in the 2 i=1
equality x−1 α = αβ + 1, we obtain n n n
= ∑ ai + ∑ bi − ∑ k
m
i=1 i=1 i=1
am+1 = ∑ ar bm−r .
r=1 n(n + 1) n(n + 1)
= + − nk
2 2
We now perform some row and column reduction on Hn (α)
without changing its determinant. Starting with Hn (α), for which simplifies to k = (n + 1)/2.
i = n, n − 1, . . . , 2 in turn, for k = 1, . . . , i − 1 subtract bi−1−k We next check that these conditions are sufficient. For
times row k from row i. In light of the recurrence relation, the this, it suffices to observe that the sequence
resulting matrix M = (mi j ) has the property that for i ≥ 2,      
n+1 n+3 n+1
i−1 1, , 2, ,..., ,n ,
2 2 2
mi j = ai+ j−1 − ∑ a j+k−1 bi−1s−k    
k=1 n+3 n−1
, 1 , . . . , n,
j−1 2 2
= ∑ ar bi+ j−2−r .
r=1
 5

of grid entries equals respectively. Since we want a convex quadrilateral, we

may assume without loss of generality that y, w > 0. The
1, 3, . . . , n, 2, . . . , n − 1. area of the quadrilateral is 21 a(y + w), which we also
want to fix; we may thus regard w as a function of y
We illustrate this for the case n = 5, k = 3 below; the (possibly restricting y to a range for which w > 0). After
selected entries are parenthesized. splitting the semicircles on which R and S lie into two
arcs each, we may also regard x and w as functions of y.
−1 0 (1) 2 3
 
2
0 1 2 (3) 4 It now suffices to observe that RS = (z − x)2 + (w − y)2
1

2 3 4 (5)
 is a nonconstant algebraic function of y, so it takes any
(2) 3 4 5 6 given value only finitely many times.
3 (4) 5 6 7 Second solution. Let ABCD be the first quadrilateral in
the sequence. Since the quadrilateral is convex, the di-
B2 No, there is no such sequence. In other words, any se- agonals AC and BD intersect. In particular they are not
quence of convex quadrilaterals with the property that parallel, so their perpendicular bisectors are not parallel
any two consecutive terms are partners must be finite. either; let O be the intersection of the bisectors.
First solution. We claim that the point O remains fixed throughout
the sequence, as do the distances OA, OB, OC, OD. To
Lemma. Given five positive real numbers a, b, c, d, K, there see this, we check this for two partners as described in
are only finitely many convex quadrilaterals with side lengths the problem statement: the diagonal BD gets reflected
a, b, c, d in that order and area K. across the perpendicular bisector of AC, so its perpen-
dicular bisector also gets reflected; the point O is the
Proof. Let PQRS be a convex quadrilateral with unique point on the perpendicular bisector of BD fixed
by the reflection. In particular, the segments OD and
PQ = a, QR = b, RS = c, SP = d. OE are mirror images across the perpendicular bisector
of AC, so their lengths coincide.
Then the congruence class of PQRS is uniquely determined by
the length of the diagonal f := PR. Moreover, as f increases, As noted in the first solution, the unordered list of
the angles ∠RPQ and ∠RPS are both strictly decreasing, so side lengths of the quadrilateral also remains invari-
∠SPQ is decreasing; by the same logic, ∠QRS is decreasing. ant throughout the sequence. Consequently, the un-
ordered list of side lengths of each of the triangles
We next recall Bretschneider’s formula: for s = (a + b + c + △OAB, △OBC, △OCD, △ODA is limited to a finite set;
d)/2, each such list uniquely determines the unoriented con-
gruence class of the corresponding triangle, and lim-
∠SPQ + ∠QRS
K 2 = (s − a)(s − b)(s − c)(s − d) − abcd cos2 . its the oriented congruence class to two possibilities.
2 Given the oriented congruence classes of the four tri-
angles we can reconstruct the quadrilateral ABCD up to
Consequently, fixing K also fixes cos2 ∠SPQ+∠QRS
2 , and thus
oriented congruence (even up to rotation around O); this
limits ∠SPQ + ∠QRS to one of two values. By the previous
proves that the sequence must be finite.
paragraph, this leaves at most two possible congruence classes
for the triangle. B3 First solution. (by Bjorn Poonen) Let tan−1 : R →
(− π2 , π2 ) be the principal branch of the arctangent func-
Returning to our original sequence, note that any two tion, and set t(x) := x − tan−1 (x). Then t(0) = 0 and
consecutive quadrilaterals in the sequence have the
same area and the same unordered list of side lengths. dt 1 x2
The latter can occur as an ordered list in at most six dif- = 1− = >0 (x ̸= 0),
dx 1 + x2 1 + x2
ferent ways (up to cyclic shift); for each of these, we
can have only finitely many distinct congruence classes so t(x) is strictly increasing. We have tan x = x if and
of quadrilaterals in our sequence with that area and or- only if x = tan−1 x + nπ for some n ∈ Z; from the previ-
dered list of side lengths. We deduce that our sequence ous analysis it follows that rn is the unique solution of
must be finite. t(x) = nπ.
Remark. Various proofs of the lemma are possible; for Let x(t) be the inverse function of t(x), so that rn =
example, here is one using Cartesian coordinates. We x(nπ). We compute that
first specify
dx 1 1
P = (0, 0), Q = (a, 0). −1 = −1 = 2
dt dt/dx x
dx 1 1 1
For two additional points R = (x, y), S = (z, w), the con- −1− 2 = 2 − 2.
ditions QR = b, SP = d restrict R and S to the circles dt t x t

