Solutions

The 12-tuple (n10 , n9 , n8 , n7 , n6 , n5 , n4 , n3 , n2 , n1 , n0 , n∅ ) following the problem num-

ber gives the performance of the top 200 or so competitors on that problem: ni is the
number who scored i, and n∅ is the number of blank papers.

51
 Solutions: The Forty-Sixth Competition (1985) 53

The Forty-Sixth William Lowell Putnam Mathematical Competition

December 7, 1985

A1. (125, 6, 0, 0, 0, 0, 0, 0, 0, 0, 61, 9)

Determine, with proof, the number of ordered triples (A1 , A2 , A3 ) of sets
which have the property that
(i) A1 ∪ A2 ∪ A3 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, and
(ii) A1 ∩ A2 ∩ A3 = ∅,
where ∅ denotes the empty set. Express the answer in the form 2a 3b 5c 7d ,
where a, b, c, and d are nonnegative integers.
Answer. The number of such triples of sets is 210 310 .

Solution. There is a bijection between triples of subsets of {1, . . . , 10} and 10 × 3

matrices with 0, 1 entries, sending (A1 , A2 , A3 ) to the matrix B = (bij ) with bij = 1 if
i ∈ Aj and bij = 0 otherwise. Under this bijection the set S of triples satisfying

A1 ∪ A2 ∪ A3 = {1, . . . , 10} and A1 ∩ A2 ∩ A3 = ∅

maps onto the set T of 10 × 3 matrices with 0, 1 entries such that no row is (000) or
(111). The number of possibilities for each row of such a matrix is 23 − 2 = 6, so
#T = 610 . Hence #S = #T = 210 310 . 
Reinterpretation. Equivalently, this problem asks for the number of ways of placing
the numbers 1 through 10 in the Venn diagram of Figure 1, where no numbers are
placed in the two regions marked with an “×”.

A1

A2 A3

FIGURE 1.
Venn diagram interpretation of the solution to 1985A1.
54 The William Lowell Putnam Mathematical Competition

A2. (29, 15, 31, 14, 38, 11, 2, 15, 6, 4, 21, 15)
Let T be an acute triangle. Inscribe a pair R, S of rectangles in T as
shown:

❏
❏
S
❏
❏
❏
❏
R ❏
❏
❏

Let A(X) denote the area of polygon X. Find the maximum value, or show
that no maximum exists, of A(R)+A(S)
A(T ) , where T ranges over all triangles
and R, S over all rectangles as above.

❆
❆
Rn−1 ❆
❆
Rn−2 ❆
.. ❆
. ❆
❆
❆
R2 ❆
❆
R1
❆
❆
❆
FIGURE 2.
A generalization of 1985A2.

A(R)+A(S)
Answer. The maximum value of A(T ) exists and equals 2/3.
Solution. In fact, for any n ≥ 2, we can find the maximum value of
A(R1 ) + · · · + A(Rn−1 )
A(T )
for any stack of rectangles inscribed in T as shown in Figure 2. The altitude of T
divides T into right triangles U on the left and V on the right. For i = 1, . . . , n − 1, let
Ui denote the small right triangle to the left of Ri , and let Un denote the small right
triangle above Rn−1 and to the left of the altitude of T . Symmetrically define V1 , . . . ,
Vn to be the right triangles on the right. Each Ui is similar to U , so A(Ui ) = a2i A(U ),
where ai is the altitude of Ui , measured as a fraction of the altitude of T . Similarly,
A(Vi ) = a2i A(V ). Hence
A(U1 ) + · · · + A(Un ) + A(V1 ) + · · · + A(Vn ) = (a21 + · · · + a2n )(A(U ) + A(V ))
= (a21 + · · · + a2n )A(T ).
Solutions: The Forty-Sixth Competition (1985) 55

Since T is the disjoint union of all the Ri , Ui , and Vi ,

A(R1 ) + · · · + A(Rn−1 ) A(U1 ) + · · · + A(Un ) + A(V1 ) + · · · + A(Vn )
=1−
A(T ) A(T )
= 1 − (a21 + · · · + a2n ).

The ai must be positive numbers with sum 1, and conversely any such ai give rise to
a stack of rectangles in T .
It remains to minimize a21 + · · · + a2n subject to the constraints ai > 0 for all i and
a1 + · · · + an = 1. That the minimum is attained when a1 = · · · = an = 1/n can be
proved in many ways:
1. The identity

n(a21 + · · · + a2n ) = (a1 + · · · + an )2 + (ai − aj )2

i<j

implies that
1 1
a21 + · · · + a2n ≥
(a1 + · · · + an )2 = ,
n n
with equality if and only if a1 = a2 = · · · = an .
2. Take b1 = · · · = bn = 1 in the Cauchy-Schwarz Inequality [HLP, Theorem 7]

(a21 + · · · + a2n )(b21 + · · · + b2n ) ≥ (a1 b1 + · · · + an bn )2 ,

which holds for arbitrary a1 , . . . , an , b1 , . . . , bn ∈ R, with equality if and only if

(a1 , . . . , an ) and (b1 , . . . , bn ) are linearly dependent.
3. Take bi = ai in Chebychev’s Inequality [HLP, Theorem 43], which states that if
a1 ≥ · · · ≥ an > 0 and b1 ≥ · · · ≥ bn > 0, then

n 
n 
n 
i=1 ai bi i=1 ai i=1 bi
≥ ,
n n n
with equality if and only if all the ai are equal or all the bi are equal.
4. Take r = 2 and s = 1 in the Power Mean Inequality [HLP, Theorem 16], which
states that for real numbers a1 , . . . , an > 0, if we define the rth power mean as

r 1/r
a1 + · · · + arn
Pr = ,
n
(and P0 = limr→0 Pr = (a1 a2 · · · an )1/n ), then Pr ≥ Ps whenever r > s, with
equality if and only if a1 = · · · = an .
5. Take f (x) = x2 in Jensen’s Inequality [HLP, Theorem 90], which states that if
f (x) is a convex (concave-up) function on an interval I, then


f (a1 ) + · · · + f (an ) a1 + · · · + a n
≥f
n n
for all a1 , . . . , an ∈ I, with equality if and only if the ai are all equal or f is linear
on a closed interval containing all the ai .
6. Let H denote the hyperplane x1 + · · · + xn = 1 in Rn . The line L through
0 = (0, . . . , 0) perpendicular to H is the one in the direction of (1, . . . , 1), which
56 The William Lowell Putnam Mathematical Competition

meets H at P = (1/n, . . . , 1/n). The quantity a21 + · · · + a2n can be viewed as the
square of the distance from 0 to the point (a1 , . . . , an ) on H, and this is minimized
when (a1 , . . . , an ) = P .
In any case, we find that the minimum value of a21 + · · · + a2n is 1/n, so the maximum
value of A(R1 )+···+A(R
A(T )
n−1 )
is 1 − 1/n. For the problem as stated, n = 3, so the
maximum value is 2/3. 
Remark. The minimum is unchanged if instead of allowing T to vary, we fix a
particular acute triangle T .
Remark. While we are on the subject of inequalities, we should also mention the
very useful Arithmetic-Mean–Geometric-Mean Inequality (AM-GM), which states that
for nonnegative real numbers a1 , . . . , an , we have
a 1 + a 2 + · · · + an 1/n
≥ (a1 a2 · · · an ) ,
n
with equality if and only if a1 = a2 = · · · = an . This is the special case P1 ≥ P0 of
the Power Mean Inequality. It can also be deduced by taking f (x) = ln x in Jensen’s
Inequality.

A3. (100, 5, 18, 0, 0, 0, 0, 0, 2, 10, 19, 47)

Let d be a real number. For each integer m ≥ 0, define a sequence {am (j)},
j = 0, 1, 2, . . . by the condition

am (0) = d/2m , and am (j + 1) = (am (j))2 + 2am (j), j ≥ 0.

Evaluate limn→∞ an (n).

Answer. The value of limn→∞ an (n) is ed − 1.

Solution. We have am (j + 1) + 1 = (am (j) + 1)2 , so by induction on j,

j
am (j) + 1 = (am (0) + 1)2 .

Hence
2n
d
lim an (n) = lim 1+ n − 1.
n→∞ n→∞ 2
If f (x) = ln(1 + dx), then f  (x) = d/(1 + dx), and in particular
ln(1 + dx)
lim = f  (0) = d.
x→0 x
Applying the continuous function ex yields

lim (1 + dx)1/x = ed ,
x→0

and taking the sequence xn = 1/2n yields

2n
d
lim 1 + n = ed ,
n→∞ 2
so limn→∞ an (n) = ed − 1. 
Solutions: The Forty-Sixth Competition (1985) 57

A4. (72, 30, 6, 23, 0, 0, 0, 0, 1, 5, 33, 31)

Define a sequence {ai } by a1 = 3 and ai+1 = 3ai for i ≥ 1. Which integers
between 00 and 99 inclusive occur as the last two digits in the decimal
expansion of infinitely many ai ?
Answer. Only 87 occurs infinitely often.
Solution. Let φ(n) denote the Euler φ-function, which equals the number of
integers between 1 and n inclusive that are relatively prime to n, or more abstractly,
the order of the multiplicative group (Z/nZ)∗ . If the prime factorization of n is
pe11 · · · pekk , then φ(n) can be computed by the formula
+
k +
k
φ(n) = φ(pei i ) = pei i −1 (pi − 1).
i=1 i=1

Euler’s Theorem [Lar1, p. 148], which is Lagrange’s Theorem (the order of an element
of a finite group divides the order of the group, [Lar1, p. 147]) applied to the group
(Z/nZ)∗ , states that aφ(n) ≡ 1 (mod n) whenever gcd(a, n) = 1.
It shows that ai = 3ai−1 modulo 100 is determined by ai−1 modulo φ(100) = 40,
for i ≥ 2. Similarly ai−1 mod 40 is determined by ai−2 mod 16, which is determined
by ai−3 mod 8, for i ≥ 4. Finally ai−3 mod 8 is determined by ai−4 mod 2 for i ≥ 5,
since 32 ≡ 1 (mod 8). For i ≥ 5, ai−4 is odd, so
ai−3 = 3ai−4 ≡ 31 ≡ 3 (mod 8)
ai−2 = 3 ai−3
≡3 3
≡ 11 (mod 16)
ai−1 = 3 ai−2
≡3 11
≡ 27 (mod 40)
ai =3 ai−1
≡3 27
≡ 87 (mod 100).
Thus the only integer that appears as the last two digits of infinitely many ai is 87. 
Remark. Carmichael’s lambda function. The exponent in Euler’s Theorem is not
always best possible. The function λ(n) giving the best possible exponent for n, i.e.,
the least positive integer λ such that aλ ≡ 1 (mod n) whenever gcd(a, n) = 1, is known
as Carmichael’s lambda function or as the reduced totient function. If n = pe11 · · · pekk ,
then
λ(n) = lcm(λ(pe11 ), . . . , λ(pekk )),
where
λ(pe ) = φ(pe ) = pe−1 (p − 1),
unless p = 2 and e ≥ 3 in which case λ(2e ) = 2e−2 instead of 2e−1 . For more
information, see [Ros1, Section 9.6].
One can simplify the computations in the above solution by using λ(n) in place of
φ(n): iteration of λ maps 100 to 20 to 4 to 2, so starting from ai−3 ≡ 1 (mod 2) we
obtain
ai−2 = 3ai−3 ≡ 31 ≡ 3 (mod 4)
ai−1 = 3 ai−2
≡3 ≡73
(mod 20)
ai =3 ai−1
≡ 3 ≡ 87
7
(mod 100).
58 The William Lowell Putnam Mathematical Competition

Stronger result. More generally, one can show that for any integer c ≥ 1, the
sequence defined by a1 = c and ai+1 = cai for i ≥ 1 is eventually constant when
reduced modulo a positive integer n. To prove this, one can use the Chinese Remainder
Theorem to reduce to the case where n is a power of a prime p. If p divides c, then
the sequence is eventually 0 modulo n; otherwise we can use Euler’s Theorem and
strong induction on n, since φ(n) < n for n > 1. See 1997B5 for a related problem,
and further discussion.

A5. (44, 2,7, 5, 0, 0, 0, 0, 10, 27, 46, 60)

2π
Let Im = 0
cos(x) cos(2x) · · · cos(mx) dx. For which integers m, 1 ≤ m ≤ 10,
is Im = 0?
Answer. The values of m between 1 and 10 for which Im = 0 are 3, 4, 7, 8.
Solution. By de Moivre’s Theorem (cos θ + i sin θ = eiθ , [Spv, Ch. 24]), we have
m
ikx
 2π +   2π
e + e−ikx −m
Im = dx = 2 ei(%1 +2%2 +···+m%m )x dx,
0 2 % =±1 0
k=1 k

where the sum ranges over the 2 m-tuples (C1 , . . . , Cm ) with Ck = ±1 for each k. For
m

integers t, ,
 2π
itx 2π, if t = 0
e dx =
0 0, otherwise.
Thus Im = 0 if and only 0 can be written as
C1 + 2C2 + · · · + mCm
for some C1 , . . . , Cm ∈ {1, −1}. If such Ck exist, then
m(m + 1)
0 = C1 + 2C2 + · · · + mCm ≡ 1 + 2 + · · · + m = (mod 2)
2
so m(m + 1) ≡ 0 (mod 4), which forces m ≡ 0 or 3 (mod 4). Conversely, if m ≡ 0
(mod 4), then
0 = (1 − 2 − 3 + 4) + (5 − 6 − 7 + 8) + · · · + ((m − 3) − (m − 2) − (m − 1) + m),
and if m ≡ 3 (mod 4),
0 = (1 + 2 − 3) + (4 − 5 − 6 + 7) + (8 − 9 − 10 + 11) + ((m − 3) − (m − 2) − (m − 1) + m).
Thus Im = 0 if and only m ≡ 0 or 3 (mod 4). The integers m between 1 and 10
satisfying this condition are 3, 4, 7, 8. 
Reinterpretation. This is a question in Fourier analysis. The function
cos(x) cos(2x) · · · cos(mx)
is continuous and periodic with period 2π, so it can be written as a Fourier series
∞ ∞
cos(x) cos(2x) · · · cos(mx) = a0 + bj cos(jx) + ck sin(kx).
j=1 k=1

The question asks: For which integers m between 1 and 10 is a0 nonzero? By similar
methods, one can show that:
 Solutions: The Forty-Sixth Competition (1985) 59

(i) ck = 0 for all k, and

(ii) bj = p/2m−1 , where p is the number of ways to express j as C1 + 2C2 + · · · + mCm ,
where C1 , . . . , Cm ∈ {1, −1}.

In particular, only finitely many bj are nonzero, and a0 + j bj = 1.

A6. (8, 2, 0, 0, 0, 0, 0, 0, 3, 19, 22, 147)

If p(x) = a0 + a1 x + · · · + am xm is a polynomial with real coefficients ai , then
set

Γ(p(x)) = a20 + a21 + · · · + a2m .

Let f (x) = 3x2 + 7x + 2. Find, with proof, a polynomial g(x) with real
coefficients such that
(i) g(0) = 1, and
(ii) Γ(f (x)n ) = Γ(g(x)n )
for every integer n ≥ 1.
Answer. One such g(x) is 6x2 + 5x + 1.
Solution. For any polynomial p(x), let γ(p(x)) = p(x)p(x−1 ), which is a Laurent
n
polynomial (an expression of the form j=m aj xj where aj are constants and m, n
are integers, not necessarily nonnegative). Then Γ(p(x)) equals the coefficient of x0 in
γ(p(x)).
We have f (x) = (3x + 1)(x + 2). Since

γ(x + 2) = (x + 2)(x−1 + 2) = (1 + 2x−1 )(1 + 2x) = γ(1 + 2x),

and γ(p(x)q(x)) = γ(p(x))γ(q(x)) for any polynomials p(x) and q(x), we find

γ(f (x)n ) = γ((3x + 1)n )γ((x + 2)n ) = γ((3x + 1)n )γ((1 + 2x)n ) = γ(g(x)n ),
where g(x) = (3x + 1)(1 + 2x) = 6x2 + 5x + 1. Taking coefficients of x0 , we obtain
Γ(f (x)n ) = Γ(g(x)n ). Moreover g(0) = 1, so we are done. 
 i
 j

Remark. One could also show by brute force that Γ ( ai x )( bj x ) is un-
changed by reversing the order of the bj , without mentioning γ or Laurent polynomials.
The coefficient of x0 in a Laurent polynomial can also be expressed as a contour
integral, via the residue theorem; hence one could begin a solution with the observation
-
1 dz
Γ(p(x)) = p(z)p(z −1 ) .
2πi |z|=1 z

Remark. The solution is not unique: for any integer k ≥ 1, the polynomials
g(x) = (3xk + 1)(2xk + 1) and g(x) = (3xk − 1)(2xk − 1) also have the desired
properties. We do not know if these are the only ones. Using the fact that the ring
R[x, 1/x] of Laurent polynomials is a unique factorization domain (UFD), we could
prove that these are the only ones if we could prove the following conjecture of Greg
Kuperberg (communicated electronically):
Suppose h1 , h2 ∈ R[x, 1/x] satisfy h1 (x) = h1 (1/x) and h2 (x) = h2 (1/x).
Then the following are equivalent:
60 The William Lowell Putnam Mathematical Competition

(a) The coefficient of x0 in h1 (x)n equals the coefficient of x0 of h2 (x)n for

all positive integers n.
(b) There exists j ∈ R[x, 1/x] such that for i = 1, 2 we have hi (x) = j(Ci xki )
for some Ci ∈ {1, −1} and ki ≥ 1.
(It is easy to prove that (b) implies (a).)

B1. (112, 29, 0, 0, 0, 0, 0, 0, 0, 30, 13, 17)

Let k be the smallest positive integer with the following property:
There are distinct integers m1 , m2 , m3 , m4 , m5 such that the
polynomial
p(x) = (x − m1 )(x − m2 )(x − m3 )(x − m4 )(x − m5 )
has exactly k nonzero coefficients.
Find, with proof, a set of integers m1 , m2 , m3 , m4 , m5 for which this
minimum k is achieved.
Answer. The minimum is k = 3, and is attained for {m1 , m2 , m3 , m4 , m5 } =
{−2, −1, 0, 1, 2}.
Solution. If k = 1, then p(x) must be x5 , but this does not have five distinct
integer zeros. If k = 2, then p(x) = x5 + axr for some nonzero a ∈ Z and 0 ≤ r ≤ 4;
this has x = 0 as double zero if r ≥ 2 and has a nonreal zero if r = 0 or r = 1. Thus
k ≥ 3. The example
x(x − 1)(x + 1)(x − 2)(x + 2) = x(x2 − 1)(x2 − 4) = x5 − 5x3 + 4x
shows that in fact k = 3. 
Remark. More generally, we can prove that given n ≥ 1, the smallest integer k for
which there exist distinct integers m1 , . . . , mn such that the polynomial
p(x) = (x − m1 ) · · · (x − mn )
has exactly k nonzero coefficients is k = (n+1)/2 = n/2+1. The key is Descartes’
Rule of Signs, which states that if p(x) = a1 xr1 + a2 xr2 + · · · + ak xrk is a polynomial
with ai ∈ R∗ and r1 > r2 > · · · > rk , then the number of positive real zeros of p(x)
counted with multiplicity is the number of sign changes in the sequence a1 , a2 , . . . , ak
minus a nonnegative even integer.
If p(x) has k nonzero coefficients, p(x) has at most k−1 positive real zeros. Applying
Descartes’ Rule of Signs to p(−x) shows that p(x) has at most k−1 negative real zeros.
Hence the total number of distinct zeros of p(x) is at most (k −1)+(k −1)+1 = 2k −1,
where the +1 is for the possibility that 0 might be a root. If p(x) has n distinct zeros
all of which are integers, then n ≤ 2k − 1, so k ≥ (n + 1)/2.
On the other hand, given n ≥ 1, we can exhibit a polynomial p(x) = (x −
m1 ) · · · (x − mn ) with distinct integer zeros m1 , . . . , mn and with at most (hence
exactly) k = (n + 1)/2 nonzero coefficients:
,
(x + 1)(x − 1)(x + 2)(x − 2) · · · (x + (k − 1))(x − (k − 1)) if n = 2k − 2
p(x) =
x(x + 1)(x − 1)(x + 2)(x − 2) · · · (x + (k − 1))(x − (k − 1)) if n = 2k − 1.
Solutions: The Forty-Sixth Competition (1985) 61

B2. (3, 89, 3, 1, 0, 0, 0, 0, 2, 11, 37, 55)

Define polynomials fn (x) for n ≥ 0 by f0 (x) = 1, fn (0) = 0 for n ≥ 1, and
d
(fn+1 (x)) = (n + 1)fn (x + 1)
dx
for n ≥ 0. Find, with proof, the explicit factorization of f100 (1) into powers
of distinct primes.
Answer. The factorization of f100 (1) is 10199 .
Solution. By induction, the given properties determine fn (x) uniquely. Comput-
ing and factoring fn (x) for the first few n suggests that fn (x) = x(x + n)n−1 . We
prove this by induction on n. The base case n = 0 is given: f0 (x) = 1. For n ≥ 0, we
indeed have fn+1 (0) = 0 and
d
fn+1 (x) = (x + n + 1)n + nx(x + n + 1)n−1
dx
= (n + 1)(x + 1)(x + n + 1)n−1
= (n + 1)fn (x + 1),
which completes the inductive step. Hence f100 (1) = 10199 . (Note that 101 is prime.)


B3. (95, 15, 15, 11, 0, 0, 0, 0, 5, 3, 23, 34)

Let
a1,1 a1,2 a1,3 ...
a2,1 a2,2 a2,3 ...
a3,1 a3,2 a3,3 ...
.. .. .. ..
. . . .
be a doubly infinite array of positive integers, and suppose each positive
integer appears exactly eight times in the array. Prove that am,n > mn for
some pair of positive integers (m, n).
Solution. Suppose not; i.e., suppose that am,n ≤ mn for all m, n ≥ 1. Let
R(k) = { (i, j) : ai,j ≤ k }.
By hypothesis, #R(k) ≤ 8k. On the other hand, R(k) contains all pairs (i, j) with
ij ≤ k, and there are
. / . / . /



k k k k k k
+ + ··· + > −1 + − 1 + ··· + − 1 > k(ln k − 1)
1 2 k 1 2 k
such pairs, since  k
1 1 1 1
+ + ··· + > dx = ln k.
1 2 k 1 x
Hence 8k > k(ln k − 1), which is a contradiction for k > e9 ≈ 8103.08. 
Remark. To solve the problem, the explicit lower bound on the rate of growth of
1
1 + 12 + · · · + k1 as k → ∞ is not really needed: it suffices to know that this sum tends
to ∞, i.e., that the harmonic series diverges.
62 The William Lowell Putnam Mathematical Competition

B4. (115, 30, 9, 1, 2, 6, 4, 0, 9, 8, 11, 6)

Let C be the unit circle x2 + y 2 = 1. A point p is chosen randomly on
the circumference of C and another point q is chosen randomly from the
interior of C (these points are chosen independently and uniformly over
their domains). Let R be the rectangle with sides parallel to the x- and
y-axes with diagonal pq. What is the probability that no point of R lies
outside of C?
Answer. The probability is 4/π 2 .

Solution. Let p = (cos θ, sin θ) and q = (a, b). The other two vertices of R are
(cos θ, b) and (a, sin θ). If |a| ≤ | cos θ| and |b| ≤ | sin θ|, then each vertex (x, y) of
R satisfies x2 + y 2 ≤ cos2 θ + sin2 θ = 1, and no points of R can lie outside of C.
Conversely, if no points of R lies outside of C, then applying this to the two vertices
other than p and q, we find

cos2 θ + b2 ≤ 1, and a2 + sin2 θ ≤ 1,

or equivalently

|b| ≤ | sin θ|, and |a| ≤ | cos θ|. (1)

These conditions imply that (a, b) lies inside or on C, so for any given θ, the probability
that the random point q = (a, b) satisfies (1) is
2| cos θ| · 2| sin θ| 2
= | sin(2θ)|,
π π
and the overall probability is
 2π  π/2
1 2 4 4
| sin(2θ)| dθ = 2 sin(2θ) dθ = 2 . 
2π 0 π π 0 π

B5. (14, 8, 2, 2, 0, 0, 0, 0, 2, 2, 61, 110) ∞

∞ −1/2 −1985(t+t−1 ) 2 √
Evaluate 0
t e dt. You may assume that −∞ e−x dx = π.
∞ −1  π −3970
Answer. The value of 0 t−1/2 e−1985(t+t ) dt is 1985 e .

Solution (adapted from [Bernau]). For a > 0, let

 ∞
−1
I(a) = t−1/2 e−a(t+t ) dt.
0

The integral converges, since the integrand is bounded by t−1/2 on (0, 1] and by e−at
on [1, ∞). Hence
0  B 1
1
−1/2 −a(t+t−1 ) −1/2 −a(t+t−1 )
I(a) = lim t e dt + t e dt .
B→∞ 1/B 1

Substitute 1/t for t in the first integral to conclude

 B
−1
I(a) = lim (t−1/2 + t−3/2 )e−a(t+t ) dt.
B→∞ 1
Solutions: The Forty-Sixth Competition (1985) 63

Now use the substitution u = a1/2 (t1/2 − t−1/2 ) to obtain

 a1/2 (B 1/2 −B −1/2 )
2
I(a) = 2a−1/2 lim e−u −2a du
B→∞ 0
 ∞
2
= 2a−1/2 e−2a e−u du
0
2
π −2a
= e ,
a
 −3970
so I(1985) = π
1985 e . 
Remark. The modified Bessel function of the second kind (also known as
Macdonald’s function) has the integral representation
 ∞
Kν (z) = e−z cosh t cosh(νt) dt
0

for Re(z) > 0 [O, p. 250]. When ν = 1/2, the substitution u = et relates this to
expressions occurring in the solution above; to be precise, I(a) = 2K1/2 (2a) for all
 π −z
a > 0. Thus K1/2 (z) = 2z e for z > 0. Similar formulas exist for Kn+1/2 (z) for
each integer n. For arbitrary ν, the function w = Kν (z) is a solution of the differential
equation
z 2 w + zw − (z 2 + ν 2 )w = 0.

B6. (5, 0, 0, 0, 0, 4, 0, 0, 9, 7, 26, 150)

Let G be a finite set of real n × n matrices {Mi }, 1 ≤ i ≤ r, which form
r
a group under matrix multiplication. Suppose that i=1 tr(Mi ) = 0, where
r
tr(A) denotes the trace of the matrix A. Prove that i=1 Mi is the n × n
zero matrix.
r
Solution 1. Let S = i=1 Mi . For any j, the sequence Mj M1 , Mj M2 , . . . , Mj Mr
is a permutation of the elements of G, and summing yields Mj S = S. Summing this
from j = 1 to r yields S 2 = rS. Therefore the minimal polynomial of S divides
x2 − rx, and every eigenvalue of S is either 0 or r. But the eigenvalues counted with
multiplicity sum to tr(S) = 0, so they are all 0. At this point, we present three ways
to finish the proof that S = 0:
1. Every eigenvalue of S − rI is −r = 0, so S − rI is invertible. Hence from
S(S − rI) = 0 we obtain S = 0.
2. The minimal polynomial p(x) of S must be x, x − r, or x(x − r). Since every zero
of the minimal polynomial is an eigenvalue, the minimal polynomial is x. By the
Cayley-Hamilton Theorem [Ap2, Theorem 7.8], p(S) = 0; that is, S = 0.
3. The Jordan canonical form of S over the complex numbers has 0’s (the eigenvalues)
on the main diagonal, possible 1’s just above the diagonal, and 0’s elsewhere. The
condition S 2 = rS implies that there are no 1’s, so the Jordan canonical form of
S is 0. Thus S = 0. 
Literature note. See [Ap2, Ch. 4] for a quick introduction to eigenvalues and
eigenvectors.
64 The William Lowell Putnam Mathematical Competition

Solution 2 (based on an idea of Dave Savitt).

Lemma. Let G be a finite group of order r. Let ρ : G → Aut(V ) be a representation

of G on some finite-dimensional complex vector space V . Then g∈G tr ρ(g) is a

nonnegative integer divisible by r, and is zero if and only if g∈G ρ(g) = 0.
Proof. Let η1 , . . . , ηs be the irreducible characters of G. Theorem 3 on p. 15 of [Se2]
s s
implies that if χ = i=1 ai ηi and ψ = i=1 bi ηi are arbitrary characters, then
s
1
χ(g)ψ(g) = ai bi . (2)
r i=1
g∈G

Applying this to the character of ρ and the trivial character 1 shows that 1r g∈G tr ρ(g)
equals the multiplicity of 1 in ρ, which is a nonnegative integer.

Now suppose that the matrix S = g∈G ρ(g) is nonzero. Choose v ∈ V with
Sv = 0. The relation ρ(h)S = S shows that Sv is fixed by ρ(h) for all h ∈ G. In
other words, Sv spans a trivial subrepresentation of ρ, so the nonnegative integer of
the previous paragraph is positive. 
We now return to the problem at hand. Unfortunately the Mi do not necessarily
define a representation of G, since the Mi need not be invertible. Instead we need to
apply the lemma to the action of G on Cn /K, for some subspace K. Given 1 ≤ i, j ≤ r,
there exists k such that Mi = Mk Mj , so ker Mj ⊆ ker Mi . This holds for all i and j,
so the Mi have a common kernel, which we call K. Then the Mi and S also act on
Cn /K. If v ∈ Cn maps to an element of Cn /K in the kernel of Mi acting on Cn /K,
then Mi Mi v ∈ Mi (K) = 0, but Mi Mi = Mj for some j, so v ∈ ker(Mj ) = K. Thus
the Mi act invertibly on Cn /K. We finish by applying the lemma to the representation
Cn /K of G, using the observation that tr S is the sum of the traces of S acting on
Cn /K and on K, with the trace on K being zero. 
Remark. One can also give a elementary variant of this solution (which somewhat
obscures the connection with representation theory). Namely, we prove by induction
on n that tr S is a nonnegative integer divisible by r, which is nonzero if S = 0. The
case n = 1 is straightforward; given the result for n − 1, choose as above a vector
v ∈ Cn with Sv = 0, so that each of the matrices preserves v. Let V be the span of v;
then the trace of S is equal to the sum of its trace on V and on the quotient Cn /V .
The former is r and the latter is a nonnegative integer divisible by r.
Literature note. See [Se2] for an introduction to representation theory. The
relations given by (2) are known as the orthogonality relations for characters.
Solutions: The Forty-Seventh Competition (1986) 65

The Forty-Seventh William Lowell Putnam Mathematical Competition

December 6, 1986

A1. (152, 23, 10, 7, 0, 0, 0, 2, 2, 3, 1, 1)

Find, with explanation, the maximum value of f (x) = x3 − 3x on the set
of all real numbers x satisfying x4 + 36 ≤ 13x2 .
Answer. The maximum value is 18.
Solution. The condition x4 +36 ≤ 13x2 is equivalent to (x−3)(x−2)(x+2)(x+3) ≤
0, which is satisfied if and only if x ∈ [−3, −2] ∪ [2, 3]. The function f is increasing
on [−3, −2] and on [2, 3], since f  (x) = 3(x2 − 1) > 0 on these intervals. Hence the
maximum value is max{f (−2), f (3)} = 18. 

A2. (155, 0, 0, 0, 0, 0, 0, 0, 0, 0, 33, 13)  

1020000
What is the units (i.e., rightmost) digit of 10100 +3 ? Here x is the
greatest integer ≤ x.
Answer. The units digit is 3.
Solution. Taking x = 10100 and y = −3 in the factorization
x200 − y 200 = (x − y)(x199 + x198 y + · · · + xy 198 + y 199 )
shows that the number
1020000 − 3200  199  100 198
I= 100
= 10100 − 10 3 + · · · + 10100 3198 − 3199 (1)
10 + 3
 20000 
10
is an integer. Moreover, I = 10 100 +3 , since

3200 9100
= < 1.
10100 + 3 10100 + 3
By (1),
I ≡ −3199 ≡ −33 (81)49 ≡ −27 ≡ 3 (mod 10),
so the units digit of I is 3. 

A3. (53, 6, 15, 1, 0, 0, 0, 1, 12, 1, 26, 86)

∞
n=0 Arccot(n + n + 1), where Arccot t for t ≥ 0 denotes the
2
Evaluate
number θ in the interval 0 < θ ≤ π/2 with cot θ = t.
Answer. The series converges to π/2.
Solution 1. If α = Arccot x and β = Arccot y for some x, y > 0, then the addition
formula
cot α cot β − 1
cot(α + β) =
cot α + cot β
shows that
xy − 1
Arccot x + Arccot y = Arccot (1)
x+y
provided that Arccot x + Arccot y ≤ π/2. The latter condition is equivalent to
Arccot x ≤ Arccot(1/y), which is equivalent to x ≥ 1/y, and hence equivalent to
66 The William Lowell Putnam Mathematical Competition

xy ≥ 1. Verifying the xy ≥ 1 condition at each step, we use (1) to compute the first
few partial sums

Arccot 1 = Arccot 1
Arccot 1 + Arccot 3 = Arccot(1/2)
Arccot 1 + Arccot 3 + Arccot 7 = Arccot(1/3),
m−1
and guess that n=0 Arccot(n2 + n + 1) = Arccot(1/m) for all m ≥ 1. This is easily
proved by induction on m: the base case is above, and the inductive step is
m m−1
Arccot(n2 + n + 1) = Arccot(m2 + m + 1) + Arccot(n2 + n + 1)
n=0


n=0

2 1
= Arccot(m + m + 1) + Arccot (inductive hypothesis)
m

2 
(m + m + 1)/m − 1
= Arccot (by (1))
m2 + m + 1 + 1/m


1
= Arccot .
m+1
Hence
∞

1
2
Arccot(n + n + 1) = lim Arccot = Arccot(0) = π/2. 
n=0
m→∞ m

Solution 2. For real a ≥ 0 and b = 0, Arccot(a/b) is the argument (between −π/2

and π/2) of the complex number a + bi. Therefore, if any three complex numbers
satisfy (a + bi)(c + di) = (e + f i), where a, c, e ≥ 0 and b, d, f = 0, then then
Arccot(a/b) + Arccot(c/d) = Arccot(e/f ). Factoring the polynomial n2 + n + 1 + i
yields
(n2 + n + 1 + i) = (n + i)(n + 1 − i).

Taking arguments, we find that Arccot(n2 + n + 1) = Arccot n − Arccot(n + 1). (This

identity can also be proved from the difference formula for cot, but then one needs
∞
to guess the identity in advance.) The series n=0 Arccot(n2 + n + 1) telescopes to
limn→∞ (Arccot(0) − Arccot(n + 1)) = π/2. 
Related question. Evaluate the infinite series:
∞
 ∞

2 8n
Arctan , Arctan .
n=1
n2 n=1
n4 − 2n2 + 5

The first is problem 26 of [WH], and both appeared in a problem due to J. Anglesio
in the Monthly [Mon3, Mon5].
Literature note. The value
 ofseries such as these have been known for a long time.
∞
In particular, n=1 Arctan n22 and some similar series were evaluated at least as
early as 1878 [Gl]. Ramanujan [Berndt, p. 37] independently evaluated this and other
 Solutions: The Forty-Seventh Competition (1986) 67

Arctan series shortly after 1903. The generalizations

∞

2xy y tanh πy
Arctan 2 − x2 + y 2
≡ Arctan − Arctan (mod π),
n=1
n x tan πx

 " πy #
∞
2xy πx
Arctan ≡ Arctan tan tanh (mod π)
n=1
(2n − 1)2 − x2 + y 2 2 2

appear in [GK]. For further history see Chapter 2 of [Berndt].

A4. (21, 3, 4, 4, 5, 7, 0, 6, 3, 7, 23, 118)

A transversal of an n×n matrix A consists of n entries of A, no two in the
same row or column. Let f (n) be the number of n × n matrices A satisfying
the following two conditions:
(a) Each entry αi,j of A is in the set {−1, 0, 1}.
(b) The sum of the n entries of a transversal is the same for all transversals
of A.
An example of such a matrix A is
 
−1 0 −1
A =  0 1 0 .
0 1 0
Determine with proof a formula for f (n) of the form

f (n) = a1 bn1 + a2 bn2 + a3 bn3 + a4 ,

where the ai ’s and bi ’s are rational numbers.

Answer. The value of f (n) is 4n + 2 · 3n − 4 · 2n + 1.

Solution (Doug Jungreis).

Lemma. Condition (b) is equivalent to the statement that any two rows of the
matrix differ by a constant vector, i.e., a vector of the form (c, c, . . . , c).
Proof. If two rows differ by a constant vector, then (b) holds. Conversely, if (b)
holds, for any i, j, k, l in {1, . . . , n}, take a transversal containing aik and ajl , and then
switch ail and ajk to get a new transversal. Since these two transversals have the
same sum, aik + ajl = ail + ajk , or equivalently, aik − ajk = ail − ajl . Thus rows i and
j differ by the constant vector with all components equal to ai1 − aj1 . Since i and j
were arbitrary, each pair of rows differs by a constant vector. 

We compute f (n) by considering four cases.

Case 1: the first row of the matrix is a constant vector. Then each row is constant
by the Lemma, so each row is (0, . . . , 0), (1, . . . , 1), or (−1, . . . , −1). Thus there are
3n such matrices.
Case 2: both 0 and 1 appear in the first row, but not −1. Then there are
2 − 2 possibilities for the first row. Each other row must differ from the first by
n

either (0, . . . , 0) or (−1, ..., −1), by the Lemma and condition (a), so there are 2n−1
possibilities for these rows. This gives a total of 2n−1 (2n − 2) possibilities in this case.
68 The William Lowell Putnam Mathematical Competition

Case 3: both 0 and −1 appear in the first row, but not 1. These are just the negatives
of the matrices in case 2, so we again have 2n−1 (2n − 2) possibilities.
Case 4: Both 1 and −1 (and possibly also 0) appear in the first row. This covers
all other possibilities for the first row, i.e., the remaining 3n − 2 · 2n + 1 possibilities.
Then every row must be equal to the first, by the Lemma and condition (a), so we
have a total of 3n − 2 · 2n + 1 possibilities in this case.
Adding the four cases gives f (n) = 4n + 2 · 3n − 4 · 2n + 1. 
Related question.
If an n × n matrix M with nonnegative integer entries satisfies condition (b),
that the sum of the n entries of a transversal is the same number m for all
transversals of A, show that M is the sum of m permutation matrices. (A
permutation matrix is a matrix with one 1 in each row and each column, and
all other entries 0.)
For a card trick related to this result, see [Kl].
This result is a discrete version of the following result.
Birkhoff-von Neumann Theorem. The convex hull of the permutation matrices
is precisely the set of doubly stochastic matrices: matrices with entries in [0, 1] with
each row and column summing to 1.

A5. (13, 4, 0, 0, 0, 0, 0, 1, 0, 2, 39, 142)

Suppose f1 (x), f2 (x), . . . , fn (x) are functions of n real variables x =
(x1 , . . . , xn ) with continuous second-order partial derivatives everywhere on
Rn . Suppose further that there are constants cij such that
∂fi ∂fj
− = cij
∂xj ∂xi
for all i and j, 1 ≤ i ≤ n, 1 ≤ j ≤ n. Prove that there is a function g(x) on
Rn such that fi + ∂g/∂xi is linear for all i, 1 ≤ i ≤ n. (A linear function is
one of the form

a0 + a1 x1 + a2 x2 + · · · + an xn .)


Solution. Note that cij = −cji for all i and j. Let hi = 1
2 j cij xj , so
1
∂hi /∂xj = 2 cij . Then
∂hi ∂hj 1 1 ∂fi ∂fj
− = cij − cji = cij = − ,
∂xj ∂xi 2 2 ∂xj ∂xi
so
∂(hi − fi ) ∂(hj − fj )
=
∂xj ∂xi
for all i and j. Hence (h1 − f1 , . . . , hn − fn ) is a gradient, i.e., there is a differentiable
function g on Rn such that ∂g/∂xi = hi − fi for each i. Then fi + ∂g/∂xi = hi is
linear for each i. 
Solutions: The Forty-Seventh Competition (1986) 69

A6. (1, 4, 1, 1, 0, 1, 0, 0, 6, 4, 64, 119)

Let a1 , a2 , . . . , an be real numbers, and let b1 , b2 , . . . , bn be distinct positive
integers. Suppose there is a polynomial f (x) satisfying the identity
n
(1 − x) f (x) = 1 +
n
ai xbi .
i=1

Find a simple expression (not involving any sums) for f (1) in terms of
b1 , b2 , . . . , bn and n (but independent of a1 , a2 , . . . , an ).
Answer. The number f (1) equals b1 b2 · · · bn /n!.

Solution 1. For j ≥ 1, let (b)j denote b(b − 1) · · · (b − j + 1). For 0 ≤ j ≤ n,

differentiating the identity j times and putting x = 1 (or alternatively substituting
x = y + 1 and equating coefficients) yields

0=1+ ai ,

0= ai bi ,

0= ai (bi )2 ,
..
.
0= ai (bi )n−1 ,

(−1)n n!f (1) = ai (bi )n .

In other words, Av = 0, where

 
−1 1 1 ··· 1  
  −1
 0 b1 b2 ··· bn   a1 
   
 0 (b1 )2 (b2 )2 ··· (bn )2  
A=  .. .. .. .. .. 
 and v =  a2  .
 . . . . .   . 
   .. 
 0 (b1 )n−1 (b2 )n−1 ··· (bn )n−1 
an
(−1)n n!f (1) (b1 )n (b2 )n ··· (bn )n

Since v = 0, det A = 0. Since (b)j is a monic polynomial of degree j in b with no

constant term, we can add a linear combination of rows 2, 3, . . . , k to row k + 1, for
2 ≤ k ≤ n, to obtain
 
−1 1 1 ··· 1
 ··· bn 
 0 b1 b2 
 
 0 b1 2
b2 2
··· b2n 
A = 
 .. .. .. .. .. .
 . . . . . 
 
 0 bn−1
1 bn−1
2 · · · bn−1
n 
(−1)n n!f (1) bn1 bn2 ··· bnn
70 The William Lowell Putnam Mathematical Competition

Expanding by minors along the first column yields 0 = det A = − det(V  ) +

n!f (1) det(V ) where
   
1 1 ··· 1 b1 b2 ··· bn
 b1 b2 ··· bn   b21 b22 ··· b2n 
   
 2 b22 ··· b2n   . .. .. ..  .
V =  b1  and V  =  .. . . . 
 . .. .. ..   
 .. . . .   n−1
b1 n−1
b2 · · · bn 
n−1

bn−1
1 bn−1
2 · · · bn−1
n bn1 bn2 ··· bnn
Factoring bi out of the ith column of V  shows that det(V  ) = b1 b2 · · · bn det(V ).
Hence −b1 b2 · · · bn det(V ) + n!f (1) det(V ) = 0. Since the bi are distinct, det(V ) = 0
(see below). Thus f (1) = b1 b2 · · · bn /n!. 
Remark (The Vandermonde determinant). The matrix V  is called the Vandermonde
matrix. Its determinant D is a polynomial of total degree n2 in the bi , and D vanishes
whenever bi = bj for i > j, so D is divisible by bi − bj whenever i > j. These bi − bj
3
have no common
 factor, so i>j (bi − bj ) divides D. But this product also has total
3
degree n2 , and the coefficient of b2 b23 · · · bn−1 in D and in i>j (bi − bj ) both equal
3 n
1, so D = i>j (bi − bj ). In particular, if the bi are distinct numbers, then D = 0.
See Problem 1941/14(ii) [PutnamI, p. 17] for an extension, and 1999B5 for another
application.
Solution 2. Subtract 1 from both sides, set x = et , and expand the left-hand side
in a power series. Since 1 − et = t + (higher order terms), we get
n
−1 + (−1)n f (1)tn + (higher order terms) = ai e b i t . (1)
i=1

The right-hand side F (t) satisfies the linear differential equation

F (n) (t) − (b1 + · · · + bn )F (n−1) (t) + · · · + (−1)n b1 b2 · · · bn F (t) = 0 (2)
with characteristic polynomial p(z) = (z − b1 )(z − b2 ) · · · (z − bn ). On the other hand,
from the left-hand side of (1) we see that F (0) = −1, F (i) = 0 for i = 1, . . . , n − 1,
and F (n) (0) = (−1)n f (1)n!. Hence taking t = 0 in (2) yields
(−1)n f (1)n! − 0 + 0 − 0 + · · · + (−1)n b1 b2 · · · bn (−1) = 0,
so f (1) = b1 b2 · · · bn /n!. 

B1. (183, 3, 7, 0, 0, 0, 0, 0, 4, 2, 1, 1)
Inscribe a rectangle of base b and height h and an isosceles triangle of
base b in a circle of radius one as shown. For what value of h do the
rectangle and triangle have the same area?†
Answer. The only such value of h is 2/5.
Solution. The radius OX (see Figure 3) has length equal to h/2 plus the altitude
of the triangle, so the altitude of the triangle is 1 − h/2. If the rectangle and triangle
have the same area, then bh = 12 b(1 − h/2). Cancel b and solve for h to get h = 2/5. 

† The figure is omitted here, since it is almost identical to Figure 3 used in the solution.
Solutions: The Forty-Seventh Competition (1986) 71

h O

FIGURE 3.

B2. (123, 31, 16, 3, 0, 0, 0, 0, 16, 5, 2, 5)

Prove that there are only a finite number of possibilities for the ordered
triple T = (x−y, y−z, z−x), where x, y, and z are complex numbers satisfying
the simultaneous equations

x(x − 1) + 2yz = y(y − 1) + 2zx = z(z − 1) + 2xy,

and list all such triples T .

Answer. The possibilities for T are (0, 0, 0), (0, −1, 1), (1, 0, −1), (−1, 1, 0).

Solution. Subtracting y(y − 1) + 2zx from x(x − 1) + 2yz, z(z − 1) + 2xy from
y(y − 1) + 2zx, and x(x − 1) + 2yz from z(z − 1) + 2xy, we find that the given system
is equivalent to

(x − y)(x + y − 1 − 2z) = 0
(y − z)(y + z − 1 − 2x) = 0
(z − x)(z + x − 1 − 2y) = 0.

If no two of x, y, z are equal, then x + y − 1 − 2z = y + z − 1 − 2x = z + x − 1 − 2y = 0,

and adding gives −3 = 0. Hence at least two of x, y, z are equal. If x = y and y = z,
then z = 2x + 1 − y = x + 1, so T = (x − x, x − z, z − x) = (0, −1, 1). By symmetry,
the only possibilities for T are (0, 0, 0), (0, −1, 1), (1, 0, −1), (−1, 1, 0). Finally we give
examples of (x, y, z) giving rise to each of the four possibilities, respectively: (0, 0, 0),
(0, 0, 1), (1, 0, 0), (0, 1, 0). 

B3. (26, 5, 4, 1, 0, 1, 0, 4, 3, 5, 33, 119)

Let Γ consist of all polynomials in x with integer coefficients. For f and g
in Γ and m a positive integer, let f ≡ g (mod m) mean that every coefficient
of f − g is an integral multiple of m. Let n and p be positive integers with
p prime. Given that f , g, h, r, and s are in Γ with rf + sg ≡ 1 (mod p) and
f g ≡ h (mod p), prove that there exist F and G in Γ with F ≡ f (mod p),
G ≡ g (mod p), and F G ≡ h (mod pn ).
72 The William Lowell Putnam Mathematical Competition

Solution. We prove by induction on k that there exist polynomials Fk , Gk ∈ Γ

such that Fk ≡ f (mod p), Gk ≡ g (mod p), and Fk Gk ≡ h (mod pk ). For the base
case k = 1, we take F1 = f , G1 = g.
For the inductive step, we assume the existence Fk , Gk as above, and try to
construct Fk+1 , Gk+1 . By assumption, h − Fk Gk = pk t, for some t ∈ Γ. We will
try Fk+1 = Fk + pk ∆1 and Gk+1 = Gk + pk ∆2 , where ∆1 , ∆2 ∈ Γ are yet to be chosen.
Then Fk+1 ≡ Fk ≡ f (mod p), Gk+1 ≡ Gk ≡ g (mod p), and

Fk+1 Gk+1 = Fk Gk + pk (∆2 Fk + ∆1 Gk ) + p2k ∆1 ∆2

≡ Fk Gk + pk (∆2 Fk + ∆1 Gk ) (mod pk+1 ).

If we choose ∆2 = tr and ∆1 = ts, then

∆2 Fk + ∆1 Gk ≡ trf + tsg = t(rf + sg) ≡ t (mod p),

so pk (∆2 Fk +∆1 Gk ) ≡ pk t (mod pk+1 ), and Fk+1 Gk+1 ≡ Fk Gk +pk t = h (mod pk+1 ),
completing the inductive step. 
Remark. This problem is a special case of a version of Hensel’s Lemma [Ei,
p. 208], a fundamental result in number theory. Here is a different version [NZM,
Theorem 2.23], which can be thought of as a p-adic analogue of Newton’s method for
solving polynomial equations via successive approximation:
Hensel’s Lemma. Suppose that f (x) is a polynomial with integral coefficients. If
f (a) ≡ 0 (mod pj ) and f  (a) ≡ 0 (mod p), then there is a unique t (mod p) such that
f (a + tpj ) ≡ 0 (mod pj+1 ).

B4. (22, 8, 6, 6, 0, 0, 0, 0, 4, 7, 59, 89)

For
√ a positive real number r, let G(r) be the minimum value of
r − m2 + 2n2 for all integers m and n. Prove or disprove the assertion
that limr→∞ G(r) exists and equals 0.
Answer. The limit exists and equals 0.
Solution (Doug Jungreis). First,
 |r2 − m2 − 2n2 | |r2 − m2 − 2n2 |
0 ≤ |r − m2 + 2n2 | = √ ≤ ,
r + m2 + 2n2 r
so it will suffice to bound the latter expression. Select the largest integer m ≥ 0 such
that r2 − m2 ≥ 0. Then m2 ≤ r2 < (m + 1)2 , so m ≤ r and r2 − m2 < 2m + 1. Next
select the largest integer n ≥0 such that r2 − m2 − 2n2 ≥ 0. Then 2n2 ≤ r2 − m2 <
2(n + 1)2 . This implies n ≤ (r2 − m2 )/2 and

|r2 − m2 − 2n2 | = r2 − m2 − 2n2

< 2(2n + 1)

≤ 2 + 4 (r2 − m2 )/2
√
≤ 2 + 2m + 1
√
≤ 2 + 2r + 1.
√
Hence G(r) ≤ (2 + 2r + 1)/r and limr→∞ G(r) = 0. 
Solutions: The Forty-Seventh Competition (1986) 73

B5. (10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 58, 133)

Let f (x, y, z) = x2 + y 2 + z 2 + xyz. Let p(x, y, z), q(x, y, z), r(x, y, z) be
polynomials with real coefficients satisfying
f (p(x, y, z), q(x, y, z), r(x, y, z)) = f (x, y, z).
Prove or disprove the assertion that the sequence p, q, r consists of some
permutation of ±x, ±y, ±z, where the number of minus signs is 0 or 2.
Solution. The assertion is false, since (p, q, r) = (x, y, −xy−z) satisfies f (p, q, r) =
f (x, y, z). 
Motivation. Take p = x, q = y, and view
x2 + y 2 + r2 + xyr = x2 + y 2 + z 2 + xyz,
as an equation to be solved for the polynomial r. It is equivalent to the quadratic
equation
r2 + (xy)r + (z 2 − xyz) = 0
and we already know one solution, namely r = z, so the quadratic is easy to factor:
(r − z)(r + (xy + z)) = 0.
Thus r = −xy − z is the other solution.
Remark. We now describe the set of all solutions (p, q, r). First, there is (x, y, z).
Second, choose a real number r with |r| > 2, factor p2 + rpq + q 2 as (p − αq)(p − βq)
for distinct real numbers α and β, choose c ∈ R∗ , and solve the system
p − αq = c
p − βq = (x2 + y 2 + z 2 + xyz − r2 )/c
for p and q as polynomials in x, y, z to obtain a solution (p, q, r). Third, consider
all triples obtainable from the two types above by iterating the following operations:
permuting p, q, r, changing the signs of an even number of p, q, r, and replacing (p, q, r)
by (−qr − p, q, r). All such triples are solutions.
We claim that all solutions arise in this way. It suffices to show that given a
solution (p, q, r), either it is of one of the first two types, or one of the operations
above transforms it to another solution with deg p + deg q + deg r lower, where degree
of a polynomial means its total degree in x, y, z, so xi y j z k has degree i + j + k.
Let (p, q, r) be a solution. We may assume deg p ≥ deg q ≥ deg r. Also we may
assume deg(p + qr) ≥ deg p, since otherwise we perform the transformation replacing
p with −qr − p. Since (p, q, r) is a solution,
p(p + qr) − (x2 + y 2 + z 2 + xyz) = −(q 2 + r2 ). (1)
Suppose deg p ≥ 2. Then deg(p + qr) ≥ deg p ≥ 2 and deg p(p + qr) ≥ 4, so (1)
implies the middle equality in
deg(p2 ) ≤ deg p(p + qr) = deg(q 2 + r2 ) ≤ deg(p2 ).
The ends are equal, so equality holds everywhere. In particular, deg(p + qr) = deg p
and deg q = deg p. Then deg(qr) ≤ max{deg(p + qr), deg p} = deg p = deg q, so r is a
74 The William Lowell Putnam Mathematical Competition

constant. The equation can be rewritten as

p2 + rpq + q 2 = x2 + y 2 + z 2 + xyz − r2 .