(x − a)2 + y2 = b2 , z2 + w2 = d 2
 6

From this we deduce that x(t) − t is strictly increasing Taking the limit as k → ∞ yields
for t > 0 (as then x(t) > 0) and x(t) − t + 1t is strictly
decreasing for t > 0 (as then tan−1 (x(t)) > 0 and so
∞
xi
f (rn + π + x) ≥ f (rn + π) + ∑ f (i) (rn + π)
t < x(t)). Evaluating at t = nπ and t = (n + 1)π, we i=1 i!
obtain k
> −π + ∑ ni+1 π i+1 xi
rn − nπ < rn+1 − (n + 1)π i=1
1 1 n2 π 2 x
rn − nπ + > rn+1 − (n + 1)π + , > −π + ;
nπ (n + 1)π 1 − nπx
which are the desired inequalities. taking x = δ yields
Second solution. Define the function  
1
0 > −π + nπ −1
f (x) := tan x − x. 1 − nπδ
1
We then have f ′ (x) = tan2 x. By induction on k, f (k) (x) and so δ < n(n+1)π as desired.
is a polynomial of degree k + 1 in tan x with leading Remark. There is a mild subtlety hidden in the proof:
coefficient k! and all coefficients nonnegative. In par- if one first bounds the finite sum as
ticular, on each of the intervals
k
π f (rn + π + x) > −π + ∑ ni+1 π i+1 xi

In := nπ, nπ + (n = 0, 1, . . . ),
2 i=1

tan x is positive and so f (k) (x) is positive for each k ≥ 1; and then takes the limit as k → ∞, the strict inequality is
replacing k with k + 1, we deduce that each f (k) (x) is not preserved. One way around this is to write f ′′ (rn ) =
strictly increasing on In for k ≥ 0. 2rn + 2rn3 , retain the extra term rn x2 in the lower bound,
take the limit as k → ∞, and then discard the extra term
We now analyze f more closely on In . As x → nπ + to get back to a strict inequality.
for n > 0, f (x) tends to f (nπ) = −nπ < 0; by contrast, 1
as x → 0+ , f (x) tends to 0 via positive values. In ei- Remark. The slightly weaker inequality δ < n2 π
fol-
ther case, as x → (nπ + π2 )− , f (x) → ∞. Since f (x) is lows at once from the inequality
strictly increasing on In , we deduce using the interme-
diate value theorem that: f ′ (rn + π) = f ′ (rn ) = tan2 rn = rn2 > n2 π 2

– f (x) has no zero in I0 ; plus the mean value theorem.

– for n > 0, f (x) has a unique zero in In . Remark. One can also reach the desired upper bound
by comparing rn+1 to rn + π using the addition formula
Since f (x) also has no zero between In and In+1 (as it for tangents:
takes exclusively negative values there), we deduce that
tan x − tan y
π tan(x + y) = .
nπ < rn < nπ + . 1 + tan x tan y
2
Namely, one then gets
This already suffices to prove the claimed lower bound:
since f (rn + π) = −π < 0 and f is strictly increasing tan rn+1 − tan(rn + π)
on In+1 , the quantity δ := rn+1 − (rn + π) is positive. δ < tan δ =
1 + tan rn+1 tan(rn + π)
To prove the upper bound, note that for k ≥ 1, for 0 < rn+1 − rn π +δ
x < nπ + π2 − rn , we have = =
1 + rn rn+1 1 + rn rn+1

f (k) (x) ≥ f (k) (rn + π) = f (k) (rn ) and hence

≥ k!rnk+1 > k!n k+1 k+1
π . π π 1
δ< < = 2 .
rn rn+1 (nπ)((n + 1)π) (n + n)π
For each k ≥ 2, we may apply the mean value theorem
with remainder to deduce that for x in the same range,
1−e−2
B4 The limit equals 2 .
k
xi First solution. We first reformulate the problem as a
f (rn + π + x) ≥ f (rn + π) + ∑ f (i) (rn + π) .
i=1 i! Markov chain. Let vk be the column vector of length n
whose i-th entry is the probability that an,k = i, so that v0
 7

is the vector (1, 0, . . . , 0). Then for all k ≥ 0, vk+1 = Avk repeatedly perform the following operation. Pick one
where A is the n × n matrix defined by light bulb uniformly at random. If it is the red bulb, do
 nothing; otherwise, switch the bulb from lit to unlit or
1

 n if i = j vice versa. After k operations of this form, the random
 j−1 if i = j − 1

variable an,k is equal to the number of lit bulbs (includ-
Ai j = n−n j ing the red bulb).
if i = j + 1
 n


We may then compute the expected value of an,n by

0 otherwise.
summing over bulbs. The red bulb contributes 1 no mat-
Let w be the row vector (1, . . . , n); then the expected ter what. Each other bulb contributes 1 if it is switched
value of an,k is the sole entry of the 1 × 1 matrix wvk = an odd number of times and 0 if it is switched an even
wAk v0 . In particular, E(n) = wAn v0 . number of times, or equivalently 12 (1 − (−1) j ) where
j is the number of times this bulb is switched. Hence
We compute some left eigenvectors of A. First, each bulb other than the red bulb contributes
w0 := (1, . . . , 1) n  
−n 1 i n
n ∑ (1 − (−1) ) (n − 1)n−i
satisfies Aw0 = w0 . Second, i=0 2 i
!
n−n n n n
   
n−i i n n−i
w1 := (n − 1, n − 3, . . . , 3 − n, 1 − n) =
2 ∑ (n − 1) − ∑ (−1) i (n − 1)
i=0 i i=0
= (n − 2 j + 1 : j = 1, . . . , n)
n−n
n−2
= ((1 + (n − 1))n − (−1 + (n − 1))n )
satisfies Aw1 = n w1 : the j-th entry of Awi equals 2
n−n 2
j−1 1 n− j = (n − (n − 2)n )
(n + 3 − 2 j) + (n + 1 − 2 j) + (n − 1 − 2 j) 2
n n n 2 n
 
1 1
n−2 = − 1− .
= (n − 2 j + 1). 2 2 n
n
−2
By the same token, we obtain This tends to 1−e2 as n → ∞. Since E(n) equals n − 1
times this contribution plus 1, E(n)
n tends to the same
n+1 1 limit.
w= w0 − w1 ;
2 2
Third solution. We compare the effect of taking an,0 =
we then have j versus an,0 = j + 1 for some j ∈ {1, . . . , n − 1}. If
mn,0 ∈ { j, j +1} then the values of an,1 coincide, as then
E(n) n + 1 1 do the subsequent values of an,k ; this occurs with prob-
= w0 An v0 − w1 An v0
n 2n 2n ability n2 . Otherwise, the values of an,1 differ by 1 and
2 n
 