The quadratic form in p, q on the left must take on negative values, since the right-hand
side is negative for x, y, and z chosen equal and sufficiently negative. Thus its
discriminant ∆ = r2 − 4 is positive, so |r| > 2. The right-hand side is irreducible
in R[x, y, z], since as a polynomial in z it is a monic quadratic with discriminant
∆ = (xy)2 − 4(x2 + y 2 − r2 ) which cannot be the square of any polynomial, since
∆ as a quadratic polynomial in x has nonzero x2 and x0 coefficients but zero x1
coefficient. Therefore the factorization (p − αq)(p − βq) of the left-hand side matches
a trivial factorization of the right-hand side, so (p, q, r) is a solution of the second type
described above.
It remains to consider the case deg p < 2. Then p, q, r are at most linear. Equating
the homogeneous degree 3 parts in

p2 + q 2 + r2 + pqr = x2 + y 2 + z 2 + xyz

shows that after permutation, (p, q, r) = (a1 x + b1 , a2 y + b2 , a3 z + b3 ) where the ai

and bi are real numbers with a1 a2 a3 = 1. Equating coefficients of xy yields b3 = 0.
Similarly b1 = b2 = 0. Equating coefficients of x2 yields a1 = ±1. Similarly a2 = ±1
and a3 = ±1. Since a1 a2 a3 = 1, (p, q, r) is obtained from (x, y, z) by an even number
of sign changes.
Remark. The preceding analysis is similar to the proof that the positive integer
solutions to the Markov equation

x2 + y 2 + z 2 = 3xyz

are exactly those obtained from (1, 1, 1) by iterations of (x, y, z) !→ (x, y, 3xy − z)
and permutations. For the connection of this equation to binary quadratic forms and
continued fractions and much more, see [CF]. For an unsolved problem about the set
of solutions, see [Guy, p. 166].
Related question. The key idea in this problem is also essential to the solution to
Problem 6 of the 1988 International Mathematical Olympiad [IMO88, p. 38]:
Let a and b be positive integers such that ab + 1 divides a2 + b2 . Show that
a2 +b2
ab+1 is the square of an integer.

B6. (3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 32, 166)

Suppose A, B, C, D are n × n matrices with entries in a field F , satisfying
the conditions that AB t and CDt are symmetric and ADt − BC t = I. Here I
is the n × n identity matrix, and if M is an n × n matrix, M t is the transpose
of M . Prove that At D − C t B = I.

Solution. The conditions of the problem are

(1) AB t = (AB t )t = BAt ,
(2) CDt = (CDt )t = DC t ,
(3) ADt − BC t = I.
Solutions: The Forty-Seventh Competition (1986) 75

Taking the transpose of (3) gives DAt −CB t = I. These four equations are the entries
in the block matrix identity


t 

A B D −B t I 0
= .
C D −C t At 0 I
(Here the matrices should be considered (2n) × (2n) matrices in the obvious way.) If
X, Y are m × m matrices with XY = Im , the m × m identity matrix, then Y = X −1
and Y X = Im too. Applying this to our product with m = 2n, we obtain

t 


D −B t A B I 0
= ,
−C t At C D 0 I
and equating the lower right blocks shows that −C t B + At D = I, as desired. 
76 The William Lowell Putnam Mathematical Competition

The Forty-Eighth William Lowell Putnam Mathematical Competition

December 5, 1987

A1. (72, 24, 22, 0, 0, 0, 0, 0, 5, 0, 45, 36)

Curves A, B, C, and D are defined in the plane as follows:†

x
A = (x, y) : x − y = 2
2 2
,
x + y2

y
B = (x, y) : 2xy + 2 = 3 ,
x + y2
 
C = (x, y) : x3 − 3xy 2 + 3y = 1 ,
 
D = (x, y) : 3x2 y − 3x − y 3 = 0 .

Prove that A ∩ B = C ∩ D.

Solution 1. Let z = x + iy. The equations defining A and B are the real and
imaginary parts of the equation z 2 = z −1 + 3i, and similarly the equations defining C
and D are the real and imaginary parts of z 3 − 3iz = 1. Hence for all real x and y,
we have

(x, y) ∈ A ∩ B ⇐⇒ z 2 = z −1 + 3i ⇐⇒ z 3 − 3iz = 1 ⇐⇒ (x, y) ∈ C ∩ D.

Thus A ∩ B = C ∩ D. 

Solution 2. Let F = x2 −y 2 − x2 +y
x
2 , G = 2xy + x2 +y 2 −3, H = x −3xy +3y −1,
y 3 2

and J = 3x y − 3x − y be the rational functions whose sets of zeros are A, B, C, and

2 3

D, respectively. The identity




x −y F H
= (1)
y x G J
shows immediately that F = G = 0
implies 
H = J = 0. Conversely if H = J = 0
x −y
at (x, y), then (x, y) = (0, 0), so det = x2 + y 2 is nonzero, so (1) implies
y x
F = G = 0 at (x, y). 
Remark. Solution 1 multiplies z 2 − z −1 − 3i by z to obtain z 3 − 3iz − 1. Solution 2
is simply doing this key step in terms of real and imaginary parts, so it is really the
same solution, less elegantly written.

A2. (117, 28, 22, 2, 0, 0, 0, 3, 10, 5, 10, 7)

The sequence of digits

123456789101112131415161718192021 . . .

is obtained by writing the positive integers in order. If the 10n th digit

in this sequence occurs in the part of the sequence in which the m-digit
numbers are placed, define f (n) to be m. For example, f (2) = 2 because the

† The equations defining A and B are indeterminate at (0, 0). The point (0, 0) belongs to neither.
Solutions: The Forty-Eighth Competition (1987) 77

100th digit enters the sequence in the placement of the two-digit integer
55. Find, with proof, f (1987).
Answer. The value of f (1987) is 1984.
Solution. Let g(m) denote the total number of digits in the integers with m or
fewer digits. Then f (n) equals the integer m such that g(m − 1) < 10n ≤ g(m).
m
There are 10r −10r−1 numbers with exactly r digits, so g(m) = r=1 r(10r −10r−1 ).
We have 1983
g(1983) ≤ 1983(10r − 10r−1 ) ≤ 1983 · 101983 < 101987
r=1
and
g(1984) ≥ 1984(101984 − 101983 ) = 1984 · 9 · 101983 > 104 · 101983 = 101987
so f (1987) = 1984. 
Motivation. Based on the growth of geometric series one might guess that g(m)
has size roughly equal to its top term, which is 9 · m · 10m−1 . Thus we seek m such
that 9 · m · 10m−1 ≈ 101987 . Using 9 ≈ 10, the condition becomes m ≈ 101987−m . This
leads to the guess m = 1984, since 1984 ≈ 103 .
Remark. There is a closed form for g(m):
m m
g(m) = r10r − r10r−1
r=1 r=1
m m−1
= r10r − (s + 1)10s
r=1 s=0
4 m−1
5 m−1
= m10 + m
r10 r
− (s + 1)10s
r=0 s=0
m−1
= m10m − 10s
s=0
= m10m − (10m − 1)/9.
b
More generally, there is a closed form for r=a P (r)xr for any integers a ≤ b, fixed
polynomial P , and number x. Rearrangement as above shows that
b b
(1 − x) P (r)xr = P (a)xa − P (b)xb+1 + (P (r) − P (r − 1))xr ,
r=a r=a+1

and the last sum is of the same type but with a polynomial P (r) − P (r − 1) of lower
degree than P , or zero if P was constant to begin with. Hence one can evaluate the
sum by induction on deg P .
b r
Alternatively, r=a P (r)x can be evaluated by repeatedly differentiating the
formula for the geometric series
b
xb+1 − xa
xr =
r=a
x−1
and taking linear combinations of the resulting identities.
 78 The William Lowell Putnam Mathematical Competition

A3. (119, 32, 1, 0, 1, 4, 2, 0, 0, 5, 11, 29)

For all real x, the real-valued function y = f (x) satisfies

y  − 2y  + y = 2ex .

(a) If f (x) > 0 for all real x, must f  (x) > 0 for all real x? Explain.
(b) If f  (x) > 0 for all real x, must f (x) > 0 for all real x? Explain.

Answers. (a) No. (b) Yes.

Solution. One solution to the differential equation is x2 ex , and the general

solution to the differential equation y  −2y  +y = 0 with characteristic equation (r−1)2
is (bx + c)ex , so
 the general solution to the original equation is f (x) = (x2 + bx + c)ex .

Then f (x) = x + (b + 2)x + (b + c) ex . Since the leading coefficient of x2 + bx + c
2

is positive, we have

f (x) > 0 for all x ⇐⇒ x2 + bx + c > 0 for all x ⇐⇒ b2 − 4c < 0.

Similarly

f  (x) > 0 for all x ⇐⇒ (b + 2)2 − 4(b + c) < 0 ⇐⇒ b2 − 4c + 4 < 0.

Clearly b2 − 4c < 0 does not imply b2 − 4c + 4 < 0. (Take b = 1, c = 1 for instance.)

But b2 − 4c + 4 < 0 does imply b2 − 4c < 0. 

A4. (14, 8, 6, 4, 0, 0, 0, 1, 28, 9, 25, 109)

Let P be a polynomial, with real coefficients, in three variables and F be
a function of two variables such that

P (ux, uy, uz) = u2 F (y − x, z − x) for all real x, y, z, u,

and such that P (1, 0, 0) = 4, P (0, 1, 0) = 5, and P (0, 0, 1) = 6. Also let A, B, C

be complex numbers with P (A, B, C) = 0 and |B − A| = 10. Find |C − A|.
√
Answer. The value of |C − A| is (5/3) 30.

Solution. Letting u = 1 and x = 0, we have that F (y, z) = P (0, y, z) is

a polynomial. Also, F (uy, uz) = P (0, uy, uz) = u2 P (0, y, z) = u2 F (y, z), so F is
homogeneous of degree 2. Therefore

P (x, y, z) = F (y − x, z − x) = a(y − x)2 + b(y − x)(z − x) + c(z − x)2

for some real a, b, c. Then 4 = P (1, 0, 0) = a + b + c, 5 = P (0, 1, 0) = a, and

6 = P (0, 0, 1) = c, so b = −7. Now

0 = P (A, B, C) = 5(B − A)2 − 7(B − A)(C − A) + 6(C − A)2 ,

so the number m = (C − A)/(B − A) satisfies 5 − 7m + 6m2 = 0.  The zeros of

6m2 − 7m + 5 are complex conjugate
√ with product 5/6, so |m| = 5/6. Thus
|C − A| = 5/6|B − A| = (5/3) 30. 
Solutions: The Forty-Eighth Competition (1987) 79

A5. (8, 0, 2, 0, 0, 0, 0, 0, 1, 1, 57, 135)

Let


- −y x
G(x, y) = , ,0 .
x2 + 4y 2 x2 + 4y 2
Prove or disprove that there is a vector-valued function
- (x, y, z) = (M (x, y, z), N (x, y, z), P (x, y, z))
F
with the following properties:
(i) M, N, P have continuous partial derivatives for all (x, y, z) = (0, 0, 0);
(ii) Curl F- = -0 for all (x, y, z) = (0, 0, 0);
(iii) F- (x, y, 0) = G(x,
- y).
-.
Answer. There is no such F
Solution. Let S be a surface not containing (0, 0, 0) whose boundary ∂S is the
ellipse x2 + 4y 2 − 4 = z = 0 parameterized by (2 cos θ, sin θ, 0) for 0 ≤ θ ≤ 2π. (For
instance, S could be the half of the ellipsoid x2 + 4y 2 + z 2 = 4 with z ≥ 0.) If F exists,
then

0= (Curl F- ) · -n dS (since Curl F- = -0 on S)
 S

= F- · d-r (by Stokes’ Theorem, e.g. [Ap2, Theorem 12.3])

 ∂S

= G- · d-r (since F- = G - on ∂S)

 2π

∂S
− sin θ 2 cos θ
= , , 0 · (−2 sin θ, cos θ, 0) dθ
0 4 4
 2π
1
= dθ
0 2
= π,
a contradiction. 

A6. (5, 10, 2, 0, 1, 0, 0, 3, 1, 4, 80, 98)

For each positive integer n, let a(n) be the number of zeros in the base 3
representation of n. For which positive real numbers x does the series
∞
xa(n)
n=1
n3
converge?
Answer. For positive real x, the series converges if and only if x < 25.
Solution. The integer n ≥ 1 has exactly k + 1 digits in base 3 if and only if
3 ≤n<3
k k+1
. Define
3k+1 −1 3k+1 −1
xa(n)
Sk = , and Tk = xa(n) .
n3
n=3k n=3k
80 The William Lowell Putnam Mathematical Competition

∞ a(n) 3
The given series n=1 x /n has all terms positive, so it will converge if and
∞
only if S converges. For 3k ≤ n < 3k+1 , we have 33k ≤ n3 < 33k+3 , so
k=0 k
∞ ∞
Tk /3 3k+3
≤ Sk ≤ Tk /3 . Therefore k=0 Sk converges if and only if k=0 Tk /33k
3k

converges.

k k+1−i
The number of n with
k
 k + 1 digits base 3 and satisfying a(n) = i is
i 2 , because there are i possibilities for the set of positions of the i zero
digits (since the leading digit cannot be zero), and then 2k+1−i ways to select 1 or 2
as each of the remaining digits. Therefore
k

k k+1−i i
Tk = 2 x = (x + 2)k .
i=0
i

Hence
k
∞ ∞
x+2
Tk /33k = ,
27
k=0 k=0

which converges if and only if |(x + 2)/27| < 1. For positive x, this condition is
equivalent to 0 < x < 25. 
Remark. More generally, let αk (n) be the number of zeros in the base k expansion
∞ αk (n) k
of n, and let Ak (x) = n=1 x /n . Then Ak (x) converges at a positive real
number x if and only if x < kk − k + 1.
Literature note. For more other convergent sums involving digits in base b
representations, see [BB]. This article contains exact formulas for certain sums, as
well as approximations by “nice” numbers that agree to a remarkable number of
decimal places.

B1. (148, 5, 4, 1, 0, 0, 0, 0, 1, 0, 15, 30)

Evaluate
 
4
ln(9 − x) dx
  .
2 ln(9 − x) + ln(x + 3)

Answer. The value of the integral is 1.

Solution. The integrand is continuous on [2, 4]. Let I be the value of the integral.
As x goes from 2 to 4, 9 − x and x + 3 go from 7 to 5, and from 5 to 7, respectively.
This symmetry suggests the substitution x = 6 − y reversing the interval [2, 4]. After
interchanging the limits of integration, this yields
 4 
ln(y + 3) dy
I=   .
2 ln(y + 3) + ln(9 − y)
Thus    
4
ln(x + 3) + ln(9 − x) 4
2I =   dx = dx = 2,
2 ln(x + 3) + ln(9 − x) 2

and I = 1. 
√
Remark. The same argument applies if ln x is replaced by any continuous function
such that f (x + 3) + f (9 − x) = 0 for 2 ≤ x ≤ 4.
Solutions: The Forty-Eighth Competition (1987) 81

B2. (47, 3, 4, 0, 0, 0, 0, 0, 1, 3, 62, 84)

Let r, s, and t be integers with 0 ≤ r, 0 ≤ s, and r + s ≤ t. Prove that
s s  s
t+1
0t  +  1t  + · · · +  st  =  .
r r+1 r+s (t + 1 − s) t−s
r
 n n(n−1)···(n+1−k)
(Note: k denotes the binomial coefficient k(k−1)···3·2·1 .)

Let F (r, s, t) be the left-hand side.

Solution 1. We prove
t+1
F (r, s, t) =  
(t + 1 − s) t−sr

by induction on s. The base case s = 0 is trivial since 00 = 1.
For s ≥ 1, r ≥ 0 and r + s ≤ t,
s−1 s−1 s−1 s−1 s−1 s−1
+ 1 +
F (r, s, t) =  t  +
0 0
 t  + · · · + s−2 t s−1
 + s−1
t

r r+1 r+s−1 r+s
= F (r, s − 1, t) + F (r + 1, s − 1, t).
Applying the inductive hypothesis to the two terms on the right gives
t+1 t+1
F (r, s, t) = t+1−s +  .
(t + 2 − s) r (t + 2 − s) t+1−s
r+1
t+1−s
The definition
t+1−s of binomial
t−s coefficients in terms of factorials lets us express r and
r+1 in terms of r ; this leads to
t+1 t+1
F (r, s, t) = " # + " # 
(t + 2 − s) t+1−s
t+1−s−r (t + 2 − s) t+1−s
t−s
r r+1
t−s
r


t+1 t+1−s−r r+1
=   +
(t + 2 − s) t−s
r
t + 1 − s t + 1−s
t+1
=  ,
(t + 1 − s) t−s
r
completing the inductive step. 
Solution 2. Writing the binomial coefficients in terms of factorials and regrouping,
we find
s


s! r! (t − r − s)! r+i t−r−i
F (r, s, t) = . (1)
t! i=0
r t−r−s
If we could prove
s



r+i t−r−i t+1
= , (2)
i=0
r t−r−s t−s+1
then substituting into (1) would yield
s! r! (t − r − s)! (t + 1)! t+1
F (r, s, t) = · =  .
t! (t − s + 1)! s! (t + 1 − s) t−s
r
82 The William Lowell Putnam Mathematical Competition

We now provide three proofs of (2):

Proof 1 (Vandermonde’s identity). We have



r+i r+i
=
r i
(r + i)(r + i − 1) · · · (r + 1)
=
i!
(−r − 1)(−r − 2) · · · (−r − i)
= (−1)i

 i!
−r − 1
= (−1)i
i
and similarly




t−r−i (t − r − s) + (s − i) −t + r + s − 1
= = · · · = (−1)s−i
t−r−s s−i s−i
and



t+1 (t − s + 1) + s s −t + s − 2
= = · · · = (−1) .
t−s+1 s s
Thus (2) can be rewritten as
s



−r − 1 −t + r + s − 1 −t + s − 2
= ,
i=0
i s−i s
by multiplying both sides by (−1)s . This is a special case of Vandermonde’s identity,
which in general states that for integers m, n, s with s ≥ 0,
s



m n m+n
= .
i=0
i s−i s

Proof 2 (generating functions). The binomial expansion gives

∞
 ∞

−(n+1) −n − 1 n+i i
(1 − x) = i
(−x) = x,
i=0
i i=0
n

so taking coefficients of xs in the identity (1 − x)−(r+1) (1 − x)−(t−r−s+1) =

(1 − x)−(t−s+2) yields (2).
Proof 3 (bijective proof ). We will show that the two sides of (2) count something in
two different ways. First set j = r + i to rewrite (2) as
r+s



j t−j t+1
= . (3)
j=r
r t−r−s s

We will show that both sides of (3) count the number of sequences of s zeros and
t − s ones, punctuated by a comma such that the number of ones occurring before the
comma is r. On one hand, the number of such
 sequences of t + 1 symbols (including
the comma) equals the right-hand side t+1s of (3), because they can be constructed
by choosing the s positions for the zeros: of the remaining positions, the (r + 1)st
must contain the comma and the others must contain ones. On the other hand, we
can count the sequences according to the position of the comma: given that there are
Solutions: The Forty-Eighth Competition (1987) 83

exactly j digits before the comma, there are rj possibilities for the digits before
 t−jthe
comma (since r of them are to be ones and the rest are to be zeros), and t−r−s
possibilities for the digits after the comma (since one needs t − r − s more ones to
bring the total number of ones to t − s). Summing over j shows that the total number
of sequences is the left-hand side of (3). 
Motivation. To find the generating function solution, first observe that the sum
in (1) looks like the coefficient of xs in a product of two series, and then figure out
what the two series must be.
Remark. Vandermonde’s identity can also be used in 1991B4.
Literature note. Generating functions are a powerful method for proving combi-
natorial identities. A comprehensive introduction to this method is [Wi].
Related question. Problem 20 of [WH] is similar:
Evaluate the sum n  n
S= 2n−1
k

k=0 k

for all positive integers n.

(Hint: the answer does not depend on n.)

B3. (27, 31, 5, 0, 0, 0, 0, 0, 3, 11, 57, 70)

Let F be a field in which 1 + 1 = 0. Show that the set of solutions to the
equation x2 + y 2 = 1 with x and y in F is given by (x, y) = (1, 0) and

2 
r −1 2r
(x, y) = , ,
r2 + 1 r2 + 1
where r runs through the elements of F such that r2 = −1.
" 2 #
Solution 1. For r2 = −1, let (xr , yr ) = rr2 −1 2r
+1 r 2 +1 . Clearly (1, 0) and (xr , yr )
,
for r = −1 are solutions.
2

Conversely suppose x2 + y 2 = 1. If x = 1, then y = 0 and we have (1, 0).

Otherwise define r = y/(1 − x). (The reason for this is that yr /(1 − xr ) = r.)
Then 1 − x2 = y 2 = r2 (1 − x)2 , but x = 1, so we may divide by 1 − x to obtain
1 + x = r2 (1 − x), and (r2 + 1)x = (r2 − 1). If r2 = −1, then this says 0 = −2,
2
contradicting 1 + 1 = 0. Thus r2 = −1, x = rr2 −1
+1 = xr , and y = r(1 − x) = r 2 +1 = yr .
2r
2 2
Hence every solution to x + y = 1 not equal to (1, 0) is of the form (xr , yr ) for some
r ∈ F with r2 = −1. 
Remark. If instead F is a field in which 1 + 1 = 0 (i.e., the characteristic of F
is 2), then x2 + y 2 = 1 is equivalent to (x + y + 1)2 = 0, and the set of solutions is
{ (t, t + 1) : t ∈ F }.

Solution 2. Essentially the same solution can be motivated by geometry. The

only solution to x2 + y 2 = 1 with x = 1 is (1, 0). For each (x, y) ∈ F 2 satisfying
x2 + y 2 = 1 with x = 1, the line through (x, y) and (1, 0) has slope y/(x − 1) in
F . Hence we can find all solutions to x2 + y 2 = 1 with x = 1 by intersecting each
nonvertical line through (1, 0) with the circle. (See Figure 4.)
84 The William Lowell Putnam Mathematical Competition

2 2
x +y =1

(1, 0)

(x, y)

FIGURE 4.
Points with coordinates in F correspond to lines through (1, 0) with slopes in F .

If Ls is the line through (1, 0) with slope s ∈ F , its intersection with x2 + y 2 = 1 can
be computed by substituting y = s(x − 1) into x2 + y 2 = 1, and solving the resulting
equation
x2 + s2 (x − 1)2 = 1.
This equation is guaranteed to have the solution x = 1, because (1, 0) is in the
intersection. Therefore we obtain a factorization,
 
(x − 1) (s2 + 1)x + (1 − s2 ) = 0,
which yields the solutions x = 1 and, if s2 = −1, also x = (s2 − 1)/(s2 + 1). (If
s2 = −1, then 1 − s2 = −2 = 0, and the second " 2 factor gives
# no solution.) Using
s −1 −2s
y = s(x − 1), we find that these give (1, 0) and s2 +1 , s2 +1 , which, as we verify, do
satisfy x2 + y 2 = 1 and y = s(x − 1). We finish by substituting s = −r. 
Remark. Let C be any nondegenerate conic over F with an F -rational point P :
this means that C is given by a polynomial f (x, y) ∈ F [x, y] of total degree 2 that does
not factor into linear polynomials over any field extension of F , and P = (a, b) ∈ F 2
is a point such that f (a, b) = 0. The same method (of drawing all lines through P
with slope in F , and seeing where they intersect f (x, y) = 0 other than at P ) lets one
parameterize the set of F -rational points of C in terms of a single parameter s. In the
language of algebraic geometry, one says that any conic over k with a k-rational point
is birationally equivalent to the line [Shaf, p. 11]. The same method works on certain
equations of higher degree [NZM, Section 5.6].
The parameterization in the preceding  paragraph is the reason why indefinite
integrals of rational functions in t and p(t) for a single quadratic polynomial p(t)
can be expressed in terms of elementary functions [Shaf, p. 7]. It also explains why
sin θ and cos θ can be expressed as rational functions of a single function:


1 − t2 2t
(cos θ, sin θ) = , where t = tan(θ/2).
1 + t2 1 + t2
This, in turn, explains why rational functions in sin θ and cos θ, such as
sin3 θ − 7 sin θ cos θ
,
1 + cos3 θ
have elementary antiderivatives.
Solutions: The Forty-Eighth Competition (1987) 85

Remark. The parameterization of solutions to x2 +y 2 = 1 over Q is closely linked to

the parameterization of primitive Pythagorean triples, i.e., positive integer solutions
to a2 + b2 = c2 with gcd(a, b, c) = 1. In any primitive Pythagorean triple, exactly one
of a and b is even; the set of primitive Pythagorean triples (a, b, c) with b even equals
the set of triples (m2 − n2 , 2mn, m2 + n2 ) with m, n ranging over positive integers of
opposite parity satisfying m > n and gcd(m, n) = 1. A more number-theoretic (less
geometric) approach to this classification is given in [NZM, Section 5.3].

B4. (13, 14, 15, 1, 0, 0, 0, 0, 11, 23, 79, 48)

Let (x1 , y1 ) = (0.8, 0.6) and let xn+1 = xn cos yn − yn sin yn and yn+1 =
xn sin yn + yn cos yn for n = 1, 2, 3, . . . . For each of limn→∞ xn and limn→∞ yn ,
prove that the limit exists and find it or prove that the limit does not
exist.
Answer. Both limits exist: limn→∞ xn = −1 and limn→∞ yn = 0.

Solution. Since (0.8)2 + (0.6)2 = 1, we have (x1 , y1 ) = (cos θ1 , sin θ1 ) where

θ1 = cos−1 (0.8). If (xn , yn ) = (cos θn , sin θn ) for some n ≥ 1 and number θn , then
by the trigonometric addition formulas, (xn+1 , yn+1 ) = (cos(θn + yn ), sin(θn + yn )).
Hence by induction, (xn , yn ) = (cos θn , sin θn ) for all n ≥ 1, where θ2 , θ3 , . . . are
defined recursively by θn+1 = θn + yn for n ≥ 1. Thus θn+1 = θn + sin θn .
For 0 < θ < π, sin θ > 0 and sin θ = sin(π − θ) < π − θ (see remark below for
explanation), so 0 < θ + sin θ < π. By induction, 0 < θn < π for all n ≥ 1. Also
θn+1 = θn + sin θn > θn , so the bounded sequence θ1 , θ2 , . . . is also increasing,
and hence has a limit L ∈ [0, π]. Since sin t is a continuous function, taking the
limit as n → ∞ in θn+1 = θn + sin θn shows that L = L + sin L, so sin L = 0.
But L ∈ [0, π] and L ≥ θ1 > 0, so L = π. By continuity of cos t and sin t,
limn→∞ xn = cos L = cos π = −1 and limn→∞ yn = sin L = sin π = 0. 
Remark. To show that sin x < x for x > 0, integrate cos t ≤ 1 from t = 0 to t = x,
and note that cos t < 1 for t ∈ (0, 2π).
Reinterpretation. This problem is about the limiting behavior of a dynamical
system. For more examples of dynamical systems, see 1992B3, 1995B4, and 1996A6.

B5. (10, 6, 3, 2, 2, 0, 0, 0, 8, 3, 16, 154)

Let On be the n-dimensional vector (0, 0, . . . , 0). Let M be a 2n × n
matrix of complex numbers such that whenever (z1 , z2 , . . . , z2n )M = On , with
complex zi , not all zero, then at least one of the zi is not real. Prove
that for arbitrary real numbers r1 , r2 , . . . , r2n , there are complex numbers
w1 , w2 , . . . , wn such that
    
w1 r1
    
Re M  ...  =  ...  .
wn r2n
(Note: if C is a matrix of complex numbers, Re(C) is the matrix whose
entries are the real parts of the entries of C.)
86 The William Lowell Putnam Mathematical Competition

Solution.
 Write M = A + iB where A and B are real 2n × n matrices.  If
z = z1 z2 · · · z2n is a row vector with real entries such  that
 z A B = 0, then
zM = zA + izB = 0, so z = 0 by hypothesis. Hence A B is an invertible real
2n × 2n matrix.
Let r be the real column vector (r1 , . . . , r2n ). If w is a complex column vector
of length n, and we write w = u + iv where u and v are the real and imaginary
 the condition Re [M w] = r is equivalent to Au − Bv = r, and to
parts of
w, then
  u  
A B = r. Since A B is invertible, we can find a real column vector
−v


u
satisfying this. 
−v

B6. (8, 1, 1, 1, 0, 0, 1, 2, 1, 3, 62, 124)

Let F be the field of p2 elements where p is an odd prime. Suppose S is
a set of (p2 − 1)/2 distinct nonzero elements of F with the property that for
each a = 0 in F , exactly one of a and −a is in S. Let N be the number of
elements in the intersection S ∩ { 2a : a ∈ S }. Prove that N is even.

We write 2S for { 2a : a ∈ S }.
Solution 1. For a ∈ S, there is a unique way to write 2a = Ca sa where Ca = ±1
3
and sa ∈ S. Then S ∩ 2S = { a ∈ S : Ca = 1 }, so a∈S Ca = (−1)#S−N = (−1)N ,
since #S = (p − 1) · (p + 1)/2 is even. In F , we have
2 + + +
2(p −1)/2 a= Ca sa = (−1)N a
a∈S a∈S a∈S

N (p2 −1)/2 p−1 (p+1)/2 (p+1)/2

so (−1) = 2 = (2 ) = 1 = 1, by Fermat’s Little Theorem
[Lar1, p. 148]. Hence N is even. 
Remark. Gauss based one of his proofs of quadratic reciprocity on the following
lemma† [NZM, Theorem 3.2], whose proof uses the same method as Solution 1:
Lemma. Let p be an odd prime, and suppose that a is an integer prime to p.
Consider the least positive residues modulo p of a, 2a, 3a, . . . " # − 1)/2)a. If n is
, ((p
the number of these that exceed p/2, then the Legendre symbol ap equals (−1)n .
# case a 2= 2, which is especially close to Solution 1, easily implies the formula
" The
2 (p −1)/8
p = (−1) , see [NZM, Theorem 3.3].
Solution 2. Let {1, x} be a basis for F over the field Fp of p elements. Let
H = {1, 2, . . . , (p − 1)/2} ⊂ Fp , and let
S0 = { a + bx : a ∈ H, b ∈ Fp or a = 0, b ∈ H }.
For each nonzero a ∈ F , exactly one of a and −a is in S0 . Also,
S0 ∩ 2S0 = { a + bx : a ∈ Q, b ∈ Fp or a = 0, b ∈ Q }

† This lemma rarely appears outside the context of the proof of quadratic reciprocity. It should not
be confused with the more important result called Gauss’s Lemma that is stated after Solution 3 to
1998B6.
Solutions: The Forty-Eighth Competition (1987) 87

where Q = H ∩ 2H, so #(S0 ∩ 2S0 ) = (#Q)p + (#Q), which is divisible by p + 1,

hence even.
Every other possible S can obtained by repeatedly replacing some α ∈ S by −α, so
it suffices to show that the parity of N = #(S ∩2S) is unchanged by such an operation
on S. Suppose S is as in the problem, and S  is the same as S except with α replaced
by −α. Define N  analogously. We will show that N  − N is even.
Note that
(i) If β ∈ S ∩ 2S, then β ∈ S  ∩ 2S  unless β = α or 2α.
(ii) If β ∈ S  ∩ 2S  , then β ∈ S ∩ 2S unless β = −α or −2α.
In other words, N  can be computed from N by subtracting 1 for each of α and 2α
that belongs to S ∩ 2S, and adding 1 for each of −α and −2α that belongs to S ∩ 2S.
These adjustments are determined according to the four cases in the table below.

Case In S ∩ 2S? N − N
α 2α −α −2α
(1) α/2 ∈ S, 2α ∈ S yes yes no no −2
(2) α/2 ∈ S, −2α ∈ S yes no no yes 0
(3) −α/2 ∈ S, 2α ∈ S no yes yes no 0
(4) −α/2 ∈ S, −2α ∈ S no no yes yes 2

In each row of the table, N  − N equals the number of times “yes” appears under the
−α and −2α headers minus the number of times “yes” appears under the α and 2α
headers. Hence N  − N is even in each case, as desired. 
Related question. The following problem, proposed by Iceland for the 1985 Inter-
national Mathematical Olympiad, is susceptible to the approach of the second solution.
Suppose x1 , . . . , xn ∈ {−1, 1}, and
x1 x2 x3 x4 + x2 x3 x4 x5 + · · · + xn−3 xn−2 xn−1 xn
+ xn−2 xn−1 xn x1 + xn−1 xn x1 x2 + xn x1 x2 x3 = 0.
Show that n is divisible by 4.
88 The William Lowell Putnam Mathematical Competition

The Forty-Ninth William Lowell Putnam Mathematical Competition

December 3, 1988

A1. (202, 0, 0, 0, 0, 0, 0, 0, 4, 2, 0, 0)
Let R be the region consisting of the points (x, y) of the cartesian plane
satisfying both |x| − |y| ≤ 1 and |y| ≤ 1. Sketch the region R and find its
area.
Remark. For this problem, the average of the scores of the top 200 or so participants
was approximately 9.76, higher than for any other problem in this volume. The second
easiest problem by this measure was 1988B1, from the same year, with an average of
9.56. At the other extreme, the average for each of 1999B4 and 1999B5 was under
0.01.
Answer. The area of R is 6.

Solution. The part of R in the first quadrant is defined by the inequalities x ≥ 0,

0 ≤ y ≤ 1, and x − y ≤ 1. This is the trapezoid T with vertices (0, 0), (1, 0), (2, 1),
(0, 1). It is the union of the unit square with vertices (0, 0), (1, 0), (1, 1), (0, 1) and the
half-square (triangle) with vertices (1, 0), (2, 1), (1, 1), so the area of T is 3/2. The
parts of R in the other quadrant are obtained by symmetry, reflecting T in both axes
(see Figure 5), so the area of R is 4(3/2) = 6. 

(–2, 1) (2, 1)

(–1, 0) (1, 0)
x

(–2, –1) (2, –1)

FIGURE 5.
The region R.

A2. (174, 1, 7, 1, 0, 0, 0, 3, 3, 0, 15, 4)

A not uncommon calculus mistake is to believe that the product rule
2
for derivatives says that (f g) = f  g  . If f (x) = ex , determine, with proof,
whether there exists an open interval (a, b) and a nonzero function g defined
on (a, b) such that this wrong product rule is true for x in (a, b).
Answer. Yes, (a, b) and g exist.
Solutions: The Forty-Ninth Competition (1988) 89

   
"  # We are asked for a solution g to f g + gf = f g 2, which is equivalent
Solution.
to g  + f −f  g = 0 if we avoid the zeros of f − f  = (1 − 2x)ex , i.e., avoid x = 1/2.
f

By the existence and uniqueness theorem for first order linear ordinary differential
equations [BD, Theorem 2.4.1], if x0 = 1/2, and y0 is any real number, then there exists
a unique solution g(x), defined in some neighborhood (a, b) of x0 , with g(x0 ) = y0 .
By taking y0 nonzero, we obtain a nonzero solution g. 
Remark. One can solve the differential equation by separation of variables, to find
all possible g. If g is nonzero, the differential equation is equivalent to
2
g f 2xex 1
=  = 2 = 1 + ,
g f −f (2x − 1)e x 2x − 1
1
ln |g(x)| = x + ln |2x − 1| + c,
2
from which one finds that the nonzero solutions are of the form g(x) = Cex |2x − 1|1/2
for any nonzero number C, on any interval not containing 1/2.
Related question. What other pairs of functions f and g satisfy (f g) = f  g  ? For
some discussion, see [Hal, Problem 2G].

A3. (42, 17, 10, 0, 0, 0, 0, 6, 5, 3, 34, 91)

Determine, with proof, the set of real numbers x for which
∞
x
1 1
csc − 1
n=1
n n
converges.
Answer. The series converges if and only if x > 1/2.
Solution. Define
1 1 1
an = csc − 1 = − 1.
n n n sin n1
Taking t = 1/n in the inequality 0 < sin t < t for t ∈ (0, π), we obtain an > 0, so each
term axn of the series is defined for any real x. Using sin t = t − t3 /3! + O(t5 ) as t → 0,
we have, as n → ∞,
1
an =  1   − 1
n n − 6n3 + O n15
1

1
=   −1
1 − 6n2 + O n14
1


1 1
= 2 +O .
6n n4
In particular, if bn = 1/n2 , then axn /bxn has a finite limit as n → ∞, so by the
∞ x
Limit Comparison Test [Spv, Ch. 22, Theorem 2], n=1 an converges if and only
∞ x ∞ −2x
if n=1 nb = n=1 n converges, which by the Integral Comparison Test [Spv,
Ch. 22, Theorem 4] holds if and only if 2x > 1, i.e., x > 1/2. 
Remark (big-O and little-o notation). Recall that O(g(n)) is a stand-in for a
function f (n) for which there exists a constant C such that |f (n)| ≤ C|g(n)| for all
sufficiently large n. (This does not necessarily imply that limn→∞ f (n)/g(n) exists.)
 90 The William Lowell Putnam Mathematical Competition

Similarly “f (t) = O(g(t)) as t → 0” means that there exists a constant C such that
|f (t)| ≤ C|g(t)| for sufficiently small nonzero t.
On the other hand, o(g(n)) is a stand-in for a function f (n) such that
f (n)
lim = 0.
n→∞ g(n)
One can similarly define “f (t) = o(g(t)) as t → 0”.

A4. (45, 29, 2, 2, 1, 25, 11, 0, 3, 3, 25, 62)

(a) If every point of the plane is painted one of three colors, do there
necessarily exist two points of the same color exactly one inch apart?
(b) What if “three” is replaced by “nine”?
Justify your answers.
Answers. (a) Yes. (b) No.
Solution.
P

A B

FIGURE 6.

(a) Suppose, for the sake of obtaining a contradiction, that we have a coloring of the
plane with three colors such that any
√ two points at distance 1 have different colors. If
A and B are two points at distance 3, then the circles of radius 1 centered at A and B
meet in two points P, Q forming equilateral triangles AP Q and BP Q. (See Figure 6.)
The three points of each equilateral triangle have different colors, and this forces
√ A
and B to have equal colors. Now consider a triangle CDE with CD = CE = 3 and
DE = 1. (See Figure 7.) We know that C, D have the same color and C, E have the
same color, so D, E have the same color, contradicting our hypothesis about points at
distance 1.
D
3

C 1

3
E

FIGURE 7.

(b) For P = (x, y) ∈ R2 , define

f (P ) = ((3/2)x mod 3, (3/2)y mod 3) ∈ {0, 1, 2}2 .
Solutions: The Forty-Ninth Competition (1988) 91

Color the nine elements of {0, 1, 2}2 with the nine different colors, and give P the
color of f (P ). (Thus the plane is covered by blocks of 3 × 3 squares, where the little
squares have side length 2/3. See Figure 8.) If P and Q are points with the √ same
color, either they belong to the same little square, in which case P Q ≤ (2/3) 2 < 1,
or to different little squares. In the latter case, either their x-coordinates differ by at
least 4/3, or their y-coordinates differ by at least 4/3, so P Q ≥ 4/3 > 1. 

(0, 0) (1, 0) (2, 0) (0, 0)

(0, 2) (1, 2) (2, 2) (0, 2)

(0, 1) (1, 1) (2, 1) (0, 1)

3/2
(0, 0) (1, 0) (2, 0) (0, 0)

FIGURE 8.

Remark. Part (a) appears in [New, p. 7], as does the following variation:
Assume that the points of the plane are each colored red or blue. Prove that
one of these colors contains pairs of points at every distance.

Remark. Figure 9 shows a well-known coloring of the plane with seven colors
such that points at distance 1 always√have different colors. Such a coloring can be
constructed as follows. Let ω = −1+2 −3 , and view the ring of Eisenstein integers
Z[ω] = { a + bω : a, b ∈ Z } as a lattice in the complex plane. Then p = (2 − ω)Z[ω]
is a sublattice of index |2 − ω|2 = 7. (In fact, p is one of the two prime ideals of Z[ω]
that divide (7).) Give each coset of p in Z[ω] its own color. Next color each point in
C according to the color of a nearest point
 in Z[ω]. This gives a coloring by hexagons.
The diameter of each hexagon equals 4/3, and if two distinct hexagons  have the
same
 color, the
 smallest distance between points of their closures is 7/3. Hence if
4/3 < d < 7/3, no two points in the plane at distance d have the same color.
Scaling the whole picture by 1/d, we find a coloring in which no two points in the
plane at distance 1 have the same color.
Remark. Let χ(n) denote the minimum number of colors needed to color the
points in Rn so that each pair of points separated by distance 1 have different colors.
Parts (a) and (b) show χ(2) > 3 and χ(2) ≤ 9, respectively. Because of the hexagonal
92 The William Lowell Putnam Mathematical Competition

✧✧❜❜✧✧❜❜✧✧❜❜✧✧❜❜✧✧❜❜✧✧❜❜
A B C D E F
✧✧❜❜✧
❜❜✧ ❜✧✧❜❜✧ ❜✧✧❜❜
❜✧✧❜❜✧ ❜✧✧✧❜❜✧
❜ ✧
F G A B C
✧✧❜❜
❜✧✧✧❜❜✧
❜✧✧❜❜
❜✧✧✧❜❜✧ ❜✧✧❜❜
❜✧✧❜❜✧
C D E F G A
❜❜✧
✧✧❜❜✧
❜✧✧❜❜✧
❜✧✧❜❜✧
❜✧✧❜❜
❜✧✧✧❜❜✧
❜ ✧
A B C D E
✧✧❜❜
❜✧ ❜✧✧❜❜
✧✧❜❜✧ ❜✧✧✧❜❜✧ ❜✧✧❜❜
❜✧✧❜❜✧
E F G A B C
❜❜✧✧❜❜✧✧❜❜✧✧❜❜✧✧❜❜✧✧❜❜✧✧

FIGURE 9.

coloring in the previous remark, we know 4 ≤ χ(2) ≤ 7. Ronald L. Graham is offering

$1000 for an improvement of either bound.
The study of χ(n) goes back at least to [Had] in 1944. As n → ∞, one has
(1 + o(1))(1.2)n ≤ χ(n) ≤ (3 + o(1))n ,
the lower and upper bounds being due to [FW] and [LR], respectively. (See the remark
after the solution to 1988A3 for the meaning of the little-o notation.) For more on
such questions, consult [Gr2] and the references listed there, and browse issues of the
journal Geombinatorics.

A5. (32, 8, 4, 0, 0, 0, 0, 1, 7, 96, 9, 51)

Prove that there exists a unique function f from the set R+ of positive
real numbers to R+ such that
f (f (x)) = 6x − f (x) and f (x) > 0 for all x > 0.

Solution. We will show that f (x) = 2x for x > 0. Fix x > 0, and consider the
sequence (an ) of positive numbers defined by a0 = x and an+1 = f (an ). The given
functional equation implies that an+2 +an+1 −6an = 0. The zeros of the characteristic
polynomial t2 + t − 6 of this linear recursion are −3 and 2, so there exist real constants
c1 , c2 such that an = c1 2n + c2 (−3)n for all n ≥ 0. If c2 = 0, then for large n, an has
the same sign as c2 (−3)n , which alternates with n; this contradicts an > 0. Therefore
c2 = 0, and an = c1 2n . In particular f (x) = a1 = 2a0 = 2x. This holds for any x.
Finally, the function f (x) = 2x does satisfy the conditions of the problem. 
Related question. The following, which was Problem 5 of the 1989 Asian-Pacific
Mathematical Olympiad [APMO], can be solved using a similar method:
Determine all functions f from the reals to the reals for which
(1) f (x) is strictly increasing,
(2) f (x) + g(x) = 2x for all real x,
Solutions: The Forty-Ninth Competition (1988) 93

where g(x) is the composition inverse function to f (x). (Note: f and g are
said to be composition inverses if f (g(x)) = x and g(f (x)) = x for all real x.)
Remark (linear recursive sequences with constant coefficients). We can describe the
general sequence x0 , x1 , x2 , . . . satisfying the linear recursion

xn+k + bk−1 xn+k−1 + · · · + b1 xn+1 + b0 xn = 0

for all n ≥ 0, where b0 , b1 , . . . , bk−1 are constants, by considering the characteristic

polynomial
f (t) = tk + bk−1 tk−1 + · · · + b1 t + b0

over a field large enough to contain all its zeros. If the zeros r1 , . . . , rk of f (t)
k
are distinct, then the general solution is xn = i=1 ci rin where the ci are arbitrary
3s
i=1 (t − ri )
mi
constants not depending on n. More generally, if f (t) = , then,
provided that we are working over a field of characteristic zero, the general solution
s
is xn = i=1 ci (n)rin where ci (n) is a polynomial of degree less than mi . For any
characteristic, the latter statement remains true if we replace the polynomial
  ci (n)

by a general linear combination of the binomial coefficients n0 , n1 , . . . , min−1 .
(These combinations are the same as the polynomials in n of degree less than mi if
the characteristic is zero, or if the characteristic is at least as large as mi .) All of
∞
these statements can be proved by showing that the generating function i=0 xi ti is
a rational function of t, and then decomposing it as a sum of partial fractions and
expanding each partial fraction in a power series to get a formula for xi . For more
discussion, see [NZM, Appendix A.4].
For example, these results can be used to show that the Fibonacci sequence, defined
by F0 = 0, F1 = 1, and Fn+2 = Fn+1 + Fn for n ≥ 0, satisfies
" √ #n " √ #n
1+ 5
2 − 1−2 5
Fn = √ .
5
The formula for the general solution to linear recursions with constant coefficients
can be thought of as the discrete analogue of the general solution to homogeneous
linear ordinary differential equations with constant coefficients.

A6. (59, 14, 10, 7, 0, 0, 0, 0, 5, 0, 21, 92)

If a linear transformation A on an n-dimensional vector space has n + 1
eigenvectors such that any n of them are linearly independent, does it
follow that A is a scalar multiple of the identity? Prove your answer.
Answer. Yes, A must be a scalar multiple of the identity.

Solution 1. Let x1 , x2 , . . . , xn+1 be the given eigenvectors, and let λ1 ,

λ2 , . . . , λn+1 be their eigenvalues. The set Bi = {x1 , . . . , xi−1 , xi+1 , . . . , xn+1 }
is a linearly independent set of n vectors in an n-dimensional vector space, so
Bi is a basis, with respect to which A is represented by the diagonal matrix
diag(λ1 , . . . , λi−1 , λi+1 , . . . , λn+1 ). Thus the trace of A equals S − λi where S =
n+1
i=1 λi . But the trace is independent of the basis chosen, so S − λi = S − λj for
all i, j. Hence all the λi are equal. With respect to the basis B1 , A is represented by
94 The William Lowell Putnam Mathematical Competition

a diagonal matrix with equal entries on the diagonal, so A is a scalar multiple of the
identity. 
Remark. One could have worked with the multiset of roots of the characteristic
polynomial, instead of their sum (the trace).
Solution 2 (Lenny Ng). Let x1 , . . . , xn+1 be the eigenvectors of A, with
eigenvalues λ1 , . . . , λn+1 . Since x1 , . . . , xn are linearly independent, they span the
vector space; hence
n
xn+1 = αi xi
i=1

for some α1 , . . . , αn . Multiply by λn+1 , or apply A to both sides, and compare:

n n
αi λn+1 xi = λn+1 xn+1 = αi λi xi .
i=1 i=1

Thus αi λn+1 = αi λi for all i between 1 and n. If αi = 0 for some i, then xn+1 can
be expressed as a linear combination of x1 , . . . , xi−1 , xi+1 , . . . , xn , contradicting the
linear independence hypothesis. Hence αi = 0 for all i, so λn+1 = λi for all i. This
implies A = λn+1 I. 

B1. (176, 25, 0, 0, 0, 0, 0, 0, 1, 1, 1, 4)

A composite (positive integer) is a product ab with a and b not necessarily
distinct integers in {2, 3, 4, . . . }. Show that every composite is expressible
as xy + xz + yz + 1, with x, y, and z positive integers.
Solution. Substituting z = 1 yields (x + 1)(y + 1), so to represent the composite
number n = ab with a, b ≥ 2, let (x, y, z) = (a − 1, b − 1, 1). 
Remark. Although the problem asks only about representing composite numbers,
all but finitely many prime numbers are representable too. Theorem 1.1 of [BC] proves
that the only positive integers not of the form xy + xz + yz + 1 for integers x, y, z > 0
are the 19 integers 1, 2, 3, 5, 7, 11, 19, 23, 31, 43, 59, 71, 79, 103, 131, 191, 211, 331,
and 463, and possibly a 20th integer greater than 1011 . Moreover, if the Generalized
Riemann Hypothesis (GRH) is true, then the 20th integer does not exist. (See [Le]
for earlier work on this problem.)
The situation is analogous to that of the class number 1 problem:
√ for many years it
was known that the squarefree integers d > 0 such that Q( −d) has class number 1
were
d = 1, 2, 3, 7, 11, 19, 43, 67, 163
and possibly one more; the existence of this tenth imaginary quadratic field of class
number 1 was eventually ruled out: see the appendix to [Se3] for the history and the
connection of this problem to integer points on modular curves.
In fact, researchers in the 19th century connected the problem of determining the
positive integers representable by xy + xz + yz + 1 to problems about class numbers of
quadratic imaginary fields, or equivalently class numbers of binary quadratic forms:
[Mord1] mentions that the connection is present in comments by Liouville, in Jour.
de maths., series 2, tome 7, 1862, page 44, on a paper by Hermite. See also [Bel],
Solutions: The Forty-Ninth Competition (1988) 95

[Wh], and [Mord2, p. 291]. The GRH implies the nonexistence of a Siegel zero for the
Dirichlet L-functions associated to these fields, and this is what is used in the proof
of Theorem 1.1 of [BC].

B2. (88, 38, 30, 0, 0, 0, 0, 0, 26, 4, 18, 4)

Prove or disprove: if x and y are real numbers with y ≥ 0 and y(y + 1) ≤
(x + 1)2 , then y(y − 1) ≤ x2 .
Solution 1. If 0 ≤ y ≤ 1, then y(y − 1) ≤ 0 ≤ x2 as desired, so assume y > 1. If
x ≤ y − 1/2 then
y(y − 1) = y(y + 1) − 2y ≤ (x + 1)2 − 2y = x2 + 2x + 1 − 2y ≤ x2 .
If x ≥ y − 1/2 then
x2 ≥ y 2 − y + 1/4 > y(y − 1). 

Solution 2.  As in Solution 1, we may assume y > 1. We are given y(y + 1) ≤ (x +

1)2 , so |x+1| ≥ y(y + 1) and |x| ≥ y(y + 1)−1 ≥ y(y− 1). (The last inequality√
follows from taking a = y − 1 and b = y in the inequality√ (a + 1)(b + 1) ≥ ab + 1
for a, b > 0, which is equivalent (via squaring) to a+b ≥ 2 ab, the AM-GM Inequality
mentioned at the end of 1985A2.) Squaring gives the result. 
Solution 3. As in Solution 1, we may assume y > 1. Let f (y) = y 2 − y and
g(x) = x2 . For y > 1, we are asked to prove that f (y + 1) ≤ g(x + 1) implies
f (y) ≤ g(x), or equivalently that f (y) > g(x) implies f (y + 1) > g(x + 1).
In this paragraph we show that for any x and y with y > 1, the inequality
f (y) ≥ g(x) implies f  (y) > g  (x). If y 2 − y ≥ x2 and y > 1, then (2y − 1)2 >
4y 2 − 4y ≥ 4x2 = (2x)2 and 2y − 1 > 0, so 2y − 1 > |2x| ≥ 2x, i.e., f  (y) > g  (x).
Now fix x and y. Let h(t) = f (y + t) − g(x + t). Given h(0) > 0, and that h(t) > 0
implies h (t) > 0, we must show that h(1) > 0. If h(1) ≤ 0, then by compactness there
exists a smallest u ∈ [0, 1] such that h(u) ≤ 0. For t ∈ (0, u), h(t) > 0, so h (t) > 0.
But h(0) > 0 ≥ h(u), so h cannot be increasing on [0, u]. This contradiction shows
h(1) > 0, i.e., f (y + 1) > g(x + 1). 
Remark. The problem is asking us to decide the truth of the first order sentence
∀x∀y((0 ≤ y) ∧ (y(y + 1) ≤ (x + 1)(x + 1))) =⇒ (y(y − 1) ≤ x · x)
in the language (R, 0, 1, +, −, ·, ≤). Roughly, a first order sentence in this language
is an expression such as the above, involving logical operations ∧ (“and”), ∨ (“or”),
¬ (“not”); binary operations +, −, ·; the relations =, ≤; variables x, y, . . . bound by
quantifiers ∃ (“there exists”) and ∀ (“for all”); and parentheses. For precise definitions,
see Chapter II of [EFT].
Tarski [Tar] proved that the first order theory of the field R is decidable; this means
that there exists a deterministic algorithm (i.e., Turing machine, computer program)
that takes as input any first order sentence and outputs YES or NO according to
whether it is true over the real numbers or not. On the other hand, the first order
theory of Z is undecidable by the work of Gödel [Göd], and J. Robinson [Robi]
combined Gödel’s result and Hasse’s work on quadratic forms to prove that the
96 The William Lowell Putnam Mathematical Competition

first order theory of Q also is undecidable. Moreover, it is known that there is no

algorithm for deciding the truth of first order sentences not involving ∀ or ¬: this was
Matiyasevich’s negative solution of Hilbert’s Tenth Problem [Mat]. The analogous
question for Q is still open. See [PZ] for more.

B3. (20, 16, 17, 2, 0, 0, 0, 0, 52, 33, 27, 41)

√ Z = {1, 2, . . . } of positive integers, let rn be the
+
For every n in the set
minimum value of |c−d 3| for all nonnegative integers c and d with c+d = n.
Find, with proof, the smallest positive real number g with rn ≤ g for all
n ∈ Z+ .
√
Answer. The smallest such g is (1 + 3)/2.
√ √
√ Let g =√(1 + 3)/2. For each fixed n, the sequence n, (n − 1) − 3,
Solution.
(n − 2) − 2 3, . . . , −n 3 is an arithmetic sequence with common difference −2g and
with terms on both sides of 0, so there exists a unique term xn in it with −g ≤ xn < g.
Then rn = |xn | ≤ g. √
For√x ∈ R, let “x mod 1” denote x − x ∈ [0, 1). Since 3 is irrational,
{ (−d 3) mod 1 : d ∈ Z+ } is dense in [0, 1). (See the√remark below.) Hence for any
C > 0, we can√find a positive integer d such that ((−d 3) mod 1) ∈ (g − 1 − C, g − 1).
Then c − d 3 ∈√(g − C, g) for some integer c ≥ 0. Let n = c + d. Then
rn = xn = c − d 3 > g − C by the uniqueness of xn above. Thus g cannot be
lowered. 
Remark. Let α be irrational, and for n ≥ 1, let an = (nα mod 1) ∈ [0, 1). Let
us explain why { an : n = 1, 2, . . . } is dense in [0, 1] when α is irrational. Given a
large integer N > 0, the Pigeonhole Principle [Lar1, Ch. 2.6] produces two integers
p, q ∈ {1, 2, . . . , N +1} such that ap and aq fall into the same subinterval [i/N, (i+1)/N )
for some 0 ≤ i ≤ N − 1. Assume q > p. Then (q − p)α is congruent modulo 1 to a
real number r with |r| < 1/N . Since α is irrational, r = 0. The multiples of (q − p)α,
taken modulo 1, will then pass within 1/N of any number in [0, 1]. This argument
applies for any N , so any nonempty open subset of [0, 1] will contain some an .
In fact, one can prove more, namely that the sequence a1 , a2 , . . . is equidistributed
in [0, 1]: this means that for each subinterval [a, b] ⊆ [0, 1],
#{ n : 1 ≤ n ≤ M and an ∈ [a, b] }
lim = b − a.
M →∞ M
One way to show this is to observe that the range {1, . . . , M } can, up to an error of
o(M ) terms if M is much larger than N (q − p), be partitioned into N -term arithmetic
sequences of the shape c, c + (q − p), . . . , c + (N − 1)(q − p) (notation as in the previous
paragraph), and the an for n in this sequence will be approximately evenly spaced
over [0, 1] when N is large.
Equidistribution can also be deduced from Weyl’s Equidistribution Theorem [Kör,
Theorem 3.1 ], which states that a sequence a1 , a2 , . . . of real numbers is equidistrib-
uted modulo 1 if and only if
n
1
lim χm (ak ) = 0 (1)
n→∞ n
k=1
Solutions: The Forty-Ninth Competition (1988) 97

for all nonzero

√
integers m, where χm (x) = e2πimx . In our application, if we set
2πi 3
ω=e , then the limit in (1) is
n

1 km 1 1 − ω (n+1)m
lim ω = lim = 0,
n→∞ n n→∞ n 1 − ωm
k=1
√
since |1 − ω (n+1)m | ≤ 2, while 1 − ω m = 0, since 3 is irrational. (See Solution 4 to
1990A2 for another application of the density result, and see Solution 2 to 1995B6 for
an application of the equidistribution of (nα mod 1) for irrational α. See 1995B6 also
for a multidimensional generalization of the equidistribution result.)

B4. (17, 1, 0, 0, 0, 0, 0, 0, 3, 2, 73, 112)

∞
Prove that if n=1 an is a convergent series of positive real numbers, then
∞
so is n=1 (an )n/(n+1) .

Solution. If an ≥ 1/2n+1 , then

an
an/(n+1)
n = 1/(n+1)
≤ 2an .
an
n/(n+1)
If an ≤ 1/2n+1 , then an ≤ 1/2n . Hence
1
an/(n+1)
n ≤ 2an + .
2n
∞ n
∞ n/(n+1)
But n=1 (2an + 1/2 ) converges, so n=1 an converges by the Comparison
Test [Spv, Ch. 22, Theorem 1]. 

B5. (9, 0, 10, 1, 0, 0, 0, 0, 1, 2, 45, 140)

For positive integers n, let Mn be the 2n + 1 by 2n + 1 skew-symmetric
matrix for which each entry in the first n subdiagonals below the main
diagonal is 1 and each of the remaining entries below the main diagonal is
−1. Find, with proof, the rank of Mn . (According to one definition, the
rank of a matrix is the largest k such that there is a k × k submatrix with
nonzero determinant.)
One may note that
 
0 −1 −1 1 1
 
0 −1 1 1 0 −1 −1 1 
 
M1 =  1 0 −1 and M2 =   1 1 0 −1 −1 .
−1 1 0  −1 1 1 0 −1
−1 −1 1 1 0

Answer. The rank of Mn equals 2n.

Solution 1. We use induction on n to prove that rank(Mn ) = 2n. We check the

n = 1 case by Gaussian elimination.
98 The William Lowell Putnam Mathematical Competition

Suppose n ≥ 2, and that rank(Mn−1 ) = 2(n − 1) is known. Adding multiples of the

first two rows of Mn to the other rows transforms Mn to a matrix of the form
 
0 −1
∗
 1 0

 
0 −Mn−1

in which 0 and ∗ represent blocks of size (2n − 1) × 2 and 2 × (2n − 1), respectively.
Thus rank(Mn ) = 2 + rank(Mn−1 ) = 2 + 2(n − 1) = 2n. 

Solution 2. Let e1 , . . . , e2n+1 be the standard basis of R2n+1 , and let v1 , . . . ,


v2n+1 be the rows of Mn . Let V = { (a1 , . . . , a2n+1 ) ∈ R2n+1 : ai = 0 }. We will
show that the row space RS(Mn ) equals V .
We first verify the following:
(a) For all m, vm ∈ V .
(b) The set { em − em−1 : 2 ≤ m ≤ 2n + 1 } is a basis of V .
(c) For 2 ≤ m ≤ 2n + 1, the vector em − em−1 is a linear combination of the vi .
n
Proof of (a): vm = i=1 (em−i − em+i ).
(All subscripts are to be considered modulo 2n + 1.)
Proof of (b): The (2n) × (2n + 1) matrix with the em − em−1 as rows is in row
echelon form (with nonzero rows).
Proof of (c): By the formula in the proof of (a),
n n
vm + vm+n = (em−i − em+i ) + (em+n−i − em+n+i )
i=1 i=1
n n
= (em−i + em+n−i ) − (em+i + em+n+i )
i=1 i=1

= (E − em+n ) − (E − em ) (where E = em )
= em − em+n

so

em − em−1 = (em − em+n ) + (em+n − em+2n )

= (vm + vm+n ) + (vm+n + vm+2n ).

Now, (a) implies RS(Mn ) ⊆ V , and (b) and (c) imply V ⊆ RS(Mn ). Thus
RS(Mn ) = V . Hence rank(Mn ) = dim V = 2n. 

Solution 3. The matrix is circulant, i.e., the entry mij depends only on j − i
modulo 2n + 1. Write aj−i = mij , where all subscripts are considered modulo 2n + 1.
(Thus ai equals 0, −1, 1 according as i = 0, 1 ≤ i ≤ n, or n + 1 ≤ i ≤ 2n.) For each
of the 2n + 1 complex numbers ζ satisfying ζ 2n+1 = 1, let vζ = (1, ζ, ζ 2 , . . . , ζ 2n ).
The vζ form a basis for C2n+1 , since they are the columns of a Vandermonde matrix
with nonzero determinant: see 1986A6. Since Mn is circulant, Mn vζ = λζ vζ where
λζ = a0 + a1 ζ + · · · + a2n ζ 2n . Thus { λζ : ζ 2n+1 = 1 } are all the eigenvalues of Mn
Solutions: The Forty-Ninth Competition (1988) 99

with multiplicity. For our Mn , λ1 = 0 and for ζ = 1,

ζ(1 − ζ n )2
λζ = ζ + · · · + ζ n − ζ n+1 − · · · − ζ 2n = = 0,
1−ζ
since gcd(n, 2n + 1) = 1. It follows that the image of Mn (as an endomorphism of
C2n+1 ) is the span of the vζ with ζ = 1, which is 2n-dimensional, so rank(Mn ) = 2n. 
Remark. This method lets one compute the eigenvalues and eigenvectors (and
hence also the determinant) of any circulant matrix. For an application of similar
ideas, see 1999B5. For introductory material on circulant matrices, see [Bar, Sect. 13.2]
and [Da]. Circulant matrices (and more general objects known as group determinants)
played an early role in the development of representation theory for finite groups:
see [Co] for a historical overview.

Solution 4. The sum of the rows of Mn is 0, so Mn is singular. (Alternatively,

this follows since Mn is skew-symmetric of odd dimension.) Hence the rank can be at
most 2n.
To show that the rank is 2n, we will prove that the submatrix A = (aij ) obtained by
deleting row 2n + 1 and column 2n + 1 of Mn has nonzero determinant. By definition,

det A = π∈S2n sgn(π)a1π(1) · · · a(2n)π(2n) , where S2n is the group of permutations of
{1, . . . , 2n}, and sgn(π) = ±1 is the sign of the permutation π. We will prove that
det A is nonzero by proving that it is an odd integer. Since aij is odd unless i = j,
the term in the sum corresponding to π is 0 if π(i) = i for some i, and odd otherwise.
Thus det A ≡ f (2n) (mod 2), where for any integer m ≥ 1, f (m) denotes the number
of permutations of {1, . . . , m} having no fixed points. We can compute f (m) using
the Inclusion-Exclusion Principle [Ros2, §5.5]: of the m! permutations, (m − 1)! fix
1, (m − 1)! fix 2, and so on, but if we subtract all these, then we must add back the
(m − 2)! permutations fixing 1 and 2 (since these have been subtracted twice), and so
on for all other pairs, and then subtract (m − 3)! for each triple, and so on; this finally
yields


m m
f (m) = m! − (m − 1)! + (m − 2)! − · · ·
1 2



m−1 m m m
+ (−1) 1! + (−1) 0!
m−1 m
≡ (−1)m−1 m + (−1)m (mod 2)

so f (2n) is odd, as desired. 

Remark. Permutations without fixed points are called derangements. The formula
for f (m) can also be written as


1 1 1 (−1)m
f (m) = m! 1 − + − + · · · + ,
1! 2! 3! m!
which is the integer nearest to m!/e.

Solution 5. As in Solution 4, Mn is singular, and hence rank(Mn ) ≤ 2n. Let A

be as in Solution 4. Then A2 is equivalent modulo 2 to the 2n × 2n identity matrix,
so det A = 0. Thus rank(Mn ) = 2n. 
100 The William Lowell Putnam Mathematical Competition

B6. (38, 9, 4, 4, 24, 8, 0, 5, 5, 1, 17, 93)

Prove that there exist an infinite number of ordered pairs (a, b) of integers
such that for every positive integer t the number at+b is a triangular number
if and only if t is a triangular number. (The triangular numbers are the
tn = n(n + 1)/2 with n in {0, 1, 2, . . . }.)
Solution 1. It is easy to check that t3n+1 = 9tn + 1, while t3n ≡ t3n+2 ≡ 0
(mod 3) for any integer n. Hence for every positive integer t, t is a triangular number
if and only if 9t + 1 is triangular. If the nth iterate of the linear map t !→ 9t + 1 is
t !→ at + b, then a chain of equivalences will show that t is a triangular number if and
only if at + b is triangular. We obtain infinitely many pairs of integers (a, b) in this
way. 
Solution 2. If t = n(n + 1)/2, then 8t + 1 = (2n + 1)2 . Conversely if t is an integer
such that 8t + 1 is a square, then 8t + 1 is the square of some odd integer 2n + 1, and
hence t = n(n + 1)/2. Thus t is a triangular number if and only if 8t + 1 is a square.
If k is an odd integer, then k 2 ≡ 1 (mod 8), and
t is a triangular number ⇐⇒ 8t + 1 is a square
⇐⇒ k 2 (8t + 1) is a square


k2 − 1
⇐⇒ 8 k t +
2
+ 1 is a square
8


k2 − 1
⇐⇒ k 2 t + is a triangular number.
8
Hence we may take (a, b) = (k 2 , (k 2 − 1)/8) for any odd integer k. 
Remark. Let us call (a, b) a triangular pair if a and b are integers with the property
that for positive integers t, t is a triangular number if and only if at + b is a triangular
number. Solution 2 showed that if k is an odd integer, then (k 2 , (k 2 − 1)/8) is a
triangular pair. In other words, the triangular pairs are of the form ((2m + 1)2 , tm ),
where m is any integer. We now show, conversely, that every triangular pair has this
form.
Suppose that (a, b) is a triangular pair. For any integer n ≥ 0, atn + b is triangular,
so 8(atn + b) + 1 = 4an2 + 4an + (8b + 1) is a square. A polynomial f (x) ∈ Z[x] taking
square integer values at all nonnegative integers must be the square of a polynomial
in Z[x]: see 1998B6 for a proof. Hence
4an2 + 4an + (8b + 1) = @(n)2 (1)
√ √
for some linear polynomial @(x). Completing the square shows that @(x) = 2 ax+ a.
Since @(n)2 is the square of an integer for any integer n, a = @(0)2 = k 2 for some integer
k, and equating constant coefficients in (1) shows that a = 8b + 1, so b = (k 2 − 1)/8.
Finally, k must be odd, in order for b to be an integer.
Solutions: The Fiftieth Competition (1989) 101

The Fiftieth William Lowell Putnam Mathematical Competition

December 2, 1989

A1. (92, 2, 6, 7, 0, 0, 0, 9, 2, 8, 41, 32)

How many primes among the positive integers, written as usual in base
10, are such that their digits are alternating 1’s and 0’s, beginning and
ending with 1?
Answer. There is only one such prime: 101.
Solution. Suppose that N = 101 · · · 0101 with k ones, for some k ≥ 2. Then
99N = 9999 · · · 9999 = 102k − 1 = (10k + 1)(10k − 1).
If moreover N is prime, then N divides either 10k + 1 or 10k − 1, and hence one of
k
10k −1
99
10k −1
= 10N+1 and 1099
k +1 = N is an integer. For k > 2, 10k − 1 and 10k + 1 are
both greater than 99, so we get a contradiction. Therefore k = 2 and N = 101 (which
is prime). 
Remark. Essentially the same problem appeared on the 1979 British Mathematical
Olympiad, reprinted in [Lar1, p. 123] as Problem 4.1.4:
Prove that there are no prime numbers in the infinite sequence of integers
10001, 100010001, 1000100010001, . . . .

A2. (141, 6, 29, 0, 0, 0, 0, 0, 5, 7, 4, 7)

a b
2
x2 ,a2 y 2 }
Evaluate emax{b dy dx, where a and b are positive.
0 0
2 2
Answer. The value of the integral is (ea b
− 1)/(ab).
Solution. Divide the rectangle into two parts by the diagonal line ay = bx to
obtain
 a b  a  bx/a  b  ay/b
max{b2 x2 ,a2 y 2 } b2 x2 2 2
e dy dx = e dy dx + ea y dx dy
0 0 0 0 0 0
 a  b
bx b2 x2 ay a2 y2
= e dx + e dy
0 a 0 b
 a2 b2  a2 b2
1 u 1 v
= e du + e dv
0 2ab 0 2ab
2 2
ea b
−1
= . 
ab
A3. (13, 4, 4, 2, 0, 0, 0, 0, 6, 8, 56, 106)
Prove that if
11z 10 + 10iz 9 + 10iz − 11 = 0,
then |z| = 1. (Here z is a complex number and i2 = −1.)
Solution 1. Let g(z) = 11z 10 + 10iz 9 + 10iz − 11. Let p(w) = −w−5 g(iw) =
11w5 + 10w4 + 10w−4 + 11w−5 . As θ increases from 0 to 2π, the real-valued function
p(eiθ ) = 22 cos(5θ) + 20 cos(4θ) changes sign at least 10 times, since at θ = 2πk/10
102 The William Lowell Putnam Mathematical Competition

for k = 0, 1, . . . , 9 its value is 22(−1)k + 20 cos(4θ), which has the sign of (−1)k . By
the Intermediate Value Theorem [Spv, Ch. 7, Theorem 5], p(eiθ ) has at least one zero
between θ = 2πk/10 and θ = 2πi(k + 1)/10 for k = 0, . . . , 9; this makes at least 10
zeros. Thus g(z) also has at least 10 zeros on the circle |z| = 1, and these are all the
zeros since g(z) is of degree 10. 

Solution 2. The equation can be rewritten as z 9 = 11−10iz

11z+10i . If z = a + bi, then

11 − 10iz 112 + 220b + 102 (a2 + b2 )
|z|9 = = .
11z + 10i 112 (a2 + b2 ) + 220b + 102
Let f (a, b) and g(a, b) denote the numerator and denominator of the right-hand side.
If |z| > 1, then a2 + b2 > 1, so g(a, b) > f (a, b), making |z 9 | < 1, a contradiction. If
|z| < 1, then a2 + b2 < 1, so g(a, b) < f (a, b), making |z 9 | > 1, again a contradiction.
Hence |z| = 1. 

Solution 3. Rouché’s Theorem [Ah, p. 153] states that if f and g are analytic
functions on an open set containing a closed disc, and if |g(z) − f (z)| < |f (z)|
everywhere on the boundary of the disc, then f and g have the same number of
zeros inside the disc. Let f (z) = 10iz − 11 and g(z) = 11z 10 + 10iz 9 + 10iz − 11, and
consider the discs |z| ≤ α with α ∈ (0, 1). Then

|g(z) − f (z)| = |11z 10 + 10iz 9 | = |z|9 |11z + 10i| < |10iz − 11| = |f (z)|

if |z| < 1, by the same calculation as in Solution 2. But f has its only zero at 11/(10i),
outside |z| = 1, so g has no zeros with |z| < α for any α ∈ (0, 1), and hence no zeros
with |z| < 1. Finally, g(−1/z) = −z −10 g(z), so the nonvanishing of g for |z| < 1
implies the nonvanishing of g for |z| > 1. Therefore, if g(z) = 0, then |z| = 1. 

A4. (54, 22, 16, 4, 0, 0, 0, 0, 4, 2, 53, 44)

If α is an irrational number, 0 < α < 1, is there a finite game with an
honest coin such that the probability of one player winning the game is
α? (An honest coin is one for which the probability of heads and the
probability of tails are both 1/2. A game is finite if with probability 1 it
must end in a finite number of moves.)
Answer. Yes, such a game exists.

Solution. Let dn be 0 or 1, depending on whether the nth toss yields heads or

∞
tails. Let X = n=1 dn /2n . Then the distribution of X is uniform on [0, 1], since
for any rational number c/2m (i.e., any dyadic rational) in [0, 1], the probability that
X ∈ [0, c/2m ] is exactly c/2m .
Say that player 1 wins the game after N tosses, if it is guaranteed at that time that
the eventual value of X will be less than α; this means that
N ∞
dn 1
+ < α.
n=1
2n 2n
n=N +1

Similarly, say that player 2 wins after N tosses, if it is guaranteed then that X will
be greater than α.
Solutions: The Fiftieth Competition (1989) 103

The game will terminate if X = α, which happens with probability 1; in fact it will
terminate at the N th toss or earlier if |X − α| > 1/2N . The probability that player 1
wins is the probability that X ∈ [0, α), which is α. 
Remark. The solution shows that the answer is yes for all real α ∈ [0, 1]: there is
no need to assume that α is irrational.
Remark. Essentially the same idea appears in [New, Problem 8]:
Devise an experiment which uses only tosses of a fair coin, but which has
success probability 1/3. Do the same for any success probability p, 0 ≤ p ≤ 1.
Related question. Show that if α ∈ [0, 1] is not a dyadic rational (i.e., not a rational
number with denominator equal to a power of 2), the expected number of tosses in
the game in the solution equals 2.
Related question. For α ∈ [0, 1], let f (α) be the minimum over all games satisfying
the conditions of the problem (such that player 1 wins with probability α) of the
expected number of tosses in the game. (For some games, the expected number
may be infinite; ignore those.) Prove that if α ∈ [0, 1] is a rational number whose
denominator in lowest terms is 2m for some m ≥ 0, then f (α) = 2 − 1/2m−1 . Prove
that if α is any other real number in [0, 1], then f (α) = 2. (This is essentially [New,
Problem 103].)

A5. (6, 1, 3, 0, 0, 0, 0, 1, 0, 2, 49, 137)

Let m be a positive integer and let G be a regular (2m + 1)-gon inscribed
in the unit circle. Show that there is a positive constant A, independent
of m, with the following property. For any point p inside G there are two
distinct vertices v1 and v2 of G such that
1 A
|p − v1 | − |p − v2 | < − .
m m3
Here |s − t| denotes the distance between the points s and t.
" #
Solution 1. The greatest distance between two vertices of G is w = 2 cos 4m+2 π
,
since these vertices with the center form an isosceles triangle with equal sides of length
1, with vertex angle 2πm/(2m + 1) and base angles π/(4m + 2). (See Figure 10.)
Hence for any vertices v1 and v2 of G, the triangle inequality gives ||p − v1 | − |p −
v2 || ≤ |v1 − v2 | ≤ w. Thus the 2m + 1 distances from p to the vertices lie in an
interval of length at most w. Let the distances be d1 ≤ d2 ≤ · · · ≤ d2m+1 . Then
2m
i=1 (di+1 − di ) = d2m+1 − d1 ≤ w, so di+1 − di ≤ w/(2m) for some i. It remains to
show that there exists A > 0 independent of m such that w/(2m) < 1/m − A/m3 . In
fact, the Taylor expansion of cos x gives


w 1 π2 −2 1 π2
= 1− + o(m ) = − + o(m−3 )
2m m 2(4m + 2)2 m 32m3
as m → ∞, so any positive A < π 2 /32 will work for all but finitely many m. We
can shrink A to make w/(2m) < 1/m − A/m3 for those finitely many m too, since
w/(2m) < 1/m for all m. 
Solution 2. We will prove an asymptotically stronger result, namely that for
a regular n-gon G inscribed in a unit circle and for any p in the closed unit disc,
104 The William Lowell Putnam Mathematical Competition

2 cos (4mp+ 2 )
1
1

FIGURE 10.

there exist vertices v1 , v2 of G such that ||p − v1 | − |p − v2 || < π 2 /n2 . Center the
polygon at (0, 0) and rotate to assume that p = (−r, 0) with 0 ≤ r ≤ 1. Let the two
vertices of G closest to (1, 0) be vi = (cos θi , sin θi ) for i = 1, 2 where θ1 ≤ 0 ≤ θ2 and
θ2 = θ1 + 2π/n. Then

||p − v1 | − |p − v2 || = |fr (θ1 ) − fr (θ2 )|,

where

fr (θ) = |(−r, 0) − (cos θ, sin θ)| = r2 + 2r cos θ + 1.

Reflecting if necessary, we may assume fr (θ1 ) ≥ fr (θ2 ). A short calculation (for

example using differentiation) shows that fr (θ) is decreasing on [0, π] and increasing
on [−π, 0]. Thus for fixed r, fr (θ1 ) − fr (θ2 ) is maximized when θ1 = 0 and θ2 = 2π/n.
Next we claim that fr (0) − fr (2π/n) is increasing with r, hence maximized at r = 1:
this is because if v2 is the point on line segment pv1 with |p − v2 | = |p − v2 |, then
as r increases, angle v2 pv2 of the isosceles triangle shrinks, making angle v2 v2 p grow,
putting v2 farther from v1 , and fr (0) − fr (2π/n) = |v2 − v1 |. See Figure 11. Hence

 " π # π2
2π
||p − v1 | − |p − v2 || ≤ f1 (0) − f1 = 2 − 2 cos < 2,
n n n

since f1 (θ) = 2 cos(θ/2) for −π ≤ θ ≤ π, and since the inequality cos x > 1 − x2 /2 for
x ∈ (0, π/3] follows from the Taylor series of cos x.
In order to solve the problem posed, we must deal with the case n = 3, i.e., m = 1,
since the bound
"π #
||p − v1 | − |p − v2 || ≤ 2 − 2 cos =1
3
is not quite good enough in that case. The proof shows, however, that equality holds
for the chosen v1 and v2 only when p is on the circle and diametrically opposite v1 ,
and in this case p is equidistant from the other two vertices. Hence the minimum of
||p−v1 |−|p−v2 || over all choices of v1 and v2 is always less than 1, and by compactness
there exists A > 0 such that it is less than 1 − A for all p in the disc, as desired. 
Solutions: The Fiftieth Competition (1989) 105

v2

fr (2π/n)

p v2 v1

FIGURE 11.
Geometric interpretation of fr (0) − fr (2π/n).

Remark. The π 2 /n2 improvement was first discovered by L. Crone and

R. Holzsager, [Mon4, Solution to 10269]. (We guess that their solution was similar
to ours.)

Stronger result. By considering the three vertices farthest from p instead of just
v1 and v2 , one can improve this to (2/3)π 2 /n2 . Moreover, (2/3)π 2 cannot be replaced
by any smaller constant, even if one insists that n be odd and that p be in G.
First let us prove the improvement. As in Solution 2, assume that p = (−r, 0). The
vertex of G closest to (1, 0) is qθ = (cos θ, sin θ) for some θ ∈ [−π/n, π/n]. Reflecting
if necessary, we may assume θ ∈ [0, π/n]. Then fr (θ − 2π/n), fr (θ), and fr (θ + 2π/n)
are among the distances from p to the vertices of G. We use a lemma that states that
for fixed θ , θ ∈ [−π, π], the function |fr (θ ) − fr (θ )| of r ∈ [0, 1] is increasing (or
zero if |θ | = |θ |): to prove this, we may assume that 0 ≤ θ < θ ≤ π and observe
that for fixed r ∈ (0, 1), the derivative

dfr (θ ) r + cos θ

=$ ,
dr
(r + cos θ )2 + sin2 θ

equals the cosine of the angle q0 pqθ , whose measure increases with θ , so

dfr (θ ) dfr (θ )

− > 0.
dr dr
106 The William Lowell Putnam Mathematical Competition

If 0 ≤ θ ≤ π/(3n), then





2π 2π 2π 2π
fr θ − − fr θ + ≤ f1 θ − − f1 θ + (by the lemma)
n n n n



θ π θ π
= 2 cos − − 2 cos +
2 n 2 n

 "π #
θ
= 4 sin sin
2 n
2
2π
< 2,
3n
since 0 < sin x < x for 0 < x < π/2. If instead π/(3n) ≤ θ ≤ π/n, then



2π 2π
fr (θ) − fr θ − ≤ f1 (θ) − f1 θ − (by the lemma)
n n

 "π #
π θ
= 4 sin − sin
2n 2 2n
2
2π
< 2.
3n
Thus there is always a pair of vertices of G whose distances to p differ by less than
(2/3)π 2 /n2 .
Now let us show that the constant (2/3)π 2 cannot be replaced by anything smaller,
even if p is required to be inside G. Without loss of generality, we may assume that the
regular n-gon G is inscribed in the circle x2 + y 2 = 1 and has one vertex at (cos θ, sin θ)
where θ = π/(3n). Take p = (−r, 0) where r = cos(π/n). Then r is the radius of the
circle inscribed in G, but since θ and −π do not differ by an odd multiple of π/n, the
point p lies strictly inside G.
Finally we show that for this choice of p, if v1 and v2 are distinct vertices of G, then


2π 2 1
||p − v1 | − |p − v2 || ≥ 2 − O .
3n n4

Since fr (θ ) is a decreasing function of |θ |, the distances from p to the vertices, in
decreasing order, are

fr (θ), fr (θ − 2π/n), fr (θ + 2π/n), fr (θ − 4π/n), fr (θ + 4π/n), . . . .

Since p is within 1 − cos(π/n) ≤ π 2 /(2n2 ) of (−1, 0), this sequence is approximated

by
f1 (θ), f1 (θ − 2π/n), f1 (θ + 2π/n), f1 (θ − 4π/n), f1 (θ + 4π/n), . . . ,

with an error of at most π 2 /(2n2 ) for each term. The latter sequence is
"π #




5π 7π 11π 13π
2 cos , 2 cos , 2 cos , 2 cos , 2 cos , ....
6n 6n 6n 6n 6n

Hence the consecutive differences in the original sequence are within π 2 /n2 of the
Solutions: The Fiftieth Competition (1989) 107

differences for the approximating sequence, which are of the form





(6k − 5)π (6k − 1)π (6k − 3)π 2π
2 cos − 2 cos = 4 sin sin
6n 6n 6n 6n




(6k − 3)π 2π 1
= 4 sin −O
6n 6n n3

or of the form



 "π #
(6k − 1)π (6k + 1)π 6kπ
2 cos − 2 cos = 4 sin sin
6n 6n 6n 6n

" #

kπ π 1
= 4 sin −O ,
n 6n n3

where 1 ≤ k ≤ n/2+O(1). If k ≥ 3, then bounding the sines from below by their values
at k = 3 and then applying sin x = x − O(x3 ) as x → 0 shows that these differences in
2π 2 π2
the approximating sequence exceed 3n 2 + n2 so the corresponding differences in the
2
2π
original sequence exceed 3n 2 . For k < 3 (the first four differences), we forgo use of the

approximating sequence and instead observe that for r = cos(π/n) and fixed integer
j, Taylor series calculations give



jπ (18 + j 2 )π 2 1
fr =2− + O ,
3n 36n2 n4

which implies that the first four differences in the original sequence (between the terms
corresponding to j = 1, 5, 7, 11, 13) are at least


2π 2 1
−O .
3n2 n4

Literature note. For an introduction to Taylor series, see [Spv, Ch. 23].

A6. (3, 1, 0, 0, 0, 0, 0, 1, 0, 4, 49, 141)

Let α = 1 + a1 x + a2 x2 + · · · be a formal power series with coefficients in
the field of two elements. Let


 1 if every block of zeros in the binary expansion of n

has an even number of zeros in the block,
an =



0 otherwise.

(For example, a36 = 1 because 36 = 1001002 , and a20 = 0 because 20 = 101002 .)

Prove that α3 + xα + 1 = 0.

Solution. It suffices to prove that α4 + xα2 + α = 0, since the ring of formal

power series over any field is an integral domain (a commutative ring with no nonzero
zerodivisors), and α = 0. Since a2i = ai for all i, and since cross terms drop out when
108 The William Lowell Putnam Mathematical Competition

we square in characteristic 2, we have

∞
α2 = an x2n
n=0
∞
α4 = an x4n , and
n=0
∞
xα2 = an x2n+1 .
n=0

Let bn be the coefficient of xn in α4 + xα2 + α. If n is odd, then the binary expansion

of n is obtained from that of (n − 1)/2 by appending a 1, an = a(n−1)/2 , and
bn = a(n−1)/2 + an = 0. If n is divisible by 2 but not 4, then bn = an = 0, since
n ends with a block of one zero. If n is divisible by 4, then the binary expansion of
n is obtained from that of n/4 by appending two zeros, which does not change the
evenness of the block lengths, so an/4 = an , and bn = an/4 + an = 0. Thus bn = 0 for
all n ≥ 0, as desired. 
Remark. The fact that the coefficients of the algebraic power series α have a
simple description is a special case of a much more general result of Christol. Let
Fq [[x]] denote the ring of formal power series over the finite field Fq with q elements.
Christol gives an automata-theoretic condition on a series β ∈ Fq [[x]] that holds if and
only if β satisfies an algebraic equation with coefficients in Fq [x]: see [C] and [CKMR].
For another application of Christol’s result, and a problem similar to this one, see the
remark following 1992A5. See Problem 1990A5 for a problem related to automata.

B1. (144, 10, 0, 0, 0, 0, 0, 0, 0, 0, 42, 3)

A dart, thrown at random, hits a square target. Assuming that any
two parts of the target of equal area are equally likely to be hit, find the
probability that the point hit is nearer
√ to the center than to any edge.
Express your answer in the form (a b + c)/d, where a, b, c, d are positive
integers.
√
Answer. The probability is (4 2 − 5)/3

FIGURE 12.
Solutions: The Fiftieth Competition (1989) 109

Solution. We may assume that the dartboard has corners at (±1, ±1). A
point
 (x, y) in the square is closer to the center than to the top edge if and only if
x + y ≤ 1−y, which is equivalent to x2 +y 2 ≤ (1−y)2 , and to y ≤ (1−x2 )/2. This
2 2

describes a region below a parabola. The region consisting of points in the board closer
to the center than to any edge is the intersection of the four symmetrical parabolic
regions inside the board: it is union of eight symmetric copies of the region A bounded
by x ≥ 0, y ≥ x, y ≤ (1 − x2 )/2. (See Figure 12.) A short calculation
√ shows
√ that the
bounding curves y = x and y = (1 − x2 )/2 intersect at (x, y) = ( 2 − 1, 2 − 1). Thus
the desired probability is
 √2−1
 √
8 Area(A) 1 − x2 4 2−5
= 2 Area(A) = 2 − x dx = . 
Area(board) 0 2 3

Related question. If a billiard table had the same shape as the region of points of
the square closer to the center than to any edge, and a ball at the center were pushed
in some direction not towards the corners, what would its path be?

B2. (49, 44, 17, 10, 0, 0, 0, 0, 3, 4, 41, 31)

Let S be a nonempty set with an associative operation that is left and
right cancellative (xy = xz implies y = z, and yx = zx implies y = z).
Assume that for every a in S the set { an : n = 1, 2, 3, . . . } is finite. Must S
be a group?
Answer. Yes, S must be a group.

Solution. Choose a ∈ S. The finiteness hypothesis implies that some term in

the sequence a, a2 , a3 , . . . is repeated infinitely often, so we have am = an for some
integers m, n ≥ 1 with m − n ≥ 2. Let e = am−n . For any x ∈ S, an ex = am x, and
cancelling an = am shows that ex = x. Similarly xe = x, so e is an identity. Now
aam−n−1 = am−n = e and am−n−1 a = am−n = e, so am−n−1 is an inverse of a. Since
S is associative and has an identity, and since any a ∈ S has an inverse, S is a group. 

B3. (36, 4, 18, 48, 4, 10, 3, 1, 4, 1, 9, 61)

Let f be a function on [0, ∞), differentiable and satisfying

f  (x) = −3f (x) + 6f (2x)

√
for x > 0. Assume that |f (x)| ≤ e− x for x ≥ 0 (so that f (x) tends rapidly to
0 as x increases). For n a nonnegative integer, define
 ∞
µn = xn f (x) dx
0

(sometimes called the nth moment of f ).

a. Express µn in terms of µ0 .
b. Prove that the sequence {µn 3n /n!} always converges, and that the limit
is 0 only if µ0 = 0.
3n −1
Answer to part a. For each n ≥ 0, µn = 3n!n ( m=1 (1 − 2−m )) µ0 .
110 The William Lowell Putnam Mathematical Competition

Solution. a. As x → ∞, f (x) tends to 0 faster than any negative power of x, so

the integral defining µn converges. Multiply the functional equation by xn for some
n ≥ 1 and integrate from 0 to some finite B > 0:
 B  B  B
xn f  (x) dx = −3 xn f (x) dx + 6 xn f (2x) dx.
0 0 0

Using integration by parts [Spv, Ch. 18, Theorem 1] on the left, and the substitution
u = 2x on the last term converts this into
B
 B  B  B/2
6
x f (x) − n
n
f (x)x n−1
dx = −3 n
x f (x) dx + n+1 un f (u) du.
0 0 0 2 0

Taking limits as B → ∞ yields

6
−nµn−1 = −3µn + µn ,
2n+1
so
n
µn = (1 − 2−n )−1 µn−1
3
for all n ≥ 1. By induction, we obtain
4 n 5−1
n! +
µn = n (1 − 2−m ) µ0 .
3 m=1
∞ 3∞
b. Since m=1 2−m converges, m=1 (1 − 2−m ) converges to some nonzero limit L,
and 4 n 5−1
3n +
−m
µn = (1 − 2 ) µ0 → L−1 µ0
n! m=1

as n → ∞. This limit L−1 µ0 is finite, and equals zero if and only if µ0 = 0. 

Remark. We show that nonzero functions f (x) satisfying the conditions of the
problem do exist. Given that f (x) tends to zero rapidly as x → ∞, one expects f (2x)
to be negligible compared to f (x) for large x, and hence one can guess that f (x) can
be approximated by a solution to y  = −3y, for example e−3x . But then the error in
the differential equation f  (x) = −3f (x) + 6f (2x) is of order e−6x as x → ∞, leading
one to guess that e−3x + a1 e−6x will be a better approximation, and so on, finally
leading one to guess

f (x) = e−3x + a1 e−6x + a2 e−12x + a3 e−24x + · · · , (1)

where the coefficients ai are to be solved for. Let a0 = 1. If we differentiate (1) term
by term, temporarily disregarding convergence issues, then for k ≥ 1, the coefficient
k
of e−3·2 x in the expression f  (x) + 3f (x) − 6f (2x) is −3 · 2k ak + 3ak − 6ak−1 . If this
is to be zero, then ak = 2k−2 a
−1 k−1
.
All this so far has been motivation. Now we define the sequence a0 , a1 , . . . by
a0 = 1 and ak = 2k−2 a
−1 k−1
for k ≥ 1, so

(−2)n
ak = ,
(2 − 1)(22 − 1) · · · (2k − 1)
Solutions: The Fiftieth Competition (1989) 111
∞
and define f (x) by (1). The series k=0 |ak | converges by the Ratio Test [Spv,
Ch. 22, Theorem 3], so (1) converges to a continuous function on [0, ∞). Moreover,
k
|ak e−3·2 x | ≤ |ak | for all complex x with Re(x) > 0, so by the Weierstrass M -test and
Weierstrass’s Theorem [Ah, pp. 37, 176], (1) converges to a holomorphic function in
this region. In particular, f (x) is differentiable on (0, ∞), and can be differentiated
term by term, so f  (x) = −3f (x) + 6f (2x) holds.
Finally, we will show that if C > 0 is sufficiently small, then
√
Cf (x), which still satisfies
the differential equation, now also satisfies |Cf (x)| ≤ e− x . Since
√
f (x) = O(e−3x + e−6x + e−12x + · · · ) = O(e−3x ) = o(e− x
)
√
as x → ∞, we have |f (x)| ≤ e− x for all x greater than some x0 . √Let M =
supx∈[0,x0 ] |f √ . If C ∈ (0, 1) is chosen so that CM ≤ 1, then |Cf (x)| ≤ e− x both for
(x)|
e− x
x ∈ [0, x0 ] and for x ∈ (x0 , ∞), as desired.

B4. (52, 7, 0, 0, 0, 0, 0, 0, 0, 1, 62, 77)

Can a countably infinite set have an uncountable collection of nonempty
subsets such that the intersection of any two of them is finite?
Answer. Yes.
Solution 1. The set Q of rational numbers is countably infinite. For each real
number α, choose a sequence of distinct rational numbers tending to α, and let Sα
be the set of terms. If α, β are distinct real numbers, then Sα ∩ Sβ is finite, since
otherwise a sequence obtained by listing its elements would converge to both α and β.
In particular, Sα = Sβ . Thus { Sα : α ∈ R } is an uncountable collection of nonempty
subsets of Q with the desired property. 
Remark. A minor variant on this solution would be to take the (countable) set of
real numbers with terminating decimal expansions, and for each of the (uncountably
many) irrational numbers a, let Sa be the set of decimal approximations to a obtained
by truncating the decimal expansion at some point.
Solution 2. Let A denote the countably infinite set of finite strings of 0’s and 1’s.
For each infinite string a = a1 a2 a3 . . . of 0’s and 1’s, let Sa = { a1 a2 . . . an : n ≥ 1 }
denote the set of finite initial substrings. There are uncountably many a, and if
a = a1 a2 . . . and b = b1 b2 . . . are distinct infinite strings, say with am = bm , then all
strings in Sa ∩ Sb have length less than m, so Sa ∩ Sb is finite. 
Solution 3. The set Z2 of lattice points in the plane is countably infinite. For
each real α, let Sα denote the set of points in Z2 whose distance to the line y = αx is
at most 1. If α, β are distinct real numbers, then Sα ∩ Sβ is a set of lattice points in
a bounded region, so it is finite. 
Solution 4. Let A be a disjoint union of a countably infinite number of countably
infinite sets, so A is countably infinite. Call a collection of subsets C of A good if it
consists of an infinite number of countably infinite subsets of A, and S ∩ T is finite for
any distinct S, T ∈ C. By construction of A, there exists a good collection (of disjoint
subsets). Order the good collections by inclusion. For any chain of good collections,
the union is also a good collection. Hence by Zorn’s Lemma, there exists a maximal
good collection Cmax .
112 The William Lowell Putnam Mathematical Competition

Suppose Cmax were countable, say Cmax = {S1 , S2 , . . . }. Because Sn is infinite while
6n−1
Si ∩ Sn is finite for i = n, there exists bn ∈ Sn − i=1 Si for each n ≥ 1. For
m < n, bm ∈ Sm but bn ∈ Sm , so bm = bn , and the set B = {b1 , b2 , . . . } is countably
infinite. Moreover, bN ∈ Sn for N > n, so B ∩ Sn is finite. Hence {B, S1 , S2 , . . . } is a
good collection, contradicting the maximality of Cmax . Thus Cmax is uncountable, as
desired. 
Remark. This question appears as [New, Problem 49], where the countably infinite
set is taken to be Z.
Remark (Zorn’s Lemma). A chain in a partially ordered set (S, ≤) is a subset
in which every two elements are comparable. Zorn’s Lemma states that if S is a
nonempty partially ordered set such that every chain in S has an upper bound in
S, then S contains a maximal element, i.e., an element m such that the only element
s ∈ S satisfying s ≥ m is m itself. Zorn’s Lemma is equivalent to the Axiom of Choice,
which states that the product of a family of nonempty sets indexed by a nonempty
set is nonempty. It is also equivalent to the Well Ordering Principle, which states
that every set admits a well ordering. (A well ordering of a set S is a total ordering
such that every nonempty subset A ⊆ S has a least element.) See pages 151 and 196
of [En].
Related question. Show that the following similar question, a restatement of [Hal,
Problem 11C], has a negative answer:
Can a countably infinite set have an uncountable collection of nonempty
subsets such that the intersection of any two of them has at most 1989
elements?

B5. (17, 17, 1, 1, 0, 0, 0, 1, 17, 29, 37, 79)

Label the vertices of a trapezoid T (quadrilateral with two parallel sides)
inscribed in the unit circle as A, B, C, D so that AB is parallel to CD
and A, B, C, D are in counterclockwise order. Let s1 , s2 , and d denote
the lengths of the line segments AB, CD, and OE, where E is the point
of intersection of the diagonals of T , and O is the center of the circle.
Determine the least upper bound of (s1 − s2 )/d over all such T for which
d = 0, and describe all cases, if any, in which it is attained.
Answer. The least upper bound of (s1 − s2 )/d equals 2. We have (s1 − s2 )/d = 2
if and only if the diagonals BD and AC are perpendicular and s1 > s2 .
Solution 1. (See Figure 13.) We may assume that AB and CD are horizontal,
with AB below CD. Then by symmetry, E = (0, e) for some e, and d = |e|. The
diagonal AC has the equation y = mx + e for some slope m > 0. Substituting
y = mx + e into x2 + y 2 = 1 results in the quadratic polynomial
q(x) = x2 + (mx + e)2 − 1 = (m2 + 1)x2 + (2me)x + (e2 − 1)
whose zeros are the x-coordinates of A and C, which also equal −s1 /2 and s2 /2,
respectively. Hence s1 − s2 is −2 times the sum of the zeros of q(x), so


−2me
s1 − s2 = −2 . (1)
m2 + 1
Solutions: The Fiftieth Competition (1989) 113

If d = 0, then


s1 − s2 2m
=2 sgn(e).
d m2 + 1

Now (m − 1)2 ≥ 0 with equality if and only if m = 1, so m2 + 1 ≥ 2m > 0. Thus

s1 − s2
≤ 2,
d
with equality if and only if m = 1 and e > 0, i.e., if the diagonals BD and AC are
perpendicular and s1 > s2 . (The latter is equivalent to e > 0 by (1).) 

Solution 2. (See Figure 13.) Again assume that AB and CD are horizontal. The
x-coordinates of B and D are s1 /2 and −s2 /2, so their midpoint M has x-coordinate
(s1 − s2 )/4. Since line OM is the perpendicular bisector of BD, ∠OM E is a right
angle, and M lies on the circle with diameter OE. For fixed d = OE, the x-coordinate
(s1 − s2 )/4 is maximized when M is at the rightmost point of this circle: then
(s1 − s2 )/4 = d/2 and (s1 − s2 )/d = 2. This happens if and only if BD has slope −1
and s1 > s2 , or equivalently if and only if the diagonals BD and AC are perpendicular
and s1 > s2 . 

s2
D C
E
M
d
A s1 B

FIGURE 13.

Literature note. For further discussion of this problem, including more solutions,
see [Lar2, pp. 33–38].

B6. (0, 1, 1, 1, 0, 0, 0, 2, 0, 11, 35, 148)

Let (x1 , x2 , . . . , xn ) be a point chosen at random from the n-dimensional
region defined by 0 < x1 < x2 < · · · < xn < 1. Let f be a continuous function
on [0, 1] with f (1) = 0. Set x0 = 0 and xn+1 = 1. Show that the expected
114 The William Lowell Putnam Mathematical Competition

value of the Riemann sum

n
(xi+1 − xi )f (xi+1 )
i=0
1
is 0 f (t)P (t) dt, where P is a polynomial of degree n, independent of f , with
0 ≤ P (t) ≤ 1 for 0 ≤ t ≤ 1.
Answer. We will show that the result holds with P (t) = 1 − (1 − t)n .

Solution 1. We may drop the i = n term in the Riemann sum, since f (xn+1 ) =
f (1) = 0. The volume of the region 0 < x1 < · · · < xn < 1 in Rn is
 1  xn  x2
··· dx1 dx2 · · · dxn = 1/n!.
0 0 0
" #
n−1
Therefore the expected value of the Riemann sum is E = i=0 Mi /(1/n!), where
 1  xn  x2
Mi = ··· (xi+1 − xi )f (xi+1 ) dx1 · · · dxn
0 0 0
 1 xn  xi+1
xi−1
= ···
(xi+1 − xi ) i f (xi+1 ) dxi dxi+1 · · · dxn
(i − 1)!
0 0
 1  xn
0
 xi+2 4 5
xii+1 xi+1
i+1
= ··· xi+1 − f (xi+1 ) dxi+1 · · · dxn
0 0 0 i! (i + 1) · (i − 1)!
 1  xn  xi+2
xi+1
i+1
= ··· f (xi+1 ) dxi+1 · · · dxn
(i + 1)!
0 0

0

1 1 1
since − =
i i+1 i(i + 1)
 1  1  xn  xi+3 i+1
xi+1
= f (xi+1 ) dxi+2 · · · dxn dxi+1
0 xi+1 xi+1 xi+1 (i + 1)!
 1 i+1
(1 − xi+1 )n−(i+1) xi+1
= f (xi+1 ) dxi+1
0 (n − (i + 1))! (i + 1)!
 1

1 n
= (1 − xi+1 )n−(i+1) xi+1
i+1 f (xi+1 ) dxi+1 .
n! 0 i + 1
Therefore
 4n−1

 5
1
n
E= (1 − xi+1 )n−(i+1) i+1
xi+1 f (xi+1 ) dxi+1
0 i+1
 i=0

 1 n

 n
= (1 − t)n−j tj  f (t) dt (we set j = i + 1 and t = xi+1 )
0 j=1
j
 1
= (1 − (1 − t)n )f (t) dt,
0

since the sum is the binomial expansion of ((1 − t) + t)n except that the j = 0 term,
1
which is (1 − t)n , is missing. Hence E = 0 f (t)P (t) dt where P (t) = 1 − (1 − t)n .
Clearly 0 ≤ P (t) ≤ 1 for 0 ≤ t ≤ 1. 
Solutions: The Fiftieth Competition (1989) 115

Solution 2. Fix n ≥ 0 and let (y1 , . . . , yn ) be chosen uniformly from [0, 1]n . Fix
t ∈ [0, 1], and let En (t) denote the expected value of
Rn = t − max ({0} ∪ ({y1 , . . . , yn } ∩ [0, t])) .
Let (x1 , . . . , xn ) be the permutation of (y1 , . . . , yn ) such that x1 ≤ · · · ≤ xn . The
distribution of (x1 , . . . , xn ) equals the distribution in the problem. Define
n−1
S= (yi+1 − zi+1 )f (yi+1 )
i=0

where zi+1 is the largest yj less than yi+1 , or 0 if no such yj exists. The terms in S
are a permutation of those in the original Riemann sum (ignoring the final term in the
Riemann sum, which is zero), so the expected  1 values of the Riemann sum and of S are
equal. Each term in S has expected value 0 En−1 (t)f (t) dt, since the expected value
of yi+1 − zi+1 conditioned on the event yi+1 = t for some t ∈ [0, 1] is the definition
 1using (yj )j=i+1 ∈ [0, 1]
n−1
of En−1 (t) . Therefore the expected value of the Riemann
sum is n 0 En−1 (t)f (t) dt.
It remains to determine En (t) for n ≥ 0. Let yn+1 be chosen uniformly in [0, 1],
independently of the y1 , . . . , yn in the definition of En (t). Then En (t) equals the
probability that yn+1 is in [0, t] and is closer to t than any other yj , since this
probability conditioned on a choice of y1 , . . . , yn equals Rn . On the other hand, this
probability equals (1 − (1 − t)n+1 )/(n + 1), since the probability that at least one of
y1 , . . . , yn+1 lies in [0, t] equals 1 − (1 − t)n+1 , and conditioned on this, the probability
that the yj in [0, t] closest to t is yn+1 is 1/(n + 1), since all possible indices for this
closest yj are equally likely. Thus En (t) = (1 − (1 − t)n+1 )/(n + 1), and we may take
P (t) = nEn−1 (t) = 1 − (1 − t)n . 
Literature note. For more on Riemann sums, see [Spv, Ch. 13, Appendix 1].
116 The William Lowell Putnam Mathematical Competition

The Fifty-First William Lowell Putnam Mathematical Competition

December 1, 1990

A1. (150, 9, 1, 0, 0, 0, 0, 0, 1, 1, 6, 33)

Let
T0 = 2, T1 = 3, T2 = 6,
and for n ≥ 3,
Tn = (n + 4)Tn−1 − 4nTn−2 + (4n − 8)Tn−3 .
The first few terms are
2, 3, 6, 14, 40, 152, 784, 5168, 40576, 363392.
Find, with proof, a formula for Tn of the form Tn = An + Bn , where (An )
and (Bn ) are well-known sequences.
Answer. We have Tn = n! + 2n .
Motivation. The hardest part of this problem is guessing the formula. There are
not many “well-known sequences” to guess. Observe that the terms are becoming
divisible by high powers of 2 (but not any other prime), and the ratio of the last two
terms given is roughly 8, and the ratio of the previous two is roughly 7.
Solution. The formula Tn = n! + 2n can be verified by induction.
Alternatively, set tn = n! + 2n . Clearly t0 = 2 = T0 , t1 = 3 = T1 and t2 = 6 = T2 .
Also,
tn − ntn−1 = 2n − n2n−1 .
Now 2n and n2n−1 are both solutions of the linear recursion
fn − 4fn−1 + 4fn−2 = 0; (1)
this follows from direct substitution. Since tn − ntn−1 is a linear combination of
solutions to (1), it must also be a solution. Hence
(tn − ntn−1 ) − 4(tn−1 − (n − 1)tn−2 ) + 4(tn−2 − (n − 2)tn−3 ) = 0,
or equivalently,
tn = (n + 4)tn−1 − 4ntn−2 + (4n − 8)tn−3 .
Thus tn = Tn , because they are identical for n = 0, 1, 2 and satisfy the same third-order
recursion (1) for n ≥ 3. 
& ' & ' & '
(1) (2) (m)
Remark. Let tn , tn , . . . , tn be sequences, each of one of
n≥1 n≥1 n≥1
the following forms:
(i) {αn }n≥1 for some α ∈ C,
(ii) {P (n)}n≥1 for some polynomial P ∈ C[n],
(iii) {(an + b)!}n≥1 for some integers a, b ≥ 0.
" #
(1) (m)
Let Q ∈ C[x1 , . . . , xm ] be a polynomial, and define un = Q tn , . . . , tn for n ≥ 1.
Then one can show that the sequences {un }n≥1 , {un+1 }n≥1 , {un+2 }n≥1 , . . . , thought
Solutions: The Fifty-First Competition (1990) 117

of as functions of n, lie in a finitely generated C[n]-submodule of the C[n]-module of all

sequences of complex numbers. Therefore {un } satisfies a nontrivial linear recursion
with polynomial coefficients. The reader may enjoy finding this recursion explicitly
for sequences such as (n!)2 + 3n or 2n n! + Fn where Fn is the nth Fibonacci number,
defined at the end of 1988A5. Problem 1984B1 [PutnamII, p. 44] is a variation on
this theme:
Let n be a positive integer, and define

f (n) = 1! + 2! + · · · + n!.