n+1 1 the situation repeats.
= w0 v0 − 1− w1 v0
2n 2n n Iterating, we see that the two sequences remain
n
n 1 apart
(in the same direction) with probability n−2

n+1 n−1 2 n and con-
= − 1− .
2n 2n n verge otherwise. Consequently, changing the start value
from j to j + 1 increases the expected value of an,n by
In the limit, we obtain n−2 n


n .
2 n Now let c be the expected value of an,n in the original
 
E(n) 1 1
lim = − lim 1 − setting where an,0 = 1. By symmetry, if we started with
n→∞ n 2 2 n→∞ n
an,0 = n the expected value would change from c to n +
1 1 −2
= − e . 1 − c; on the other hand, by the

n previous paragraph it
2 2 would increase by (n − 1) n−2 . We deduce that
n

Remark. With a bit more work, one can show that A 1

 
n−2 n
 
has eigenvalues n−2 j
n for j = 0, . . . , n − 1, and find the
c= n + 1 − (n − 1)
2 n
corresponding left and right eigenvectors. In particular,
it is also possible (but much more complicated) to ex- and as above this yields the claimed limit.
press v0 as a linear combination of right eigenvectors
and use this to calculate An v0 . B5 For convenience, we extend the problem to allow non-
Second solution. We reinterpret the Markov chain in negative values for k and m.
combinatorial terms. Consider an apparatus consisting First solution. Let R(n, k) denote the number of subsets
of one red light bulb, which is initially lit, plus n − 1 of {1, ..., n} of size k where repetitions are allowed. The
white light bulbs, which are initially unlit. We then
 8

“sticks and stones” argument shows that which we prove by induction on k. The base case k = 0
  is evident; given (4) with k replaced by k − 1, we apply
n+k−1 (3) to obtain
R(n, k) = :
k
fk,m (n)
there is a bijection of these subsets with linear arrange-
ments of k (unlabeled) sticks and z − 1 (unlabeled) R(n, m + 1) k−1
= R(n, k)R(n, m + 1) + ∑ R(m + 1, i)R(n, i)
stones, where we recover the subset by counting the R(m + 2, k − 1) i=0
number of stones to the left of each stick. R(n, m + 2) k−1
Let fk,m (n) := ∑nz=1 R(z, k)R(z, m).
It is known that for − ∑ R(m + 2, i)R(n, i)
R(m + 3, k − 1) i=0
any positive integer k, the sum of the k-th powers of all
positive integers less than or equal to n is a polynomial R(n, m + 1) k
= ∑ R(m + 1, i)R(n, i)
in n (given explicitly in terms of Bernoulli numbers via R(m + 2, k) i=0
Faulhaber’s formula); hence fk,m (n) is a polynomial in
n. We wish to show that this polynomial has nonnega- yielding (4) as written.
tive coefficients. Since R(n, i) = n(n + 1)(n + 2) · · · (n + i − 1)/i! clearly
Using the recursion for binomial coefficients, we obtain has positive coefficients for all i, the explicit formula
(4) implies that fk,m (n) also has positive coefficients for
R(n, k)R(n, m) = fk,m (n) − fk,m (n − 1) all k and m.
n
Second solution. (by an anonymous Putnam partici-
= ∑ (R(z, k)R(z, m) − R(z − 1, k)R(z − 1, m)) pant) As in the first solution, we deduce that fk,m (n)
z=1
n is a polynomial in n of degree k + m + 1 satisfying
= ∑ (R(z, k)R(z, m) − R(z − 1, k)R(z, m) fk,m (0) = 0 and fk,m (n) − fk,m (n − 1) = R(n, k)R(n, m).
z=1 Since fk,m (n) > 0 for n ≫ 0, this polynomial has posi-
+R(z − 1, k)R(z, m) − R(z − 1, k)R(z − 1, m)) tive leading coefficient. To prove that it has nonnegative
n coefficients, it will suffice to prove the stronger asser-
= ∑ (R(z, k − 1)R(z, m) + R(z − 1, k)R(z, m − 1)) tion that the roots of fk,m (x) are all real and nonpositive,
z=1 as then this will imply that fk,m (x) = c ∏k+m j=0 (x + r j ) for
n some r j ≥ 0.
= ∑ (R(z, k − 1)R(z, m) Since R(n, m) = 0 for m = 0, −1, . . . , −m+1, we deduce
z=1
that fk,m (n) = 0 for n = 0, −1, . . . , −m. Consequently,
+(R(z, k) − R(z, k − 1))R(z, m − 1))
fk,m (x) can be written as x(x + 1) · · · (x + m)Q(x) for
= fk−1,m (n) + fk,m−1 (n) − fk−1,m−1 (n). some polynomial Q(x) of degree k, and it will suffice
to check that Q(x) has k distinct negative real roots.
It follows from the latter equation (replacing the index
m by m + 1) that From the equality fk,m (n) − fk,m (n − 1) =
R(n, k)R(n, m), if we substitute in for Q(x) and
fk,m (n) = R(n, k)R(n, m + 1) + fk−1,m (n) − fk−1,m+1 (n); (3) divide out common factors, we obtain