Find polynomials P (x) and Q(x) such that

f (n + 2) = P (n)f (n + 1) + Q(n)f (n),

for all n ≥ 1.

A2. √(63, 25, 16, 4, 0, 0, 0, 4, 5, 6, 21, 57)

√ √
Is 2 the limit of a sequence of numbers of the form 3
n − 3 m, (n, m =
0, 1, 2, . . . )?
√
3
√
Answer. Yes. In fact, every real number r is a limit of numbers of the form n− 3 m.
Solution 1. By the binomial expansion,

1/3


√ √ 1 1
3
n + 1 − 3 n = n1/3 1 + − n1/3 = n1/3 1 + O − n1/3 = O(n−2/3 )
n n
√ √
so 3 n + 1 − 3 n → 0 as n → ∞. (Alternatively, one could use
√ √ 1
3
n+1− 3n=  3
 √
3
= O(n−2/3 )
(n + 1)2 + 3 n(n + 1) + n2
or 
√ √ 1 n+1 −2/3
3
n+1− 3n= x dx = O(n−2/3 ),
3 n
√
3 √
3
n+1− n
or the Mean Value Theorem applied to the difference quotient (n+1)−n .)
3
If m > r , then the numbers
√ √ √ √ √ √ √
0 = 3 m − 3 m < 3 m + 1 − 3 m < · · · < 3 m + 7m − 3 m = 3 m
√
partition the interval [0, 3 m], containing r, in such a way that the largest subinterval
is of size O(m−2/3 ). By taking m sufficiently large, one can find a difference among
these that is arbitrarily close to r. 
Remark. We show more generally, that for any sequence {an } with an → +∞ and
an+1 − an → 0, the set S = { an − am : m, n ≥ 0 } is dense in R. Given r ≥ 0 and
C > 0, fix m such that |aM +1 − aM | < C for all M ≥ m. If n is the smallest integer
≥ m with an ≥ am + r, then an < am + r + C, so an − am is within C of r. This shows
that S is dense in [0, ∞), and by symmetry S is dense also in (−∞, 0].
Remark. Let f (x) be a function such that f (x) → +∞ and f  (x) → 0 as x → +∞.
The Mean Value Theorem shows that the hypotheses of the previous remark are
satisfied by the sequence an = f (n).
118 The William Lowell Putnam Mathematical Competition

Solution 2. Fix r ∈ R and C > 0. Then for sufficiently large positive integers n,
(n + r)3 − (n + r − C)3 = 3n2 C + O(n) > 1,

so (n + r − C)3 ≤ (n + r)3  ≤ (n + r)3 , and 3 (n + r)3  is within C of n + r. Hence
" √ #
lim 3 (n + r)3  − 3 n = r. 
n→∞
√ √
Solution 3. As in Solution 1, 3 n + 1 − 3 n → 0 as n → ∞, so the set
√ √
S = { 3 n − 3 m} contains arbitrarily small positive numbers. √ Also S√is closed under
√ √ 3 3
multiplication by positive integers k since k( 3 n − 3 m) = k 3 n − k 3 m. Any set
with the preceding two properties is dense in [0, ∞), because any finite open interval
(a, b) in [0, ∞) contains a multiple of any element of S ∩ (0, b − a). By symmetry S is
dense in (−∞, 0] too. 
√
Solution 4. Let bn = (n 3 5 mod 1) ∈ [0, 1]. As in the remark √ following 1988B3,
{bn } is dense in [0, 1]. Thus given
√ C > 0,
√ we can√ find n such that n 3
5 − r is within C
3 3
of some integer m ≥ 0. Then 5n − m = n 5 − m is within C of r.
3 3 3


A3. (4, 4, 4, 0, 0, 0, 0, 0, 22, 36, 76, 55)

Prove that any convex pentagon whose vertices (no three of which are
collinear) have integer coordinates must have area ≥ 5/2.
Solution. A lattice polygon is a plane polygon whose vertices are lattice points,
i.e., points with integer coordinates. The area of any convex lattice polygon has area
equal to half an integer: this follows from Pick’s Theorem mentioned in the second
remark below; alternatively, by subdivision one can reduce to the case of a triangle,
in which case the statement follows from the first remark below.
Consider a convex lattice pentagon ABCDE of minimal area. Since the area is
always half an integer, the minimum exists. If the interior of side AB contains a lattice
point F , then AF CDE is a convex lattice pentagon with smaller area, contradicting
the choice of ABCDE. (As is standard, vertices are listed in order around each
polygon.) Applying this argument to each side, we may assume that all boundary
lattice points are vertices.
Separate the vertices into four classes according to the parity of their coordinates.
By the Pigeonhole Principle, one class must contain at least two vertices. The midpoint
M between two such vertices has integer coordinates. By the previous paragraph, these
two vertices cannot form a side of the polygon. Also, the pentagon is convex, so M
is in the interior of the pentagon. Connecting M to the vertices divides the polygon
into 5 triangles, each of area at least 1/2, so the whole polygon has area at least 5/2. 
Remark. The bound 5/2 cannot be improved: the polygon with vertices (0, 0),
(1, 0), (2, 1), (1, 2), (0, 1) is convex and has area 5/2 (see Figure 14).
Remark. The area of a plane triangle with vertices (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) equals
 
x1 y1 1
1
det x2 y2 1
2
x3 y3 1
In particular, if xi , yi ∈ Z, the area is half an integer.
Solutions: The Fifty-First Competition (1990) 119

FIGURE 14.
A convex pentagon with area 5/2 whose vertices have integer coordinates.

For a formula for the volume of an n-dimensional simplex in Rn , see Solution 5 to

1993B5.
Remark (Pick’s Theorem). Given a lattice polygon, let i be the number of
internal lattice points and let b be the number of boundary lattice points. Pick’s
Theorem [Lar1, p. 68] states that the area of the polygon equals i + b/2 − 1.
In the previous solution, we could have used Pick’s Theorem in two places: in the
first paragraph to prove that a lattice polygon has half-integer area (even if it is not
convex), and as a substitute for the last sentence, using i ≥ 1 and b = 5.
Related question. Problem 1981A6 [PutnamII, p. 37] is similar:


Suppose that each of the vertices of ABC is a lattice point in the (x, y)-plane
and that there is exactly one lattice point P in the interior of the triangle.
The line AP is extended to meet BC at E. Determine the largest possible
value for the ratio of the lengths of segments |AP |/|P E|.
Here is a solution to 1981A6 using Pick’s Theorem. (See [PutnamII, p. 118] for a
non-Pick solution.)
We may reduce to the case where BC has no lattice points on it other than B, C,
and possibly E, by replacing the base BC with the shortest segment along it with
lattice endpoints and containing E in its interior.
Case 1: E is a lattice point. Then the reflection of E across P is also a lattice point,
so it must coincide with A, and |AP |/|P E| = 1.
Case 2: E is not a lattice point. Without loss of generality (by applying an affine
transformation preserving the lattice), we may assume B = (0, 0) and C = (1, 0). If
A = (x, y), y > 0, then x = 0, 1, and by Pick’s Theorem,
y = 2 + #{boundary lattice points} − 2
=d+e+1
where d = gcd(x, y), e = gcd(1 − x, y). Since d and e are relatively prime, de divides
y, so de ≤ d + e + 1, or equivalently (d − 1)(e − 1) ≤ 2. We have several subcases:
120 The William Lowell Putnam Mathematical Competition

• If d = 1, then y = 2 + e is divisible by e, so e = 1 or 2. If e = 1, then y = 2, and

the ratio |AP |/|P E| is 2. If e = 2, then y = 3, and the ratio is 3.
• If e = 1, then essentially the same argument gives a ratio of 2 or 3.
• The case d = e = 2 is not allowed, since d and e are coprime.
• If d = 3 and e = 2, then y = 6, giving a ratio of 5.
• If d = 2 and e = 3, then the ratio is also 5.
Hence the maximum is 5. This argument can be refined to show that equality is
achieved only for one triangle (up to automorphisms of the plane preserving lattice
points).
Remark. The group of linear transformations of the plane preserving the lattice
points and fixing the origin is the group GL2 (Z) defined in the introduction. It is
important in number theory and related fields.
Remark. Fix integers d ≥ 2 and k ≥ 1. It follows easily from [He] that up to the
action of GLd (Z) and translation by lattice points, there are only finitely many convex
lattice polytopes in Rn having exactly k interior lattice points. For d = 2 and k = 1,
we get 16 polygons; see [PRV] for a figure showing all of them. Five of these 16 are
triangles, and checking each case gives another proof that |AP |/|P E| ≤ 5, and that
equality is possible for only one of the five triangles.

A4. (44, 7, 6, 6, 0, 0, 0, 0, 32, 15, 30, 61)

Consider a paper punch that can be centered at any point of the plane
and that, when operated, removes from the plane precisely those points
whose distance from the center is irrational. How many punches are needed
to remove every point?
Answer. Three punches are needed.

Solution. Punches at two points P and Q are not enough to remove all points,
because if r is any rational number exceeding P Q/2, the circles of radius r centered
at P and Q intersect in at least one point R, and R is not removed by either punch.
We next show that three carefully chosen punches suffice.
Proof 1: Existential. Punch twice, at distinct centers. Since each punch leaves
countably many circles, and any two distinct circles intersect in at most two points,
the two punches leave behind a countable set. Consider all circles with rational radii
centered at points of this set. Their intersections with a fixed line L form a countable
set S. A point of L − S is at an irrational distance from all unpunched points; apply
the third punch there. 
Remark. The fact that the plane is not a countable union of circles can also be
deduced from measure theory: a circle (without its interior) in the plane has measure
zero, and a countable union of measure zero sets still has measure zero, but the entire
plane has infinite measure. See [Ru] for more on measure theory.
√
Proof 2: Constructive. Choose α ∈ R such that α2 is irrational, for example α = 3 2.
Use punches centered at A = (−α, 0), B = (0, 0), and C = (α, 0). If P = (x, y) is any
Solutions: The Fifty-First Competition (1990) 121

point,
AP 2 + CP 2 − 2BP 2 = (x + α)2 + y 2 + (x − α)2 + y 2 − 2(x2 + y 2 ) = 2α2
is irrational, so AP , BP , CP cannot all be rational. Hence all P get removed. 
Remark. The motivation for taking AP + CP − 2BP is that it is the linear
2 2 2

combination which eliminates the terms involving x or y.

Remark. Both proofs easily generalize to prove the same result where the punch
removes only those points whose distance from the center is transcendental. Recall
that a real or complex number α is said to be algebraic if α is a zero of a nonzero
polynomial with rational coefficients, and α is said to be transcendental otherwise. In
Proof 1, observe that the set of real algebraic numbers is countable. In Proof 2, simply
take α transcendental.
Remark. Essentially the same question appeared as [New, Problem 28].
Related question. There are many interesting questions concerning distances
between points in a subset of the plane. For example, for any set of n points, in
the plane, Erdős [Er], [Hon2, Ch. 12] proved that

• the number of different distances produced must be at least n − 3/4 − 1/2,
• the smallest distance produced cannot occur more often than 3n − 6 times,
• the greatest distance produced cannot occur more often than n times,
• no distance produced can occur more than 2−1/2 n3/2 + n/4 times.
Also, Problem 5 on the 1987 International Mathematical Olympiad [IMO87] asks
Let n be an integer greater than or equal to 3. Prove that there is a set of n
points in the plane such that the distance between any two points is irrational
and each set of three points determines a nondegenerate triangle with rational
area.

A5. (29, 5, 0, 0, 0, 0, 0, 0, 1, 0, 58, 108)

If A and B are square matrices of the same size such that ABAB = 0,
does it follow that BABA = 0?
Answer. No.
Solution 1. Direct multiplication shows that the 3 × 3 matrices
   
0 0 1 0 0 1

A= 0 0 0 , B= 1  0 0 (1)
0 1 0 0 0 0
give a counterexample. 
Solution 2. A more enlightening way to construct a counterexample is to use
a transition diagram, as in the following example. Let e1 , e2 , e3 , e4 be a basis of a
four-dimensional vector space. Represent the matrices as in Figure 15. For example,
the arrow from e4 to e3 labelled B indicates that Be4 = e3 ; the arrow from e1 to 0
indicates that Ae1 = 0. Then it can be quickly checked that ABAB annihilates the
four basis vectors, but BABAe4 = e1 . (Be careful with the order of multiplication
when checking!) 
122 The William Lowell Putnam Mathematical Competition

A B A B A

#✥#✥#✥#✥#✥
B A B A
e4 e3 e2 e1 0
B
✧✦✧✦✧✦✧✦✧✦
FIGURE 15.
Schematic representation of counterexample in Solution 2 to 1990A5. Alternatively, a
transition diagram for a finite automaton.

Remark. The counterexample of Solution 1 also can be obtained from a transition

diagram.
Remark. There are no 1 × 1 or 2 × 2 counterexamples. The 1 × 1 case is clear. For
the 2 × 2 case, observe that ABAB = 0 implies B(ABAB)A = 0, and hence BA is
nilpotent. But if a 2 × 2 matrix M is nilpotent, its characteristic polynomial is x2 , so
M2 = 0 by the Cayley-Hamilton Theorem [Ap2, Theorem 7.8]. Thus BABA = 0.
Remark. For any n ≥ 3, there exist n × n counterexamples: enlarge the matrices
in (1) by adding rows and columns of zeros.
Stronger result. Here we present a conceptual construction of a counterexample,
requiring essentially no calculations. Define a word to be a finite sequence of A’s and
B’s. (The empty sequence ∅ is also a word.) Let S be a finite set of words containing
BABA and its “right subsequences” ABA, BA, A, ∅, but not containing any word
having ABAB as a subsequence. Consider a vector space with basis corresponding
to these words (i.e., eBABA , eABA , e∅ , etc.). Let A be the linear transformation
mapping ew to eAw if Aw ∈ S and to 0 otherwise. Define a linear transformation
B similarly. Then ABAB = 0 but BABA = 0. (This gives a very general way of
dealing with any problem of this type.)
Remark. To help us find a counterexample, we imposed the restriction that each of
A and B maps each standard basis vector ei to some ej or to 0. With this restriction,
the problem can be restated in terms of automata theory:
Does there exist a finite automaton with a set of states Σ = {0, e1 , e2 , . . . , en }
in which all states are initial states and all but 0 are final states, and a set
of two productions {A, B} each mapping 0 to 0, such that the language it
accepts contains ABAB but not BABA?
See Chapter 3 of [Sa] for terminology. The language accepted by such a finite
automaton is defined as the set of words in A and B that correspond to a sequence of
productions leading from some initial state to some final state. Technically, since our
finite automaton does not have a unique initial state, it is called nondeterministic,
even though each production maps any given state to a unique state. (Many
authors [HoU, p. 20] do not allow multiple initial states, even in nondeterministic
finite automata; we could circumvent this by introducing a new artificial initial state,
with a new nondeterministic production mapping it to the desired initial states.) One
theorem of automata theory [Sa, Theorem 3.3] is that any language accepted by a
Solutions: The Fifty-First Competition (1990) 123

nondeterministic finite automaton is also the language accepted by some deterministic

finite automaton.
Many lexical scanners, such as the UNIX utility grep [Hu], are based on the theory
of finite automata. See the remark in 1989A6 for the appearance of automata theory
in a very different context.

A6. (6, 6, 54, 0, 0, 0, 0, 4, 0, 45, 85)

If X is a finite set, let |X| denote the number of elements in X. Call an
ordered pair (S, T ) of subsets of {1, 2, . . . , n} admissible if s > |T | for each
s ∈ S, and t > |S| for each t ∈ T . How many admissible ordered pairs of
subsets of {1, 2, . . . , 10} are there? Prove your answer.
Answer. The number of admissible ordered pairs of subsets of {1, 2, . . . , 10} equals
the 22nd Fibonacci number F22 = 17711.
Solution 1. Let Am,n be the set of admissible pairs (S, T ) with S ⊆ {1, 2, . . . , m}
and T ⊆ {1, 2, . . . , n}, and let am,n = |Am,n |. Suppose m ≥ n ≥ 1. Then
Am−1,n ⊆ Am,n . We now show that the maps

Am,n − Am−1,n ↔ Am−1,n−1

(S, T ) !→ (S − {m}, { t − 1 : t ∈ T })
(U ∪ {m}, { v + 1 : v ∈ V }) ← (U, V )

are well-defined inverse bijections.

If (S, T ) ∈ Am,n −Am−1,n , then m ∈ S. Let S  = S −{m} and T − = { t−1 : t ∈ T }.
Then |S| ≥ 1, and t > |S| ≥ 1 for all t ∈ T , so T − ⊆ {1, 2, . . . , n − 1}. Since (S, T )
is admissible, each element of S  is greater than |T | = |T − |. Also, each element of
T is greater than |S|, so each element of T − is greater than |S| − 1 = |S  |. Hence
(S  , T − ) ∈ Am−1,n−1 .
If (U, V ) ∈ Am−1,n−1 , let U  = U ∪ {m} ⊆ {1, 2, . . . , m} and V + = { v + 1 : v ∈
V } ⊆ {1, 2, . . . , n}. Since (U, V ) is admissible, each element of U is greater than |V |,
but m ≥ n > |V | also, so each element of U  is greater than |V |. Moreover, each
element of V is greater than |U |, so each element of V + is greater than |U | + 1 = |U  |.
Hence (U  , V + ) ∈ Am,n − Am−1,n .
Composing the two maps just defined in either order gives the identity, so both
are bijections. Hence am,n = am−1,n + am−1,n−1 for m ≥ n ≥ 1. In particular,
an,n = an,n−1 + an−1,n−1 (because ai,j = aj,i ), and an,n−1 = an−1,n−1 + an−1,n−2 , so
each term of
a0,0 , a1,0 , a1,1 , a2,1 , a2,2 , a3,2 , a3,3 , . . .

is the sum of the two preceding terms. Starting from a0,0 = 1 and a1,0 = 2, we find
that the 21st term in the sequence

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377,

610, 987, 1597, 2584, 4181, 6765, 10946, 17711, . . .

is a10,10 = 17711. (The sequence is the Fibonacci sequence defined at the end of
1988A5, but starting with F2 = 1.) 
124 The William Lowell Putnam Mathematical Competition

Solution 2 (Jeremy Rouse). Let am,n be as in Solution 1. If S is an i-element

subset of {j + 1, j + 2, . . . , m} and T is a j-element subset of {i + 1, i + 2, . . . , n},
then (S, T ) is an admissible pair; conversely, each admissible pair (S, T ) with S ⊆
{1, 2, . . . , m}, T ⊆{1, 2,
 . . . , n}, |S| = i, and |T | = j arises in this way. Hence
 m−j n−i
am,n = i,j i j , where the sum ranges over pairs of nonnegative integers
(i, j) satisfying i + j ≤ min{m, n} (so that the binomial coefficients are nonzero). Let
Fn be the nth Fibonacci number. We will give a bijective proof that



n−j n−i
= F2n+2 (1)
i,j
i j

for all n ≥ 0.
For m ≥ 1, let Rm mean “1 × m rectangle,” and let Nm denote the number of ways
to tile an Rm with 1 × 1 squares and 1 × 2 dominos (rectangles). We now prove (1) by
showing that both sides equal N2n+1 . A tiling of an Rm ends either with a square or
a domino, so Nm = Nm−1 + Nm−2 for m ≥ 3. Together with N1 = 1 and N2 = 2, this
proves Nm = Fm+1 by induction. In particular N2n+1 equals F2n+2 , the right-hand
side of (1).
On the other hand, if we start with a pair of tilings, one a tiling of an Rn+i−j by
n − i − j squares and i dominos, and the other a tiling of an Rn+j−i by n − i − j
squares and j dominos, we may form a tiling of an R2n+1 by appending the two, with
a square inserted in between. Conversely, any tiling of an R2n+1 arises from such
a pair: every tiling of an R2n+1 contains an odd number of squares, so there is a
“median” square, and the pieces to the left and right of this square
  constitute
 a pair
of tilings. The number of such pairs for fixed i and j equals n−j n−i
, so N
  n−i i j 2n+1
equals i,j n−j i j , the left-hand side of (1).
This proves (1). The desired value a10,10 = F22 is then found by calculating
F0 , F1 , . . . , F22 successively, using Fn+2 = Fn+1 + Fn . 

B1. (114, 2, 52, 0, 0, 0, 0, 11, 5, 3, 10, 4)

Find all real-valued continuously differentiable functions f on the real
line such that for all x
 x
 
(f (x))2 = (f (t))2 + (f  (t))2 dt + 1990.
0

√
Answer. There are two such functions, namely f (x) =
√ 1990ex , and f (x) =
− 1990ex .

Solution. For a given f , the functions on the left- and right-hand sides are equal
if and only if their values at 0 are equal, i.e., f (0)2 = 1990, and their derivatives are
equal for all x, i.e.,

2f (x)f  (x) = (f (x))2 + (f  (x))2 for all x.

The latter condition is equivalent to each of the following: (f (x) − f  (x))2 = 0,

f  (x) = f (x), f (x) = Ce√
x
for some constant C. Combining this condition
√ with
f (0) = 1990 yields C = ± 1990, so the desired functions are f (x) = ± 1990ex . 
2
Solutions: The Fifty-First Competition (1990) 125

B2. (23, 5, 4, 9, 0, 0, 3, 0, 0, 32, 125)

Prove that for |x| < 1, |z| > 1,
∞
(1 − z)(1 − zx)(1 − zx2 ) · · · (1 − zxj−1 )
1+ (1 + xj ) = 0.
j=1
(z − x)(z − x2 )(z − x3 ) · · · (z − xj )

Solution. Let S0 = 1, and for n ≥ 1, let

n
(1 − z)(1 − zx)(1 − zx2 ) · · · (1 − zxj−1 )
Sn = 1 + (1 + xj ) .
j=1
(z − x)(z − x2 )(z − x3 ) · · · (z − xj )

Since S1 = (1 − zx)/(z − x) and S2 = (1 − zx)(1 − zx2 )/(z − x)(z − x2 ), we suspect

that
(1 − zx)(1 − zx2 ) · · · (1 − zxn )
Sn = ,
(z − x)(z − x2 )(z − x3 ) · · · (z − xn )
which is easily proved by induction.
It remains to prove that limn→∞ Sn = 0. If Sn = 0 for some n, then SN = 0 for all
N ≥ n, so limn→∞ Sn = 0. Otherwise
Sn+1 1 − zxn+1 1
= →
Sn z − xn+1 z
as n → ∞, since xn+1 → 0. By the Ratio Test, limn→∞ Sn = 0. 

B3. (97, 7, 4, 2, 0, 0, 0, 0, 12, 2, 54, 23)

Let S be a set of 2×2 integer matrices whose entries aij (1) are all squares
of integers, and, (2) satisfy aij ≤ 200. Show that if S has more than 50387
(= 154 − 152 − 15 + 2) elements, then it has two elements that commute.
Solution. Let U be the set of 2 × 2 matrices satisfying (1) and
(2). Let
 D be the
1 1
set of diagonal matrices in U , and let J be the set of multiples of in U . The
1 1
numbers less than or equal to 200 that are squares of integers are the 15 numbers 02 ,
12 , . . . , 142 , so |U | = 154 , |D| = 152 , and |J| = 15. Now
(i) any two matrices from D commute,
(ii) any two matrices from J commute, and



1 1 1 4
(iii) and commute.
0 1 0 1
Suppose that no two elements of S commute. Write
   
S = S ∩ (D ∪ J) ∪ S ∩ (D ∪ J)c .
(Here X c denotes the complement of X.) By (i) and (ii), S can contain at most one
element of D and at most one element of J, so |S ∩ (D ∪ J)| ≤ 2. By (iii),
|S ∩ (D ∪ J)c | < |U ∩ (D ∪ J)c |
= |U | − |D| − |J| + |D ∩ J|
= 154 − 152 − 15 + 1.
Hence |S| ≤ 2 + (154 − 152 − 15) = 50387. 
126 The William Lowell Putnam Mathematical Competition

Remark. The number 50387 is far from the best possible. Liu and Schwenk have
shown, using an inclusion-exclusion argument, that the maximum number of elements
in U in which no two elements commute is 32390 [LS]. Because the bound given in
the problem is so far from being optimal, there are many possible solutions.

B4. (9, 5, 2, 1, 0, 0, 0, 0, 3, 2, 63, 116)

Let G be a finite group of order n generated by a and b. Prove or disprove:
there is a sequence

g1 , g2 , g3 , . . . , g2n

such that
(1) every element of G occurs exactly twice, and
(2) gi+1 equals gi a or gi b, for i = 1, 2, . . . , 2n. (Interpret g2n+1 as g1 .)

Solution. We use graph theory terminology; see the remark below. Construct
a directed multigraph D whose vertices are the elements of G, and whose arcs are
indexed by G × {a, b}, such that the arc corresponding to the pair (g, x) goes from
vertex g to vertex gx. (See Figure 16 for an example, with G equal to the symmetric
group S3 , a equal to the transposition (12), and b equal to the 3-cycle (123).) At
the vertex g, there are two arcs going out (to ga and to gb), and two arcs coming in
(from ga−1 and from gb−1 ). Also, D is weakly connected, since a and b generate G.
Hence, by the first theorem in the remark below, D has an Eulerian circuit. Take g1 ,
g2 , . . . , gm to be the the startpoints of the arcs in this circuit, in order. Each element
of G occurs exactly twice in this sequence, since each vertex of D has outdegree 2; in
particular m = 2n. Also, for 1 ≤ i ≤ 2n, the element gi+1 equals either gi a or gi b,
because the two outgoing arcs from gi end at gi a and gi b. 
Remark (Eulerian paths and circuits). A directed multigraph D consists of a set V
(whose elements are called vertices), and a set E (whose elements are called arcs or
sometimes edges or directed edges), with a map E → V × V (thought of as sending
an arc to the pair consisting of its startpoint and the endpoint). Typically one draws
each element of V as a point, and each arc of E as an arc from the startpoint to the
endpoint, with an arrow to indicate the direction. What makes it a multigraph is that
for some vertices v, w ∈ V , there may be more than one arc from v to w. Also, there
may be loops: arcs from a vertex to itself. Call D finite if V and E are both finite
sets.
The outdegree of a vertex v is the number of arcs in E having v as startpoint.
Similarly the indegree of v is the number of arcs in E having v as endpoint. If v, w ∈ V
then a path from v to w in D is a finite sequence of arcs in E, not necessarily distinct,
such that the startpoint of the first arc is v, the endpoint of each arc (other than the
last) is the startpoint of the next arc, and the endpoint of the last arc is w. Such a
path is called a circuit or cycle if v = w. An Eulerian path in D is a path in which
each arc in E occurs exactly once. An Eulerian circuit is a circuit in which each arc
in E occurs exactly once. We say that D is strongly connected if for every two distinct
vertices v, w ∈ V , there is a path from v to w in D. On the other hand, D is weakly
connected if for every two distinct vertices v, w ∈ V , there is a path from v to w in
Solutions: The Fifty-First Competition (1990) 127

(12)

a a

b (1) b
b b

(231) (123)
a a
b
a a

(23) (13)
b

FIGURE 16.
Cayley graph of the group S3 with generators a = (12) and b = (123)

the corresponding undirected graph, that is, a path consisting of arcs some of which
may be the reverses of the arcs in E.
Then one can prove the following two theorems.
• A finite directed multigraph with at least one arc has an Eulerian circuit if and only
if it is weakly connected and the indegree and outdegree are equal at each vertex.
• A finite directed multigraph with at least one arc has an Eulerian path but not an
Eulerian circuit if and only if it is weakly connected and the indegree and outdegree
are equal at each vertex, except at one vertex at which indegree is 1 larger than
outdegree, and one other vertex at which outdegree is 1 larger than indegree.
See Chapter 7 of [Ros2], especially §7.4 and §7.5.
Remark. The directed multigraph constructed in the solution is called the Cayley
digraph or Cayley diagram associated to G and its set of generators {a, b}. See [Fr,
pp. 87–91].

B5. (16, 11, 15, 12, 0, 0, 0, 11, 5, 5, 48, 78)

Is there an infinite sequence a0 , a1 , a2 , . . . of nonzero real numbers such
that for n = 1, 2, 3, . . . the polynomial

pn (x) = a0 + a1 x + a2 x2 + · · · + an xn

has exactly n distinct real roots?

Answer. Yes, such an infinite sequence exists.

Solution 1. Take a0 = 1, a1 = −1, and for n ≥ 1 construct an+1 inductively as

follows. Suppose pn (x) has n distinct real zeros: x1 < x2 < · · · < xn . Choose α0 , . . . ,
αn so that
α0 < x1 < α1 < · · · < xn < αn .
128 The William Lowell Putnam Mathematical Competition

Then the signs of pn (α0 ), pn (α1 ), . . . , pn (αn ) alternate. Define an+1 = −C sgn(pn (αn )),
where C is positive and small enough that
   
sgn pn+1 (αi ) = sgn pn (αi )
for all i. Let
pn+1 (x) = pn (x) + an+1 xn+1 .
By the Intermediate Value Theorem, pn+1 has a zero between αi and αi+1 for
0 ≤ i ≤ n − 1, and a zero greater than αn since
   
sgn pn+1 (αn ) = lim sgn pn+1 (x) .
x→∞

Because pn+1 (x) is of degree n + 1, it cannot have more than these n + 1 zeros, so
pn+1 (x) has n + 1 distinct real zeros, as desired. 
2
Solution 2. For n ≥ 0, let an = (−1)n 10−n . For 0 ≤ k ≤ n,
n
2 2
(−1)k 10−k pn (102k ) = (−1)i−k 10−(i−k)
i=0
n−k
2
= (−1)j 10−j
j=−k
∞
2
>1−2 10−j
j=1

> 0,
so pn (1), pn (102 ), pn (104 ), . . . , pn (102n ) alternate in sign. By the Intermediate Value
Theorem, it follows that pn (x) has at least n distinct real zeros. Since pn (x) has degree
n, there cannot be more than n zeros. 
Remark. Solution 2 is motivated by the theory of Newton polygons for polynomials
with coefficients in the field Qp of p-adic numbers. Let | · |p denote the p-adic absolute
value on Qp . If {ai }i≥0 is a sequence of nonzero p-adic numbers such that the lower
convex hull of { (i, − ln |ai |p ) : 0 ≤ i ≤ n } consists of n segments with different slopes,
n 2
then i=0 ai xi has n distinct zeros in Qp ; in particular this holds for ai = pi . The
analogous statement over R, with | · |p replaced by the standard absolute value, is not
true in general, but it is true if the differences between the slopes are sufficiently large
relative to n. For an introduction to p-adic numbers and Newton polygons, see [Kob].

B6. (5, 0, 0, 0, 0, 0, 0, 0, 38, 6, 29, 123)

Let S be a nonempty closed bounded convex set in the plane. Let K be a
line and t a positive number. Let L1 and L2 be support lines for S parallel
to K, and let L be the line parallel to K and midway between L1 and L2 .
Let BS (K, t) be the band of points whose distance from L is at most (t/2)w,
where w is the distance between L1 and L2 . What is the smallest t such
that !
S∩ BS (K, t) = ∅
K

for all S? (K runs over all lines in the plane.)

Solutions: The Fifty-First Competition (1990) 129

Support line L1 w
tw

L S
B S( K,t ) K

Support line L2

Answer. The smallest t is 1/3.

Motivation. To approach this problem, it is natural to experiment with simple
shapes. The triangle suggests that the answer should be t = 1/3, and that for any
t ≥ 1/3, the intersection !
S∩ BS (K, t)
K

should contain the centroid, and is hence nonempty.

Solution 1. We first show that the intersection can be empty for t < 1/3. Suppose
that S is a triangle. Dissect the triangle into 9 congruent subtriangles, as shown in
Figure 17. If K is parallel to one of the sides of the triangle, S ∩ BS (K, t) is contained
in the three subtriangles in the “middle strip” (and does not meet the boundary of
the strip). Hence if t < 1/3, and K1 , K2 , K3 are parallel to the three sides of the
triangle,
!3
S∩ BS (Ki , t)
i=1

is empty. This is illustrated in Figure 18. (Also, if t = 1/3, then this intersection
consists of just the centroid. This motivates the rest of the solution.)

L1

1/3

1/6
L

1/3

L2

FIGURE 17.
130 The William Lowell Putnam Mathematical Competition

BS (CA,t) BS (AB,t)

BS (BC,t)

B C

FIGURE 18.
S∩ K BS (K, t) may be empty for t < 1/3.

Recall
 that the centroid of a measurable region S in R2 is the unique point P such
−→
that Q∈S  P Q dA = 0. Equivalently, the coordinates
 of the centroid (x, y) are given
by x = S x dA/ S 1 dA, and y = S y dA/ S 1 dA. If S is convex, then the centroid
lies within S.
We now show that the intersection of the problem is nonempty for t ≥ 1/3 for any
S, by showing that each strip BS (K, t) contains the centroid of S. By symmetry, it
suffices to show that the centroid of S is at most 2/3 of the distance from L1 to L2 .
Think of L1 as the upper support line. (See Figure 19.) Let Pi be a point of contact
of Li with S, for i = 1, 2. For a variable point Q to the left of P2 on L2 (possibly
←→
Q = P2 ), let A be the intersection of S with the open half-plane to the left of QR1 , and
let B be the part of (possibly degenerate) 4QP1 P2 lying outside S. As Q moves to a
nearby point Q , Area(A) and Area(B) each change by at most Area(4QQ P1 ), which
can be made arbitrarily small by choosing Q close to Q; hence Area(A) and Area(B)
vary continuously as functions of Q. The difference δ(Q) = Area(A) − Area(B) is also
a continuous function of Q. At Q = P2 , Area(A) ≥ 0 and Area(B) = 0, so δ(P2 ) ≥ 0.
But as Q tends to infinity along L1 , Area(A) is bounded by Area(S) and Area(B)
grows without bound, so δ(Q) < 0 for some Q. By the Intermediate Value Theorem,
there is some position of Q for which δ(Q) = 0, i.e., for which Area(A) = Area(B).
Fix such a Q.

P1
L1

A
A A¢

B
L2
Q P2

FIGURE 19.
Solutions: The Fifty-First Competition (1990) 131

We claim that if A ∈ A and B ∈ B, then B lies below A. To show this, let A

be the intersection of P1 P2 with the horizontal line through A. Since S is convex,
A ∈ P1 P2 ⊆ S and 4AA P1 ⊆ S. Then B ∈ 4AA P1 by the definition of B. But
4AA P1 contains all points of 4QP1 P2 lying above or at the same level as A, so B
must lie below A.
Let S̃ denote the region obtained from S by removing A and adding B, and
performing the corresponding operations to the right of P1 P2 . By the previous
paragraph, the centroid of S̃ lies at least as low as the centroid of S. But S̃ is a
triangle, with base on L2 and opposite vertex at P1 , so the centroid of S̃ lies exactly
2/3 of the way from L1 to L2 . Hence the centroid of S lies at most 2/3 of the way
from L1 to L2 .
Thus the minimal t for which the intersection is nonempty is 1/3. 
Solution 2. As in the first paragraph of Solution 1, t ≥ 1/3. We now show that
t = 1/3 works, by proving that the centroid of S is in BS (K, t) for all K. Without loss
of generality, we may rotate, rescale, and translate to assume that L2 is the x-axis and
L1 is the line y = 3. Let P be a point where S meets L2 . Let A be the area of S. It
suffices to show that the centroid (x, y) satisfies y ≤ 2, since then y ≥ 1 by symmetry.
Partially cover S with nonoverlapping inscribed triangles each having one vertex
at P , as in Figure 20, and let CA be the area of the part of S not covered. Each
triangle has vertex y-coordinates 0, y1 , y2 where 0 ≤ y1 , y2 ≤ 3, so the centroid of the
triangle has y-coordinate at most 2. Let y  denote the y-coordinate of the centroid
of the triangle-tiled portion of S, let y % denote the y-coordinate of the centroid of the
remainder, and let y S denote the y-coordinate of the centroid of S. Then y  ≤ 2,
y % ≤ 3, and
Ay S = CAy % + (A − CA)y 
y S = Cy % + (1 − C)y 
≤ 3C + 2(1 − C)
≤ 2 + C.
But C can be made arbitrarily small by choosing the triangles appropriately, so y S ≤ 2,
as desired. 

L1
✧ ❧
✧ ❈ ❧
✧ ❈
✧ ❧
✧ ❈ ❧
❅ ❈ ❧
❅ ❈ #✁
#✁
❅ ❈ #
❜ ❅ ❈ # ✁
❆❜ ❅ ❈ # ✁
❆P❜❜ ❅ ❈ # ✁
PP❜ #
P❜
P❅ ❜ ❈ # ✘✘✘✁
P❜ P❈#
❅ ✘✘✘
L2
P

FIGURE 20.
Partially covering S by triangles.
132 The William Lowell Putnam Mathematical Competition

Remark. We sketch a justification of the last sentence. Using the convexity of

S, one can prove that there exist continuous functions f (x) ≤ g(x) defined on an
interval [a, b] such that S is the region
 bbetween the graphs of f and g on [a, b]. The
b
approximations to A = a g(x) dx − a f (x) dx given by the Trapezoid Rule can be
made arbitrarily close to A by taking sufficiently fine subdivisions of [a, b]. We may
assume that the x-coordinate of P is one of the sample points; then the approximation
is represented by the area of an inscribed polygon with one vertex at P . Cut the
polygon into triangles by connecting P to the other vertices with line segments.
Remark. A more analytic approach to proving that the centroid is in the central
1/3 of the strip is the following. Choose coordinates so that L1 and L2 are the lines
y = 1 and y = 0 respectively. Let f (t) be the length of the intersection of S and the
line y = t. Convexity of S implies that f (t) is a nonnegative concave-down continuous
function on [0, 1], and the desired result would follow from
 1 
1 1
tf (t) dt ≥ f (t) dt. (1)
0 3 0
(Geometrically, this states that the centroid is at least 1/3 of the way from L2 to L1 .
Along with the corresponding statement with the roles of L1 and L2 reversed, this
shows that the centroid is in BS (K, 1/3).)
For any function f with continuous second derivative, integration by parts twice
yields
 1

2  1  1
2 
1 t t t t
t− f (t) dt = − f (t) − − f  (t) dt (2)
0 3 2 3 0 0 2 3

3  1  1
3 
f (1) t t2 t t2
= − − f  (t) + − f  (t) dt
6 6 6 0 6 6
 1
3  0
f (1) t t2 
= + − f (t) dt.
6 0 6 6
If in addition f (1) ≥ 0 and f is concave-down, then this implies (1) since the final
integrand is everywhere nonnegative.
To prove
 1

1 f (1)
t− f (t) dt ≥ (3)
0 3 6
for all concave-down continuous functions, including those that are not twice differen-
tiable, it remains to prove that any concave-down continuous function f on [0, 1] is a
uniform limit of concave-down functions with continuous second derivatives. Adjusting
f by a linear function, we may assume f (0) = f (1) = 0. By continuity of f at 0 and
1, the function f (t) is the uniform limit of the concave-down continuous functions
min{f (t), ct, c(1 − t)} on [0, 1] as c → +∞. Hence we may replace f by such an
approximation to assume that f (t) ≤ min{ct, c(1 − t)} for some c > 0. We can then
extend f to a concave-down continuous function on R by setting f (t) = ct for t < 0
and f (t) = c(1−t) for t > 1. We now show that it is the uniform limit of concave-down
smooth functions. Define a sequence of smooth nonnegative functions δn supported
Solutions: The Fifty-First Competition (1990) 133
∞
on [−1/n, 1/n] with δ (t) dt
−∞ n
= 1, and define the convolution
 ∞
fn (t) = f (u)δn (t − u) du.
−∞

Then f is the uniform limit of fn on [0, 1], and fn is smooth. Finally, fn is also concave-
down, because the convolution can be viewed as a weighted average of translates of
f.
Remark (Eric Wepsic). Alternatively, one can prove (3) for all concave-down
continuous functions by discretizing (2). For n ≥ 1, define

n


i 1 i 1
An = − f , and
i=1
n 3 n n
n


f (1) i f ((i + 1)/n) − 2f (i/n) + f ((i − 1)/n) 1
Bn = + g ,
6 i=1
n 1/n2 n

where g(t) = t3 /6 − t2 /6. (These are supposed to be approximations to the two ends
of (2). The ratio
f ((i + 1)/n) − 2f (i/n) + f ((i − 1)/n)
1/n2

is an approximation of f  (i/n) in the same spirit as the formula in Proof 4 of the

lemma in 1992A4.)
It is not quite true that An = Bn , but we can bound the difference. First collect
terms in Bn with the same value of f , and use g(0) = g(1) = 0, to obtain

n
Bn = bin f (i/n)
i=1

with
      

 n g n1 − g n0 + 0 , if i = 0
   i+1  i  
bin = n g n − 2g n + g i−1 , if 1 ≤ i ≤ n − 1

 1 + n 0 − g  n  + g  n−1  ,
n

6 n n if i = n.

Since g is infinitely differentiable, Taylor’s Theorem [Spv, Ch. 19, Theorem 4] centered
at x shows




1 1 g  (x) 1
g x+ − 2g(x) + g x − = 2
+ O
n n n n3

and



1 g  (x) 1
g x± − g(x) = ± +O
n n n2
134 The William Lowell Putnam Mathematical Competition

as n → ∞, where the constants implied by the O’s are uniform for x ∈ [0, 1]. Thus
 1

 
g (0) + O n ,  if i = 0
bin = g (i/n) + O n12 , if 1 ≤ i ≤ n − 1

 1 − g  (1) + O  1  ,
n

6 n if 1 ≤ i ≤ n − 1
  

 1
O ,
 i n 1 1 1
if i = 0
n − 3 n + O n2 , if 1 ≤ i ≤ n − 1
=

O  1  ,

if 1 ≤ i ≤ n − 1.
n

where again the implied constants are uniform. Except for the O’s, these are the same
as the coefficients of f (i/n) in An . Thus, if M = sup{ |f (t)| : t ∈ [0, 1] }, then
4n−1 5
An − Bn = O(1/n)M + O(1/n2 )M + O(1/n)M = O(1/n).
i=1

In particular, limn→∞ (An −Bn ) = 0. If f is concave-down, then for all i, f ((i+1)/n)−

2f (i/n) + f ((i − 1)/n) ≤ 0 and g(i/n) ≤ 0, so the definition of Bn implies Bn ≥ f (1)/6
1
for all n. If f is continuous, then An is the nth Riemann sum for 0 (t − 1/3)f (t) dt, so
1
limn→∞ An = 0 (t − 1/3)f (t) dt. Combining the three previous sentences yields (3),
whenever f is concave-down and continuous.
Remark. A similar result is that for every compact convex set S in the plane, there
exists at least one point P ∈ S such that every chord AB of S containing P satisfies
AP
1/2 ≤ ≤ 2.
BP
This result can be generalized to an arbitrary number of dimensions [Berg, Corol-
lary 11.7.6].
Solutions: The Fifty-Second Competition (1991) 135

The Fifty-Second William Lowell Putnam Mathematical Competition

December 7, 1991

A1. (189, 0, 3, 0, 0, 0, 0, 0, 0, 1, 20, 0)

A 2 × 3 rectangle has vertices at (0, 0), (2, 0), (0, 3), and (2, 3). It rotates
90◦ clockwise about the point (2, 0). It then rotates 90◦ clockwise about
the point (5, 0), then 90◦ clockwise about the point (7, 0), and finally, 90◦
clockwise about the point (10, 0). (The side originally on the x-axis is now
back on the x-axis.) Find the area of the region above the x-axis and below
the curve traced out by the point whose initial position is (1, 1).
Answer. The area of the region is 7π/2 + 6.
Solution.

(6, 2)

(3, 1) (9, 1) (11, 1)

(1, 1)

(2, 0) (5, 0) (7, 0) (10, 0)

FIGURE 21.

The point (1, 1) rotates around (2, 0) to (3, 1), then around (5, 0) to (6, 2), then
around (7, 0) to (9, 1), then around (10, 0) to (11, 1). (See Figure 21.) The area
of concern consists of four 1 × 1 right triangles
√ of area 1/2, four 1 × 2 triangles of
area 1,√two quarter circles of area (π/4)( 2)2 = π/2, and two quarter circles of area
(π/4)( 5)2 = 5π/4, for a total area of 7π/2 + 6. 

A2. (150, 17, 2, 1, 0, 0, 0, 0, 3, 5, 15, 20)

Let A and B be different n × n matrices with real entries. If A3 = B3 and
A B = B2 A, can A2 + B2 be invertible?
2

Answer. No.
Solution. We have

(A2 + B2 )(A − B) = A3 − B3 − A2 B + B2 A = 0,

and A − B = 0, so A2 + B2 is not invertible. 

A3. (42, 35, 29, 0, 0, 0, 0, 0, 6, 5, 63, 33)

Find all real polynomials p(x) of degree n ≥ 2 for which there exist real
numbers r1 < r2 < · · · < rn such that
(i) p(ri ) = 0, i = 1, 2, . . . , n, and
" #
(ii) p ri +r2 i+1 = 0 i = 1, 2, . . . , n − 1,
where p (x) denotes the derivative of p(x).
136 The William Lowell Putnam Mathematical Competition

Answer. The real polynomials with the required property are exactly those that are
of degree 2 with 2 distinct real zeros.

Solution.
All degree 2 polynomials with 2 distinct real zeros work. If p(x) = ax2 + bx + c has
two real zeros r1 < r2 (i.e., b2 − 4ac > 0), then

p(x) = a(x − r1 )(x − r2 );

comparing coefficients of x, we get −b/a = r1 + r2 , from which

p (x) = 2a(x − (r1 + r2 )/2).

Geometrically, this is clearer: y = p(x) is a parabola, symmetric about some vertical

axis x = d, and p (d) = 0. The zeros x = r1 , x = r2 must also be symmetric about
the axis, so d = (r1 + r2 )/2.
No polynomial of higher degree works. Suppose r1 < · · · < rn are all real and n > 2,
so
p(x) = a(x − r1 )(x − r2 ) · · · (x − rn ).

Let r = (rn−1 + rn )/2.

From here, we can follow two (similar) approaches.
Approach 1 uses the following exercise: If p(x) is a degree n polynomial with zeros
r1 , . . . , rn , then


 1 1
p (x) = p(x) + ··· + (1)
x − r1 x − rn
/ {r1 , . . . , rn }. (This can be shown using the product rule for derivatives, or
for x ∈
d
more directly by computing dx ln p(x) in two ways. This useful equation also comes
up in 1992A2.) Since p(r) = 0,
p (r) 1 1 1 1
= + ··· + + +
p(r) r − r1 r − rn−2 r − rn−1 r − rn
1 1
= + ··· + (since r − rn = −(r − rn−1 ))
r − r1 r − rn−2
> 0,

so p (r) = 0.
Approach 2 is the following: let q(x) = (x − r1 ) · · · (x − rn−2 ) so

p(x) = (x − r1 )(x − r2 ) · q(x),

and apply the product rule to obtain

p (x) = a · 2(x − r)q(x) + a(x − rn−1 )(x − rn )q  (x).

Rolle’s Theorem (see remark below) implies that all the zeros of q  (x) lie between r1
and rn−2 . Hence (rn−1 + rn )/2 is not a zero of q  (x), so p(x) does not satisfy the
hypotheses of the problem. 
Remark. Rolle’s Theorem is the following:
Solutions: The Fifty-Second Competition (1991) 137

Rolle’s Theorem. Let [a, b] be a closed interval in R. Let f (t) be a function that
is continuous on [a, b] and differentiable on (a, b), and suppose that f (a) = f (b). Then
there exists c ∈ (a, b) such that f  (c) = 0.
In particular, if f (t) ∈ R[t] is a polynomial of degree n with n distinct real zeros,
then the real zeros of f  (t) are contained in the interval spanned by the zeros of f (t).
The previous sentence has a generalization to complex polynomials, known as Lucas’
Theorem, or sometimes as the Gauss-Lucas Theorem:
Lucas’ Theorem. If p(z) ∈ C[z] is a polynomial of degree at least 1, then the zeros
of p (z) are contained in the convex hull of the set of zeros of p(z).
See [Mar] for this and many related results, as well as the following open question:
Let p(z) ∈ C[z] be a polynomial whose complex zeros all lie in the disc
|z| ≤ 1, and let z1 be any one of the zeros. Must p (z) have a zero in the
disc |z − z1 | ≤ 1?
B. Sendov conjectured in 1962 that the answer is yes, but in the literature it is
sometimes called the “Ilyeff Conjecture,” because of a misattribution. The statement
has been proved for deg p ≤ 8 [BX].
Related question. Let n > 2, and let p(x) ∈ R[x] be a polynomial of degree n
with n distinct real zeros r1 < · · · < rn . Rolle’s Theorem implies that p (x) has n − 1
distinct real zeros, interlaced between the zeros of p(x). Prove that the largest real
zero of p (x) is closer to rn than to rn−1 , and that the smallest real zero of p (x) is
closer to r1 than to r2 .
See [An] for related ideas.

A4. (86, 33, 43, 0, 0, 0, 0, 0, 12, 3, 21, 15)

Does there exist an infinite sequence of closed discs D1 , D2 , D3 , . . . in the
plane, with centers c1 , c2 , c3 , . . . , respectively, such that
(i) the ci have no limit point in the finite plane,
(ii) the sum of the areas of the Di is finite, and
(iii) every line in the plane intersects at least one of the Di ?
Answer. Yes, such a sequence of closed discs exists.
Solution. Let ai = 1/i for i ≥ 1 (or choose any other sequence of positive numbers
∞ ∞
ai satisfying i=1 ai = ∞ and i=1 a2i < ∞). For n ≥ 1, let An = a1 + a2 + · · · + an .
Let U be the union of the discs of radius an centered at (An , 0), (−An , 0), (0, An ),
(0, −An ), for all n ≥ 1. Then U covers the two coordinate axes, and has finite total
area. Every line in the plane meets at least one axis, and hence meets U . Finally,
the centers have no limit point, since every circle C centered at the origin encloses at
most finitely many centers: if C has radius R, we can choose n such that An > R,
and then less than 4n centers lie inside C. 
Motivation. The essential idea is as follows. This construction covers a one-
dimensional set (the union of the two axes). Any line must meet this set, and it is pos-
sible to cover it with an infinite number of circles of finite area. Then a little care must
be taken to make sure the centers do not have a limit point. Another one-dimensional
set that would work is the union of the circles of integer radius centered at the origin.
138 The William Lowell Putnam Mathematical Competition

Related question. Gabriel’s Horn is the surface of revolution G obtained by

revolving the graph of y = 1/x, x ≥ 1 around the x-axis. Prove that G encloses a
finite volume but has infinite surface area. (This leads to the following fun “paradox”:
one cannot paint the inside of the horn because it has infinite surface area, yet one
can paint it by filling it with a finite amount of paint and then emptying it!)

A5. (23, 4, 5, 0, 0, 0, 0, 0, 3, 6, 82, 90)

Find the maximum value of
 y
x4 + (y − y 2 )2 dx
0
for 0 ≤ y ≤ 1.
Answer. The maximum value of the integral is 1/3.
Motivation. To find the maximum of I(y), one naturally looks at I  (y); if it is
never 0, then the maximum must occur at an endpoint of the interval in question.
y
Solution 1. For 0 ≤ y ≤ 1 let I(y) = 0 x4 + (y − y 2 )2 dx.
1
Note that I(1) = 0 x2 dx = 1/3. We will prove I  (y) > 0 for 0 < y < 1. Since I(y)
is continuous on [0, 1], this will prove that 1/3 is the maximum.
By the remark following this solution,
  y
1
I  (y) = y 4 + (y − y 2 )2 + (y − y 2 )(1 − 2y)  dx,
0 x + (y − y 2 )2
4

so we must check that

  y
1
2y 2 − 2y + 1 > (1 − y)(2y − 1)  dx. (1)
0 x4 + (y − y 2 )2
If y < 1/2, (1) holds because the right side is negative. If y ≥ 1/2, (1) would follow
from
  y
1
2y 2 − 2y + 1 > (1 − y)(2y − 1)  dx
0 (y − y 2 )2
= (1 − y)(2y − 1)y/(y − y 2 )
= 2y − 1,
which is equivalent to 2y > 2y 2 , hence true, since 0 < y < 1. 
Remark. We discuss how to differentiate certain integrals depending on a
parameter. Fix M > 0, and define the triangle
T = { (x, y) ∈ R2 : 0 < y < M, 0 ≤ x ≤ y }.
Suppose that f (x, y) is a continuous function on T such that the partial derivative fy
(alternative notation for ∂f /∂y) exists on theinterior of T and extends to a continuous
y
function on T . For 0 < y < M , let I(y) = 0 f (x, y) dx. We wish to compute I  (y)
for 0 < y < M .

 tWe will express I (y) in terms of partial derivatives of the auxiliary function J(t, u) =
0
f (x, u) dx defined for (t, u) ∈ T . By the Fundamental Theorem of Calculus,
Jt (t, u) = f (t, u) on T . (If t = 0 or t = u, we consider Jt as only a one-sided derivative.)
Solutions: The Fifty-Second Competition (1991) 139

Since fy extends to a continuous

t function on T , we may differentiate under the integral
sign to obtain Ju (t, u) = 0 fy (x, u) dx on T . Then by the Multivariable Chain Rule
[Ru, p. 214],
dJ(y, y)
I  (y) =
dy
∂y ∂y
= Jt (y, y) + Ju (y, y)
∂y ∂y
 y
= f (y, y) + fy (x, y) dx.
0

The same method can be used to differentiate integrals of the form

 h(y)
I(y) = f (x, y) dx.
g(y)

√ √
Solution 2. If u, v ≥ 0, then u2 + v 2 ≤ u2 + 2uv + v 2 = u + v. Taking u = x2
and v = y − y 2 we obtain
 y  y
 2  
x4 + (y − y 2 )2 dx ≤ x + y − y 2 dx
0 0
2
= y − y3
2
3
1
≤ ,
3
with
 2equality
 everywhere if y = 1: the last inequality uses the positivity of
d
dy y − 2 3
3 y = 2y(1 − y) on (0, 1). Thus the maximum value is 1/3. 

A6. (8, 21, 8, 1, 0, 0, 0, 0, 6, 7, 40, 122)

Let A(n) denote the number of sums of positive integers a1 + a2 + · · · + ar
which add up to n with a1 > a2 + a3 , a2 > a3 + a4 , . . . , ar−2 > ar−1 + ar ,
ar−1 > ar . Let B(n) denote the number of b1 + b2 + · · · + bs which add up to
n, with
(i) b1 ≥ b2 ≥ · · · ≥ bs ,
(ii) each bi is in the sequence 1, 2, 4, . . . , gj , . . . defined by g1 = 1, g2 = 2, and
gj = gj−1 + gj−2 + 1, and
(iii) if b1 = gk then every element in {1, 2, 4, . . . , gk } appears at least once as
a bi .
Prove that A(n) = B(n) for each n ≥ 1.
(For example, A(7) = 5 because the relevant sums are 7, 6 + 1, 5 + 2, 4 + 3,
4 + 2 + 1, and B(7) = 5 because the relevant sums are 4 + 2 + 1, 2 + 2 + 2 + 1,
2 + 2 + 1 + 1 + 1, 2 + 1 + 1 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1 + 1 + 1.)

Solution 1. First, given a sum counted by A(n), we construct a “tableau” of

Fibonacci numbers. Start with two rows, the lower row with ar ones and the upper
140 The William Lowell Putnam Mathematical Competition

with ar−1 ones, as shown below.

ar−1 : 1 1 1 1 1 1 1
ar : 1 1 1 1 1
The top row is longer than the bottom row since ar−1 > ar .
Since ar−2 > ar−1 + ar , we may add a new row on top, such that each new entry
directly above two entries is the sum of those two, each entry above a single entry is
equal to that entry, and the rest are enough ones (at least one) to make the sum of
the new row equal ar−2 .
ar−2 : 2 2 2 2 2 1 1 1 1 1
ar−1 : 1 1 1 1 1 1 1
ar : 1 1 1 1 1
Next, ar−3 > ar−2 + ar−1 , so we can add another row on top, such that each entry
above at least two entries is the sum of the two below it, each entry above one entry
is equal to that entry, and the rest are enough ones (at least one) to make the row
sum ar−3 , as shown below.
ar−3 : 3 3 3 3 3 2 2 1 1 1 1
ar−2 : 2 2 2 2 2 1 1 1 1 1
ar−1 : 1 1 1 1 1 1 1
ar : 1 1 1 1 1
Continue this construction to make a tableau of r rows, in which the total of all
entries is n. Let s be the number of columns, and let bi denote the sum of the ith
column. Then bi = F1 + F2 + · · · + Fj , where j is the length of the ith column. The
sequence gj = F1 + F2 + · · · + Fj satisfies g1 = 1 = g1 , g2 = 2 = g2 and for j ≥ 3,
 
gj−1 + gj−2 + 1 = F1 + F2 + · · · + Fj−1
+ F1 + · · · + Fj−2 + 1
= F2 + F3 + · · · + Fj + F1 (since F1 = F2 = 1)
= gj ,
so gj = gj for all j. Thus the bi satisfy condition (i) in the definition of B(n). The
sum of the bi equals the sum of all entries of the tableau, namely n. Each row is at
least one longer than the next row, so each column is either equal in length or one
longer than the next column. Thus the bi satisfy conditions (ii) and (iii) as well.
Conversely, given a sum b1 + · · · + bs counted by B(n), construct a (top and left
justified) tableau in which the entries of the ith column from the top down are Fk ,
Fk−1 , . . . , F1 , where k is the positive integer such that bi = gk ; then let ai denote the
sum of the ith row.
It is straightforward to check that these constructions define inverse bijections
between the partitions counted by A(n) and those counted by B(n). Thus A(n) =
B(n) for all n. 
Solution 2. Let A(n, r) denote the number of sums counted by A(n) with exactly
r terms. Let B(n, r) denote the number of sums counted by B(n) in which b1 = gr .
Solutions: The Fifty-Second Competition (1991) 141

Fix r ≥ 1. Form the r × r matrix

 
1 −1 −1 0 · · · 0
0 1 −1 −1 · · · 0
 
 
0 0 1 −1 · · · 0
M = (mij ) = 
0 0 0 1 ··· 0

 
 .. .. .. .. .. .. 
. . . . . .
0 0 0 0 ··· 1
with 

 if j − i = 0,
1
mij = −1 if j − i = 1 or j − i = 2,


0 otherwise.

Let a and c denote column vectors (a1 , a2 , . . . , ar ) and (c1 , c2 , . . . , cr ), respectively.

Since M is an integer matrix with determinant 1, the relation M a = c is a bijection
between the set of a in Zr and the set of c in Zr . Under this bijection, the inequalities
a1 > a2 + a3 , . . . , ar−2 > ar−1 + ar , ar−1 > ar , ar > 0 correspond to c1 > 0, . . . ,
cr−2 > 0, cr−1 > 0, cr > 0. Also, applying the identity
   
g1 g2 · · · gr M = 1 1 · · · 1
 
to a shows that the condition ai = n corresponds to gi ci = n. Thus A(n, r),
which counts the number of a ∈ Zr satisfying the inequalities and sum condition,

equals the number of c ∈ Zr satisfying ci > 0 for all i and gi ci = n. Such a c may
be matched with the sequence b1 , b2 , . . . , bs consisting of cr copies of gr , . . . , and c1
copies of g1 ; the b-sequences arising in this way are exactly the sequences summing to
n that satisfy conditions (i)–(iii) of the problem and b1 = gr . Hence the number of
such c equals B(n, r).
Thus A(n, r) = B(n, r). Summing over r yields A(n) = B(n). 
Related question. The most famous result along these lines is the following theorem
of Euler [NZM, Theorem 10.2]:
Let O(n) be the number of (nonincreasing) partitions of a positive integer n into
odd parts, and let D(n) be the number of (nonincreasing) partitions of n into different
parts. Then O(n) = D(n) for all n.

B1. (192, 6, 2, 0, 6, 0, 0, 0, 0, 5, 0, 2)
For each integer n ≥ 0, let S(n) = n − m2 , where m is the greatest integer
with m2 ≤ n. Define a sequence (ak )∞ k=0 by a0 = A and ak+1 = ak + S(ak ) for
k ≥ 0. For what positive integers A is this sequence eventually constant?
Answer. This sequence is eventually constant if and only if A is a perfect square.

Solution. If ak is a perfect square, then ak+1 = ak , and the sequence is constant

thereafter.
Conversely, if ak is not a perfect square, then suppose r2 < ak < (r + 1)2 . Then
S(ak ) = ak − r2 is in the interval [1, 2r], so ak+1 = r2 + 2S(ak ) is greater than r2 but
less than (r + 2)2 , and not equal to (r + 1)2 by parity. Thus ak+1 is also not a perfect
142 The William Lowell Putnam Mathematical Competition

square, and is greater than ak . Hence if A is not a perfect square, then no ak is a

perfect square, and the sequence diverges to infinity. 

B2. (93, 30, 8, 0, 0, 0, 0, 0, 7, 1, 57, 17)

Suppose f and g are nonconstant, differentiable, real-valued functions on
R. Furthermore, suppose that for each pair of real numbers x and y,

f (x + y) = f (x)f (y) − g(x)g(y),

g(x + y) = f (x)g(y) + g(x)f (y).

If f  (0) = 0, prove that (f (x))2 + (g(x))2 = 1 for all x.

Motivation. We can use calculus to reach the intermediate goal of proving that
the H(x) = f (x)2 + g(x)2 is constant. In other words, we try to prove H  (x) = 0 by
differentiating the given functional equations. This motivates the first solution.
Alternatively, the hypotheses and conclusion in the problem are recognizable as
identities satisfied by f (x) = cos x and g(x) = sin x. If f (x) were cos x and g(x) were
sin x, then h(x) = f (x) + ig(x) would satisfy h(x + y) = h(x)h(y). This suggests the
approach of the second solution.
Solution 1. Differentiate both sides with respect to y to obtain

f  (x + y) = f (x)f  (y) − g(x)g  (y),

g  (x + y) = f (x)g  (y) + g(x)f  (y).

Setting y = 0 yields

f  (x) = −g  (0)g(x) and g  (x) = g  (0)f (x).

Thus
2f (x)f  (x) + 2g(x)g  (x) = 0,

and therefore
f (x)2 + g(x)2 = C

for some constant C. Since f and g are nonconstant, C = 0. The identity

  
f (x + y)2 + g(x + y)2 = f (x)2 + g(x)2 f (y)2 + g(y)2 ,

implies C = C 2 . But C = 0, so C = 1. 
Solution 2. Define h : R → C by h(x) = f (x) + ig(x). Then h is differentiable,
and h (0) = bi for some b ∈ R. The two given functional equations imply h(x + y) =
h(x)h(y). Differentiating with respect to y and substituting y = 0 yields h (x) =
h(x)h (0) = bi · h(x), so h(x) = Cebix for some C ∈ C. From h(0 + 0) = h(0)h(0) we
obtain C = C 2 . If C = 0, then h would be identically zero, and f and g would be
constant, contradiction. Thus C = 1. Finally, for any x ∈ R,
f (x)2 + g(x)2 = |h(x)|2 = |ebix |2 = 1. 

Remark. It follows that f (x) = cos(bx) and g(x) = sin(bx) for some nonzero b ∈ R.
Solutions: The Fifty-Second Competition (1991) 143

Remark. Solution 1 is very close to what one gets if one writes out Solution 2 in
terms of real and imaginary parts.
Remark. Suppose we change the problem by dropping the assumption that f
and g are differentiable, and assume only that f and g are continuous, and that
f  (0) exists and is zero. Then we could still conclude f (x)2 + g(x)2 = 1 by following
Solution 2, because it is true that any continuous function h : R → C satisfying
h(x + y) = h(x)h(y) is either identically zero, or of the form h(x) = ebx for some
b ∈ C. This is a consequence of the following lemma, sometimes attributed to Cauchy.
Lemma. Suppose f : R → R is a continuous function such that
f (x + y) = f (x) + f (y). (1)
Then f (x) = cx for some c ∈ R.
Proof. By substituting x = y = 0 into (1), we find f (0) = 0. Let c = f (1). It is not
hard to show that f (x) = cx for rational x (by using (1) repeatedly). Since f (x) − cx
is a continuous function that vanishes on a dense set (the rationals), it must be 0. 
Reinterpretation. Solution 2 can be understood in a more sophisticated context.
If h is not identically zero, h : R → C∗ is a continuous homomorphism between real
Lie groups, so by [War, Theorem 3.38], it is a homomorphism of Lie groups. (In
other words, it is automatically analytic.) Since homomorphisms of Lie groups are
determined by their induced Lie algebra homomorphisms [War, Theorem 3.16], h must
be identically zero, or of the form h(x) = ebx for some b ∈ C.

B3. (38, 11, 4, 0, 0, 0, 0, 0, 0, 5, 7, 49, 99)

Does there exist a real number L such that, if m and n are integers
greater than L, then an m × n rectangle may be expressed as a union of
4 × 6 and 5 × 7 rectangles, any two of which intersect at most along their
boundaries?
Answer. Yes, such a real number L exists.
The solution will use a generalization of the following well-known result:
Theorem 1. If a and b are positive integers, then there exists a number g such that
every multiple of gcd(a, b) greater than g may be written in the form ra + sb, where r
and s are nonnegative integers.
This is sometimes called the “Postage Stamp Theorem” because if gcd(a, b) = 1,
then every amount of postage greater than g cents can be paid for with a-cent and
b-cent stamps. In this case, g may be taken to be ab − a − b, but no smaller: ab − a − b
is not of the form ra + sb with r, s ≥ 0. For further discussion, see [NZM, Section 5.1];
the theorem appears as Problem 16.
Proof. Suppose first that gcd(a, b) = 1. Then 0, a, 2a, . . . , (b − 1)a is a complete
set of residues modulo b. Thus, for any integer k greater than (b − 1)a − 1, k − qb = ja
for some q ≥ 0, j = 0, 1, 2, . . . , b − 1, hence the claim for this special case.
For general a and b, write a = da0 and b = db0 , where d = gcd(a, b) and
gcd(a0 , b0 ) = 1. We showed that all sufficiently positive integers are expressible as
ra0 + sb0 . Multiplying by d, we find that all sufficiently positive multiples of d are
expressible as ra + sb. 
144 The William Lowell Putnam Mathematical Competition

Solution. We begin by forming 20 × 6 and 20 × 7 rectangles. From Theorem 1,

we may form 20 × n rectangles for n sufficiently large. We may also form 35 × 5 and
35 × 7 rectangles, hence 35 × n rectangles for n sufficiently large. We may further form
42 × 4 and 42 × 5 rectangles, hence 42 × n rectangles for n sufficiently large.
Since gcd(20, 35) = 5, there exists a multiple m0 of 5, relatively prime to 42
and independent of n, for which we may form an m0 × n rectangle. Finally, since
gcd(m0 , 42) = 1, we may use m0 × n and 42 × n rectangles to form m × n rectangles
for all m and n sufficiently large. 
Related question. By using the explicit value of g in the discussion after the
statement of the Theorem 1, this approach shows that one can construct an m × n
rectangle if m ≥ 41 · 54 = 2214 and n ≥ 30. Hence one can take L = 2213 in the
original problem. Dave Savitt has shown that the conditions m ≥ 65 and n ≥ 80
suffice. Hence one can take L = 79. How much better can one do? We do not know
the smallest possible L.
Related question. Problem 1971A5 [PutnamII, p. 15] is related. See also [Hon2,
Ch. 13]:
A game of solitaire is played as follows. After each play, according to the
outcome, the player receives either a or b points (a and b are positive integers
with a greater than b), and his score accumulates from play to play. It has
been noticed that there are thirty-five non-attainable scores and that one of
these is 58. Find a and b.
Remark. Let a1 , . . . , an be positive integers with greatest common divisor 1.
The problem of determining the greatest integer g = g(a1 , . . . , an ) not expressible
n
as i=1 ri ai with ri nonnegative integers is called the Frobenius Problem, because
according to A. Brauer [Br], Frobenius mentioned it in his lectures. There is an exten-
sive literature on the problem, dating back at least to the nineteenth century [Shar].
For each fixed n the problem can be solved in polynomial time, but the problem as a
whole is NP-hard [Ka]. For the solution in some special cases, see [Sel]. The Frobenius
Problem is relevant to the running time analysis of the sorting algorithm Shellsort [P1].
The reader may enjoy proving that g exists (say by induction on n).
Related question. Another generalization of Theorem 1 is Problem 3 from the
1983 International Mathematical Olympiad [IMO79–85, p. 6] In the language of the
preceding remark, it asks for a proof that g(bc, ca, ab) = 2abc − ab − bc − ca, where a,
b, and c be positive integers, no two of which have a common divisor greater than 1.
Remark. Suppose that we insist that the length 4 side of each 4 × 6 rectangle be
parallel to the length m side of the big m × n rectangle, and suppose we similarly
restrict the orientation of the 5 × 7 rectangles. Then the answer to the problem
becomes NO!
Let us explain why. Label the 1 × 1 squares in the m × n rectangle in the obvious
way with (i, j) where i, j are integers satisfying 0 ≤ i ≤ m−1 and 0 ≤ j ≤ n−1. Write
the monomial xi y j in the square (i, j). Then the sum of the monomials inside an a × b
rectangle equals xi y j (1+x+· · ·+xa−1 )(1+y+· · ·+y b−1 ) for some nonnegative integers
i, j. If the m×n rectangle is a disjoint union of 4×6 and 5×7 rectangles (disjoint except
at boundaries), then the polynomial f (x, y) = (1 + x + · · · + xm−1 )(1 + y + · · · + y n−1 )
Solutions: The Fifty-Second Competition (1991) 145

is in the ideal of C[x, y] generated by (1 + x + · · · + x4−1 )(1 + y + · · · + y 6−1 ) and

(1 + x + · · · + x5−1 )(1 + y + · · · + y 7−1 ). In particular,
(xm − 1)(y n − 1)
f (x, y) =
(x − 1)(y − 1)

must vanish at (e2πi/4 , e2πi/7 ). But if 4 does not divide m and 7 does not divide
n, then f (e2πi/4 , e2πi/7 ) is nonzero and hence it is impossible to dissect the m × n
rectangle in the specified way.
This argument is related to a beautiful proof [Wag, Proof 1] of the following
remarkable fact.
Theorem 2. If a rectangle is tiled by rectangles each of which has at least one
integer side, then the tiled rectangle has at least one integer side.
Remark. Theorem 1 also comes up in other advanced contexts. For example, it
corresponds to the fact that two measures of the “nonsmoothness” of the complex
singularity given by y p = xq in C2 are the same [ACGH, p. 60].

B4. (21, 1, 7, 0, 0, 0, 0, 0, 23, 1, 37, 123)

Suppose p is an odd prime. Prove that
p


p p+j
≡ 2p + 1 (mod p2 ).
j=0
j j

p  p+j 
p
Solution 1. The sum j=0 j j is the coefficient of xp in
 
p
 p

p p
(1 + x)p+j = (1 + x)j  (1 + x)p
j=0
j j=0
j
 p
= (1 + x) + 1 (1 + x)p

= (2 + x)p (1 + x)p ,

so this coefficient equals

p


p p
2k .
k p−k
k=0
p
But p divides k for k = 0. Thus,
p






p p+j p p 0 p p p
≡ 2 + 2
j=0
j p 0 p p 0

≡ 1 + 2p (mod p2 ). 

Remark. If at each step of Solution 1, one writes the coefficient of xp explicitly,

one obtains a solution that does not use generating functions, but instead uses
Vandermonde’s identity (mentioned in 1987B2) at the first equality sign.
146 The William Lowell Putnam Mathematical Competition

Solution 2. Modulo p, the sets {1, 2, . . . , p − 1} and {1−1 , 2−1 , . . . , (p − 1)−1 } are
the same (where the inverses are modulo p). Thus


2p p+1 p+2 p + (p − 1) 2p
= · ··· ·
p 1 2 p−1 p
≡ (1−1 p + 1)(2−1 p + 1) · · · ((p − 1)−1 p + 1) · 2
≡ 2(1p + 1)(2p + 1) · · · ((p − 1)p + 1)
44p−1 5 5
≡2 k p+1
k=1

≡2 (mod p2 ).

Although 1−1 , . . . , (p − 1)−1 are only defined modulo p, they appear in the equations
above multiplied by p, so the terms make sense modulo p2 .
p

 p−1


p p+j p p+1p+2 p+j 2p
=1+ ··· +
j=0
j j j=1
j 1 2 j p
p−1

p
≡1+ (1−1 p + 1)(2−1 p + 1) · · · (j −1 p + 1) + 2
j=1
j
p−1

p  
≡1+ +2 (since p | pj for 1 ≤ j ≤ p − 1)
j=1
j
p

p
≡ +1
j=0
j
≡ 2p + 1 (mod p2 ). 
2p
Remark. Here is another way to prove p ≡ 2 (mod p2 ) for odd primes p. We
have


2p (p + 1)(p + 2) · · · (p + p − 1)
=2 (1)
p 1 · 2 · · · (p − 1)
and
+
(p−1)/2
(x + 1)(x + 2) · · · (x + p − 1) = (x + j)(x + p − j)
j=1

+
(p−1)/2
= (x2 + px + j(p − j))
j=1

= (p − 1)! + λpx + x2 f (x),

for some λ ∈ Z and f (x) ∈ Z[x]. Substituting x = p yields

(p + 1)(p + 2) · · · (p + p − 1) ≡ (p − 1)! (mod p2 ),

 
which we substitute into (1). Since (p−1)! is prime to p, we deduce 2p p ≡ 2 (mod p ).
2
Solutions: The Fifty-Second Competition (1991) 147
 
Remark. In fact, if p > 3, then 2p p ≡ 2 (mod p3 ); this is Wolstenholme’s
Theorem. Another version of this theorem is that if p is a prime greater than 3,
then the numerator of 1 + 12 + · · · + p−1
1
is divisible by p2 . (See [Bak1, p. 26] or the
remarks following 1997B3.) It is not hard to derive one version from the other; see
below.
Remark. Wolstenholme’s Theorem can be generalized. For example, for p ≥ 5,



2p 2
≡ (mod p3 )
p 1

2

2p 2p
≡ (mod p6 ),
p2 p

3
2
2p 2p
3
≡ (mod p9 ),
p p2
and so on. Like Wolstenholme’s Theorem, these assertions can be recast in terms of
a sum of reciprocals. Namely, write

n + pn
n 
2p p +i
=
pn i=1
i

and divide up the terms into those with i divisible by p and those with i not divisible
by p. This gives
 n−1   n−1 

n p+ n p+ p−1
+ pn + pj + i
2p p + pj
=  .
pn j=1
pj j=1 i=1
pj + i
2pn−1 
The term in the first parentheses is precisely pn−1 . Therefore
2pn 
+

n−1
p+ p−1
pn pn
2pn−1  = 1+
pj + i
pn−1 j=1 i=1
pn−1 p−1
1 1
≡1+p n
+ p2n (mod p3n ).
pj + i (pj + i)(pj  + i )
j=1 i=1 pj+i=pj  +i

The last sum is divisible by pn . Thus the assertions follow from the congruence
pn−1 p−1
1
≡0 (mod p2n ),
j=1 i=1
pj + i

which is discussed in remarks following 1997B3.

Remark. Congruences between binomial coefficients can be proved using analytic
properties of Morita’s p-adic gamma function Γp (x). See [Robe, p. 380] for such a
proof of the following theorem of Kazandzidis, which generalizes the congruences of
the previous remark: for all primes p ≥ 5 we have



pn n
≡ (mod pm ),
pk k
148 The William Lowell Putnam Mathematical Competition

 
where pm is the largest power of p dividing p3 nk(n − k) nk . The same congruence
holds for p = 3, but with p3 replaced by 32 . (In other words, one loses a factor of 3.)

B5. (38, 4, 3, 0, 3, 0, 1, 0, 9, 3, 50, 102)

Let p be an odd prime and let Zp denote (the field of ) integers modulo
p.
How many elements are in the set

{ x2 : x ∈ Zp } ∩ { y 2 + 1 : y ∈ Zp }?

Warning. In current mathematics, especially in number theory, the notation Zp is

usually reserved for the ring of p-adic integers. If one intends the additive group of
integers modulo p, it is safer to write Z/pZ or its abbreviation Z/p, or Fp if, as in this
problem, one wishes to consider the set as a field. More generally, for any prime power
q = pm with p prime and m ≥ 1, there is a unique finite field with q elements up to
isomorphism [Se1, Theorem 1], and it is denoted Fq , or GF(q) in older literature.
Answer. The number of elements in the intersection is p/4 = (p+3)/4. In other
words, the answer is
p+3 p+1
if p ≡ 1 (mod 4), and if p ≡ 3 (mod 4).
4 4

Solution 1. Let S be the set of solutions to x2 = y 2 + 1 over
Fp . The

 linear
1 1
change of coordinates (u, v) = (x + y, x − y) is invertible since det = −2 is
1 −1
nonzero in Fp . Hence |S| equals the number of solutions to uv = 1 over Fp . There is
one possible v for each nonzero u, and no v for u = 0, so |S| = p − 1.
The problem asks for the size of the image of the map φ : S → Fp taking (x, y) to
x2 . If z = x2 for some (x, y) ∈ S, then φ−1 (z) = {(±x, ±y)}, which has size 4, except
in the cases z = 1 (in which case x = ±1 and y = 0, making φ−1 (z) of size 2) and
z = 0 (in which case x = 0 and y 2 = −1, again making φ−1 (z) of size 2); the latter
exception occurs if and only if −1 is a square in Fp . Hence |S| = 4|φ(S)| − 2 − 2c,
where c is 1 or 0 according as −1 is a square in Fp or not. Thus the answer to the
problem is
(p − 1) + 2 + 2c p + 1 + 2c
|φ(S)| = = .
4 4
This should be an integer, so c = 1 if p ≡ 1 (mod 4) and c = 0 if p ≡ 3 (mod 4). In
either case, |φ(S)| = p/4 as claimed. 
Remark. Alternatively, one can count the solutions to x2 = y 2 + 1 over Fp by
explicit parametrization:
r + r−1 r − r−1
x= , y=
2 2
for r ∈ F∗p . (Given a solution (x, y), take r = x + y, so that r−1 = x − y.) This
parametrization shows that the conic x2 = y 2 + 1 is birationally equivalent to the line
over Fp ; this is not surprising in light of the remark after Solution 2 to 1987B3.
Solutions: The Fifty-Second Competition (1991) 149

Solution 2. We use the Legendre symbol, defined by



  
1, if a ≡ k 2 (mod p) for some k ≡ 0 (mod p)
a
= 0, if a ≡ 0 (mod p)
p 

−1, otherwise.

The number of nonzero squares in Fp equals the number of nonsquares, so

p−1

a−k
=0 (1)
a=0
p

for any k ∈ Z.
In this solution only, if P is a statement, let [P ] be 1 if P is true, and 0 if P is false.
Let F2p denote the set of squares in Fp , including 0. The problem asks us to compute
p−1
N= [a ∈ F2p ] · [a − 1 ∈ F2p ].
a=0

Substituting the identity

 
1 a
[a ∈ F2p ] = 1+ + [a = 0]
2 p
we obtain
4

 p−1



5
1 −1 1 a a−1
N= 1+ +1+ + 1+ 1+
4 p p a=0
p p
7
 8 p−1





1 −1 1 1 a a−1 a a−1
= =1 + + 1+ + +
2 p 2 4 a=0 p p p p
7
 8 p−1


1 −1 1 p 1 a a−1
= =1 + + + ,
2 p 2 4 4 a=0 p p
p−1 " # " a−k #
by (1) twice. For k ∈ Z, let S(k) = a=0 ap p . We want S(1). For k not
divisible by p, the substitution a = kb, with b running over the residue classes modulo
p, shows that
p−1




k b k b−1
S(k) =
p p p p
b=0
p−1

 " #2
b b−1 k
= (since p = 1)
p p
b=0

= S(1).

Also,
p−1 p−1
 p−1

a a−k
S(k) = = 0,
a=0
p p
k=0 k=0
150 The William Lowell Putnam Mathematical Competition

since each inner sum is zero by (1). Thus

S(0) p−1
S(1) = − =− = −1.
p−1 p−1
Hence 7

8 
1 −1
p+1
N= =1 + .
2 4p
" #
But N is an integer. Thus if p ≡ 1 (mod 4), then −1 = 1 and N = (p + 3)/4; if
" # p

p ≡ 3 (mod 4), then −1p = −1 and N = (p + 1)/4. 

Remark. In each solution, we proved:
Corollary. The number −1 is a square modulo an odd prime p if and only if p ≡ 1
(mod 4).
From this and unique factorization in the ring of Gaussian integers
Z[i] = { a + bi : a, b ∈ Z },
one can prove:
Theorem. Every prime p ≡ 1 (mod 4) is a sum of two squares.
Hint: There is some x such that x2 ≡ −1 (mod p), i.e., there is some integer k such
that kp = (x − i)(x + i); consider prime factorizations of both sides, and note that p
is not a factor of x ± i.
A remarkable one-sentence proof of the theorem, due to D. Zagier [Z], is the
following:
The involution on the finite set S = { (x, y, z) ∈ Z≥0 : x2 + 4yz = p } defined by

 (x + 2z, z, y − x − z) if x < y − z
(x, y, z) !→ (2y − x, y, x − y + z) if y − z < x < 2y

(x − 2y, x − y + z, y) if x > 2y
has exactly one fixed point (1, 1, (p − 1)/4), so #S is odd and the involution defined
by (x, y, z) !→ (x, z, y) also has a fixed point.
Using the corollary and theorem above, one can prove the well-known theorem that
a positive integer can be expressed as a sum of two squares if and only if every prime
congruent to 3 modulo 4 appears with even power in the factorization of n. One can
even count the number of ways to express a positive integer as a sum of two squares
[IR, pp. 278–280].
Remark (Gauss and Jacobi sums). In Solution 1 we counted the number of solutions
to x2 = y 2 + 1 in Fp . Much more generally, given a1 , . . . , ar ∈ F∗p , @1 , . . . , @r ≥ 1, and
b ∈ F∗p one can express the number of solutions over Fp to
a1 x!11 + a2 x!22 + · · · + ar x!rr = b (2)
in terms of Jacobi sums, which we will define here. First, a multiplicative character
on Fp is a homomorphism χ : F∗p → C∗ , extended to a function on Fp by defining
χ(0) = 0. If χ1 , . . . , χ! are multiplicative characters, the Jacobi sum is defined by
J(χ1 , . . . , χ! ) = χ1 (t1 ) χ2 (t2 ) · · · χ! (t! ),
t1 +···+t =1
Solutions: The Fifty-Second Competition (1991) 151

where the sum is taken over all @-tuples (t1 , . . . , t! ) with the ti in Fp summing
to 1. Jacobi sums are related to Gauss sums, which are sums of the form ga (χ) =
p−1
t=0 χ(t)ζ
at
where χ is a multiplicative character on Fp , ζ = e2πi/p , and a ∈ Z. See
Chapters 6 and 8 of [IR] for more details. See also [Lan2, Chapter IV,§3] for more
general Gauss sums, over Z/nZ for n not necessarily prime, and over finite fields not
necessarily of prime order. Theorem 5 in Chapter 8 of [IR] gives the formula for the
number of solutions to (2) in terms of Jacobi sums.
Remark (the Weil Conjectures). The formula just mentioned led Weil [Weil2] to
formulate his famous conjectures (now proven) about the number of points on varieties
over finite fields. He proved the conjectures himself in the case of curves [Weil1]. Their
proof for arbitrary algebraic varieties uses ideas of Grothendieck and Deligne, among
others. For a survey, see [Har, Appendix C]. See Proof 3 of the stronger result following
1998B6 for one application of the conjectures.

B6. (2, 0, 0, 0, 1, 0, 1, 2, 0, 4, 30, 173)

Let a and b be positive numbers. Find the largest number c, in terms of
a and b, such that
sinh ux sinh u(1 − x)
ax b1−x ≤ a +b
sinh u sinh u
for all u with 0 < |u| ≤ c and for all x, 0 < x < 1. (Note: sinh u = (eu −e−u )/2.)
Answer. The largest c for which the inequality holds for 0 < |u| ≤ c is c = | ln(a/b)|.
Motivation. The key idea here is to strip away the inessential details. First, since
the inequality is even in u, we may assume u > 0. Also, we may assume a ≥ b without
loss of generality. A second step is to realize that the substitution v = eu will remove
the exponentials, leaving
v x − v −x v 1−x − v −(1−x)
ax b1−x ≤ a −1
+b .
v−v v − v −1
Multiplying by v − v −1 (which is positive) yields

ax b1−x v − ax b1−x v −1 ≤ av x − av −x + bv 1−x − bv −1+x .

Since this equation is homogeneous in {a, b}, we can divide by b and set r = a/b,
yielding
rx v − rx v −1 ≤ rv x − rv −x + v 1−x − v −1+x .

At this point, some inspiration is necessary. Notice that the terms involving x are
of the form r±x or v ±x . This suggests the substitution v = r±1 ; since v, r ≥ 1 we
substitute v = r. Then after this substitution, the inequality becomes 0 ≥ 0, i.e.,
equality holds for all x! This strongly suggests that c = ln r, and that a good strategy
would be to check that the inequality only “gets better” as |u| decreases from c, and
“gets worse” as |u| increases from c.
This intuition then leads directly to the following argument.

Solution 1. We will show that c = | ln(a/b)| by proving that the inequality is

satisfied if and only if 0 < |u| ≤ | ln(a/b)|. The right-hand side is an even function of
152 The William Lowell Putnam Mathematical Competition

u; hence it suffices to consider u > 0. Replacing x by 1 − x and interchanging a and

b preserves the inequality, so we may assume a ≥ b.
We will show that for each x ∈ (0, 1), the function

sinh ux sinh u(1 − x)

F (u) = a +b − ax b1−x
sinh u sinh u

of u is decreasing for u > 0. For this, it suffices to show that for each α ∈ (0, 1), the
function
sinh αu
f (u) =
sinh u

is decreasing for u > 0, since F (u) is a constant plus a sum of positive multiples of
two such functions f . Differentiating with respect to u yields

α cosh(αu) sinh u − sinh(αu) cosh u

f  (u) = ,
sinh2 u

which for u > 0 is negative if and only if the value of the function

g(u) = α tanh u − tanh(αu)

is negative. The negativity of g(u) for u > 0 follows from g(0) = 0 and the negativity
of
α α
g  (u) = − ,
cosh u cosh2 αu
2

which in turn follows since cosh is increasing on R+ . Thus F (u) is decreasing for
u > 0.
If a > b then F (u) is zero at u = ln(a/b). If a = b then limu→0+ F (u) = 0 by
L’Hôpital’s Rule [Spv, Ch. 11, Theorem 9]. Since F (u) is decreasing for u > 0, in both
cases we have F (u) ≥ 0 for 0 < u ≤ ln(a/b) and F (u) < 0 for u > ln(a/b), as desired. 

Solution 2. As before, we may assume u > 0 and a ≥ b. Taking ln of each side

and rearranging, we find that the inequality is equivalent to

x(ln(a sinh u)) + (1 − x)(ln(b sinh u)) ≤ ln(a sinh ux + b sinh u(1 − x)),

which, if we define
f (x) = a sinh ux + b sinh u(1 − x),

can be written as
x ln f (1) + (1 − x) ln f (0) ≤ ln f (x).

This will hold if u is such that ln f (x) is concave for x ∈ (0, 1), and will fail if ln f (x)
is strictly convex on (0, 1).
Solutions: The Fifty-Second Competition (1991) 153

We compute
d f
ln f =
dx f
d2 f f  − (f  )2
ln f =
dx2 f2
f  = au cosh ux − bu cosh u(1 − x)
f  = au2 sinh ux + bu2 sinh u(1 − x)
f f  − (f  )2 = (a2 + b2 )u2 (sinh2 ux − cosh2 ux)
 
+ 2abu2 sinh ux sinh u(1 − x) + cosh ux cosh u(1 − x)
 
= (a2 + b2 )u2 (−1) + 2abu2 cosh ux + u(1 − x)
= u2 (−a2 − b2 + 2ab cosh u)
d2
so dx2 ln f has constant sign for x ∈ (0, 1), and the following are equivalent:
d2
ln f ≤ 0 for x ∈ (0, 1)
dx2
−a2 − b2 + 2ab cosh u ≤ 0
−a2 − b2 + abeu + abe−u ≤ 0
−(a − eu b)(a − e−u b) ≤ 0
−(a − eu b) ≤ 0 (since u > 0 and a ≥ b > 0)
u ≤ ln(a/b). 

Remark. Taking the limit as u → 0 in the inequality of the problem yields

ax + b(1 − x) ≥ ax b1−x ,
a weighted version of the AM-GM Inequality mentioned at the end of 1985A2.
154 The William Lowell Putnam Mathematical Competition

The Fifty-Third William Lowell Putnam Mathematical Competition

December 5, 1992

A1. (31, 82, 42, 10, 0, 0, 0, 7, 23, 6, 2, 0)

Prove that f (n) = 1 − n is the only integer-valued function defined on the
integers that satisfies the following conditions:
(i) f (f (n)) = n, for all integers n;
(ii) f (f (n + 2) + 2) = n for all integers n;
(iii) f (0) = 1.

Solution. If f (n) = 1 − n, then f (f (n)) = f (1 − n) = 1 − (1 − n) = n, so (i) holds.

Similarly, f (f (n + 2) + 2) = f ((−n − 1) + 2) = f (1 − n) = n, so (ii) holds. Clearly
(iii) holds, and so f (n) = 1 − n satisfies the conditions.
Conversely, suppose f satisfies the three given conditions. By (i), f (0) = 1 and
f (1) = 0. From condition (ii), f (f (f (n + 2) + 2)) = f (n), and applying (i) yields
f (n + 2) + 2 = f (n), or equivalently f (n + 2) = f (n) − 2. Easy inductions in both
directions yields f (n) = 1 − n. 

A2. (157, 1, 0, 0, 0, 0, 0, 0, 2, 14, 14, 15)

Define C(α) to be the coefficient of x1992 in the power series expansion
about x = 0 of (1 + x)α . Evaluate
 1

1 1 1 1
C(−y − 1) + + + ··· + dy.
0 y+1 y+2 y+3 y + 1992

Answer. The value of the integral is 1992.

Solution. From the binomial theorem, we see that
α − 1991
C(α) = α(α − 1) · · · ,
1992!
so C(−y − 1) = (y + 1) · · · (y + 1992)/1992!. Therefore



1 1 d (y + 1) · · · (y + 1992)
C(−y − 1) + ··· + = .
y+1 y + 1992 dy 1992!
(The same formula for the derivative of a factored polynomial came up in 1991A3.)
Hence the integral in question is
 1
 1
d (y + 1) · · · (y + 1992) (y + 1) · · · (y + 1992)
dy =
0 dy 1992! 1992! 0
1993! − 1992!
= = 1992. 
1992!
A3. (55, 20, 7, 0, 0, 0, 0, 0, 16, 7, 45, 53)
For a given positive integer m, find all triples (n, x, y) of positive integers,
with n relatively prime to m, which satisfy (x2 + y 2 )m = (xy)n .
Answer. There are no solutions if m is odd. If m is even, the only solution is
(n, x, y) = (m + 1, 2m/2 , 2m/2 ).
Solutions: The Fifty-Third Competition (1992) 155

Solution. Note that x2 + y 2 > xy, so n > m. Let d = gcd(x, y), so x = ad

and y = bd where gcd(a, b) = 1. Then (a2 + b2 )m = d2(n−m) (ab)n . If p is a prime
factor of a, then p divides the right side of this equation, but not the left. Hence
a = 1, and similarly b = 1. Thus 2m = d2(n−m) . Now m must be even, so say
m = 2k, from which 2k = dn−2k , so n − 2k divides k. Since gcd(m, n) = 1, we have
n − 2k = gcd(n − 2k, k) = gcd(n, k) = 1, so n = 2k + 1. Thus d = x = y = 2k , and
the solutions are as claimed. 

A4. (17, 6, 7, 0, 0, 0, 2, 0, 73, 18, 47, 33)

Let f be an infinitely differentiable real-valued function defined on the
real numbers. If


1 n2
f = 2 , n = 1, 2, 3, . . . ,
n n +1
compute the values of the derivatives f (k) (0), k = 1, 2, 3, . . . .
Answer. We have ,
(k) (−1)k/2 k! if k is even;
f (0) =
0 if k is odd.

Solution. Let g(x) = 1/(1 + x2 ) and h(x) = f (x) − g(x). The value of g (k) (0) is
∞
k! times the coefficient of xk in the Taylor series 1/(1 + x2 ) = m=0 (−1)m x2m , and
the value of h(k) (0) is zero by the lemma below (which arguably is the main content
of this problem). Thus
,
(−1)k/2 k! if k is even;
(k) (k)
f (0) = g (0) + h (0) =(k)

0 if k is odd.

Lemma. Suppose h is an infinitely differentiable real-valued function defined on

the real numbers such that h(1/n) = 0 for n = 1, 2, 3, . . . . Then h(k) (0) = 0 for all
nonnegative integers k.
This lemma can be proved in many ways (and is a special case of a more general
result stating that if h : R → R is infinitely differentiable, if k ≥ 0, and if a ∈ R,
then h(k) (a) is determined by the values of h on any sequence of distinct real numbers
tending to a).
Proof 1 (Rolle’s Theorem). Since h(x) = 0 for a sequence of values of x strictly
decreasing to 0, h(0) = 0. By Rolle’s Theorem, h (x) has zeros between the zeros of
h(x); hence h (x) = 0 for a sequence strictly decreasing to 0, so h (0) = 0. Repeating
this argument inductively, with h(n) (x) playing the role of h(x), proves the lemma. 

Proof 2 (Taylor’s Formula). We prove that h(n) (0) = 0 by induction. The n = 0

case follows as in the previous proof, by continuity. Now assume that n > 0, and
h(k) (0) = 0 is known for k < n. Recall Taylor’s Formula (Lagrange’s form, see
e.g. [Ap1, Section 7.7]) which states that for any x > 0 and integer n > 0, there exists
θn ∈ [0, x] such that

h(x) = h(0) + h (0)x + · · · + h(n−1) (0)xn−1 /(n − 1)! + h(n) (θn )xn /n!.
156 The William Lowell Putnam Mathematical Competition

By our inductive hypothesis,

h(0) = · · · = h(n−1) (0) = 0.
Hence by taking x = 1, 1/2, 1/3, . . . , we get h(n) (θm ) = 0, where 0 ≤ θm ≤ 1/m. But
limm→∞ θm = 0, so by continuity h(n) (0) = 0. 
Proof 3 (J. P. Grossman). By continuity, h(0) = 0. Let k be the smallest
nonnegative integer such that h(k) (0) = 0. We assume h(k) (0) > 0; the same argument
applies if h(k) (0) < 0. Then there exists C such that h(k) (x) > 0 on (0, C]. Repeated
integration shows that h(x) > 0 on (0, C], a contradiction. 
Proof 4 sketch (explicit computation). By definition,
f (C) − f (0)
h (0) = lim .
%→0 C
More generally, if h is infinitely differentiable in a neighborhood of 0, then
k k k−j
(k) j=0 j (−1) h(jC)
h (0) = lim k
. (1)
%→0 C
(This can be proved by applying L’Hôpital’s Rule k times to the expression on the
right.) Then choose C = 1/n where n runs over the multiples of lcm(1, . . . , k), to
obtain h(k) (x) = 0. 

Remark. The formula (1) holds under the weaker assumption that h(k) (0) exists.
To prove this, apply L’Hôpital’s Rule k − 1 times, and then write the resulting
expression as a combination of limits of the form
h(k−1) (jC) − h(k−1) (0)
lim ,
%→0 jC
each of which equals h(k) (0), by definition.
Remark. Note that h(x) need not be the zero function! An infinitely differentiable
function need not be represented by its Taylor series at a point, i.e., it need not be
analytic. For example, consider
, 2
e−1/x sin(π/x) for x = 0,
h(x) =
0 for x = 0.
It is infinitely differentiable for all x, and all of its derivatives are 0 at x = 0. (It
satisfies the hypotheses of the lemma!)

A5. (1, 9, 1, 0, 0, 0, 0, 0, 5, 3, 72, 112)

For each positive integer n, let
0 if the number of 1’s in the binary representation of n is even,
an =
1 if the number of 1’s in the binary representation of n is odd.
Show that there do not exist positive integers k and m such that
ak+j = ak+m+j = ak+2m+j ,
for 0 ≤ j ≤ m − 1.
Solutions: The Fifty-Third Competition (1992) 157

Solution 1. The sequence begins

0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, . . . .
The problem is to show that there are not “three identical blocks in a row”.
The definition of an implies that a2n = an = 1 − a2n+1 .
Suppose that there exist k, m as in the problem; we may assume that m is minimal
for such examples.
Suppose first that m is odd. We’ll suppose ak = ak+m = ak+2m = 0; the case ak = 1
can be treated similarly. Since either k or k + m is even, ak+1 = ak+m+1 = ak+2m+1 =
1. Again, since either k+1 or k+m+1 is even, ak+2 = ak+m+2 = ak+m+2 = 0. By this
means, we see that the terms ak , ak+1 , . . . , ak+m−1 alternate between 0 and 1. Then
since m − 1 is even, ak+m−1 = ak+2m−1 = ak+3m−1 = 0. But, since either k + m − 1
or k + 2m − 1 is even, that would imply that ak+m = ak+2m = 1, a contradiction.
Thus m must be even. Extracting the terms with even indices in
ak+j = ak+m+j = ak+2m+j , for 0 ≤ j ≤ m − 1,
and using the fact that ar = ar/2 for even r, we get
ak/2+i = ak/2+(m/2)+i = ak/2+m+i , for 0 ≤ i ≤ m/2 − 1.
(The even numbers ≥ k are 2k/2, 2k/2 + 2, . . . .) This contradicts the minimality
of m.
Hence there are no such k and m. 
Solution 2 (J.P. Grossman).
Lemma. Let c(i, j) be the number of carries when i is added to j in binary. Then
ai+j ≡ ai + aj + c(i, j) (mod 2).
Proof. Perform the addition column by column as taught in grade school, writing
a “1” above the next column every time there is a carry. Then modulo 2, the total
number of 1’s appearing “above the line” is ai + aj + c(i, j), and the total number
of 1’s in the sum is ai+j . But in each column the digit of the sum is congruent
modulo 2 to the sum of the 1’s that appear above it. Summing over all columns yields
ai+j ≡ ai + aj + c(i, j) (mod 2). 
We are asked to show that there is no block of 2m consecutive integers such that
ak = ak+m for all k in the block. Suppose such a block exists.
If 2n ≤ m < 2n+1 , then m has n + 1 bits (binary digits), and since 2m ≥ 2n+1 , any
2m consecutive integers will exhibit every possible pattern of n + 1 low bits. Let κ be
the integer in this block with the n + 1 low bits zero. Then c(κ, m) = 0, so the lemma
gives aκ+m ≡ aκ + am (mod 2). Since aκ+m = aκ , we obtain am = 0. Running this
argument in reverse, we have ak+m ≡ ak + am (mod 2) and c(k, m) ≡ 0 (mod 2) for
all k in the block.
In particular, c(k, m) = 0 for the k with the n + 1 low bits equal to 0 except for one
1, so m cannot contain the pattern 01. Thus the binary expansion of m has the form
1 · · · 10 · · · 0; that is, m = 2n+1 − 2s for some s ≥ 0.
Suppose we can find k1 and k2 in the block such that k1 /2n+1  = k2 /2n+1 ,
k1 ≡ 2s (mod 2n+1 ), and k2 ≡ 2s+1 (mod 2n+1 ). Then c(k1 , m) = c(k2 , m) + 1,
158 The William Lowell Putnam Mathematical Competition

contradicting the fact that c(k, m) is even for each k in the block. Thus no such pair
(k1 , k2 ) exists.
On the other hand, since 2m ≥ 2n+1 , there does exist k in the block such that
k ≡ 2s (mod 2n+1 ). The smallest element of the block must be at least k − 2n+1 + 1,
or else we could take k1 = k − 2n+1 and k2 = k1 + 2s . The largest element must be at
most k + 2s − 1, or else we could take k1 = k and k2 = k1 + 2s . Thus the block has
length at most
2n+1 + 2s − 1 < 2n+1 + 2s+2 − 2s+1 − 1 ≤ 2n+2 − 2s+1 − 1 = 2m − 1,
which is a contradiction.
Related question. Show that if two adjacent blocks are identical, then the length
of each is a power of 2. Show that any power of 2 is achievable.
Remark. This sequence is sometimes called the Thue-Morse sequence; see [AS] for
examples of its ubiquity in mathematics, including its use by Machgielis Euwe (chess
world champion 1935–37) to show that infinite games of chess may occur despite the
so-called “German rule” which states that a draw occurs if the same sequence of moves
occurs three times in succession.
Related question. Another fact discussed in [AS] is the following (see there for
further references). The result of Christol mentioned in the remark following 1989A6
∞
implies that the generating function F (x) = n=0 an xn considered as a power series
with coefficients in F2 , is algebraic over the field F2 (x) of rational functions with
coefficients in F2 . We leave it to the reader to verify that
(x + 1)3 F (x)2 + (x + 1)2 F (x) + x = 0.
On the other hand, if the coefficients of F (x) are considered to be rational, then
one can show that F (x) is transcendental over Q(x), and even that F (1/2) is a
transcendental number.
Related questions. The Thue-Morse sequence has a self-similarity property: if one
replaces each “0” with “0, 1”, and each “1” with “1,0”, one recovers the original
sequence. The “extraction of terms with even indices” in Solution 1 is just reversing
this self-similarity.
Here are a few more problems and results relating to “arithmetic self-similarity”.
(a) Show that the sequence s(0), s(1), s(2), . . . , defined by s(3m) = 0, s(3m + 1) =
s(m), s(3m+2) = 1, also has the property that there are not “three identical blocks in
a row”. This problem is given in [Hal, p. 156], along with a discussion of its motivation
in chess.
(b) 1993A6.
(c) Problem 3 on the 1988 International Mathematical Olympiad [IMO88, p. 37] is:
A function f is defined on the positive integers by
f (1) = 1, f (3) = 3, f (2n) = f (n),
f (4n + 1) = 2f (2n + 1) − f (n), f (4n + 3) = 3f (2n + 1) − 2f (n),
for all positive integers n. Determine the number of positive integers n, less
than or equal to 1988, for which f (n) = n.
Solutions: The Fifty-Third Competition (1992) 159

(d) Also, [YY, p. 12] has a series of similar problems. For example, this Putnam
problem answers the question Yaglom and Yaglom pose in Problem 124(b). The
subsequent problem is harder:
Show that there exist arbitrarily long sequences consisting of the digits 0, 1,
2, 3, such that no digit or sequence of digits occurs twice in succession. Show
that there are solutions in which the digit 0 does not occur; thus three digits
is the minimum we need to construct sequences of the desired type.
For a link to a discrete version of dynamical systems, [YY] suggests [MH].

A6. (9, 3, 4, 0, 0, 0, 0, 0, 0, 10, 32, 22, 123)

Four points are chosen at random on the surface of a sphere. What is
the probability that the center of the sphere lies inside the tetrahedron
whose vertices are at the four points? (It is understood that each point is
independently chosen relative to a uniform distribution on the sphere.)
Answer. The probability is 1/8.

Solution 1. Set up a coordinate system so that the sphere is centered at the

origin of R . Identify points with vectors in R3 .
3

Let P1 , P2 , P3 , P4 be the four random points on the sphere. We may suppose that
the choice of each Pi is made in two steps: first a random point Qi is chosen, then a
random sign Ci ∈ {−1, 1} is chosen and we set Pi = Ci Qi .
The probability that Q3 is in the linear subspace spanned by Q1 and Q2 is zero.
Similar statements hold for any three of the Qi , so we may assume that every three of
the Qi are linearly independent. The probability that Q4 is in the plane through Q1 ,
Q2 , Q3 is zero, so we may assume that Q1 Q2 Q3 Q4 is a nondegenerate tetrahedron
TQ . We may also assume that for any choices of the Ci , the tetrahedron TP with the
Pi as vertices is nondegenerate.

The linear map LQ : R4 → R3 sending (w1 , w2 , w3 , w4 ) to wi Qi is surjective, so
ker LQ is 1-dimensional, generated by some (w1 , w2 , w3 , w4 ). Since every three Qi are
linearly independent, every wi is nonzero.
We claim that 0 lies in the interior of TQ if and only if all the wi have the same sign.
If 0 lies in the interior of TQ , then 0 is a convex combination of the vertices in which
4
each occurs with nonzero coefficient; i.e., 0 = i=1 ri Qi for some r1 , r2 , r3 , r4 > 0
with sum 1. Then (w1 , w2 , w3 , w4 ) is a multiple of (r1 , r2 , r3 , r4 ), so the wi have the
same sign. Conversely if the wi have the same sign, then w1 + w2 + w3 + w4 = 0
4 1
and 0 = i=1 ri Qi where (r1 , r2 , r3 , r4 ) = w1 +w2 +w 3 +w4
(w1 , w2 , w3 , w4 ) is such that
r1 , r2 , r3 , r4 > 0 sum to 1, so 0 lies in the interior of TQ .
Fix Q1 , Q2 , Q3 , Q4 and (w1 , w2 , w3 , w4 ). Then (C1 w1 , C2 w2 , C3 w3 , C4 w4 ) is a generator
of ker LP , where LP is defined using the Pi instead of the Qi . The numbers C1 w1 ,
C2 w2 , C3 w3 , C4 w4 have the same sign (all plus or all minus) for exactly 2 of the 24 = 16
choices of (C1 , C2 , C3 , C4 ). Thus, conditioned on a choice of the Qi , the probability
that TP contains the origin in its interior is 2/16 = 1/8. This is the same for any
(Q1 , . . . , Q4 ), so the overall probability that 0 lies inside TP also is 1/8. 
Remark. As noted in [HoS], the same argument proves the following generalization.
160 The William Lowell Putnam Mathematical Competition

Let Rn be endowed with a probability measure µ that is symmetric with respect to

the map x !→ −x, and such that, when n + 1 points are chosen independently with
respect to µ, with probability one their convex hull is a simplex. Then the probability
that the origin is contained in the simplex generated by n + 1 such random points is
1/2n .
The original problem is the case n = 3 with µ equal to the uniform distribution.
The case n = 2 with the uniform distribution appears, along with a strategy for
approaching it, in [Hon1, Essay 1].
Remark. The article [HoS] mentions also the following result of J. G. Wendel [Wen],
which generalizes the original result in a different direction: if N points are chosen at
random from the surface of the unit sphere in Rn , the probability that there exists a
hemisphere containing all N points equals
n−1

−N +1 N −1
2 .
k
k=0

Remark. Here is a bizarre related fact: if five points are chosen at random from a
ball in R3 , the probability that one of them is contained in the tetrahedron generated
by the other four is 9/143 [So].