this can also be recovered by applying Abel summation 1

(x + m)Q(x) − (x − 1)Q(x − 1) = R(x, k).
(summation by parts) to ∑nz=1 R(z, k)R(z, m). m!
Using (3), we can evaluate fk,m by induction on k: for Substituting x = 0, −1, . . . , −k + 1 in turn, we obtain
the first few values we obtain
j+1
Q(− j) = − Q(− j − 1) ( j = 0, . . . , k − 1).
f0,m (n) = R(n, m + 1) m− j
f1,m (n) = R(n, 1)R(n, m + 1) + R(n, m + 1) − R(n, m + 2)
In particular, if any of Q(0), . . . , Q(−k) were zero, then
= R(n, m + 1)((m + 1)n + 1)/(m + 2) all of them would be zero and Q would have too many
R(m + 1, 1)R(n, 1) + 1 roots for its degree. Consequently, Q(0), . . . , Q(−k) are
= R(n, m + 1)
m+2 all nonzero and alternating in sign. By the intermediate
value theorem, Q has a root r j in the interval (− j −
and similarly 1, − j) for j = 0, . . . , k − 1; this completes the proof.
f2,m (n) = R(n, m + 1)(R(m + 1, 2)R(n, 2) + R(m + 1, 1)R(n, 1) B6 The claim holds with c = − 21 . Set t := 1/(1 − x), so that
+ R(m + 1, 0)R(n, 0))/R(m + 2, 2). x = 1 − 1/t and

This leads us to conjecture that 1 1 1

− − 2 ≤ log x ≤ − .
t t t
R(n, m + 1) k
fk,m (n) = ∑ R(m + 1, i)R(n, i), (4)
R(m + 2, k) i=0
 9

Set also m := ⌊t⌋. In the following arguments, we use c Suppose next that a < − 21 . Then
to refer to some positive constant independent of n and
∞
t, but a different such constant at each appearance. 2
Fa (x)e−t = ∑ na e2n−t xn
Suppose first that a > − 21 . Then n=1
∞
2
−t
Fa (x)e =
∞
a 2n−t n2
≤ ∑ na e−t(1−n/t) .
∑n e x n=1
n=1
∞
2 2 2 Fix ε > 0 such that a + ε < − 12 . For the summands
≥ ∑ na e2n−t−n /t−n /t with t − t 1/2+ε < n < t + t 1/2+ε , we may bound the
n=1
∞ summand from above by ct a ; this range of the sum is
2 2 2
= ∑ na e−n /t e−t(1−n/t) . then dominated by ct a+1/2+ε . For the summands with
2ε
n=1 n < t − t 1/2+ε , we may bound the summand by na e−t ;
√ this range of the sum is then dominated by te−t . For
2ε
If we restrict the sum to the range t < n < t + t, we 1/2+ε
a
may bound the summand from below by ct ; we then the summands with n > t −t , we may again bound
a −t 2ε
have Fa (x)e−t > ct a+1/2 and this tends to ∞ as t → ∞. the summand by n e ; this range of the sum is then
2ε
dominated by ct a+1 e−t . Since all three bounds tends
to 0 as t → ∞, so then does Fa (x)e−t .