We end with a variant of the solution to the original problem.

Solution 2. Pick the first three points first; call them A, B, C. Consider the
spherical triangle T defined by those points. The center of the sphere lies in the convex
hull of A, B, C, and another point P if and only if it lies in the convex hull of T and
P . This happens if and only if P is antipodal to some point of T . So the desired
probability is the expected fraction of the sphere’s surface area which is covered by T .
Denote the antipode to a point P by P  . We consider the eight spherical triangles
ABC, A BC, A B  C, ABC  , AB  C  , A B  C  . Denote these by T0 , T1 , . . . , T7 ; we
regard Ti as a function of the random variables A, B, C. There is an automorphism
of our probability space defined by (A, B, C) !→ (A , B, C), so T0 and T1 have the
same distribution. By choosing similar automorphisms, T0 and Ti have the same
distribution for all i. In particular, the expected fraction of the sphere covered by Ti
is independent of i. On the other hand, the triangles T0 , . . . , T7 cover the sphere (with
overlap of measure zero), since they are the eight regions formed by the three great
circles obtained by extending the sides of spherical triangle ABC, so the probability
we seek is 1/8. 

B1. (145, 15, 4, 0, 0, 0, 0, 0, 6, 14, 11, 8)

Let S be a set of n distinct real numbers. Let AS be the set of numbers
that occur as averages of two distinct elements of S. For a given n ≥ 2,
what is the smallest possible number of elements in AS ?
Answer. The smallest possible number of elements of AS is 2n − 3.
Solution. Let x1 < x2 < · · · < xn represent the elements of S. Then
x1 + x2 x1 + x3 x1 + xn x2 + xn x3 + xn xn−1 + xn
< < ··· < < < < ··· <
2 2 2 2 2 2
Solutions: The Fifty-Third Competition (1992) 161

represent 2n − 3 distinct elements of AS , so AS has at least 2n − 3 distinct elements.

On the other hand, if we take S = {1, 2, . . . , n}, the elements of AS are 32 , 42 , 52 ,
2 . There are only 2n − 3 such numbers; thus there is a set AS with at most
. . . , 2n−1
2n − 3 distinct elements. 
Related question. This is a generalization of Problem 2 of the 1991 Asian-Pacific
Mathematical Olympiad [APMO],
Suppose there are 997 points given on a plane. If every two points are joined
by a line segment with its midpoint coloured in red, show that there are at
least 1991 red points on the plane. Can you find a special case with exactly
1991 red points?

B2. (159, 10, 7, 0, 0, 0, 0, 0, 1, 4, 13, 9)

For nonnegative integers n and k, define Q(n, k) to be the coefficient of
x in the expansion of (1 + x + x2 + x3 )n . Prove that
k

k


n n
Q(n, k) = ,
j=0
j k − 2j
 
where ab is the standard
  binomial coefficient. (Reminder:
  For integers a
and b with a ≥ 0, ab = b! (a−b)!
a!
for 0 ≤ b ≤ a, with ab = 0 otherwise.)
Solution. Write (1 + x + x2 + x3 )n as (1 + x2 )n (1 + x)n . The coefficient
 n of xk
2j
gets contributions from the x term in the first factor (with coefficient j ) times the
 n 
xk−2j term in the second factor (with coefficient k−2j ). 
Remark. This solution can also be stated in the language of generating functions:
Q(n, k)xk = (1 + x + x2 + x3 )n
k≥0

= (1 + x2 )n (1 + x)n



n 2j n i
= x x
j i
j≥0 i≥0



2j+i n n
= x
j i
j≥0 i≥0



n n
= xk .
j k − 2j
k≥0 j≥0

B3. (23, 11, 10, 0, 0, 0, 0, 0, 27, 24, 71, 37)

For any pair (x, y) of real numbers, a sequence (an (x, y))n≥0 is defined as
follows:
a0 (x, y) = x,
(an (x, y))2 + y 2
an+1 (x, y) = , for n ≥ 0.
2
Find the area of the region { (x, y)|(an (x, y))n≥0 converges }.
Answer. The area is 4 + π.
162 The William Lowell Putnam Mathematical Competition

Solution. The region of convergence is shown in Figure 22; it is a (closed) square

{ (x, y) : −1 ≤ x, y ≤ 1 } of side 2 with (closed) semicircles of radius 1 centered at
(±1, 0) described on two opposite sides.

(–1, 1) (1, 1)

(–1, –1) (1, –1)

FIGURE 22.
The region of convergence.

Note that (x, y), (−x, y), (x, −y), and (−x, −y) produce the same sequence after the
first step, so will restrict attention to the first quadrant (x, y ≥ 0), and use symmetry
to deal with the other three.
Fix y; we will determine for which x (x, y) is in the region. Let f (w) = (w2 + y 2 )/2,
so an (x, y) = f n (x). If the limit exists and is L, then f (L) = L, so L = 1 ± 1 − y 2
(call these values L+ and L− ). It will be useful to observe that

f (w) − L = (w − L)(w + L)/2. (1)

In particular, if y > 1, the limit cannot exist; we now assume y ≤ 1.

If w >L+ , then f (w) > w (from (1) for L = L+ , and w + L+ > 2). Hence if
x > 1 + 1 − y 2 , the sequence x, f (x), f (f (x)), . . . , cannot converge.
If x = L+ , then x, f (x), f (f (x)), . . . is the constant sequence (L+ ), and hence
converges.
If 0 ≤ w < L+ , then
(i) f (w) satisfies the same inequality, i.e., 0 ≤ f (w) < L+ (the left inequality is
immediate, and the right follows from (1) for L = L+ ),
(ii) |f (w) − L− | < |w − L− | (from (1) for L− and w + L− < L+ + L− = 2) , and
(iii) w − L− and f (w) − L− have the same sign (from (1) for L = L− ).
Then (i)–(iii) imply that if 0 ≤ x < L+ , then

x, f (x), f (f (x)), . . .

converges to L− . 
Solutions: The Fifty-Third Competition (1992) 163

Remark. Problem 1996A6 involves a similar dynamical system.

Related question. For another problem involving a dynamical system, observe what
happens when you repeatedly hit the cosine button on a calculator, and explain this
phenomenon. In essence, this is Problem 1952/7 [PutnamI, p. 37]:
Given any real number N0 , if Nj+1 = cos Nj , prove that limj→∞ Nj exists
and is independent of N0 .
Literature note. For introductory reading on dynamical systems, see [De] or [Mi].

B4. (35, 11, 13, 0, 0, 0, 0, 0, 12, 5, 48, 79)

Let p(x) be a nonzero polynomial of degree less than 1992 having no
nonconstant factor in common with x3 − x. Let


d1992 p(x) f (x)
=
dx 1992 x −x
3 g(x)
for polynomials f (x) and g(x). Find the smallest possible degree of f (x).
Answer. The smallest possible degree of f (x) is 3984.
Solution. By the Division Algorithm, we can write p(x) = (x3 − x)q(x) + r(x),
where q(x) and r(x) are polynomials, the degree of r(x) is at most 2, and the degree
of q(x) is less than 1989. Then



d1992 p(x) d1992 r(x)
= 1992 .
dx1992 x3 − x dx x3 − x
Write r(x)/(x3 − x) in the form
A B C
+ + .
x−1 x x+1
Because p(x) and x3 − x have no common factor, neither do r(x) and x3 − x, and thus
A, B, and C are nonzero. Thus


d1992 r(x)
dx1992 x3 − x


A B C
= 1992! + 1993 +
(x − 1)1993 x (x + 1)1993

1993 
Ax (x + 1)1993 + B(x − 1)1993 (x + 1)1993 + C(x − 1)1993 x1993
= 1992! .
(x3 − x)1993
Since A, B, C are nonzero, the numerator and denominator have no common factor.
Expanding the numerator yields
(A + B + C)x3986 + 1993(A − C)x3985 + 1993(996A − B + 996C)x3984 + · · · .
From A = C = 1, B = −2 we see that the degree can be as low as 3984. A lower
degree would imply A + B + C = 0, A − C = 0, 996A − B + 996C = 0, implying that
" #
A = B = C = 0, a contradiction.
d1992 p(x)
Expressing dx 1992 x3 −x in other than lowest terms can only increase the degree of
the numerator. (Nowhere in the problem does it say that f (x)/g(x) is to be in lowest
terms.) 
164 The William Lowell Putnam Mathematical Competition

Remark (Noam Elkies). Although the proposer of this problem presumably had
polynomials with real or complex coefficients in mind, the same solution works for
"
polynomials over a field of characteristic greater #than 1992. When the characteristic
d1992 p(x)
is less than 1992, one can show that dx 1992 x −x is always zero, so the problem does
3

not make much sense.

B5. (62, 4, 4, 0, 0, 0, 0, 3, 6, 2, 49, 73)

Let Dn denote the value of the (n − 1) × (n − 1) determinant
3 1 1 1 ··· 1
1 4 1 1 ··· 1
1 1 5 1 ··· 1
1 1 1 6 ··· 1 .
.. .. .. .. .. ..
. . . . . .
1 1 1 1 ··· n + 1
Is the set {Dn /n!}n≥2 bounded?
Answer. No. Each of the solutions below shows that


1 1
Dn = n! 1 + + · · · + .
2 n
Thus the sequence {Dn /n!}n≥2 is the nth partial sum of the harmonic series, which
is unbounded as n → ∞.

Solution 1. Subtract the first row from each of the other rows, to get
 
3 1 1 1 ··· 1
−2 3 0 0 · · · 0
 
 
−2 0 4 0 · · · 0
Dn = det −2 0 0 5

· · · 0 . (1)
 
 .. .. .. .. .. .
 . . . . . .. 
−2 0 0 0 · · · n
Then for 2 ≤ i ≤ n − 1, add 2/(i + 1) times the ith column to the first column to
obtain  
3 + 23 + 24 + · · · + n2 1 1 1 ··· 1
 ··· 0
 0 3 0 0 
 
 0 0 4 0 ··· 0
Dn = det 
 0 0 0 5 ···

0 .
 
 .. .. .. .. .. .. 
 . . . . . .
0 0 0 0 ··· n
The resulting matrix is upper triangular, so the determinant is the product of the
diagonal elements, which is


1 1
n! 1 + + · · · + . 
2 n
Solutions: The Fifty-Third Competition (1992) 165

Solution 2 (J.P. Grossman). We derive a recursion for Dn+1 . Expand (1)

along the last row to get Dn = nDn−1 + (n − 1)!. Divide by n! to obtain
Dn Dn−1 1
= + ,
n! (n − 1)! n
from which the result follows. 
Solution 3. Define  
1 + a1 1 1 1 ··· 1
 1 ··· 
 1 + a2 1 1 1 
 
 1 1 1 + a3 1 ··· 1 
Dn+1 (a1 , . . . , an ) = det 
 1 1 1 1 + a4 ··· 1
.

 
 .. .. .. .. .. .. 
 . . . . . . 
1 1 1 1 ··· 1 + an
The problem asks for Dn (2, 3, . . . , n). We will prove the identity
+
n n +
n
Dn+1 (a1 , . . . , an ) = ai + aj . (2)
i=1 i=1 j=1
j=i

This formula follows immediately from the recursion

Dn+1 (a1 , . . . , an ) = an Dn (a1 , . . . , an−1 ) + an−1 Dn (a1 , . . . , an−2 , 0).

To prove this recursion, subtract the (n − 1)st column from the nth column, and then
expand along the nth column.
If all the ai are nonzero, we can write the polynomial Dn (a1 , . . . , an−1 ) in the form


1 1 1
Dn (a1 , . . . , an−1 ) = a1 a2 · · · an−1 1 + + + ··· + .
a1 a2 an−1
This problem is the special case ai = i + 1. 
Remark. This formula can also be used for a generalization of Problem 1993B5;
see page 188.
Solution 4. The following result is Problem 7 of Part VII of [PS], and appeared
as Problem 1978A2 [PutnamII, p. 31].
Let a, b, p1 , p2 , . . . , pn be real numbers with a = b. Define

f (x) = (p1 − x)(p2 − x)(p3 − x) · · · (pn − x).

Then
 
p1 a a a ··· a a
b p ··· a
 2 a a a 
b b p ··· a
 3 a a 
 

det  b b b p 4 ··· a a  = bf (a) − af (b) . (3)
 .. .. .. .. .. .. .. 
 b−a
. . . . . . .
 
b b b b ··· pn−1 a
b b b b ··· b pn
166 The William Lowell Putnam Mathematical Competition

The left side of (3) is a polynomial in b, and hence continuous in b. Letting a = 1,

b → 1, and using L’Hôpital’s Rule, we find
 
p1 1 1 1 · · · 1 1
1 p 1 1 ··· 1
 2 1 
1 1 p 1 ··· 1
 3 1 
 
det  1 1 1 p4 · · · 1 1  = f (1) − f  (1).
. .. 
. .. .. .. . . .. 
. . . . . . .
 
 1 1 1 1 · · · pn−1 1 
1 1 1 1 ··· 1 pn
Letting pi = ai + 1, we recover (2), using for example equation (1) on page 136. 
Remark. There are many other approaches that proceed by applying row and
column operations to find a recursion for Dn . Here we sketch one more.
Subtract the next-to-last column from the last column, and then subtract the next-
to-last row from the last row, and expand along the last row to obtain
Dn+1 = (2n + 1)Dn − n2 Dn−1 .
If we set rn = Dn /n!, this becomes rn+1 − rn = n
n+1 (rn − rn−1 ). One proves
rn+1 − rn = 1/(n + 1) by induction, and finally
1 1 1
rn = 1 + + + ··· + .
2 3 n
B6. (0, 0, 0, 0, 0, 0, 0, 0, 5, 4, 39, 155)
Let M be a set of real n × n matrices such that
(i) I ∈ M, where I is the n × n identity matrix;
(ii) if A ∈ M and B ∈ M, then either AB ∈ M or −AB ∈ M, but not both;
(iii) if A ∈ M and B ∈ M, then either AB = BA or AB = −BA;
(iv) if A ∈ M and A = I, there is at least one B ∈ M such that AB = −BA.
Prove that M contains at most n2 matrices.
Solution 1 (Noam Elkies). Suppose A, B are in M. By (iii), AB = CBA,
where C = ±1, for any B in M. Then
AAB = ACBA = C2 BAA = BAA,
so A2 commutes with any B in M; of course the same is true of −A2 . On the other
hand, by (ii), A2 or −A2 is in M. Let C be the one that is in M.
If C is not I, then by (iv) we can find a B in M such that CB = −BC. But we
know CB = BC for any B in M. Thus CB = 0, which is impossible by (ii).
We conclude that C = I. In other words, for any A in M, A2 = ±I.
Now suppose M has more than n2 matrices. The space of real n × n matrices has

dimension n2 , so we can find a nontrivial linear relation D∈M xD D = 0. Pick such a
relation with the smallest possible number of nonzero xD . We will construct a smaller
relation, obtaining a contradiction and finishing the proof.

Pick an A with xA nonzero, and multiply by it on the right: D∈M xD DA = 0.
In light of (ii) the matrices DA run over M modulo sign; so we have a new relation
Solutions: The Fifty-Third Competition (1992) 167

E∈M yE E = 0. The point of this transformation is that now the coefficient yI of I
is ±xA , which is nonzero.
Pick any D other than I such that yD is nonzero. By (iv), we can pick B in M such

that DB = −BD. Multiply E∈M yE E = 0 by B on both the left and the right, and
add:
yE (BE + EB) = 0.
E∈M

Now by (iii) we have BE + EB = (1 + CBE )BE, where CBE = ±1. In particular,

CBI = 1 (clear) and CBD = −1 (by construction of B). So we get

yE (1 + CBE )BE = 0,
E∈M

where at least one term does not disappear and at least one term does disappear. As
before, the matrices BE run over M modulo sign. So we have a relation with fewer
terms, as desired. 
Remark. This proof did not need (i) in an essential way. The only modification to
the proof needed to avoid using (i), is that instead of arranging for yI to be nonzero,
we might have y−I nonzero instead.
In fact, if M is nonempty, (i) follows from (ii), (iii), and (iv), as we now show.
Choose A ∈ M. Then ±A2 ∈ M, so ±I ∈ M by the first half of Solution 1. But
−I ∈ M contradicts (iv).
Remark. The result holds not only for real matrices, but also for matrices with
entries in any field k: if the characteristic of k is not 2, the proof proceeds as before;
if the characteristic is 2, then (ii) implies that M is empty.
Remark. An argument similar to that in the latter half of the solution proves
the following result of field theory (“independence of characters”): if L is a finite
extension of a field K, then the automorphisms of L over K are linearly independent
in the K-vector space of K-linear maps L → L [Ar, Theorem 12].

Solution 2. We prove the result more generally for complex matrices, by induction
on n.
If n = 1, then the elements of M commute so (iv) cannot be satisfied unless
M = {I}. Suppose that n > 1 and that the result holds for sets of complex matrices
of smaller dimension.
We may assume |M| > 1, so by (iv), there exist C, D ∈ M with CD = −DC. Fix
such C, D. As in Solution 1, C 2 = ±I, so the eigenvalues of C are ±λ where λ = 1
or i. Furthermore, Cn = Vλ ⊕ V−λ , where Vλ , V−λ are the eigenspaces corresponding
to λ and −λ (i.e., the nullspaces of (C − λI), (C + λI)) respectively. (Here we follow
the convention that the matrices are acting on Cn by right-multiplication.) Observe
that if X ∈ M then

CX = XC =⇒ (C ± λI)X = X(C ± λI) =⇒ V±λ X = V±λ ;

CX = −XC =⇒ (C ± λI)X = (−1)X(C ∓ λI) =⇒ V±λ X = V∓λ .

In particular, since Vλ D = V−λ , dim(Vλ ) = dim(V−λ ) = n/2.

168 The William Lowell Putnam Mathematical Competition

Let N = { X ∈ M : CX = XC, DX = XD }. If Y ∈ M then exactly one of Y ,

Y C, Y D, Y CD is in N . It follows that |N | = |M|/4.
For X ∈ N , let φ(X) be the (n/2) × (n/2) matrix representing, with respect to
a basis of Vλ , the linear transformation given by v !→ vX for v ∈ Vλ . Then φ is
injective. To see this, assume φ(X) = φ(Y ), so vX = vY for v ∈ Vλ ; but if v ∈ V−λ
then vD ∈ Vλ , so vXD = vDX = vDY = vY D, which again implies vX = vY ; since
X and Y induce the same transformations of both Vλ and V−λ , it follows that X = Y .
It suffices finally to show that φ(N ), a set of (n/2)×(n/2) complex matrices, satisfies
(i), (ii), (iii), (iv), for then, by induction, |φ(N )| ≤ (n/2)2 , whence |M| = 4|N | =
4|φ(N )| ≤ n2 .
Conditions (i), (ii), (iii) for φ(N ) are inherited from those of M. To show (iv),
let φ(A) ∈ φ(N ), with φ(A) not the (n/2) × (n/2) identity matrix. Then A = I
(since φ is injective) and AB = −BA for some B ∈ M. Let B  be the element of
{B, BC, BD, BCD} belong to N . Since AB  = −B  A, φ(A)φ(B  ) = −φ(B  )φ(A). 

Solution 3. Again we prove the result more generally for complex matrices. We
will use the following facts about the set S of irreducible complex representations of
a finite group G up to equivalence:
1. The number of conjugacy classes of G is |S|. [Se2, Theorem 7]
2. The number of one-dimensional representations in S is |G/G |, where G is the
commutator subgroup of G. (This is a consequence of the previous fact applied
to G/G , since all one-dimensional representations must be trivial on G .)
3. The sum of the squares of the dimensions of the representations in S equals
|G|. [Se2, Corollary 2]
As in Solution 1, we have A2 = ±I for any A ∈ M. Thus any finite subset
{A1 , . . . , Ak } ⊆ M generates a finite group G0 , whose elements are of the form

±Ai1 Ai2 · · · Aim

where i1 < i2 < · · · < im . If A = ±I is in the center of G0 , then A or −A belongs to

M, so by (iv) some B in M does not commute with A. Let G1 be the group generated
by A1 , . . . , Ak , B. If there were some A in G0 such that A B is central in G1 , then

A BA = AA B = A AB = −A BA,

giving a contradiction. Hence G1 has a strictly smaller center than G0 . By repeating

this enlargement process, we can find a finite set A1 , . . . , Ak of elements of M
(k  > k) generating a finite group G with center Z = {±I}. Note that |G| ≥ 2k.
If X ∈ G − Z, then X has precisely two conjugates, namely itself and −X. Thus G
has 1 + |G|/2 conjugacy classes, and therefore G has 1 + |G|/2 inequivalent irreducible
representations over C. The number of inequivalent representations of dimension 1
is |G/G |. Since G = {±I} =Z, this number is |G|/2. The remaining irreducible
representation η has dimension |G|/2, since the sum of the squares of the dimensions
of the irreducible representations equals |G|. Then η must occur in the representation
G N→GLn (C), since Z is in the kernel of all the 1-dimensional representations. Hence
n ≥ |G|/2, or equivalently 2n2 ≥ |G|.
Solutions: The Fifty-Third Competition (1992) 169

Thus k ≤ |G|/2 ≤ n2 . Since all finite subsets of M have cardinality at most n2 , we

have |M| ≤ n2 . 

Remark. If we knew in advance that M were finite, the proof would be cleaner:
we could let G be the group { ±A : A ∈ M }.
Group theory interlude. Here we collect some facts from group theory that will be
used in the next remark to classify all possible M. First recall some definitions, from
pages 4, 8, and 183 of [Gor]. Let Z denote the center of a group G. If p is a prime,
then G is called a p-group if G is finite of order a power of p. An elementary abelian
p-group is a finite abelian group killed by p, hence isomorphic to (Z/pZ)k for some
integer k ≥ 0. A special p-group is a p-group G such that either G is elementary
abelian, or {1}  G = Z  G with Z and G/Z elementary abelian. Finally an
extraspecial p-group is a nonabelian special p-group G with |Z| = p.
There is a relatively simple classification of extraspecial p-groups [Gor, p. 204]. We
reproduce the classification for p = 2 here. First one has the dihedral group D of order
8 and the quaternion group Q of order 8. Identify the center of each of these with
{±1}. For integers a, b ≥ 0 not both zero, we let Ga,b denote the quotient of Da × Qb
by H ⊂ {±1}a+b , where H is the kernel of the multiplication map {±1}a+b → {±1}.
(This quotient is called a central product.) A group G is an extraspecial 2-group if
and only if G ∼ = Ga,b for some a, b as above. Moreover, Ga,b ∼ = Ga ,b if and only
if a + b = a + b and b ≡ b (mod 2). For convenience, we also define G0,0 = Z/2,


although it is not extraspecial. Then |Ga,b | = 22a+2b+1 for all a, b ≥ 0.

The argument of Solution 3 shows that over C, Ga,b has 22a+2b representations of
dimension 1 that are trivial on the center, and one other irreducible representation η,
of dimension 2a+b . We can construct η explicitly as follows. Let H = R + Ri + Rj + Rk
denote Hamilton’s ring of quaternions, in which i2 = j 2 = k 2 = −1, ij = −ji = k,
jk = −kj = i, and ki = −ik = j. The 2-dimensional representations ηD and ηQ of D
and Q are obtained by complexifying the real representation D → GL2 (R) given by
symmetries of the square and the inclusion Q = {±1, ±i, ±j, ±k} → H∗ , respectively,
noting that H ⊗R C ∼ = M2 (C). The tensor product of these irreducible representations
gives an irreducible representation of Da × Qb whose kernel is exactly H; this induces
η. The corresponding simple factor of the group algebra CGa,b is the complexification
of a simple factor of RGa,b isomorphic to M2 (R)⊗a ⊗R H⊗b . The latter is isomorphic to
M2a+b (R) or M2a+b−1 (H) according as b is even or odd, since the Brauer group Br(R)
is of order 2, generated by the class of H. (For the definition of the Brauer group, see
Section 4.7 of [J].) Hence η is the complexification of a representation η  over R if and
only if b is even, i.e., if and only if Ga,b ∼
= Gα,0 for some α ≥ 0. If b is odd, we instead
let η  be the unique faithful R-irreducible representation; it is of dimension 2 dim η,
and is obtained by identifying the Cn for η with R2n .

Remark. We are now ready to classify all possible M. The argument of Solution 3
shows that if M satisfies (i), (ii), (iii), (iv), then G = { ±A : A ∈ M } is a finite group
such that
(1) the center Z has order 2 and contains the commutator subgroup G ; and
(2) there is a faithful representation ρ : G → GLn (R) identifying Z with {±I}.
170 The William Lowell Putnam Mathematical Competition

Conversely, given a group G with ρ : G → GLn (R) satisfying (1) and (2), any set M
of coset representatives for {±I} in ρ(G) with I ∈ M satisfies (i), (ii), (iii), (iv).
Suppose that G and ρ satisfy (1) and (2). The argument at the beginning of
Solution 1 shows that every element of G/Z has order dividing 2; hence inversion on
G/Z is a homomorphism, and G/Z is an elementary abelian 2-group. Thus G is an
extraspecial 2-group or G ∼ = Z/2Z. (In particular |M| = |G|/2 is always a power of
4.) Since ρ is nontrivial on Z, η  must occur in ρ. Conversely, if G = Ga,b for some
a, b ≥ 0, and ρ : Ga,b → Mn (R) is a real representation containing η  , then G and ρ
satisfy (1) and (2), and hence we obtain all possibilities for M by choosing a set of
coset representatives for {±I} in ρ(G) with I ∈ M, for such G and ρ.
Remark. We can also determine all situations in the original problem in which
equality holds, i.e., in which M is a set of n2 matrices in Mn (R) satisfying (i), (ii),
(iii), (iv). In this case, G ∼
= Ga,b for some a ≥ 0 and b ∈ {0, 1}, |G| = 2n2 = 22a+2b+1 ,
and ρ is n-dimensional. Then dim η = 2a+b = n, and dim η  equals n or 2n according
as b = 0 or b = 1. But ρ contains η  , so ρ ∼ = η  and b = 0. It follows that equality
is possible if and only if n = 2a for some a ≥ 0, and in that case M is uniquely
determined up to a change of signs of its elements not equal to I and up to an overall
conjugation.
Similarly, if we relax the conditions of the problem to allow M to contain complex
matrices, then the equality cases arise from n = 2a , G ∼ = Ga,0 or G ∼ = Ga−1,1 (the
∼
latter being possible only if a ≥ 1), and ρ = η.
Solutions: The Fifty-Fourth Competition (1993) 171

The Fifty-Fourth William Lowell Putnam Mathematical Competition

December 4, 1993

A1. (185, 2, 0, 0, 0, 0, 0, 0, 1, 0, 14, 5)

The horizontal line y = c intersects the curve y = 2x − 3x3 in the first
quadrant as in the figure.† Find c so that the areas of the two shaded
regions are equal.
Answer. The area of the two regions are equal when c = 4/9.

(b, c)
y=c

3
y = 2x – 3x

FIGURE 23.

Solution. Let (b, c) denote the rightmost intersection point. (See Figure 23.) We
wish to find c such that  b
 
(2x − 3x3 ) − c dx = 0.
0

This leads to b − (3/4)b − bc = 0. After substituting c = 2b − 3b3 and solving, we find

2 4

a unique positive solution, namely b = 2/3. Thus c = 4/9. To validate the solution,
we check that (2/3, 4/9) is rightmost among the intersection points of y = 4/9 and
y = 2x − 3x√ : the zeros of 2x − 3x − 4/9 = (2/3 − x)(3x + 2x − 2/3) other than 2/3
3 3 2

are (−1 ± 3)/3, which are less than 2/3. 

A2. (146, 21, 6, 0, 0, 0, 0, 0, 8, 1, 17, 8)

Let (xn )n≥0 be a sequence of nonzero real numbers such that

x2n − xn−1 xn+1 = 1 for n = 1, 2, 3, . . . .

Prove there exists a real number a such that xn+1 = axn −xn−1 for all n ≥ 1.

Solution 1. Since the terms are nonzero, we can define an = (xn+1 + xn−1 )/xn
for n ≥ 1. It suffices to show a1 = a2 = · · · . But x2n+1 −xn xn+2 = 1 = x2n −xn−1 xn+1 ,

† The figure is omitted here, since it is almost identical to Figure 23 used in the solution.
172 The William Lowell Putnam Mathematical Competition

so

xn+1 (xn+1 + xn−1 ) = xn (xn+2 + xn )

xn+1 + xn−1 xn+2 + xn
=
xn xn+1
an = an+1

since xn , xn+1 = 0. The result follows by induction. 

Solution 2. Given x0 , x1 , x2 , let a = (x0 + x2 )/x1 . From x21 − x0 x2 = 1, we get

x21 − ax1 x0 + x20 = 1.
Now

x22 − x1 (ax2 − x1 ) = (ax1 − x0 )2 − x1 (a(ax1 − x0 ) − x1 )

= x21 − ax1 x0 + x20 = 1,

so x3 = ax2 − x1 .
By essentially the same algebra (and a simple induction), xn+1 = axn − xn−1 for
all n ≥ 1. 

Solution 3. Let a = (x0 + x2 )/x1 . Consider a sequence defined by yn+1 − ayn +

yn−1 = 0 where yi = xi for i = 0, 1, 2; we wish to show that yi = xi for all nonnegative
integers i.
If a = ±2, then by the theory of linear recursive sequences (discussed after the
solution to 1988A5), the general solution is

yn = Arn + Bsn ,

where A and B are constants, and r and s are the roots of the characteristic equation
t2 − at + 1 = 0. Then rs = 1, and (r − s)2 = a2 − 4 by the quadratic formula or by
the identity (r − s)2 = (r + s)2 − 4rs. Now

yn2 − yn+1 yn−1 = (Arn + Bsn )2 − (Arn−1 + Bsn−1 )(Arn+1 + Bsn+1 )

= −AB(rs)n−1 (r − s)2
= −AB(a2 − 4),

which is independent of n. Thus yn2 − yn+1 yn−1 = y12 − y2 y0 = 1 for all n, so yn = xn

by induction on n.
If a = ±2, then the general solution is

yn = (A + Bn)(±1)n .

Then
yn2 − yn+1 yn−1 = B 2 ,

which is independent of n. Hence as before, yn2 − yn+1 yn−1 = y12 − y2 y0 = 1 for all n,
so yn = xn for all n as well. 
Solutions: The Fifty-Fourth Competition (1993) 173

Solution 4. For all n,




x + xn+1 xn + xn+2 x xn x xn+2
det n−1 = det n−1 + det n+1
xn xn+1 xn xn+1 xn xn+1
= −1 + 1 = 0.
Thus (xn−1 + xn+1 , xn + xn+2 ) = an (xn , xn+1 ) for some scalar an . Hence
xn+1 + xn−1 xn+2 + xn
=
xn xn+1
for all n ≥ 1, since xn , xn+1 = 0, so we are done by induction. 
Remark. In a similar manner, one can prove that if (xn )n≥0 is a sequence of
nonzero real numbers such that
 
xn xn+1 · · · xn+k
xn+1 xn+2 · · · xn+k+1 
  n
det  . .. .. ..  = cr
 .. . . . 
xn+k xn+k+1 · · · xn+2k
for n = 1, 2, 3, . . . , then there exist real numbers a1 , . . . , ak such that
xn+k+1 = a1 xn+k + a2 xn+k−1 + · · · + ak xn+1 + (−1)k rxn .

Related question. The Fibonacci numbers have a similar structure: Fn+1 − Fn −

Fn−1 = 0, and Fn2 − Fn−1 Fn+1 = (−1)n+1 . How do the solutions generalize to this
case? Can they be generalized even further?

A3. (2, 11, 24, 0, 0, 0, 0, 0, 27, 8, 54, 81)

Let Pn be the set of subsets of {1, 2, . . . , n}. Let c(n, m) be the number of
functions f : Pn → {1, 2, . . . , m} such that f (A ∩ B) = min{f (A), f (B)}. Prove
that
m
c(n, m) = jn.
j=1

Solution 1. Let S = {1, 2, . . . , n} and Si = S − {i}. First we show that

f !→ (f (S), f (S1 ), f (S2 ), . . . , f (Sn )) defines a bijection between the set of allowable
functions f and the set of (n + 1)-tuples (j, a1 , a2 , . . . , an ) of elements of {1, 2, . . . , m}
such that j ≥ ai for i = 1, 2, . . . , n. This map is well defined, since for any allowable
f , we have f (Si ) = f (Si ∩ S) = min{f (Si ), f (S)} ≤ f (S). The inverse map takes the
tuple (j, a1 , a2 , . . . , an ) to the function f such that f (S) = j and f (T ) = mini∈T ai
for T  S. This function satisfies f (A ∩ B) = min{f (A), f (B)}, and these two
constructions are inverse to each other.
Now we count the (n + 1)-tuples. For fixed j, the possibilities for each ai are
1, 2, . . . , j, so there are j n possibilities for (a1 , a2 , . . . , an ). Summing over j, which can
m
be anywhere from 1 to m, yields c(n, m) = j=1 j n . 
Solution 2. We will use induction on m. The base case, m = 1, states that
c(n, 1) = 1 for all n, which is obvious.
174 The William Lowell Putnam Mathematical Competition

Suppose m ≥ 2. Given f : Pn → {1, 2, . . . , m} not identically 1 such that

f (A ∩ B) = min{f (A), f (B)}, let Sf be the intersection of all A ∈ Pn such that
f (A) ≥ 2. The property implies that f (Sf ) ≥ 2, and that if T ⊇ Sf then f (T ) ≥ 2.
Hence the sets T for which f (T ) ≥ 2 are exactly those that contain Sf .
Given S ∈ Pn , how many f ’s have Sf = S? To give such an f is the same as
specifying a function

g : { T ∈ Pn : T ⊇ S } → {2, 3, . . . , m}

satisfying g(A ∩ B) = min{g(A), g(B)}. There is an intersection-preserving bijection

from { T ∈ Pn : T ⊇ S } to the set of subsets of {1, 2, . . . , n} − S (the bijection maps
T to T − S), and an order-preserving bijection from {2, 3, . . . , m} to {1, 2, . . . , m − 1},
so the number of such functions g equals c(n − #S, m − 1).
Thus, remembering to count the identically 1 function, we have

c(n, m) = 1 + c(n − #S, m − 1)

S

n
n
=1+ c(n − k, m − 1)
k
k=0
n
 m−1
n
=1+ j n−k (inductive hypothesis)
k j=1
k=0
m−1 n

n n−k
=1+ j
j=1
k
k=0
m−1
=1+ (1 + j)n
j=1
m
= jn,
j=1

completing the inductive step. 

A4. (3, 2, 0, 0, 0, 0, 0, 0, 0, 0, 44, 158)

Let x1 , x2 , . . . , x19 be positive integers each of which is less than or equal
to 93. Let y1 , y2 , . . . , y93 be positive integers each of which is less than or
equal to 19. Prove that there exists a (nonempty) sum of some xi ’s equal
to a sum of some yj ’s.
Solution 1. We move a pebble among positions numbered −18, −17, . . . , 0, 1,
2, . . . , 93, until it revisits a location. The pebble starts at position 0. Thereafter, if
the pebble is at position t, we move it as follows. If t ≤ 0, choose some unused xi ,
move the pebble to t + xi , and then discard that xi . If t > 0, choose some unused yj ,
move the pebble to t − yj , and then discard that yj . Since xi ≤ 93 and yj ≤ 19, the
pebble’s position stays between −18 and 93.
In order to continue this process until a location is revisited, we must show that
there is always an unused xi or yj as needed. If t ≤ 0 and a revisit has not yet
occurred, then one xi has been used after visiting each nonpositive position except the
Solutions: The Fifty-Fourth Competition (1993) 175

current one, so the total number of xi ’s used so far is at most 19 − 1 = 18, and at
least one xi remains. Similarly, if t > 0 and a revisit has not yet occurred, then one
yj has been used after visiting each positive position except the current one, so the
total number of yj ’s used so far is at most 93 − 1 = 92, and at least one yj remains.
Since there are only finitely many xi ’s and yj ’s to be used, the algorithm must
eventually terminate with a revisit. The steps between the two visits of the same
position constitute a sum of some xi ’s equal to a sum of some yj ’s. 
Our next solution is similar to Solution 1, but we dispense with the algorithmic
interpretation.

Solution 2. For the sake of generality, replace 19 and 93 in the problem statement
k !
by m and n respectively. Define Xk = i=1 xi and Y! = j=1 yj . Without loss of
generality, assume Xm ≥ Yn . For 1 ≤ @ ≤ n, define f (@) by

Xf (!) ≤ Y! < Xf (!)+1 ,

so 0 ≤ f (@) ≤ m. Let g(@) = Y! −Xf (!) . If g(@) = 0 for some @, we are done. Otherwise,

g(@) = Y! − Xf (!) < xf (!)+1 ≤ n,

so 0 < g(@) ≤ n − 1 whenever 1 ≤ @ ≤ n. Hence by the Pigeonhole Principle, there

exist @0 < @1 such that g(@0 ) = g(@1 ). Then
f (!1 ) !1
xi = Xf (!1 ) − Xf (!0 ) = Y!1 − Y!0 = yj . 
i=f (!0 )+1 j=!0 +1

Solution 3 (based on an idea of Noam Elkies). With the same notation as

in the previous solution, without loss of generality Xm ≥ Yn . If equality holds, we are
done, so assume Xm > Yn . By the Pigeonhole Principle, two of the (m + 1)(n + 1)
sums Xi + Yj (0 ≤ i ≤ m, 0 ≤ j ≤ n) are congruent modulo Xm , say

Xi1 + Yj1 ≡ Xi2 + Yj2 (mod Xm ).

But the difference

(Xi1 − Xi2 ) + (Yj1 − Yj2 ) (1)

lies strictly between −2Xm and +2Xm , so it equals 0 or ±Xm . Clearly i1 = i2 ;

without loss of generality i1 > i2 , so (1) equals 0 or Xm . If j1 < j2 , then (1) must be
0, so Xi1 − Xi2 = Yj2 − Yj1 are two equal subsums. If j1 > j2 , then (1) must be Xm ,
and Yj1 − Yj2 = Xm − (Xj1 − Xj2 ) are two equal subsums. 

A5. (3, 3, 0, 0, 0, 0, 0, 0, 1, 16, 37, 147)

Show that
 −10
2 1

 11 2  11
2
x2 − x x2 − x 10 x2 − x
dx + dx + dx
−100 x3 − 3x + 1 1
101
x3 − 3x + 1 101
100
x3 − 3x + 1

is a rational number.
176 The William Lowell Putnam Mathematical Competition

Solution 1. The polynomial x3 − 3x + 1 changes sign in each of the intervals

[−2, −1], [1/3, 1/2], [3/2, 2], so it has no zeros outside these intervals. Hence in the
problem, the integrand is continuous on the three ranges of integration.
By the substitutions x = 1/(1 − t) and x = 1 − 1/t, the integrals over [1/101, 1/11]
and [101/100, 11/10] are respectively converted into integrals over [−100, −10]. The
integrand

2
x2 − x
Q(x) = ,
x3 − 3x + 1
is invariant under each of the substitutions x → 1/(1 − x) and x → 1 − 1/x. Hence
the sum of the three given integrals is expressible as
 −10
2

x2 − x 1 1
1 + + dx. (1)
−100 x3 − 3x + 1 x2 (1 − x)2
But
2
1 1 1
= x+1− − ,
Q(x) x x−1

so the last integral is of the form u−2 du. Hence its value is
−10
x2 − x
− ,
x3 − 3x + 1 −100

which is rational. 

Solution 2. Set
 t
2  1/(1−t)
2
x2 − x x2 − x
f (t) = dx + dx
−100 x3 − 3x + 1 1
101
x3 − 3x + 1
 1−1/t
2
x2 − x
+ dx
101
100
x3 − 3x + 1

for −100 ≤ t ≤ −10. We want f (−10). By the Fundamental Theorem of Calculus,



 1 1 1 1
f (t) = Q(t) + Q +Q 1− .
1 − t (1 − t)2 t t2
We find that Q(1/(1−x)) = Q(1−1/x) = Q(x) (in fact x !→ 1/(1−x) and x !→ 1−1/x
are inverses of each other), so
 −10
2

x2 − x 1 1
f (−10) = 1+ 2 + dx.
−100 x3 − 3x + 1 x (1 − x)2
The integrand equals
x4 − 2x3 + 3x2 − 2x + 1
.
(x3 − 3x + 1)2
We guess that this is the derivative of a quotient
Ax2 + Bx + C
x3 − 3x + 1
Solutions: The Fifty-Fourth Competition (1993) 177

and solve for the undetermined coefficients, obtaining A = −1, B = 1, C = 0. Thus

−10
−x2 + x
f (−10) = 3 ,
x − 3x + 1 −100

which is rational. 
Remark. Both solutions reached the formula (1) in the middle, so we could
construct two more solutions by matching the first half of Solution 1 with the second
half of Solution 2, or vice versa.
11131110
Remark. The actual value of the sum of the three integrals is .
107634259
Related question. Let k be a positive integer, and suppose f (x) ∈ Q[x] has degree
at most 3k − 2. Then
 −10  111  11
f (x) f (x) 10 f (x)
dx + dx + dx (1)
−100 (x − 3x + 1) (x − 3x + 1) 101 (x − 3x + 1)k
3 k 1 3 k 3
101 100

is rational.
Proof. We will use the theory of differentials on the projective line P1 . Let
τ (x) = 1 − 1/x. Then τ , as an automorphism of P1 , permutes the three zeros of
p(x) = x3 − 3x + 1 cyclically, since a calculation shows p(1 − 1/x) = −x−3 p(x). In
particular, τ 3 is the identity, since an automorphism of P1 is determined its action on
three distinct points, or by direct calculation.
f (x)
k dx on P . Because deg f ≤ 3k − 2, the
1
Let ω be the differential (x3 −3x+1)
substitution x = 1/t, dx = −t−2 dt shows that ω has no pole at ∞, so ω is regular on
all of P1 except possibly at the zeros of x3 − 3x + 1.  −10
The substitutions used in Solution 1 transform (1) into −100 η where η = ω + τ ∗ ω +
(τ 2 )∗ ω. Then η is regular outside the zeros of x3 − 3x + 1. Since τ ∗ η = η, the residues
of η at the three zeros are equal, but the sum of the residues of any differential is
zero, so all the residues are zero. Hence η = dg for some rational function g on P1
having poles at most at the zeros of x3 − 3x + 1. Adding a constant to g, we may
assume g(0) = 0. We want to show that g has rational coefficients. Write g in lowest
terms, with monic denominator. Any automorphism σ of C over Q applied to the
coefficients of g results in another rational function g1 with dg1 = η and g1 (0) = 0.
Then d(g − g1 ) = 0, so g − g1 is constant, and evaluating
 −10 at 0 shows that g = g1 . This
holds for all σ, so g ∈ Q(x). Finally, (1) equals −100 dg = g(−10) − g(−100), which
is rational. 

Remark. We prove that if instead f (x) ∈ Q[x] has degree exactly 3k − 1,

then (1) is irrational. Such f (x) can be expressed as a nonzero rational multiple
of f1 (x) = (3x2 − 3)(x3 − 3x + 1)k−1 plus a polynomial in Q[x] of degree at most
3k − 2, so it suffices to consider f (x) = f1 (x). Then (1) becomes


−10 1/11 11/10 34719358188367000
ln p(x) −100 + ln p(x) 1/101 + ln p(x) 101/100 = ln ,
49251795138277608547
which is irrational, since the only rational number q > 0 with ln q ∈ Q is q = 1.
Remark. One can strengthen “irrational” to “transcendental” in the previous
remark, since Lindemann proved in 1882 that all nonzero numbers that are logarithms
178 The William Lowell Putnam Mathematical Competition

of algebraic numbers are transcendental [FN, Corollary 2.2]. (Recall that a complex
number is algebraic if it is the zero of some nonzero polynomial with rational
coefficients, and transcendental otherwise.)
Remark. We can explain why for the polynomial p(x) = x3 − 3x + 1 there was a
fractional linear transformation with rational coefficients permuting its zeros cyclically.
An automorphism of P1 is uniquely determined by its action on any three distinct
points, so given any cubic polynomial p(x) ∈ Q[x] with distinct zeros, there is a unique
fractional linear transformation τ ∈ L(x) permuting the zeros, where L is the splitting
field of p(x). If σ ∈ Gal(L/Q), then σ acts on τ , and the action of στ on the three zeros
is obtained by conjugating the action of τ by the permutation of the zeros given by σ.
In particular, στ = τ for all σ ∈ Gal(L/Q) if and only if the Gal(L/Q) is contained in
the cyclic subgroup of order 3 of the permutations of the three zeros. Hence τ ∈ Q(x)
if and only if the discriminant of p(x) is a square. In the problem, the discriminant of
p(x) = x3 − 3x + 1 is 81.
Remark. This action of the group of order three on the projective line appeared
in Problem 1971B2 [PutnamII, p. 15]:
Let F (x) be a real valued function defined for all real x except for x = 0 and
x = 1 and satisfying the functional equation F (x) + F ((x − 1)/x) = 1 + x.
Find all functions F (x) satisfying these conditions.

A6. (0, 1, 0, 0, 0, 0, 1, 3, 3, 11, 36, 152)

The infinite sequence of 2’s and 3’s

2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 2, 3, 3, 3, 2, . . .

has the property that, if one forms a second sequence 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+rm for some nonnegative
integer m. (Note: x denotes the largest integer less than or equal to x.)
Motivation. Assuming the result, we derive the value of r. Fix a large integer m.
The result implies that the mth 2 in the sequence occurs at the nth term, where
n ≈ rm. Interleaved among these m 2’s are n − m 3’s. By the self-generation
property of the sequence, these 2’s and 3’s describe an initial segment with n 2’s
separated by blocks containing a total of 2m + 3(n − m) 3’s. But the length of
this segment should also approximate r times the number of 2’s in the segment, so
n + 2m + 3(n − m) ≈ rn. Substituting n ≈ rm, dividing by m, and taking the limit√as
m → ∞ yields r + 2 + 3(r − 1) = r · r, so r2 − 4r + 1 = 0. Clearly r ≥ 1, so r = 2 + 3.
Note that 3 < r < 4.
√
Solution. Let r = 2 + 3. The sequence is uniquely determined by the
self-generation property and its first few terms. Therefore it suffices to show that
the sequence a0 , a1 , a2 , . . . has the self-generation property when we define an = 2 if
n = rm for some m, and an = 3 otherwise. (Note that the first term is a0 .)
Solutions: The Fifty-Fourth Competition (1993) 179

The self-generation property for (an ) is equivalent to

3 if n = rm for some m,
r(n + 1) − rn =
4 otherwise.
Since r(n + 1) − rn is always 3 or 4, we seek to prove:
n is of the form rm ⇐⇒ r(n + 1) − rn = 3.
This is a consequence of the following series of equivalences, in which we use the
identity
z/r = 4z − rz = 4z − rz (1)
for integer z:
n is of the form rm.
⇐⇒ There is an m for which n + 1 > rm > n.
⇐⇒ There is an m for which (n + 1)/r > m > n/r.
⇐⇒ (n + 1)/r − n/r = 1.
⇐⇒ 4(n + 1) − (n + 1)r − 4n + nr = 1 (by (1) with z = n and z = n + 1).
⇐⇒ (n + 1)r − nr = 3. 
Remark. A similar argument
√ shows that an = 3 if and only if n has the
form sm where s = (1 + 3)/2 and m is a positive integer. Hence the positive
integers are disjointly partitioned into the two sequences (rm)m>0 and (sm)m>0 .
Two sequences forming a partition of the positive integers are called complementary
sequences. These are discussed in [Hon1, Essay 12].
Complementary sequences of the form (rm) and (sm) are called Beatty se-
quences. A short argument shows that if (rm) and (sm) are complementary, then
r and s are irrational and 1/r + 1/s = 1. Conversely, Problem 1959A6 [PutnamI,
p. 57] asks:
Prove that, if x and y are positive irrationals such that 1/x + 1/y = 1, then
the sequences x, 2x, . . . , nx, . . . and y, 2y, . . . , ny, . . . together
include every positive integer exactly once.
This result is known as Beatty’s Theorem. For further discussion, see [PutnamI,
p. 514]. For another Beatty-related problem, see Problem 1995B6.
Related question. How far can you generalize this? For example, the infinite
sequence of 1’s and 2’s starting 1, 2, 1, 2, 2, 1, . . . (where the leading term is considered
the 0th term) has the property that, if one forms a second sequence that records the
number of 2’s between successive 1’s, the result is identical to the identical to the given
sequence.√Then the nth term of the sequence is 1 if and only if n = mτ 2 , where
τ = (1 + 5)/2 is the golden mean.
Related questions. This problem is reminiscent of Problem 3 on the 1978 Interna-
tional Mathematical Olympiad [IMO79–85, p. 1]:
The set of all positive integers is the union of two disjoint subsets
{f (1), f (2), . . . , f (n), . . . }, {g(1), g(2), . . . , g(n), . . . },
180 The William Lowell Putnam Mathematical Competition

where
f (1) < f (2) < · · · < f (n) < · · · ,

g(1) < g(2) < · · · < g(n) < · · · ,

and g(n) = f (f (n)) + 1 for all n ≥ 1. Determine f (240).

The two subsets again turn out to form a Beatty sequence.
Another related result, due to E. Dijkstra [Hon3, p. 13], is the following.
Suppose f (n), n = 0, 1, . . . is a nondecreasing sequence of nonnegative integers
which is unbounded above. Let the “converse function” g(n) (n ≥ 0) be the
number of values m for which f (m) ≤ n. Then the converse function of g is f ,
and the sequences F (n) = f (n) + n and G(n) = g(n) + n are complementary.
Yet another problem, related to the previous two, is the following [New, Problem 47]:
Suppose we “sieve” the integers as follows: take a1 = 1 and then delete
a1 + 1 = 2. The smallest untouched integer is 3, which we call a2 , and then
we delete a2 + 2 = 5. The next untouched integer is 4 = a3 , and we delete
a3 + 3 = 7, etc. The sequence a1 , a2 , . . . is then

1, 3, 4, 6, 8, 9, 11, 12, 14, 16, 17, . . . .

Find a formula for an .

The sieve just described arises in the analysis of Wythoff’s game with two heaps [Wy],
[BCG, pp. 62, 76], [Hon1].

B1. (83, 29, 9, 0, 0, 0, 0, 0, 6, 10, 38, 32)

Find the smallest positive integer n such that for every integer m, with
0 < m < 1993, there exists an integer k for which
m k m+1
< < .
1993 n 1994

Answer. The smallest positive integer n satisfying the condition is n = 3987.

a
Lemma 1. Suppose a, b, c, and d are positive numbers, and b < dc . Then
a a+c c
< < .
b b+d d
This lemma is sometimes called the mediant property or the règle des nombres
moyens [Nel, pp. 60–61]. It appears without proof in the Triparty en la Science
des Nombres, which was written by the French physician Nicholas Chuquet in 1484
(although his work went unpublished until 1880). It is not hard to verify this inequality
directly, but there is an even easier way of remembering it: If a sports team wins a
of b games in the first half of a season, and c of d games in the second half, then its
overall record ((a + c)/(b + d)) is between its records in its two halves (a/b and c/d).
Alternatively, Figure 24 gives a “proof without words.” For three more proofs without
words of this result, see [Nel, pp. 60–61].
Solutions: The Fifty-Fourth Competition (1993) 181

(c, d ) (a + c, b + d )

(a, b)

FIGURE 24.
A proof without words of Lemma 1.

Solution 1. By Lemma 1,
m 2m + 1 m+1
< < .
1993 3987 1994
We will show that 3987 is best possible. If
1992 k 1993
< < (1)
1993 n 1994
then
1 n−k 1
> > ,
1993 n 1994
so
n
1993 < < 1994.
n−k
Clearly n − k = 1, so n − k ≥ 2. Thus n > 1993(n − k) ≥ 3986, and n ≥ 3987. 
Solution 2. Subtracting everything in the desired inequality from 1, and using
the change of variables M = 1993 − m, K = n − k, the problem becomes: determine
the smallest positive integer n such that for every integer M with 1993 > M > 0,
there exists an integer K for which
M K M
> > , or equivalently, 1993K < nM < 1994K.
1993 n 1994
For M = 1, K cannot be 1 and hence is at least 2, so n > 1993 · 2 = 3986. Thus
n ≥ 3987. On the other hand n = 3987 works, since then for each M , K = 2M
satisfies the inequalities. 
Solution 3 (Naoki Sato).
Lemma 2. Let a, b, c, d, p, and q be positive integers such that a/b < p/q < c/d,
and bc − ad = 1. Then p ≥ a + c and q ≥ b + d.
Proof. Since bp − aq > 0, bp − aq ≥ 1. Also, cq − dp > 0, so cq − dp ≥ 1. Hence
d(bp − aq) + b(cq − dp) ≥ b + d, which simplifies to (bc − ad)q ≥ b + d. But bc − ad = 1,
so q ≥ b + d. The proof of p ≥ a + c is similar. 
Now
1992 k 1993
< <
1993 n 1994
for some k, and 1993·1993−1992·1994 = 1, so n ≥ 1993+1994 = 3987. And n = 3987
works, by Lemma 1. 
182 The William Lowell Putnam Mathematical Competition

Remark. Looking at (1) was key. What clues suggest that it would be helpful to
look at large m?
Remark. This problem and Lemma 2 especially are related to the beautiful topic
of Farey series; see [Hon1, Essay 5].

B2. (89, 0, 1, 0, 0, 0, 0, 0, 7, 3, 42, 65)

Consider the following game played with a deck of 2n cards numbered
from 1 to 2n. The deck is randomly shuffled and n cards are dealt to each of
two players, A and B. Beginning with A, the players take turns discarding
one of their remaining cards and announcing its number. The game ends as
soon as the sum of the numbers on the discarded cards is divisible by 2n+1.
The last person to discard wins the game. Assuming optimal strategy by
both A and B, what is the probability that A wins?
Answer. The probability that A wins is 0, i.e., B can always win.

Solution. Player B can always win, because B can always guarantee that A will
not win on the next move: B holds one more card than A, and each of A’s cards
causes at most one of B’s cards to be a fatal play. Hence B has at least one safe play.
Player B wins on the last move if not earlier, since the sum of the numbers on all the
cards is n(2n + 1). 
Remark. David Savitt gives the following ‘alternate solution’: “I maintain that A,
playing optimally, will decline to participate!”

B3. (111, 16, 13, 0, 0, 0, 0, 0, 4, 20, 10, 33)

Two real numbers x and y are chosen at random in the interval (0, 1)
with respect to the uniform distribution. What is the probability that the
closest integer to x/y is even? Express the answer in the form r + sπ, where
r and s are rational numbers.
Answer. The limit is (5 − π)/4. That is, r = 5/4 and s = −1/4.

Solution. The probability that x/y is exactly half an odd integer is 0, so we may
safely ignore this possibility.
The closest integer to xy is even if and only if 0 < xy < 12 or 4n−1 2 < xy < 4n+12 for
some integer n ≥ 1. The former occurs inside the triangle with vertices (0, 0), (0, 1),
( 12 , 1), whose area is 14 . The latter occurs inside the triangle (0, 0), (1, 4n−1
2 2
), (1, 4n+1 ),
whose area is 4n−1 − 4n+1 . These regions are shown in Figure 25.
1 1

Hence the total area is

1 1 1 1
+ − + + ··· .
4 3 5 7
Comparing this with Leibniz’s formula
π 1 1 1
= 1 − + − + ···
4 3 5 7
shows that the total area is (5 − π)/4. 
Solutions: The Fifty-Fourth Competition (1993) 183

(1/2, 1)

(1, 2/3)

(1, 2/5)
(1, 2/7)
(1, 2/9)

…
FIGURE 25.

Remark. Leibniz’s formula can be derived as follows:

π
= arctan 1
4
 1
dx
= 2
0 1+x
 1
 
= (1 − x2 ) + (x4 − x6 ) + (x8 − x10 ) + · · · dx
0
 1  1  1
= (1 − x2 ) dx + (x4 − x6 ) dx + (x8 − x10 ) dx + · · ·

0 
0 
 0
1 1 1 1 1
= 1− + − + − + ··· .
3 5 7 9 11

The interchange of integral and summation in the penultimate step is justified by the
Monotone Convergence Theorem [Ru, p. 319], since each integrand x4k − x4k+2 is
nonnegative on [0, 1].
Remark. This gives an interesting connection between π and number theory (and
gives an “experimental” way to estimate π). Here are more links. In fact, 6/π 2 is both
the probability that a pair of positive integers are relatively prime, and the probability
that a positive integer is squarefree (where these probabilities are appropriately defined
as limits). More generally, for any k ≥ 2, the probability that a set of k positive
integers has greatest common divisor 1, and the probability that a positive integer is
kth-power-free, is 1/ζ(k), where

1 1 1
ζ(k) = + k + k + ···
1k 2 3
is the Riemann zeta function. (See [NZM, Theorem 8.25] for a proof in the case n = 2;
the general case is similar.) It can be shown that ζ(2n) is a rational multiple of π 2n
184 The William Lowell Putnam Mathematical Competition

for all positive integers n, and in fact

22n−1 Bn 2n
ζ(2n) = π
(2n)!
where Bn is the nth Bernoulli number [HW, Section 17.2]. (See 1996B2 for another
appearance of Bernoulli numbers.) The special case ζ(2) = π 2 /6 can be computed
using Fourier analysis [Fo, Exercise 8.17 or 8.21], trigonometry [NZM, Appendix A.3],
or multivariable change of variables [Sim, p. 748].

B4. (0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 61, 144)

The function K(x, y) is positive and continuous for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1,
and the functions f (x) and g(x) are positive and continuous for 0 ≤ x ≤ 1.
Suppose that for all x, 0 ≤ x ≤ 1,
 1  1
f (y)K(x, y) dy = g(x) and g(y)K(x, y) dy = f (x).
0 0

Show that f (x) = g(x) for 0 ≤ x ≤ 1.

Solution. For convenience of notation, define the linear operator T by
 1
(T h)(x) = h(y)K(x, y) dy
0

for any continuous function h on [0, 1]. Then T f = g and T g = f .

By the Extreme Value Theorem [Ap1, Section 3.16], the continuous function
f (x)/g(x) attains a minimum value on [0, 1], say r. Thus f (x) − rg(x) ≥ 0 on [0, 1],
with equality somewhere. Suppose that f − rg is not identically zero on [0, 1]. Then
by continuity f − rg is positive on some interval, so T (f − rg) is positive on [0, 1],
and so is T 2 (f − rg). But by linearity, T 2 (f − rg) = T (g − rf ) = f − rg, which
is zero somewhere. This contradiction shows that f − rg = 0. Thus T g = rg, and
g = T 2 g = r2 g, so r2 = 1. Also r ≥ 0, so r = 1. Hence f = g. 
Remark. This result also follows from a continuous analogue of the Perron-
Frobenius Theorem, that every square matrix with positive entries has a unique
positive eigenvector. Namely, if F (x, y) is a continuous function from [0, 1] × [0, 1]
to R+ , then
 there exists a continuous function f (x) from [0, 1] to R+ such that
cf (x) = 0,1 F (x, y)f (y) dy for some c > 0, and f is unique up to a scalar multiple. The
article of Birkhoff [Bi] cited in the remarks following 1997A6 proves this continuous
analogue as well as the original theorem.
To apply this result to the given problem, note that T 2 has eigenvectors f and g
which are both positive of eigenvalue 1. The result implies then that f and g are
scalar multiples of each other, so f is an eigenvector of T of eigenvalue ±1. Clearly
the plus sign is correct, so f = T f = g.
Stronger result. For further discussion along these lines, see [Hol], where a short
proof of the following generalization is given.
Let X be a normed linear space, and let C be a strict cone in X. Let T be a
linear operator on X mapping C − {0} into the interior of C. If x ∈ C and
T k x = x for some integer k ≥ 1, then T x = x.
Solutions: The Fifty-Fourth Competition (1993) 185

B5. (6, 1, 0, 0, 0, 0, 0, 0, 1, 7, 65, 127)

Show there do not exist four points in the Euclidean plane such that the
pairwise distances between the points are all odd integers.

All of the following solutions involve finding a contradiction modulo some power
of 2. The first solution is long, but uses little more than coordinate geometry.
Solution 1. For real numbers x and y, and for a positive integer n, let “x ≡ y
(mod n)” mean that x − y is an integer divisible by n. Choose a coordinate system in
which the four points are (0, 0), (a, 0), (r, s), (x, y). Here a is an odd integer, and we
may assume a > 0. The square of an odd integer is congruent to 1 modulo 8, so if all
the pairwise distances are odd integers, we have
r2 + s2 ≡ 1 (mod 8)
(r − a) + s ≡ 1
2 2
(mod 8)
x +y ≡1
2 2
(mod 8)
(x − a) + y ≡ 1
2 2
(mod 8)
(x − r) + (y − s) ≡ 1
2 2
(mod 8).
Subtracting the first two yields 2ar ≡ a2 (mod 8). Thus r is a rational number whose
denominator is a multiple of 2 and a divisor of 2a. The same is true of x. Therefore
we can multiply all coordinates by the odd integer a, to reduce to the case where the
denominators of r and x are both 2. Then the congruence 2ar ≡ a2 (mod 8) between
integers implies r ≡ a/2 (mod 4). If r = a/2 + 4b, then
r2 = a2 /4 + 4ab + 16b2 ≡ a2 /4 (mod 4),
so the first congruence implies
s2 ≡ 1 − r2 ≡ 1 − a2 /4 (mod 4).
Similarly x ≡ a/2 (mod 4) and y 2 ≡ 1 − a2 /4 (mod 4). Also
x − r ≡ a/2 − a/2 ≡ 0 (mod 4),
so (x − r)2 ∈ 16Z, and the last of the five congruences yields
(y − s)2 ≡ 1 − (x − r)2 ≡ 1 (mod 8).
We will derive a contradiction from the congruences s2 ≡ y 2 ≡ 1 − a2 /4 (mod 4)
and (y − s)2 ≡ 1 (mod 8) obtained above. First,
(y + s)2 ≡ 2y 2 + 2s2 − (y − s)2
≡ 2(1 − a2 /4) + 2(1 − a2 /4) − 1
≡ 3 − a2
≡2 (mod 4).
Multiplying this integer congruence by (y − s)2 ≡ 1 (mod 8) yields (y 2 − s2 )2 ≡ 2
(mod 4). But y 2 and s2 are rational by the beginning of this paragraph, so y 2 − s2 is
a rational number with square congruent to 2 modulo 4. This is impossible. 
186 The William Lowell Putnam Mathematical Competition

Solution 2.
Lemma. If r = cos α, s = cos β, and t = cos(α+β), then 1−r2 −s2 −t2 +2rst = 0.
Proof. We have

cos(α + β) = cos α cos β − sin α sin β,

(cos(α + β) − cos α cos β)2 = sin2 α sin2 β,
(t − rs)2 = (1 − r2 )(1 − s2 ),

from which the result follows. 

Suppose that O, A, B, C are four points in the plane such that a = OA, b = OB,
c = OC, x = BC, y = CA, z = AB are all odd integers. Using the Law of Cosines,
let
a2 + b2 − z 2
r = cos ∠AOB =
2ab
b2 + c2 − x2
s = cos ∠BOC =
2bc
c2 + a2 − y 2
t = cos ∠AOC = .
2ca
But ∠AOB + ∠BOC = ∠AOC as directed angles, so the lemma implies 1 − r2 − s2 −
t2 + 2rst = 0. Substituting the values of r, s, t, and multiplying by 4a2 b2 c2 yields

4a2 b2 c2 − c2 (a2 + b2 − z 2 )2 − a2 (b2 + c2 − x2 )2 − b2 (c2 + a2 − y 2 )2

+ (a2 + b2 − z 2 )(b2 + c2 − x2 )(c2 + a2 − y 2 ) = 0.

The square of an odd integer is 1 modulo 4, so we obtain

4−1−1−1+1≡0 (mod 4),

a contradiction. 

Solution 3 (Manjul Bhargava). Suppose -v1 , -v2 , -v3 are vectors in 3-space. Then
the volume V of the parallelepiped spanned by the vectors is given by |-v1 · (-v2 ×-v3 )| =
| det M |, where M = (-v1-v2-v3 ) is the 3 × 3 matrix with entries given by the vectors -v1 ,
-v2 , -v3 . Since det M = det M T ,
 
-v1 · -v1 -v1 · -v2 -v1 · -v3
V 2 = M · M T = det -v2 · -v1 -v2 · -v2 -v2 · -v3  . (1)
-v3 · -v1 -v3 · -v2 -v3 · -v3
The volume of the tetrahedron spanned by the vectors is V /6. If the edges of the
tetrahedron are a = |-v1 |, b = |-v2 |, c = |-v3 |, x = |-v2 − -v3 |, y = |-v3 − -v1 |, z = |-v1 − -v2 |,
then by the Law of Cosines, written as

2-vi · -vj = |-vi |2 + |-vj |2 − |-vi − -vj |2 , (2)

 
2a2 a2 + b2 − z 2 a2 + c2 − y 2
8V 2 = det a2 + b2 − z 2 2b2 b2 + c2 − x2  .
a +c −y b +c −x
2 2 2 2 2 2
2c2
Solutions: The Fifty-Fourth Competition (1993) 187

Suppose now that there were 4 points as described in the problem. Locate one point
at the origin, and let -v1 , -v2 , -v3 be the vectors from the origin to the other three; they
are coplanar, so V = 0. Since squares of odd integers are congruent to 1 modulo 8,
 
2 1 1
8V 2 ≡ det 1 2 1 ≡ 4 (mod 8),
1 1 2

which gives a contradiction. 

Remark. Suppose -v1 , -v2 , . . . , -vn are vectors in Rm . Then the n-dimensional
volume V of the parallelepiped spanned by the vectors is given by
 
-v1 · -v1 · · · -v1 · -vn
2  .. .. ..  .
V = det  . . .  (3)
-vn · -v1 · · · -vn · -vn

(Note that m need not equal n.) This generalizes (1).

Solution 4. Suppose as before there were 4 such points; define -v1 , -v2 , -v3 as in
the previous solution. Then -vi · -vi ≡ 1 (mod 8), and from (2), 2-vi · -vj ≡ 1 (mod 8) for
i = j as well.
No three of the points can be collinear. Hence -v3 = x-v1 + y-v2 for some scalars x
and y. Then

2-v1 · -v3 = 2x-v1 · -v1 + 2y-v1 · -v2

2-v2 · -v3 = 2x-v2 · -v1 + 2y-v2 · -v2 (4)
2-v3 · -v3 = 2x-v3 · -v1 + 2y-v3 · -v2 .

Since -v1 is not a scalar multiple of -v2 ,


-v1 · -v1 -v1 · -v2
det >0
-v2 · -v1 -v2 · -v2

(by the two-dimensional version of (3)) so the first two equations in (4) have a unique
rational solution for x, y, say x = X/D, y = Y /D, where X, Y , and D are integers.
We may assume gcd(X, Y, D) = 1. Then multiplying (4) through by D we have

D ≡ 2X + Y (mod 8)
D ≡ X + 2Y (mod 8)
2D ≡ X + Y (mod 8).

Adding the first two congruences and subtracting the third gives 2X+2Y ≡ 0 (mod 8),
so, by the third congruence, D is even. But then the first two congruences force X
and Y to be even, giving a contradiction. 
188 The William Lowell Putnam Mathematical Competition

Solution 5. If P0 , . . . , Pn are the vertices of a simplex in Rn , and dij = (Pi Pj )2 ,

then its volume V satisfies
 
0 1 1 ··· 1
1 d00 d01 · · · d0n 
n+1  
(−1) 1 d10 d11 · · · d1n 
V2 = n det  .
2 (n!)2 . .. .. .. .. 
 .. . . . . 
1 dn0 dn1 · · · dnn
This is proved on page 98 of [Bl], where this determinant is called a Cayley-Menger
determinant. In particular, the formula for n = 3 (which goes back to Euler) implies
that the edge lengths a, b, c, x, y, z (labelled as in Solution 2) of a degenerate
tetrahedron satisfy  
0 1 1 1 1
 1 0 z 2 y 2 a2 
 
det 
1 z
2
0 x2 b2   = 0.
1 y 2 x2 0 c2 
1 a2 b2 c2 0
If a, b, c, x, y, z are odd integers, we get a contradiction modulo 8. 
Remark. See 1988B5 for another example of showing a matrix is invertible by
reducing modulo some number.
Remark. Either Solution 3 or 5, along with a formula from Solution 3 of 1992B5,
can be used to prove the following generalization:
Suppose n is an integer not divisible by 8. Show that there do not exist n points
in Rn−2 such that the pairwise distances between the points are all odd integers.
On the other hand, we do not know whether or not there exist 8 points in R6 such
that the pairwise distances between the points are all odd integers.

B6. (2, 0, 0, 0, 0, 0, 0, 0, 1, 2, 59, 143)

Let S be a set of three, not necessarily distinct, positive integers. Show
that one can transform S into a set containing 0 by a finite number of
applications of the following rule: Select two of the three integers, say x
and y, where x ≤ y and replace them with 2x and y − x.
The following three solutions seem related only by the central role played by powers
of 2.
Solution 1 (attributed to Garth Payne). It suffices to show that (a, b, c) with
0 < a ≤ b ≤ c can be transformed into (b , a , c ), where a is the remainder when b
is divided by a. (Hence any set with no zeros can be transformed into a set with a
smaller element.)
Let b = qa + r with r < a; let q = q0 + 2q1 + 4q2 + · · · + 2k qk be the binary
representation of q (so qi = 0 or 1, and qk = 1). Define g0 (d, e, f ) = (2d, e, f − d) and
g1 (d, e, f ) = (2d, e − d, f ). Then
gqk (· · · (gq1 (gq0 (a, b, c))) · · · ) = (b , r, c )
describes a sequence of legal moves. 
Solutions: The Fifty-Fourth Competition (1993) 189

Solution 2. Let a, b, c be the elements of S. We prove the result by strong

induction on a + b + c, a quantity that is preserved by applications of the rule. For
the sake of obtaining a contradiction, assume that S cannot be transformed to a triple
containing 0.
First we reduce to the case that exactly one of a, b, c is odd. If two are odd, apply
the rule with those two, and then none are odd. If none are odd, divide all numbers
by 2 and apply the inductive hypothesis. If three are odd, apply the rule once, and
exactly one is odd. Once exactly one is odd, this will remain so.
Without loss of generality, a is odd and b and c are even. Let 2n be the highest
power of 2 dividing b + c. We will describe a series of moves leaving us with a , b ,
c , with a odd and 2n+1 dividing b + c . Repeating such a series of moves results in
a triple a , b , c with b + c divisible by 2m where 2m > a + b + c. Then since
b , c > 0, we have b + c ≥ 2m > a + b + c = a + b + c , contradicting a > 0.
We first apply a series of moves so that b and c are divisible by different powers of 2,
and the one divisible by the smaller power of 2 (b, say) is also smaller. If b and c have
the same number of factors of 2, then applying the rule to those two will yield both
divisible by a higher power of 2, or one will have fewer factors of 2 than the other.
Since b + c is constant here, after a finite number of applications of the rule, b and c
will not have the same number of factors of 2. Also, possibly after some additional
moves (on b and c), the one of b and c divisible by the smaller power of 2 is also
smaller.
Now if a > b, then apply the rule to (a, b); a remains odd, b is doubled, and b + c is
divisible by a higher power of 2 as desired.
On the other hand, if a < b, apply the rule first to (a, b), and then to (b − a, c).
(Note that c > b > b − a.) The result is the triple (2a, 2b − 2a, c − b + a). Now the
odd number is c − b + a, and the sum of the even numbers is 2b, which has one more
factor of 2 than b + c. 
Solution 3 (Dylan Thurston). As in Solution 2, we assume there is some triple
a, b, c that cannot be transformed to a triple containing 0, and that exactly one of
a, b, c is odd. We give a recipe showing that if two of the triples are divisible by 2n ,
then we can reach a state where two are divisible by 2n+1 ; since the sum of the triple
is constant, this eventually gives a contradiction.
The result is trivial if both of the multiples of 2n are already divisible by 2n+1 , or if
neither are (just apply the rule once to that pair). So assume one is a multiple of 2n+1 ,
and the other is not, so the triple is (after renaming) (a, b, c) ≡ (0, 2n , x) (mod 2n+1 )
(where x is odd).
We can assume a > c. (Otherwise, apply the rule to (a, c) repeatedly until this is so.)
Then apply the rule to (a, c), giving the triple (a , b , c ) ≡ (−x, 2n , 2x) (mod 2n+1 ).
If a < b , then apply the rule to (a , b ), giving (−2x, 2n + x, 2x) (mod 2n+1 ).
Repeated applications of the rule to (2x, −2x) eventually produce (0, 0) (mod 2n+1 )
(regardless of which is bigger at each stage).
If on the other hand a > b , apply the rule to (a , b ) giving (2n − x, 0, 2x)
(mod 2n+1 ). Apply the rule repeatedly to the 0 and 2n − x terms until the 0 is
bigger, and then apply it once more to get (−2x, 2n + x, 2x) (mod 2n+1 ). Again apply
the rule repeatedly to (2x, −2x) to eventually produce (0, 0) (mod 2n+1 ). 
190 The William Lowell Putnam Mathematical Competition

Related question. Problem 3 on the 1986 International Mathematical Olympiad

[IMO86] is similar, although the solution is quite different:
To each vertex of a regular pentagon an integer is assigned in such a way
that the sum of all five numbers is positive. If three consecutive vertices
are assigned the numbers x, y, z respectively and y < 0 then the following
operation is allowed: the numbers x, y, z are replaced by x + y, −y, z + y
respectively. Such an operation is performed repeatedly as long as at least one
of the five numbers is negative. Determine whether this procedure necessarily
comes to an end after a finite number of steps.
Solutions: The Fifty-Fifth Competition (1994) 191

The Fifty-Fifth William Lowell Putnam Mathematical Competition

December 3, 1994

A1. (59, 59, 54, 21, 0, 0, 0, 0, 8, 0, 3, 2)

Suppose that a sequence a1 , a2 , a3 , . . . satisfies 0 < an ≤ a2n + a2n+1 for all
∞
n ≥ 1. Prove that the series n=1 an diverges.
2m −1
Solution. For m ≥ 1, let bm = i=2m−1 ai . Summing an ≤ a2n + a2n+1 from
n = 2m−1 to n = 2m − 1 yields bm ≤ bm+1 for all m ≥ 1. For any t ≥ 1,
2t −1 t
an = bm ≥ tb1 = ta1 ,
n=1 m=1
∞
which is unbounded as t → ∞ since a1 > 0, so n=1 an diverges.
Alternatively, assuming the series converges to a finite value L, we obtain the
contradiction
L = b1 + (b2 + b3 + · · · )
≥ b1 + (b1 + b2 + · · · )
= b1 + L. 
Related question. An interesting variant is the following [New, Problem 87]:
(An )n∈Z+ is a sequence of positive numbers satisfying An < An+1 + An2 for
∞
all n ∈ Z+ . Prove that n=1 An diverges.

A2. (169, 3, 2, 0, 0, 0, 0, 0, 1, 3, 22, 6)

Let A be the area of the region in the first quadrant bounded by the line
y = 12 x, the x-axis, and the ellipse 19 x2 + y 2 = 1. Find the positive number m
such that A is equal to the area of the region in the first quadrant bounded
by the line y = mx, the y-axis, and the ellipse 19 x2 + y 2 = 1.
Answer. To make the areas equal, m must be 2/9.
Solution 1. The linear transformation given by x1 = x/3, y1 = y transforms the
region R bounded by y = x/2, the x-axis, and the ellipse x2 /9 + y 2 = 1 into the region
R bounded by y1 = 3x1 /2, the x1 -axis, and the circle x21 + y12 = 1; it also transforms

y y1

y = x/2 y = 3x1 /2

x x1

FIGURE 26.
192 The William Lowell Putnam Mathematical Competition

the region S bounded by y = mx, the y-axis, and x2 /9 + y 2 = 1 into the region S 
bounded by y1 = 3mx1 , the y1 -axis, and the circle. Since all areas are multiplied by
the same (nonzero) factor under the linear transformation, R and S have the same
area if and only if R and S  have the same area. However, we can see by symmetry
about the line y1 = x1 that this happens if and only if 3m = 2/3, that is, m = 2/9. 

Solution 2 (Noam Elkies). Apply the linear transformation (x, y) → (3y, x/3).
This preserves area, and the ellipse x2 /9 + y 2 = 1. It switches the x and y axes, and
takes y = x/2 to the desired line, x/3 = (3y/2), i.e. y = (2/9)x. Thus m = 2/9. 
Remark. There are, of course, less enlightened solutions. Setting up the integrals
for the two areas yields the equation
 3/√13 " #  3/√1+9m2 " #
9 − 9y 2 − 2y dy = 1 − x2 /9 − mx dx.
0 0

At this point, one might guess that a substitution y = cx will transform one integral
into the other, if c and m satisfy
3 3
√ = c√ , 3c = 1, 2c2 = m,
13 1 + 9m2
and in fact, c = 1/3 and m = 2/9 work. If this shortcut is overlooked, then as a last
resort one could use trigonometric substitution to evaluate both sides: this yields



3 3 3 1
Arcsin √ = Arcsin √ .
2 13 2 1 + 9m2
Solving yields m = 2/9.

A3. (0, 10, 67, 0, 0, 0, 0, 0, 30, 31, 40, 28)

Show that if the points of an isosceles right triangle of side length 1 are
each colored with one of four colors, then there
√ must be two points of the
same color which are at least a distance 2 − 2 apart.
Motivation. The standard approach to √ such a problem would be to find five points
in the triangle, no two closer than 2 − 2; then two of them would have the same
color. Unfortunately, such a set of five points does not exist, so a subtler argument is
required.

Solution. Suppose that it is possible to √ color the points of the triangle with four
colors so that any two points at least 2 − 2 apart receive different colors. Suppose
the vertices √
of the triangle are A =
√ (0, 0), B = (1, 0), C = (0, 1). Define √ also
√ the
points D =√( 2 − 1, 0)√ and E = (0, 2 − 1) on the two sides, and F = (2 − 2, 2 − 1)
and G = ( 2 − 1, 2 − 2) on the diagonal; see Figure 27. Note that
√
BD = BF = CE = CG = DE = DG = EF = 2 − 2.

The color of B must be different from the colors of the other named points. The same
holds for C. Thus the vertices of the pentagram AGDEF must each be painted one
of the two remaining colors. Then two adjacent vertices
√ of the pentagram must have
the same color. But they are separated by at least 2 − 2, so this is a contradiction. 
Solutions: The Fifty-Fifth Competition (1994) 193

E F

A D B

FIGURE 27.

√
Remark. The 2 − 2 in the problem is best possible. If we give triangles ADE,
BDF , CEG, and quadrilateral DEGF each their own color, coloring √ common edges
arbitrarily, then no two points at distance strictly greater than 2− 2 receive the same
color.
Related questions. There are many problems of this sort. Other typical examples
are Problems 1954M2 and 1960M2 [PutnamI, pp. 41, 58]:
Consider any five points P1 , P2 , P3 , P4 , P5 in the interior of a square S of
side-length 1. Denote by dij the distance between√the points √ Pi and Pj . Prove
that at least one of the distances dij is less than 2/2. Can 2/2 be replaced
by a smaller number in this statement?
Show that if three points
√ are √
inside a closed square of unit side, then some
pair of them are within 6 − 2 units apart.
Problem 1988A4 also is similar.

A4. (12, 17, 20, 0, 0, 0, 0, 0, 15, 3, 43, 96)

Let A and B be 2 × 2 matrices with integer entries such that A, A + B,
A + 2B, A + 3B, and A + 4B are all invertible matrices whose inverses have
integer entries. Show that A + 5B is invertible and that its inverse has
integer entries.

Solution. A square matrix M with integer entries has an inverse with integer
entries if and only if det M = ±1: if N is such an inverse, (det M )(det N ) =
det(M N ) = 1 so det M = ±1; conversely, if det M = ±1, then ±M  is an inverse
with integer entries, where M  is the classical adjoint of M . Let f (x) = det(A + xB).
Then f (x) is a polynomial of degree at most 2, such that f (x) = ±1 for x = 0, 1, 2,
3, and 4. Thus by the Pigeonhole Principle f takes one of these values three or more
times. But the only polynomials of degree at most 2 that take the same value three
times are constant polynomials. In particular, det(A + 5B) = ±1, so A + 5B has an
inverse with integer entries. 
194 The William Lowell Putnam Mathematical Competition

A5. (20, 13, 4, 0, 0, 0, 0, 0, 6, 2, 57, 104)

Let (rn )n≥0 be a sequence of positive real numbers such that limn→∞ rn = 0.
Let S be the set of numbers representable as a sum

ri1 + ri2 + · · · + ri1994 ,

with i1 < i2 < · · · < i1994 . Show that every nonempty interval (a, b) contains
a nonempty subinterval (c, d) that does not intersect S.

Solution 1. We may permute the ri to assume r0 ≥ r1 ≥ · · · . This does not

change S or the convergence to 0. If b ≤ 0, the result is clear, so we assume b > 0.
Since rn → 0, only finitely many rn exceed b/2. Thus we may choose a positive
number a1 so that a < a1 < b and rn ∈ / [a1 , b) for all n. Then for an element of
S ∩ (a1 , b), there are only a finite number of possibilities for i1 (since 0 < a1 /1994 ≤
ri1 < a1 ); let I1 be the set of such i1 .
Choose a2 so that a1 < a2 < b and ri1 + rn ∈ / [a2 , b) for all i1 ∈ I1 and
n ≥ 0. Then for each i1 , there are only a finite number of possibilities for i2 (since
0 < (a2 − ri1 )/1993 ≤ ri2 < a2 − ri1 ); let I2 be the set of ordered pairs (i1 , i2 ) of
possibilities.
Similarly, inductively choose am so that am−1 < am < b and ri1 + · · · + rim−1 + rn ∈ /
[am , b) for all (i1 , . . . , im−1 ) ∈ Im−1 and n ≥ 0. The set Im of ordered m-tuples
(i1 , . . . , im ) of possibilities is finite.
Then (c, d) = (a1994 , b) does not intersect S. 
Remark. A cleaner variation is to show that if A is a nowhere dense subset of
R, and (rn )n≥0 converges to 0, then A + {rn } is also nowhere dense (where A + {rn }
denotes the set of numbers of the form a + rn , where a ∈ A). Then the result follows
by induction.

Solution 2. It suffices to show that any sequence in S contains a monotonically

nonincreasing subsequence. For then, letting (tn )n≥0 be any strictly increasing
sequence within (a, b), some (in fact, all but a finite number) of the intersections
S ∩ (tn , tn+1 ) would have to be empty. Otherwise, one could form a strictly increasing
sequence (sn )n≥0 by taking Sn ∈ S ∩ (tn , tn+1 ).
Let (sn )n≥0 be a sequence in S. For n = 0, 1, 2, . . . , write

sn = rf (n,1) + rf (n,2) + · · · + rf (n,1994) with f (n, 1) < f (n, 2) < · · · < f (n, 1994).

The sequence (rf (n,1) )n≥0 has a monotonically nonincreasing subsequence (since
(rn )n≥0 is a positive sequence converging to 0). Thus we may replace (sn )n≥0 by a
subsequence for which (rf (n,1) )n≥0 is monotonically nonincreasing. In a similar fash-
ion, we pass to subsequences so that, successively, each of (rf (n,2) )n≥0 , (rf (n,3) )n≥0 ,
. . . , (rf (n,1994) )n≥0 may be assumed to be monotonically nonincreasing. The resulting
(sn )n≥0 is monotonically nonincreasing. 
Remark. A totally ordered set T is well-ordered if and only if every nonempty
subset has a least element. This condition is equivalent to the condition that every
sequence in T contain a nondecreasing subsequence. Hence −S is well-ordered under
the ordering induced from the reals. Solution 2 can be interpreted as proving that
Solutions: The Fifty-Fifth Competition (1994) 195

a finite sum of well-ordered subsets of the reals is well-ordered. See the remark on
Zorn’s Lemma in 1989B4 for more on well orderings.

Solution 3. Let C be the set { rn : n ≥ 0 } ∪ {0}. Then C is closed and bounded,

so C is compact. Hence C 1994 is compact, and its image S  under the “sum the
coordinates” map R1994 → R is compact. Clearly S ⊂ S  .
Let (a, b) be a nonempty open interval. Since S  is countable, (a, b)−S  is nonempty;
it is open since S  is closed. Hence (a, b) − S  includes a nonempty open interval. 
Remark. The proofs above generalize to give the same conclusion for any
convergent sequence (rn )n≥0 .

A6. (5, 8, 10, 0, 0, 0, 0, 0, 7, 4, 34, 138)

Let f1 , f2 , . . . , f10 be bijections of the set of integers such that for each
integer n, there is some composition fi1 ◦ fi2 ◦ · · · ◦ fim of these functions
(allowing repetitions) which maps 0 to n. Consider the set of 1024 functions

F = {f1e1 ◦ f2e2 ◦ · · · ◦ f10

e10
},

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 of integers, then at most 512 of the
functions in F map A to itself.

Solution. We say that a bijection of the integers to itself preserves a subset A if

it restricts to a bijection of A.
Let G be the group of permutations of Z generated by the fi . Then G has elements
mapping any integer m to 0, and elements mapping 0 to any integer n, so G acts
transitively on Z. Hence no nonempty proper subset A can be preserved by all the fi .
It remains to prove the following lemma.
Lemma. Let f1 , . . . , fn be bijections Z → Z, and let A be a subset of Z. Suppose
that some fi does not preserve A. Then at most 2n−1 elements of

Fn = { f1e1 ◦ f2e2 ◦ · · · ◦ fnen : ei = 0 or 1 }

preserve A.
Proof 1 of Lemma (inductive). The lemma is true for n = 1, so assume it it is true
for all n < k (for some k > 1), and false for n = k. Then by the Pigeonhole Principle,
there are e1 , . . . , ek−1 ∈ {0, 1} such that both
e e
f1e1 ◦ f2e2 ◦ · · · ◦ fk−1
k−1
and f1e1 ◦ f2e2 ◦ · · · ◦ fk−1
k−1
◦ fk

fix A. Hence fk preserves A as well. By the inductive hypothesis, at most 2k−2 ele-

ments of Fk−1 fix A; thus at most 2k−1 elements of Fk fix A, giving a contradiction. 

Proof 2 of Lemma (noninductive). Let k be the largest integer such that fk does
not map A to itself, and suppose that more than 2n−1 of the functions Fn map A to
itself. By the Pigeonhole Principle, there are

e1 , . . . , ek−1 , ek+1 , . . . , en ∈ {0, 1}

196 The William Lowell Putnam Mathematical Competition

such that both

e e e e
f1e1 ◦ · · · ◦ fk−1
k−1
◦ fk+1
k+1
◦ · · · ◦ fnen and f1e1 ◦ · · · ◦ fk−1
k−1
◦ fk ◦ fk+1
k+1
◦ · · · ◦ fnen
e e
both fix A. Hence both F1 = f1e1 ◦ · · · ◦ fk−1
k−1
and F1 = f1e1 ◦ · · · ◦ fk−1
k−1
◦ fk both map
−1
A to itself. But then F1 ◦ F2 = fk also maps A to itself, giving a contradiction. 
Remark. The solution shows that the problem statement remains true even if the
condition that A be a nonempty finite subset of Z is relaxed to the condition that A
be a nonempty proper subset of Z.

B1. (45, 26, 57, 0, 0, 0, 0, 0, 42, 28, 6, 2)

Find all positive integers that are within 250 of exactly 15 perfect
squares.
Answer. An integer N is within 250 of exactly 15 perfect squares if and only if
either 315 ≤ N ≤ 325 or 332 ≤ N ≤ 350.
Solution. The squares within 250 of a positive integer N form a set of consecutive
squares. If N is such that there are 15 such squares, then they are m2 , (m + 1)2 , . . . ,
(m + 14)2 for some m ≥ 0. If m = 0, then 142 ≤ N + 250 < 152 , contradicting N > 0.
Now, given N, m > 0, the following two conditions are necessary and sufficient for
m2 , (m + 1)2 , . . . , (m + 14)2 to be the squares within 250 of N :
(m + 14)2 ≤ N + 250 ≤ (m + 15)2 − 1,

m2 ≥ N − 250 ≥ (m − 1)2 + 1.
Subtraction shows that these imply
28m + 196 ≤ 500 ≤ 32m + 222,
which implies m = 9 or 10.
If m = 9, the two conditions 232 ≤ N + 250 ≤ 242 − 1, 92 ≥ N − 250 ≥ 82 + 1 are
equivalent to 315 ≤ N ≤ 325. If m = 10, the two conditions 242 ≤ N + 250 ≤ 252 − 1,
102 ≥ N − 250 ≥ 92 + 1 are equivalent to 332 ≤ N ≤ 350. 

B2. (28, 8, 49, 0, 0, 0, 0, 0, 56, 10, 39, 16)

For which real numbers c is there a straight line that intersects the curve
y = x4 + 9x3 + cx2 + 9x + 4
in four distinct points?
Answer. There exists such a line if and only if c < 243/8.
Remark. Since all vertical lines meet the curve in one point, we need only consider
lines of the form y = mx + b.
Solution 1 (geometric). The constant and linear terms of
P (x) = x4 + 9x3 + cx2 + 9x + 4
are irrelevant to the problem; y = P (x) meets the line y = mx + b in four points if
and only if y = P (x) + 9x + 4 meets the line y = (m + 9)x + (b + 4) in four points.
Solutions: The Fifty-Fifth Competition (1994) 197

Also, y = P (x) meets the line y = mx + b in four points if and only if y = P (x − α)

meets the line y = m(x − α) + b in four points, so we may replace the given quartic
with P (x − 9/4) = x4 + (c − 243/8)x2 + · · · (where we ignore the linear and constant
terms).
The problem is then to determine the values of c for which there is a straight line
that intersects y = x4 + (c − 243/8)x2 in four distinct points. The result is now
apparent from the shapes of the curves y = x4 + ax2 . For example, when a < 0, this
“W-shaped” curve has a relative maximum at x = 0, so the horizontal lines y = −C for
small positive C intersect the curve in four points, while for a ≥ 0, the curve is always
concave upward, so no line can intersect it in more than two points; see Figure 28. 

a<0 a≥0

FIGURE 28.
Graphs of y = x4 + ax2 for a < 0 and a ≥ 0.

Solution 2 (algebraic). We wish to know if we can choose m and b so that

q(x) = x4 + 9x3 + cx2 + 9x + 4 − (mx + b)

has four distinct real solutions α1 , α2 , α3 , α4 . If we can find four distinct real numbers
such that

α1 + α2 + α3 + α4 = −9 (1)

α1 α2 + α1 α3 + α1 α4 + α2 α3 + α2 α4 + α3 α4 = c, (2)

then we can choose m and b appropriately so that q(x) has these four zeros (by the
34
expansion of i=1 (x − αi )).
Then from

0< (αi − αj )2 = 3(α1 + α2 + α3 + α4 )2 − 8c (3)

i<j

we get c < 243/8. Conversely, we must show that if c < 243/8, then we can find
distinct real α1 , α2 , α3 , α4 satisfying (1) and (2). This is indeed possible; try
(α1 , α2 , α3 , α4 ) = (9/4 − µ, 9/4 + µ, 9/4 − ν, 9/4 + ν) where µ and ν are distinct
positive numbers. Then (1) is automatically satisfied, and we can choose µ and ν so
that the right side of (3), which is 8(µ2 + ν 2 ), is any desired positive number. 
198 The William Lowell Putnam Mathematical Competition

Remark. The previous two solutions highlight two ways of constructing a line for
given c. Other possible tools to do this include Rolle’s Theorem, the Mean Value
Theorem [Spv, Ch. 11, Theorem 4], Descartes’ Rule of Signs, and more. Also, another
quick way to see that c < 243/8 is necessary is to consider the discriminant of the
quadratic P  (x).
Related question. A related problem is 1977A1 [PutnamII, p. 29]:
Consider all lines which meet the graph of
y = 2x4 + 7x3 + 3x − 5
in four distinct points, say (xi , yi ), i = 1, 2, 3, 4. Show that
x1 + x2 + x3 + x4
4
is independent of the line and find its value.

B3. (27, 10, 8, 5, 0, 0, 0, 2, 45, 49, 15, 45)

Find the set of all real numbers k with the following property: For any
positive, differentiable function f that satisfies f  (x) > f (x) for all x, there
is some number N such that f (x) > ekx for all x > N .
Answer. The desired set is (−∞, 1).
Solution. Let h(x) = ln f (x) − x. Then the problem becomes that of determining
for which k the following holds: if a real-valued function h(x) satisfies h (x) > 0 for
all x, then there exists a number N such that h(x) > (k − 1)x for all x > N .
The function −e−x is always negative, but has positive derivative, so no number
−x
k ≥ 1 is in the set. (This corresponds to the function f (x) = ex−e .) On the other
hand any k < 1 is in the set: choose N such that (k − 1)N < h(0); then for x > N ,
h(x) > h(0) > (k − 1)N > (k − 1)x. 

B4. (15, 1, 4, 0, 0, 0, 0, 0, 5, 22, 71, 88)

For n ≥ 1, let dn be the greatest common divisor of the entries of An − I,
where


3 2 1 0
A= and I = .
4 3 0 1
Show that limn→∞ dn = ∞.
Solution 1. Experimentation suggests and induction on n proves that there exist
integers an , bn > 0 such that

n an bn
A = .
2bn an
Since det An = 1, we have a2n − 1 = 2b2n . Thus an − 1 divides 2b2n . By definition,
dn = gcd(an − 1, bn ), so 2d2n = gcd(2(an − 1)2 , 2b2n ) ≥ an − 1. From An+1 = A · An we
have an+1 > 3an , so limn→∞ an = ∞. Hence limn→∞ dn = ∞. 


a b
Solution 2 (Robin Chapman). The set of matrices of the form with
2b a


1 1
a, b ∈ Z is closed under left multiplication by . It follows by induction on n
2 1
Solutions: The Fifty-Fifth Competition (1994) 199

n

1 1 a b
that = for some a, b ∈ Z depending on n. Taking determinants
2 1 2b a

2

1 1 3 2
shows a2 − 2b2 = (−1)n . But = , so
2 1 4 3

n

2

3 2 1 0 a b 1 0
− = −
4 3 0 1 2b a 0 1

2 
a + 2b − 1
2
2ab
= .
4ab a2 + 2b2 − 1

If n is odd then a2 − 2b2 = −1, so a2 + 2b2 − 1 = 2a2 and all entries are divisible by
a. If n is even a2 − 2b2 = 1, so a2 + 2b2 − 1 = 4b2 and all entries are divisible by b.
Both a and b increase as n → ∞ (by the same argument as in Solution 1), so we are
done. 

Solution 3. Define the sequence r0 , r1 , r2 , . . . by r0 = 0, r1 = 1, and

rk = 6rk−1 − rk−2 for k > 1. Then rn > 5rn−1 for n ≥ 1, so limn→∞ rn = ∞.
We first show by induction on k that

An − I = rk+1 (An−k − Ak ) − rk (An−k−1 − Ak+1 ) for k ≥ 0. (1)

This is clear for k = 0 and, for the inductive step, using A2 − 6A + I = 0 (the
characteristic equation), we have

rk+1 (An−k − Ak ) − rk (An−k−1 − Ak+1 )

 
= rk+1 (6An−k−1 − An−k−2 ) − (6Ak+1 − Ak+2 ) − rk (An−k−1 − Ak+1 )
= (6rk+1 − rk )(An−k−1 − Ak+1 ) − rk+1 (An−k−2 − Ak+2 )
= rk+2 (An−k−1 − Ak+1 ) − rk+1 (An−k−2 − Ak+2 ).

Applying (1) with k = n/2, we obtain

rn/2 (An/2+1 − An/2−1

 ) if n is even,
An − I =
r(n+1)/2 + r(n−1)/2 (A(n+1)/2 − A(n−1)/2 ) if n is odd.

In either case, the entries of An − I have a common factor that goes to ∞ since
limn→∞ rn = ∞. 

Solution √4. The entries of√ An are each of the form α1 λn1 + α2 λn2 , where
λ1 = 3 + 2 2 and λ2 = 3 − 2 2 are the eigenvalues of A. This follows from
diagonalization (as suggested by our hint), or from the theory of linear recursive
relations and the Cayley-Hamilton Theorem: the latter yields A2 − 6A + I = 0,
so An+2 − 6An+1 + An = 0 for all n, and each entry of An satisfies the recursion
xn+2 − 6xn+1 + xn = 0. Using the entries for n = 1, 2, we derive
 n n n n
λ1 +λ2 λ1 −λ2
√
An =  λn −λn
2 2 2 
λn n .
1√ 2 1 +λ2
2 2
200 The William Lowell Putnam Mathematical Competition

√ √
Since λi = µ2i where µ1 = 1 + 2 and µ2 = 1 − 2, we see

n 
λ1 + λn2 λn − λn
dn = gcd − 1, 1 √ 2
2 2 2

n 
(µ1 − µ2 ) (µ1 − µn2 )(µn1 + µn2 )
n 2 n
= gcd , √
2 2 2

n 
n 
(µ1 − µ2 )n
(µ1 − µn2 ) (µn1 + µn2 )
= √ gcd √ ,
2 2 2 2
√
since (µn1 −µn2 )/ 2 and (µn1 +µn2 )/2 are (rational) integers. Since |µ1 | > 1 and |µ2 | < 1,
we conclude limn→∞ (µn1 − µn2 ) = ∞. Hence limn→∞ dn = ∞. 
Solution 5. The characteristic polynomial of A is x2 − 6x + 1, so
A has distinct

√ λ 0
eigenvalues λ, λ−1 , where λ = 3+2 2. Hence A = CDC −1 where D = and
√ 0 λ−1
C is an invertible matrix with entries√in Q( 2). Choose an integer k ≥ 1 such that
the entries of kC and kC
−1 are in Z[ 2]. Then k (A − I) = (kC)(D − I)(kC )
2 n n −1

1 0 √
and Dn − I = (λn − 1) so λn − 1 divides k 2 dn in Z[ 2]. Taking norms,
0 λ−n
we find that the (rational) integer (λn − 1)(λ−n − 1) divides k 4 d2n . But |λ| > 1, so
|(λn − 1)(λ−n − 1)| → ∞ as n → ∞. Hence limn→∞ dn = ∞. 
Remark.

Each solution generalizes to establish the same result for integral matrices
a b
A= of determinant 1 and | trace(A)| > 1 (the latter to guarantee rn → ∞
c d
where rn = trace(A)rn−1 − rn−2 ).

B5. (11, 4, 4, 0, 0, 0, 0, 0, 32, 10, 15, 130)

For any real number α, define the function fα (x) = αx. Let n be a
positive integer. Show that there exists an α such that for 1 ≤ k ≤ n,‡
fαk (n2 ) = n2 − k = fαk (n2 ).

Solution 1. We will show that α satisfies the conditions of the problem if and
only if

1/n
1 n2 − n + 1
1− ≤α< ,
n2 n2
and then show that this interval is nonempty.
We have fαk (n2 ) = n2 − k for k = 1, . . . , n if and only if α(n2 − k + 1) = n2 − k
for k = 1, . . . , n, which holds if and only if
n2 − k
≤α<1 for k = 1, . . . , n.
n2 −k+1
Since
n2 − k 1
=1− 2
n2 −k+1 n −k+1

‡ Here fαk (n2 ) = fα (· · · (fα (n2 )) · · · ), where fα is applied k times to n2 .

Solutions: The Fifty-Fifth Competition (1994) 201

decreases with k, these hold if and only if 1 − n12 ≤ α < 1. We assume this from now
on.
Next we consider the conditions fαk (n2 ) = n2 − k for k = 1, . . . , n. Since fαk (n2 ) is
an integer less than αk n2 , we have fαk (n2 ) ≤ fαk (n2 ). We have already arranged that
fαk (n2 ) = n2 − k, so fαk (n2 ) = n2 − k if and only if αk n2 < n2 − k + 1. Moreover, if
the latter holds for k = n, then it holds for k = 1, . . . , n by reverse induction, since
multiplying αk n2 < n2 − k + 1 by
n2 1 1 n2 − (k − 1) + 1
α−1 ≤ = 1 + ≤ 1 + ≤
n2 − 1 n2 − 1 n2 − k + 1 n2 − k + 1
yields the inductive step. Hence all the conditions hold if and only if

2 1/n
1 n −n+1
1− 2 ≤α < .
n n2
It remains to show that this interval is nonempty, or equivalently, that

n
1 1 1
1− 2 < 1 − + 2. (1)
n n n
First we will prove


n 2
(1 − x)n ≤ 1 − nx + x for 0 ≤ x ≤ 1. (2)
2
The two sides are equal at x = 0, so it suffices to show that their derivatives satisfy

−n(1 − x)n−1 ≤ −n + n(n − 1)x for 0 ≤ x ≤ 1.

Again the two sides are equal at x = 0, so, differentiating again, it suffices to show

n(n − 1)(1 − x)n−2 ≤ n(n − 1) for 0 ≤ x ≤ 1.

This is obvious. Taking x = 1/n2 in (2) yields

n
1 1 n(n − 1)/2 1 1
1− 2 ≤1− + < 1 − + 2. 
n n n4 n n

Remark. Inequality (2) is a special case of the fact that for real a > b > 0 and
integers 0 < k < n, the partial binomial expansion



n n−1 k n
a −
n
a b + · · · + (−1) an−k bk (3)
1 k
is greater than or less than (a − b)n , according to whether k is even or odd. It can
be proved by the same argument used to prove (2). This is sometimes called the
Inclusion-Exclusion Inequality, because if a, b are sizes of sets A B ∅, then (3)
is the overestimate or underestimate for #(A − B) obtained by terminating the
n

inclusion-exclusion argument prematurely.

2
Solution 2 (Dave Rusin). Let α = e−1/n . The same method used to prove (2)
shows that

1 − r + r2 /2 > e−r > 1 − r. (4)

202 The William Lowell Putnam Mathematical Competition

Substituting r = k/n2 (0 < k ≤ n) and simplifying, we find

2
1 k
n −k+
2
> αk n2 > n2 − k.
2 n
The right side is an integer, and the left side is at most 1/2 more, so αk n2  = n2 − k.
Since α > 1 − 1/n2 (again by (4)), fαk (n2 ) = n2 − k for 1 ≤ k ≤ n by the same
argument as in the previous solution. 

B6. (14, 11, 1, 0, 0, 0, 0, 0, 16, 10, 50, 104)

For any integer a, set
na = 101a − 100 · 2a .
Show that for 0 ≤ a, b, c, d ≤ 99, na + nb ≡ nc + nd (mod 10100) implies
{a, b} = {c, d}.

The following lemma states that 2 is a primitive root modulo the prime 101.
Lemma. 2a ≡ 1 (mod 101) if and only if a is divisible by 100.
Proof. We need to show that the order m of the image of 2 in the group (Z/101Z)∗
is 100. Since the group has order 100, m divides 100. If m were a proper divisor of
100, then m would divide either 100/2 = 50 or 100/5 = 20, so either 250 or 220 would
be 1 modulo 101. But we compute
210 = 1024 ≡ 14 (mod 101)
220 ≡ 142 ≡ −6 (mod 101)
240 ≡ (−6)2 ≡ 36 (mod 101)
2 50
≡ 36 · 14 ≡ −1 (mod 101).

Corollary 1. If a and b are nonnegative integers such that 2a ≡ 2b (mod 101),
then a ≡ b (mod 100).
Proof. Without loss of generality, assume a ≥ b. If 101 divides 2a −2b = 2b (2a−b −1),
then 2a−b ≡ 1 (mod 101), so a − b ≡ 0 (mod 100) by the lemma. 

Solution. By the Chinese Remainder Theorem, na + nb ≡ nc + nd (mod 10100)

is equivalent to
a + b ≡ c + d (mod 100) (1)
and
2a + 2b ≡ 2 c + 2d (mod 101). (2)
By Fermat’s Little Theorem, (1) implies 2a+b ≡ 2c+d (mod 101), or equivalently
2a 2b ≡ 2c 2d (mod 101) (3)
Solve for 2b in (2) and substitute into (3) to obtain
2a (2c + 2d − 2a ) ≡ 2c 2d (mod 101),
Solutions: The Fifty-Fifth Competition (1994) 203

or equivalently
0 ≡ (2a − 2c )(2a − 2d ) (mod 101).
(That such a factorization exists could have been guessed from the desired conclusion
that a = c or a = d.) Hence 2a ≡ 2c (mod 101) or 2a ≡ 2d (mod 101). By the
corollary above, a is congruent to c or d modulo 100. Then by (1), b is congruent to
the other. But 0 ≤ a, b, c, d ≤ 99, so these congruences are equalities. 
Remark. A more conceptual way to get from (2) and (3) to the conclusion that 2a ,
2 are congruent to 2c , 2d in some order, modulo 101, is to observe that (x−2a )(x−2b )
b

and (x − 2c )(x − 2d ) are equal as elements of the polynomial ring F101 [x] over the finite
field F101 . Then take the multiset of zeros of each.
204 The William Lowell Putnam Mathematical Competition

The Fifty-Sixth William Lowell Putnam Mathematical Competition

December 2, 1995

A1. (115, 64, 20, 0, 0, 0, 0, 0, 1, 0, 2, 2)

Let S be a set of real numbers which is closed under multiplication (that
is, if a and b are in S, then so is ab). Let T and U be disjoint subsets of
S whose union is S. Given that the product of any three (not necessarily
distinct) elements of T is in T and that the product of any three elements
of U is in U , show that at least one of the two subsets T, U is closed under
multiplication.

Solution. Suppose on the contrary that there exist t1 , t2 ∈ T with t1 t2 ∈ U and

u1 , u2 ∈ U with u1 u2 ∈ T . Then (t1 t2 )u1 u2 ∈ U while t1 t2 (u1 u2 ) ∈ T , contradiction. 
Remark. It is possible for T to be closed and U not to be: for example, let T and
U be the sets of integers congruent to 1 and 3, respectively, modulo 4.

A2. (26, 10, 14, 0, 0, 0, 0, 0, 14, 3, 56, 81)

For what pairs (a, b) of positive real numbers does the improper integral
 ∞
$ $ 
√ √ √ √
x+a− x− x − x − b dx
b

converge?
Answer. The integral converges if and only if a = b.
√
Solution. Using 1 + t = 1 + t/2 + O(t2 ) repeatedly, we obtain
%2
$√
√ a
x + a − x = x1/4 1+ −1
x
2
a
=x 1/4
+ O(x−2 )
2x
2
a −1/4  
= x 1 + O(x−1 )
2
% 2
$ √
√ b
x− x−b=x 1/4
1− 1−
x
2
b
= x1/4 + O(x−2 )
2 2x
b −1/4  
= x 1 + O(x−1 ) .
2
   ∞
Thus the original integrand is a/2 − b/2 x−1/4 + O(x−5/4 ). Since b x−5/4 dx
 ∞   
converges, the original
 ∞ −1/4integral converges if and only if b a/2 − b/2 x−1/4 dx
converges. But b x dx diverges, so the original integral converges if and only if
a = b. 
Remark. See 1988A3 for a similar argument.
Solutions: The Fifty-Sixth Competition (1995) 205

Remark. In the case a = b, one can show the integral converges by a “telescoping”
argument. Write
 d
$ $ 
√ √ √ √
x+a− x− x − x − a dx
c
 $√  $√
d √ d−a √
= x + a − x dx − x+a− x dx
c c−a
 $√  $√
d √ c−a √
= x + a − x dx − x + a − x dx.
d−a c

As d → ∞ for fixed c, the second term is constant, while the first term tends to 0. (It
is an integral over an interval of fixed length, and the integrand tends to 0 uniformly.)

A3. (95, 44, 39, 0, 0, 0, 0, 0, 12, 5, 3, 6)

The number d1 d2 . . . d9 has nine (not necessarily distinct) decimal digits.
The number e1 e2 . . . e9 is such that each of the nine 9-digit numbers formed
by replacing just one of the digits di in d1 d2 . . . d9 by the corresponding digit
ei (1 ≤ i ≤ 9) is divisible by 7. The number f1 f2 . . . f9 is related to e1 e2 . . . e9
is the same way: that is, each of the nine numbers formed by replacing one
of the ei by the corresponding fi is divisible by 7. Show that, for each i,
di − fi is divisible by 7. [For example, if d1 d2 . . . d9 = 199501996, then e6 may
be 2 or 9, since 199502996 and 199509996 are multiples of 7.]

Solution. Let D and E be the numbers d1 . . . d9 and e1 . . . e9 , respectively. We

are given that

(ei − di )109−i + D ≡ 0 (mod 7)

(fi − ei )109−i + E ≡ 0 (mod 7)

for i = 1, . . . , 9. Sum the first relation over i = 1, . . . , 9 to get E − D + 9D ≡ 0

(mod 7), or equivalently E + D ≡ 0 (mod 7). Now add the first and second relations
for any particular value of i to get (fi − di )109−i + E + D ≡ 0 (mod 7). Since E + D
is divisible by 7, and 10 is coprime to 7, we conclude di − fi ≡ 0 (mod 7). 

A4. (39, 3, 13, 0, 0, 0, 0, 0, 19, 1, 83, 40)

Suppose we have a necklace of n beads. Each bead is labelled with an
integer and the sum of all these labels is n − 1. Prove that we can cut the
necklace to form a string whose consecutive labels x1 , x2 , . . . , xn satisfy
k
xi ≤ k − 1 for k = 1, 2, . . . , n.
i=1

Solution 1. Number the beads 1, 2, . . . starting from some arbitrary position,

and let zi be the label of bead i, with the convention that zn+i = zi . Let Sj =
z1 + · · · + zj − j(n − 1)/n, so Sn+j = Sj . Then the sum of the labels on beads
m + 1, m + 2, . . . , m + k is Sm+k − Sm + k(n − 1)/n.
206 The William Lowell Putnam Mathematical Competition

This suggests choosing m such that Sm is maximal, and cutting between m and
m + 1. Indeed, if we do so, then
k(n − 1) k
Sm+k − Sm + ≤k− ,
n n
but the left side is an integer, so we can replace the right side by k − 1. 
Remark. In fact, this argument shows that the location of the cut is unique,
assuming that the orientation of the necklace is fixed. Namely, for any other choice
of m, we would have Sm+k − Sm + k(n−1) n ≥ k(n−1)
n for some k, but the left side is
an integer, so we may replace the right side by k. (This implies in particular that the
maximal Sm is unique up to replacing m by m + in, but this was already evident: the
Sj have distinct fractional parts, so no two are equal.)
For a more visual reinterpretation, plot the points (i, z1 + · · · + zi ) for i ≥ 1. The
set of these points is mapped into itself by translation by (n, n − 1), and the upper
support line of slope (n − 1)/n of this set passes through (m, z1 + · · · + zm ), where m
is as chosen above.

Solution 2. Replace each label x with 1 − x, to obtain a necklace Nn with sum

of labels 1. It suffices to prove that we can cut Nn so that the consecutive labels
k
y1 , . . . , yn satisfy i=1 yi ≥ 1 for k = 1, . . . , n. We prove this by induction on n, the
case n = 1 being trivial. Suppose n > 1; then some bead b in Nn has a positive label,
since otherwise the sum of labels would not be positive. Form the new necklace Nn−1
by merging b with its successor, replacing the two labels by their sum. The induction
hypothesis provides a cut point in Nn−1 such that the partial sums are positive; the
corresponding cut point for Nn has the same property. 
Literature note. In the reformulation used in Solution 2, this problem appears
in [GKP, Section 7.5], but it is much older: see [Tak] and the references listed
there for some generalizations and applications. It is closely related to that of the
“ballot problem”: if m votes are cast for A and n votes are cast for B in some order,
find the number of orderings of the m + n votes in which A is never behind during
the tally. When m = n, the solution to the ballot problem involves the Catalan
numbers [St, Corollary 6.2.3]; see [St, Ex. 6.19 a.-nnn.] for more occurrences of the
Catalan numbers.

A5. (1, 1, 2, 1, 0, 0, 0, 6, 18, 14, 43, 118)

Let x1 , x2 , . . . , xn be differentiable (real-valued) functions of a single vari-
able t which satisfy
dx1
= a11 x1 + a12 x2 + · · · + a1n xn
dt
dx2
= a21 x1 + a22 x2 + · · · + a2n xn
dt
..
.
dxn
= an1 x1 + an2 x2 + · · · + ann xn
dt
Solutions: The Fifty-Sixth Competition (1995) 207

for some constants aij > 0. Suppose that for all i, xi (t) → 0 as t → ∞. Are
the functions x1 , x2 , . . . , xn necessarily linearly dependent?
Answer. Yes, the functions must be linearly dependent.
Solution. If (v1 , . . . , vn ) is a (complex) eigenvector of the matrix (aij ) with
(complex) eigenvalue λ, then the function y = v1 x1 +· · ·+vn xn satisfies the differential
equation
dy
= vi ai,j xj = λvj xj = λy.
dt i,j j

Therefore y = ceλt for some c ∈ C.

Now the trace of the matrix (aij ) is a11 + · · · + ann , which is positive. Since the
trace is the sum of the eigenvalues of the matrix, there must be at least one eigenvalue
λ = α + iβ with positive real part. Setting y = v1 x1 + · · · + vn xn , we have that
y = ceλt = ceαt eiβt .
We are given that |xi | → 0 as t → ∞ for i = 1, . . . , n, which implies that |y| → 0 as
well. On the other hand, for α > 0, eαt eiβt does not tend to 0 as t → ∞. Therefore we
must have c = 0, and hence v1 x1 + · · · + vn xn = 0. Since the xi are linearly dependent
over C, they are also linearly dependent over R, as desired. 
Remark. We did not require the eigenvalue λ to be real. It turns out, however, that
a matrix with nonnegative entries always has an eigenvector with nonnegative entries,
and its corresponding eigenvalue is then also nonnegative. Moreover, if the matrix has
all positive entries, it has exactly one positive real eigenvector: this assertion is the
Perron-Frobenius Theorem, which is important in the theory of Markov processes and
random walks. See the remarks following 1997A6 for further discussion.
Literature note. See Chapter 7 of [BD] for a treatment of systems of linear
differential equations with constant coefficients, such as the one occurring here.

A6. (1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 61, 141)

Suppose that each of n people writes down the numbers 1, 2, 3 in random
order in one column of a 3 × n matrix, with all orders equally likely and
with the orders for different columns independent of each other. Let the
row sums a, b, c of the resulting matrix be rearranged (if necessary) so that
a ≤ b ≤ c. Show that for some n ≥ 1995, it is at least four times as likely
that both b = a + 1 and c = a + 2 as that a = b = c.
Solution 1. For integers x, y, z with x + y + z = 0 and n a positive integer, let
f (x, y, z, n) be the number of 3 × n matrices whose columns are permutations of 1, 2,
3, and whose row sums p, q, r (in that order) satisfy p − q = x, q − r = y, r − p = z.
Then we have the recursion
f (x, y, z, n + 1) = f (x + 1, y − 1, z, n) + f (x + 1, y, z − 1, n) + f (x − 1, y + 1, z, n)
+ f (x, y + 1, z − 1, n) + f (x − 1, y, z + 1, n) + f (x, y − 1, z + 1, n).
When x = y = z = 0, the right-hand side counts the number of possible 3 × n
matrices satisfying both b = a + 1 and c = a + 2. Thus the problem requires us to
208 The William Lowell Putnam Mathematical Competition

show that 4f (0, 0, 0, n) ≤ f (0, 0, 0, n + 1) for some n ≥ 1995. Suppose this were not
the case; then we would have f (0, 0, 0, n) = O(4n ) as n → ∞. On the other hand,
f (0, 0, 0, 6m) is at least the number of 3 × 6m matrices such that each permutation
of 1, 2, 3 appears
 in exactly
 m columns. The latter number
 equals6mthe multinomial
6m 6m
coefficient m,m,m,m,m,m , so we would have m,m,m,m,m,m = O(4 ). This rate of
growth contradicts the fact that
 6(m+1)

m+1,m+1,m+1,m+1,m+1,m+1
lim  6m

m→∞
m,m,m,m,m,m
(6m + 6)(6m + 5)(6m + 4)(6m + 3)(6m + 2)(6m + 1)
= lim
m→∞ (m + 1)(m + 1)(m + 1)(m + 1)(m + 1)(m + 1)
= 66 . 

Remark. Instead of estimating the ratio of consecutive multinomial coefficients,

we could obtain a direct contradiction from Stirling’s approximation (see the remarks
following 1996B2), which implies that


6m
∼ (2πm)−5/2 66m+1/2
m, m, m, m, m, m
as m → ∞.

Solution 2. In fact, we can prove 4f (x, y, z, n) ≤ f (x, y, z, n + 1) for all n ≥ 3

and all x, y, and z. Using the recursion in Solution 1, we can deduce this from the
claim for n = 3. The verification for n = 3 can be made from the following table,
which shows the results of computing f (x, y, z, n) for small values of x, y, z, n from
the recursion. (We list only those (x, y, z) with x ≥ 0 ≥ z ≥ y and f (x, y, z, 3) = 0,
since these suffice by symmetry.)
(x, y, z) f (x, y, z, 0) f (x, y, z, 1) f (x, y, z, 2) f (x, y, z, 3) f (x, y, z, 4)
(0, 0, 0) 1 0 6 12 60
(1, −1, 0) 0 1 2 15 60
(2, −2, 0) 0 0 1 6 34
(2, −1, −1) 0 0 2 6 48
(3, −3, 0) 0 0 0 1 12
(3, −2, −1) 0 0 0 3 16

Reinterpretation. Consider a random walk on the triples of integers (x, y, z) with
x + y + z = 0, where a move increases one integer by 1 and decreases another one by 1.
(More geometrically, one can think of the triples as the points of a triangular lattice,
where a move leaves a lattice point and goes to an adjacent one.) Then f (x, y, z, n)/6n
is the probability of ending up at (x, y, z) after n steps, and for fixed x, y, z, one can
show that the ratio between these probabilities for successive n tends to 1. Greg
Lawler points out that this is a special case of the Local Central Limit Theorem for
two-dimensional walks. For more on random walks on lattice points of Euclidean
space, see [Spt].
Solutions: The Fifty-Sixth Competition (1995) 209

A related fact is that the random walk is recurrent: the expected number of returns
∞
to the origin is infinite, or equivalently, n=1 f (x, y, z, n)/6n diverges. This is not
true of the corresponding random walk on a lattice of three (or more) dimensions.

B1. (124, 26, 7, 0, 0, 0, 0, 0, 4, 10, 11, 22)

For a partition π of {1, 2, 3, 4, 5, 6, 7, 8, 9}, let π(x) be the number of elements
in the part containing x. Prove that for any two partitions π and π  , there
are two distinct numbers x and y in {1, 2, 3, 4, 5, 6, 7, 8, 9} such that π(x) = π(y)
and π  (x) = π  (y). [A partition of a set S is a collection of disjoint subsets
(parts) whose union is S.]
Solution. For a given π, no more than three different values of π(x) are possible.
(Four would require one part each of size at least 1,2,3,4, which is already too many
elements.) If no such x, y exist, each pair (π(x), π  (x)) occurs for at most one x, and
since there are at most 3 × 3 possible pairs, each must occur exactly once. Moreover,
there must be three different values of π(x), each occurring 3 times. However, any
given value of π(x) occurs kπ(x) times, where k is the number of distinct parts of
that size. Thus π(x) can occur 3 times only if π(x) equals 1 or 3, but we have three
distinct values occurring 3 times, contradiction. 
Remark. Noam Elkies points out that the only n for which all pairs of partitions
of {1, . . . , n} satisfy the statement of the problem are n = 2, 5, 9.

B2. (2, 54, 26, 0, 0, 0, 0, 0, 3, 13, 51, 55)

An ellipse, whose semi-axes
  have lengths a and b, rolls without slipping
on the curve y = c sin xa . How are a, b, c related, given that the ellipse
completes one revolution when it traverses one period of the curve?
Answer. The given condition forces b2 = a2 + c2 .
Solution 1. The assumption of rolling without slipping means that any point,
the length of the curve already traversed equals the perimeter of the arc of the ellipse
that has already touched the curve. For the ellipse to complete one revolution in one
period of the sine curve, the perimeter of the ellipse must then equal the length of a
period of the sine curve. Stating this in terms of integrals, we have
 2π   2πa 
(−a sin θ)2 + (b cos θ)2 dθ = 1 + (c/a cos x/a)2 dx.
0 0

Let θ = x/a in the second integral and write 1 as sin2 θ + cos2 θ; the result is
 2π   2π $
2
2 2 2
a sin θ + b cos θ dθ = a2 sin2 θ + (a2 + c2 ) cos2 θ dθ.
0 0
Since the left side is increasing as a function of b, we have equality if and only if
b2 = a2 + c2 . 
Solution 2. As in Solution 1, it suffices to find the relation between a, b, c that
makes the length of one period of the sine curve equal to the perimeter of the ellipse.
We will show that b2 = a2 + c2 . By scaling a, b, c, we may assume a = 1.
In the strip 0 ≤ θ ≤ 2π in the (θ, z)-plane, graph one period S of the curve
z = c sin θ. Curl the strip and glue its left and right edges to form the cylinder
210 The William Lowell Putnam Mathematical Competition

x2 + y 2 = 1 in (x, y, z)-space so that a point (θ, z) on the strip maps to (cos θ, sin θ, z)
on the cylinder. Then S is mapped to the curve E described parametrically by
(x, y, z) = (cos θ, sin θ, c sin θ) for 0 ≤ θ ≤ 2π, and S and E have the same length.
The parameterization is of the form (cos θ)-v + (sin θ)w - where -v = (1, 0, 0) and
w- = (0, 1, c) are orthogonal
√ vectors, so E is an ellipse with semi-axes of lengths
|-v | = 1 and |w| 2
- = 1 + c . (Alternatively, this can be seen geometrically, since E is
the intersection of the plane z = cy with the cylinder.) Thus when b2 = 1 + c2 , the
length of S equals the perimeter of the ellipse with semi-axes of lengths 1 and b.
On the other hand, the length of S is an increasing function of |c|. (If this is not
clear geometrically, consider the arc length formula from calculus.) Hence for fixed b,
the only values of c for which the length of S equals the perimeter of the ellipse with
semi-axes of lengths 1 and b are those for which b2 = 1 + c2 . 
Remark. Noam Elkies points out that it is not obvious that the condition
b2 = a2 + c2 is sufficient for the rolling to be physically possible: it is conceivable
that the ellipse could be too fat to touch the bottom of the sine curve. We show that
the rolling is in fact physically possible if b2 = a2 + c2 , for an appropriately chosen
starting point, by showing that the radius of curvature of the ellipse at any point is
always less than that of the sine curve at the corresponding point.
The radius of curvature of the curve y = c sin(x/a) at (x, y) is

(1 + (y  )2 )3/2 (a2 + c2 cos2 (x/a))3/2


= ,
|y | ac| sin(x/a)|
while the radius of curvature of the ellipse x = a cos θ, y = b sin θ at a given θ is

((x )2 + (y  )2 )3/2 (a2 sin2 θ + b2 cos2 θ)3/2

= .
|x y  − y  x | ab

(See [Ap2, pp. 538–539] for the basic properties of the radius of curvature.) If we
start with the points θ = 0 of the ellipse and x = 0 of the sine curve in contact, the
parameters will be related by the equation θ = x/a. Since ac| sin(x/a)| ≤ ac < ab,
the radius of curvature of the sine curve will always be larger than that of the ellipse
at the corresponding point, so the rolling will be physically possible.
Remark. The result of the previous calculation can also be explained by the
geometry of Solution 2. Since we obtain the ellipse by rolling up one period of the
sine curve, it suffices to prove the following geometrically believable claim: Rolling
up a curve does not increase the radius of curvature at any point. Here “rolling up a
curve” means taking the image of a smooth curve in the (θ, z)-plane under the map
(θ, z) !→ (cos θ, sin θ, z).
To prove the claim, parameterize the smooth curve by (θ(t), z(t)) so that the speed
for all t is 1. The radius of curvature is 1/|-a|, where -a is the acceleration [Ap2, p. 538].
Hence it remains to show that the magnitude of the acceleration is not decreased by
rolling up.
-
Let -a(t) and A(t) be the accelerations of the original and rolled-up curves, respec-
 
tively. For the original curve, the velocity is (θ , z ) and the acceleration is -a = (θ , z  ),
so the magnitude of the acceleration is |-a| = (θ )2 + (z  )2 .
Solutions: The Fifty-Sixth Competition (1995) 211

For the image curve, the velocity is (−θ sin θ, θ cos θ, z  ). The speed is (θ )2 + (z  )2
= 1. (We expect this, since “rolling up” is an isometric embedding: it does not involve
stretching the paper.) The acceleration is
- = (−θ sin θ − (θ )2 cos θ, θ cos θ − (θ )2 sin θ, z  ),
A

so
- 2 = (θ )2 + (z  )2 + (θ )4 = |-a|2 + (θ )4 ≥ |-a|2 ,
|A|

as desired. Equality holds precisely when the original curve is moving only in the
z-direction (i.e., when the curve has a vertical tangent vector); this never happens for
the sine curve.

B3. (54, 15, 11, 0, 0, 0, 0, 0, 33, 15, 49, 27)

To each positive integer with n2 decimal digits, we associate the determi-
nant of the matrix obtained by writing the digits in order across

the rows.
8 6
For example, for n = 2, to the integer 8617 we associate det = 50.
1 7
Find, as a function of n, the sum of all the determinants associated with
n2 -digit integers. (Leading digits are assumed to be nonzero; for example,
for n = 2, there are 9000 determinants.)
Answer. The sum is 45 for n = 1, 20250 for n = 2, and 0 for n ≥ 3.

Solution. The set of matrices in question can be constructed as follows: choose

the first row from one set of possibilities, choose the second row from a second set,
and so on. By the multilinearity of the determinant, the answer is the determinant of
the matrix whose kth row is the sum of the possibilities for that row.
For n = 1, this sum matrix is the 1 × 1 matrix
with entry
 45, so the answer
450 405
is 45 in this case. For n = 2, the sum matrix is , whose determinant
450 450  
is 20250. (More explicitly, the first row is the sum of the 90 vectors a b  with
a ∈ {1, . . . , 9} and b ∈ {0, . . . , 9}; the second row is the sum of the 100 vectors a b
with a ∈ {0, 1, . . . , 9} and b ∈ {0, . . . , 9}.) For n ≥ 3, all but the first row of the sum
matrix are equal, so its determinant is 0. 
Remark. Alternatively, for n ≥ 3, any matrix either has its last two columns equal,
or cancels with the matrix obtained by switching these two columns. For n = 2, this
argument eliminates all matrices from consideration except those with a 0 in the top
right, and these can be treated directly.

B4. (9, 18, 62, 4, 0, 0, 0, 2, 21, 33, 10, 45)

Evaluate
%
1
8 2207 − .
2207 − 2207−···
1

√
a+b c
Express your answer in the form d , where a, b, c, d are integers.
√
Answer. The expression equals (3 + 5)/2.
 212 The William Lowell Putnam Mathematical Competition

Solution. The infinite continued fraction is defined as the limit L of the sequence
defined by x0 = 2207 and xn+1 = 2207 − 1/xn ; the limit exists because the sequence
is decreasing (by a short induction). Also, L must satisfy L = 2207 − 1/L. Moreover,
xi > 1 for all i by induction, so L ≥ 1. Let r = L1/8 , so r8 + r18 = 2207. Then

2
4 1 1
r + 4 = 2207 + 2 = 472 , so r4 + 4 = 47;
r r

2
2 1 1
r + 2 = 49, so r2 + 2 = 7; and
r r

2
1 1
r+ = 9, so r + = 3.
r r
√ √
Thus r2 − 3r + 1 = 0, so r = 3± 5
2 . But r = L1/8 ≥ 1, so r = 3+ 5
2 . 
Remark. This question deals with the asymptotic behavior of the dynamical system
given by x0 = 2207, xn+1 = 2207 − 1/xn . See 1992B3 for another dynamical system.
Stronger result. Let Ln denote the nth Lucas number: that is, L0 = 2, L1 = 1,
and Ln+2 = Ln+1 + Ln for n ≥ 0. Then it can be shown [Wo] that for any n > 0,
% √
1 3 + 5
2n −
n L = .
L2n − L −···
1
2n
2

B5. (72, 11, 13, 0, 0, 0, 0, 0, 9, 6, 42, 51)

A game starts with four heaps of beans, containing 3, 4, 5, and 6 beans.
The two players move alternately. A move consists of taking either
(a) one bean from a heap, provided at least two beans are left behind in
that heap, or
(b) a complete heap of two or three beans.
The player who takes the last heap wins. To win the game, do you want
to move first or second? Give a winning strategy.
Answer. The first player has a winning strategy, described below.
Solution. The first player wins by removing one bean from the pile of 3, leaving
heaps of size 2, 4, 5, 6. Regarding heaps of size 2 and heaps of odd size as “odd”,
and heaps of even size other than 2 as “even”, the total parity is now even. As long
as neither player removes a single bean from a heap of size 3, the parity will change
after each move. Thus the first player can always ensure that after his move, the total
parity is even and there are no piles of size 3. (If the second player removes a heap of
size 2, the first player can move in another odd heap; the resulting heap will be even
and so cannot have size 3. If the second player moves in a heap of size greater than
3, the first player can move in the same heap, removing it entirely if it was reduced
to size 3 by the second player.) 
Remark. Because 3, 4, 5, 6 are not so large, it is also possible to describe a winning
strategy with a state diagram, where a state consists of an unordered tuple of heap
sizes, together with a label (1st or 2d) saying which player is to move next. The
diagram would need to show, for each state labelled 1st, an arrow indicating a valid
Solutions: The Fifty-Sixth Competition (1995) 213

move to a state labelled 2nd, and for each state labelled 2nd, arrows to all states that
can be reached in one move. If all paths eventually lead to the state in which there
are no heaps left and the second player is to move, then the first player has a winning
strategy. Figure 29 shows a state diagram for a smaller instance of the problem: it
gives a winning strategy for the first player in the game with heaps of size 3, 3, and 5.

(3, 3, 5 : 1st)


(3, 3, 4 : 2nd)
o PPP
ooooo PPP
PPP
oo PPP
wooo  '
(3, 4 : 1st) (2, 3, 4 : 1st) (3, 3, 3 : 1st)

 
(2, 2, 4 : 2nd) (3, 3 : 2nd)
o
ooooo
oo
wooo  
(2, 4 : 1st) (2, 2, 3 : 1st) (2, 3 : 1st)
PPP
PPP
PPP
PPP
 ' 
(4 : 2nd) (2, 2 : 2nd)

 x 
(3 : 1st) (2 : 1st)
OOO nn
OOO
OOO nnnnn
OO' nn
wnnn
(∅ : 2nd)

FIGURE 29.
An example of a state diagram.

Remark. In the language of game theory, this game is normal (the last player to
move wins) and impartial (the options in any given position are the same for both
players). The classic example of such a game is Nim: the game starts with some piles
of sticks, and each player in turn can remove any number of sticks from a single pile.
It has long been known that one can determine who wins in a given Nim position by
adding the sizes of the piles in base 2 without carries: the position is a second player
win if the sum is zero. The point is that every position with nonzero sum has a move
to a zero position, but every zero position has moves only to nonzero positions. So
from a nonzero position, the first player can arrange to move only to zero positions;
when the game ends, it must do so after a move by the first player.
214 The William Lowell Putnam Mathematical Competition

The main result of Sprague-Grundy theory [BCG, Volume 1, Ch. 3, p. 58] is that
every normal impartial game works the same way! Explicitly, to each position one
assigns a “nim-value” which is the smallest nonnegative integer not assigned to any
position reachable from the given position in one move. In particular, the final position
has value 0, and a position is a second player win if and only if it has value 0. The
result then is that the value of a disjunction of two positions (in which a player may
move in one position or the other) is obtained by adding the values of the positions
in base 2 without carries.
In the original problem, piles of size 0, 2, 3, 4, 5, 6 have values 0, 1, 2, 0, 1, 0, so
the initial position has value 2 ⊕ 0 ⊕ 1 ⊕ 0 = 3, and the winning first move creates a
position of value 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0, as expected.

B6. (3, 2, 0, 0, 0, 0, 0, 0, 2, 3, 61, 133)

For a positive real number α, define
S(α) = { nα : n = 1, 2, 3, . . . }.
Prove that {1, 2, 3, . . . } cannot be expressed as the disjoint union of three
sets S(α), S(β) and S(γ).
Solution 1. Suppose on the contrary that S(α), S(β), S(γ) form a partition of
the positive integers. Then 1 belongs to one of these sets, say S(α), and so nα < 2 for
some n ≥ 1. In particular α < 2; moreover, α > 1 or else S(α) contains every positive
integer, contradicting our hypothesis.
Let m ≥ 2 be the integer satisfying 1+1/m ≤ α < 1+1/(m−1). Then nα = n for
n = 1, . . . , m − 1, but mα = m + 1, so m is the smallest integer in the complement
S(β)∪S(γ) of S(α). Moreover, any two consecutive elements of this complement differ
by m or m + 1.
Without loss of generality, suppose m ∈ S(β), so β = m. Let n be an element of
S(γ); by the previous paragraph, the nearest elements of S(β) on either side of n each
differ from n by at least m, so they differ from each other by at least 2m. This is a
contradiction, because consecutive elements of S(β) differ by at most m + 1 < 2m. 
Related question. A similar argument arises in Problem 3 from the 1999 USA
Mathematical Olympiad [USAMO]:
Let p > 2 be a prime and let a, b, c, d be integers not divisible by p, such that
{ra/p} + {rb/p} + {rc/p} + {rd/p} = 2
for any integer r not divisible by p. Prove that at least two of the numbers
a + b, a + c, a + d, b + c, b + d, c + d are divisible by p. (Note: {x} = x − x
denotes the fractional part of x.)

Solution 2 (attributed to John H. Lindsey II). Let a = 1/α, b = 1/β, c = 1/γ.

For n ≥ 1, let f (n, x) = #(S(x) ∩ {1, 2, . . . , n}). Since S(α) = {1, 2, . . . }, α > 1. Now
kα ≤ n is equivalent to kα < n + 1 and to k < (n + 1)a, so f (n, α) = (n + 1)a − 1.
Hence if {1, 2, 3, . . . } is the disjoint union of S(α), S(β) and S(γ), then
n = f (n, α) + f (n, β) + f (n, γ) = (n + 1)a + (n + 1)b + (n + 1)c − 3. (1)
Solutions: The Fifty-Sixth Competition (1995) 215

Dividing by n + 1 and taking the limit as n → ∞ shows 1 = a + b + c. Multiplying

this by n + 1 and subtracting the result from (1) yields
2 = g(n, a) + g(n, b) + g(n, c), (2)
where g(n, x) = (n + 1)x − (n + 1)x.
We will contradict (2) by proving that for any x > 0, the average value of g(i, x)
for 1 ≤ i ≤ n tends to a limit less than or equal to 1/2 as n → ∞. If x is rational,
say m/d in lowest terms, then any block of d consecutive integers contains exactly
one solution z to mz ≡ j (mod d) for each integer j, so the values of g(i, x) for i in
this block are 0/d, 1/d, 2/d, . . . , (d − 1)/d in some order; hence the average value of
g(i, x) for 1 ≤ i ≤ n tends to


1 0 1 d−1 d−1 1
+ + ··· + = < .
d d d d 2d 2
If instead x is irrational, then the average value of g(i, x) for 1 ≤ i ≤ n tends to 1/2
by the equidistribution result mentioned in 1988B3. 
Stronger result. Solution 2 generalizes to show that if k > 2 and α1 , α2 , . . . , αk
are real numbers, then {1, 2, . . . } cannot be the disjoint union of S(α1 ), S(α2 ), . . . ,
S(αk ). This fact also follows from the following paragraph.
Stronger result. The sets S(α), S(β), S(γ) cannot be disjoint. Proof (John
Rickard): Let N be an integer larger than max{α, β, γ} and consider the N 3 + 1
triples vn = ({n/α}, {n/β}, {n/γ}) ∈ R3 for n = 0, . . . , N 3 , where {x} = x − x
denotes the fractional part of x. If we divide the unit cube 0 ≤ x, y, z ≤ 1 in R3 into
N 3 subcubes of side length 1/N , then by the Pigeonhole Principle, two of the vn , say
vi and vj , lie in the same subcube. Let k = |i − j|. Then k/α is within 1/N of some
integer m, and mα equals k − 1 or k, depending on whether k/α > m or not. In
other words, S(α) contains k or k − 1. Similarly, S(β) and S(γ) each contain k − 1
or k. We conclude that one of k − 1 or k lies in at least two of S(α), S(β), S(γ), so
these sets are not disjoint.
Warning. Not all authors use the notation {x} for the fractional part of x. In fact,
the same notation is used in Problem 1997B1 to denote the distance from x to the
nearest integer. And sometimes it denotes the set whose only element is x!
Stronger result. If S(α) and S(β) are disjoint, then there exist positive integers
m and n such that m/α + n/β = 1. This can be deduced from the following fact: if
x1 , . . . , xk are real numbers, then the closure of the set of points ({nx1 }, . . . , {nxk }) in
the closed k-dimensional unit cube is the set of points (y1 , . . . , yk ) in the cube with the
property that for every sequence of integers n1 , . . . , nk such that n1 x1 +· · ·+nk xk ∈ Z,
we have n1 y1 + · · · + nk yk ∈ Z. (Hint: reduce to the special case where this set is the
entire cube.)
The k = 1 case of this fact is the denseness result proved in the remark in 1988B3.
One can also generalize the equidistribution result given there: if 1, x1 , . . . , xk are
linearly independent over Q, the k-tuples ({nx1 }, . . . , {nxk }) are equidistributed in
the k-dimensional unit cube. (That is, for any subcube of volume µ, the number of
n ≤ N such that the k-tuple ({nx1 }, . . . , {nxk }) lies in the subcube is asymptotic
to µN as N → ∞.) This can be deduced from a multidimensional version of
216 The William Lowell Putnam Mathematical Competition

Weyl’s Equidistribution Theorem, which can be proved in the same way as the one-
dimensional case in [Kör, Theorem 3.1 ].
Reinterpretation. The multidimensional version of Weyl’s Equidistribution The-
orem can be deduced from the following fact: every continuous function on the n-
dimensional unit cube can be uniformly approximated by finite linear combinations
of characters. (A character is a continuous group homomorphism from Rn /Zk to C∗ ;
each such homomorphism is
(x1 , . . . , xk ) !→ exp(2πi(a1 x1 + · · · + ak xk ))
for some a1 , . . . , ak ∈ Z.) The approximation assertion follows from basic properties of
Fourier series, which are ubiquitous in physics and engineering as well as mathematics.
See [Kör, Theorem 3.1 ] for the proof in the k = 1 case, and [DMc, Section 1.10] for
the general case.
Related question. Problem 1979A5 [PutnamII, p. 33] is related:
Denote by [x] the greatest integer less than or equal to x and by S(x) the
sequence [x], [2x], [3x], . . . . Prove that there are distinct real solutions α and
β of the equation x3 − 10x2 + 29x − 25 = 0 such that infinitely many positive
integers appear both in S(α) and S(β).
Remark. If α and β are irrational and 1/α+1/β = 1, then S(α) and S(β) partition
the positive integers (Beatty’s Theorem). See 1993A6 for another instance of this fact,
and the solution to that problem for further discussion.
Solutions: The Fifty-Seventh Competition (1996) 217

The Fifty-Seventh William Lowell Putnam Mathematical Competition

December 7, 1996

A1. (87, 26, 47, 0, 0, 0, 0, 0, 6, 9, 31, 0)

Find the least number A such that for any two squares of combined area
1, a rectangle of area A exists such that the two squares can be packed in
the rectangle (without the interiors of the squares overlapping). You may
assume that the sides of the squares will be parallel to the sides of the
rectangle.
√
Answer. The least such A is (1 + 2)/2.
Solution 1. If x and y are the sides of two squares with combined area 1, then
x2 + y 2 = 1. Suppose without loss of generality that x ≥ y. Then the shorter side of a
rectangle containing both squares without overlap must be at least x, and the longer
side must be at least x + y. Hence the desired value of A is the maximum of x(x + y)
subject to the constraints x2 + y 2 = 1 and x ≥ y > 0.
To find this maximum, we make the trigonometric substitution x = cos θ, y = sin θ
with θ ∈ (0, π/4]. Then
1
cos2 θ + sin θ cos θ =(1 + cos 2θ + sin 2θ)
2 √
1 2
= + cos(2θ − π/4)
2 √2
1+ 2
≤ ,
2
√
with equality for θ = π/8. Hence the least A is (1 + 2)/2. 
Solution 2. As in Solution 1, we find the maximum value of x(x + y) on the arc
defined by the constraints x2 + y 2 = 1 and x ≥ y > 0. By the theory of Lagrange
multipliers [Ap2, Ch. 9.14], the maximum on the closure of the arc occurs either at
an endpoint, or at a point on the arc where the gradient of x(x + y) is a multiple of
the gradient of x2 + y 2 − 1, i.e., where there exists λ such that (2x + y, x) = λ(x, y).
If such a λ exists, we may substitute x = λy into 2x + y = λx to obtain (λ2 −
2λ − 1)y = 0. If y = 0, then 2 2
√ x = λy = 0, contradicting x + y = 1. Therefore
λ − 2λ − 1 = 0 and λ = 1 ± 2. Only the plus sign gives λ > 0, and for this choice,
2

there is a unique pair (x, y) with x = λy > y > 0 and x2 + y 2 = 1. For this pair, we
have 1 = x2 + y 2 = (λ2 + 1)y 2 = (2λ + 2)y 2 , so
x(x + y) = (λ2 + λ)y 2
λ2 + λ
=
2λ + 2
λ
=
2 √
1+ 2
= .
2
√ √
At the endpoints (1, 0) and
√ ( 2/2, 2/2) of the closure of the arc, we have x(x+y) = 1,
so the maximum is (1 + 2)/2. 
218 The William Lowell Putnam Mathematical Competition

A2. (6, 11, 27, 0, 0, 0, 0, 0, 65, 23, 46, 28)

Let C1 and C2 be circles whose centers are 10 units apart, and whose radii
are 1 and 3. Find, with proof, the locus of all points M for which there
exists points X on C1 and Y on C2 such that M is the midpoint of the line
segment XY .
Answer. Let O1 and O2 be the centers of C1 and C2 , respectively. (We are assuming
C1 has radius 1 and C2 has radius 3.) Then the desired locus is an annulus centered
at the midpoint O of O1 O2 , with inner radius 1 and outer radius 2.
Solution. If X lies on C1 , Y lies on C2 , and M is the midpoint of XY , then
" #
−−→ 1 −−→ −→
OM = OX + OY
2
1 "−−→ −−→ −−→ −−→#
= OO1 + O1 X + OO2 + O2 Y
2
1 −−→ −−→#
"
= O1 X + O2 Y .
2
−−→ −−→
By the triangle inequality, the length of O1 X + O2 Y lies between 3 − 1 and 3 + 1.
Conversely, every vector of length @ between 3−1 and 3+1 can be expressed as the sum
of a vector of length 1 and a vector of length 3, by building a (possibly degenerate)
triangle of side lengths @, 1, 3 having the given vector as one side. Thus the set of
possible M is precisely the closed annulus centered at O with inner radius 1 and outer
radius 2. 
Related question. Given C1 , C2 , and a point M in the annulus, construct points
X on C1 and Y on C2 using compass and straightedge such that M is their midpoint.

A3. (63, 18, 6, 0, 0, 0, 0, 0, 0, 0, 47, 72)

Suppose that each of 20 students has made a choice of anywhere from 0
to 6 courses from a total of 6 courses offered. Prove or disprove: there are
5 students and 2 courses such that all 5 have chosen both courses or all 5
have chosen neither course.
Answer. There need not exist 5 such students and 2 such courses.
 
Solution. The number of ways to choose 3 of the 6 courses is 63 = 20. Suppose
each student chooses a different set of 3 courses. Then given any pair of courses, four
students have chosen both and four students have chosen neither course. Thus there
do not exist 5 students who have chosen both courses, nor do there exist 5 students
who have chosen neither course. 
Remark. It can be shown that in any example contravening the assertion of the
problem, each student must choose 3 courses. To see this, note that for each pair of
courses, thereare
 at most 4 + 4 = 8 students who choose both or neither. Thus there
are at most 8 62 = 120 pairs (s, {c, d}), where s is a student and {c, d} is a set of two
distinct courses, such that s either chooses both courses or chooses kneither
6−kcourse.

On the other hand, if a student chooses k courses, then there are 2 + 2 pairs
of courses such that the student chose both or chose neither. This sum is minimized
for k = 3, where it equals 6. Thus there are at least 6 × 20 = 120 pairs (s, {c, d});
 Solutions: The Fifty-Seventh Competition (1996) 219

if any students choose more or fewer than 3 courses, we get strict inequality and a
contradiction.
Remark. A consequence of the previous observation is that the example given
above is unique if we assume no two students choose the same set of courses. On the
other hand, there exist other counterexamples to the assertion of the problem in which
more than one student selects the same set of courses. Here is an elegant construction
attributed to Robin Chapman. Take an icosahedron, and label its vertices with the
names of the 6 courses so that two vertices have the same label if and only if they
are antipodal. For each of the 20 faces of the icosahedron, have one student select the
three courses labelling the vertices of the face; then antipodal faces give rise to the
same set of courses. Concretely, if the courses are A, B, C, D, E, F , the sets could be
ABC, ACD, ADE, AEF , AF B, BCE, CDF , DEB, EF C, F BD.

Literature note. The study of problems such as this one, asking for combinatorial
structures not containing certain substructures, is known as Ramsey theory. See [Gr1]
for an introduction.

A4. (3, 7, 7, 0, 0, 0, 0, 0, 7, 19, 67, 96)

Let S be a set of ordered triples (a, b, c) of distinct elements of a finite set
A. Suppose that
(1) (a, b, c) ∈ S if and only if (b, c, a) ∈ S;
(2) (a, b, c) ∈ S if and only if (c, b, a) ∈
/ S [for a, b, c distinct];
(3) (a, b, c) and (c, d, a) are both in S if and only if (b, c, d) and (d, a, b) are
both in S.
Prove that there exists a one-to-one function g from A to R such that
g(a) < g(b) < g(c) implies (a, b, c) ∈ S.
Solution. We may assume that A is nonempty. Pick some x in A. Let us define a
binary relation on A as follows: we define b < c if and only if b = x = c or (x, b, c) ∈ S.
For x, b, c distinct, we have
(x, b, c) ∈ S ⇐⇒ (c, x, b) ∈ S ⇐⇒ (b, x, c) ∈ S ⇐⇒ (x, c, b) ∈ S
by (1),(2),(1), respectively. Thus b < c if and only if c < b. We also have b < x for
any b = x, since (x, b, x) ∈ S.
We now show that if b < c and c < d, then b < d. These conditions imply c, d = x;
if b = x, we are done, so assume b = x. Then (x, b, c) and (x, c, d) are in S; by (1),
(c, d, x) is also in S, and then (3) implies (d, x, b) ∈ S. By (1) again, (x, b, d) ∈ S, and
so b < d.
Thus < is a total ordering of A, and so we can find the desired function g. For
example, let g(a) be the number of y ∈ A such that y < a. 
Remark. The construction of the total ordering did not use the finiteness of A.
The finiteness was only used to construct the order-preserving injection of A into R.
In fact, such an injection also exists if A is countable: list the elements of A in an
arbitrary fashion (without regard to the total ordering), and choose their images in R
one at a time, making sure that the total ordering is preserved.
220 The William Lowell Putnam Mathematical Competition

A5. (4, 4, 2, 0, 0, 0, 0, 0, 0, 13, 40, 143)

If p is a prime number greater than 3 and k = 2p/3, prove that the sum




p p p
+ + ··· +
1 2 k
of binomial coefficients is divisible by p2 .
 
Solution. For 1 ≤ n ≤ p − 1, p divides np and


1 p 1 p−1 p−2 p−n+1 (−1)n−1
= · · ··· ≡ (mod p),
p n n 1 2 n−1 n
where the congruence x ≡ y (mod p) means that x − y is a rational number whose
numerator, in reduced form, is divisible by p. Therefore
k
 k
1 p (−1)n−1
≡
n=1
p n n=1
n
k k/2
1 1
= −2
n=1
n n=1
2n
k p−1
1 1
≡ + (mod p).
n=1
n n
n=p− k/2

Any prime p > 3 is of one of the forms 6r + 1 or 6s + 5. In the former case, k = 4r and
p−k/2 = 4r +1 = k +1. In the latter case, k = 4s+3 and p−k/2 = 4s+4 = k +1.
In either case, we conclude that
k
 p−1 p/2 p/2
1 p 1 1 1
≡ ≡ + ≡ 0 (mod p),
n=1
p n n=1
n n=1
n n=1
p−n
which completes the proof. 
Related question. Compare with Problem 1 of the 1979 International Mathematical
Olympiad [IMO79–85, p. 2]:
Let m and n be positive integers such that
m 1 1 1 1
= 1 − + − ··· − + .
n 2 3 1318 1319
Prove that m is divisible by 1979.

A6. (4, 4, 11, 0, 0, 0, 0, 0, 18, 13, 46, 110)

Let c ≥ 0 be a constant. Give a complete description, with proof, of the
set of all continuous functions f : R → R such that f (x) = f (x2 + c) for all
x ∈ R.
Answer. If 0 ≤ c ≤ 1/4 then f is constant. If c > 1/4, then f is uniquely determined
(as described below) by its restriction g to [0, c], which may be any continuous function
on [0, c] with g(0) = g(c).
Solution. The functional equation implies f (x) = f (x2 + c) = f ((−x)2 + c) =
f (−x). Conversely, if f satisfies the functional equation for x ≥ 0 and also satisfies
Solutions: The Fifty-Seventh Competition (1996) 221

f (x) = f (−x), then it satisfies the functional equation for all x. So we will consider
only x ≥ 0 hereafter.
Case 1: 0 ≤ c ≤ 1/4. In this case, the zeros a, b of x2 − x + c are real and satisfy
0 ≤ a ≤ b. We will show that f (x) = f (a) for all x ≥ 0. First suppose 0 ≤ x < a.
Define x0 = x and xn+1 = x2n + c for n ≥ 0. Then we have xn < xn+1 < a for all n,
by induction: if xn < a, then on one hand xn+1 − xn = x2n − xn + c > 0, and on the
other hand xn+1 = x2n + c < a2 + c = a. We conclude that xn converges to a limit
L not exceeding a, which must satisfy L2 + c = L. Since the only real roots of this
equation are a and b, we have L = a. Now since f (xn ) = f (x) for all n and xn → a,
we have f (x) = f (a) by continuity.
If a < x < b, we argue similarly. Define xn as above; then we have a < xn+1 < xn
for all n, by induction: if xn < b, then on one hand xn+1 − xn = x2n − xn + c < 0, and
on the other hand xn+1 = x2n + c > a2 + c = a. So again f (x) = f (a) for a < x < b.
By continuity, f (b) = f (a).
√
Finally, suppose x > b. Now define x0 = x and xn+1 = xn − c for n ≥ 0. Then
we have b < xn+1 < xn for all n,√by induction: on one hand, x2n+1 + c = xn < x2n + c,
and on the other√hand xn+1 > b − c = b. Thus xn converges to a limit L not less
than b, and L = L − c, so L = b. So analogously, f (x) = f (b) = f (a).
We conclude that f (x) = f (a) is constant.
Case 2: c > 1/4. In this case, x2 + c > x for all x. Let g be any continuous function
on [0, c] with g(0) = g(c). Then g extends to a continuous functionf by setting
√ √
f (x) = g(x) for x < c, f (x) = g( x − c) for c ≤ x < c2 + c, f (x) = g( x − c − c)
for c2 + c ≤ x < (c2 + c)2 + c, and so on. Conversely, any f satisfying the functional
equation is determined by its values on [0, c] and satisfies f (0) = f (c), and so arises
in this fashion. 
Stronger result. More generally, let p be a polynomial not equal to x or to c − x
for any c ∈ R. One can imitate the above solution to classify continuous functions
f : R → R such that f (x) = f (p(x)) for all x ∈ R. The classification hinges on
whether p has any fixed points.
Case 1: p(x) has a fixed point. In this case, we show that f must be constant.
We first reduce to the case where p has positive leading coefficient. If this is not the
case and p has odd degree, we work instead with p(p(x)); if p has even degree, then
g(x) = f (−x) satisfies g(x) = g(−p(−x)), so we work instead with −p(−x).
Let r be the largest fixed point of p; then p(x) − x does not change sign for x > r,
and it tends to +∞ as x → +∞, so it is positive for all x > r. In particular, p maps
[r, ∞) onto itself. For any x > r, define the sequence {xn } by setting x0 = x and
xn+1 is the smallest element of [r, ∞) such that p(xn+1 ) = xn . The sequence {xn } is
decreasing, so it converges to a limit L ≥ r, and L = lim xn = lim p(xn+1 ) = p(L).
Thus L = p(L), which can only occur if L = r. Since f (xn ) = f (x0 ) for all n and
xn → r, we have f (x0 ) = f (r). Consequently, f is constant on [r, ∞).
Suppose, in order to obtain a contradiction, that f is not constant on all of R. Let

T = inf{ t ∈ R : f is constant on [t, ∞)}.

Then f is constant on [T, ∞). It is also constant on [p(T ), ∞), so p(T ) ≥ T . It is

222 The William Lowell Putnam Mathematical Competition

also constant on p−1 ([T, ∞)); if p(T ) > T , this set also includes T − C for small C,
contradiction. Thus p(T ) = T .
Since p has finitely many fixed points, either p(T − C) > T − C for small C, or
p(T − C) < T − C for small C. In the former case, the sequence with x0 = T − C and
xn+1 = p(xn ) converges to T , so f (x0 ) = f (T ). In the latter case, the sequence with
x0 = T − C and xn+1 the largest element of (xn , T ) with q(xn+1 ) = xn converges to
T , so g(x0 ) = T . In either case, g is constant on a larger interval than [T, ∞), a
contradiction.
Case 2: p has no fixed point. In this case, p must have even degree, so as in Case
1, we may assume p has positive leading coefficient, which implies that p(x) > x for
all x ∈ R. There exists a0 ∈ R such that p is strictly increasing on [a0 , ∞), since any
a0 larger than all zeros of p has this property. For n ≥ 0, set an+1 = p(an ), so {an }
is an increasing sequence. It cannot have a limit, since that limit would have to be a
fixed point of p, so an → ∞ as n → ∞.
We claim f is determined by its values on [a0 , a1 ]. Indeed, these values determine
f on [a1 , a2 ], on [a2 , a3 ], and so on, and the union of these intervals is [a0 , ∞).
Furthermore, for any x ∈ R, if we set x0 = x and xn+1 = p(xn ), then the sequence
{xn } is increasing and unbounded, so some xn lies in [a0 , ∞), and f (xn ) = f (x) is
determined by the values of f on [a0 , a1 ].
Conversely, let g(x) be any continuous function on [a0 , a1 ] with g(a0 ) = g(a1 ); we
claim there exists a continuous f : R → R with f (x) = f (p(x)) whose restriction to
[a0 , a1 ] is g. For x ∈ R, again set x0 = x and xn+1 = p(xn ). If n is large enough, there
exists t ∈ [a0 , a1 ) such that xn = pk (t) for some integer k, and this t is independent
of n. Hence we may set f (x) = f (t).
It remains to show that f is continuous. We can define a continuous inverse p−1 of
p mapping [a1 , ∞) to [a0 , ∞), and f is given on [a0 , a1 ] by f (x) = g(x), on [a1 , a2 ] by
f (x) = g(p−1 (x)), on [a2 , a3 ] by f (x) = g(p−2 (x)), and so on. Thus f is continuous
on each of the intervals [an , an+1 ], and hence on their union [a0 , ∞). For x arbitrary,
we can find a neighborhood of x and an integer k such that pk maps the neighborhood
into [a0 , ∞), and writing f (t) = f (pk (t)) for t in the neighborhood, we see that f is
also continuous there.
We conclude that the continuous functions f satisfying f (x) = f (p(x)) correspond
uniquely to the continuous functions g on [a0 , a1 ] such that g(a0 ) = g(a1 ). Otherwise
put, the quotient of R by the map p is isomorphic to the interval [a0 , a1 ] with the
endpoints identified.
Remark. The ideas in this problem come from the theory of dynamical systems.
For a problem involving a related dynamical system, see 1992B3.

B1. (113, 72, 17, 0, 0, 0, 0, 0, 3, 1, 0, 0)

Define a selfish set to be a set which has its own cardinality (number
of elements) as an element. Find, with proof, the number of subsets of
{1, 2, . . . , n} which are minimal selfish sets, that is, selfish sets none of
whose proper subsets is selfish.
Answer. The number of subsets is Fn , the nth Fibonacci number. (See the definition
at the end of 1988A5.)
Solutions: The Fifty-Seventh Competition (1996) 223

Solution 1. Let fn denote the number of minimal selfish subsets of {1, . . . , n}.
We have f1 = 1 and f2 = 2. For n > 2, the number of minimal selfish subsets of
{1, . . . , n} not containing n is equal to fn−1 . On the other hand, for any minimal
selfish set containing n, by removing n from the set and subtracting 1 from each
remaining element, we obtain a minimal selfish subset of {1, . . . , n − 2}. (Note that
1 could not have been an element of the set because {1} is itself selfish.) Conversely,
any minimal selfish subset of {1, . . . , n − 2} gives rise to a minimal selfish subset of
{1, . . . , n} containing n by the inverse procedure. Hence the number of minimal selfish
subsets of {1, . . . , n} containing n is fn−2 . Thus we obtain

fn = fn−1 + fn−2 ,

which together with the initial values f1 = f2 = 1 implies that fn = Fn . 

Solution 2. A set is minimal selfish if and only if its smallest member is equal
to its cardinality. (If a selfish set contains a member smaller than its cardinality, it
contains a subset of that cardinality which is also selfish.) Thus we can compute the
number Sn of minimal selfish subsets of {1, . . . , n} by summing, over each cardinality
k, the number of subsets of {k + 1, . . . , n} of size k − 1 (since each
 of these, plus {k},

is minimal selfish, and vice versa). In other words, Sn = k n−k k−1 . We directly verify
that S1 = S2 = 1, and that



n−k n−1−k
Sn−1 + Sn = +
k−1 k−1
k

 k


n−k n−k
= +
k−1 k−2
k

k
n+1−k
=
k−1
k
= Sn+1 .

Thus by induction, Sn = Fn for all n. 

 n−k
Remark. The identity Fn = k k−1 can be interpreted as the fun fact
that certain diagonal sums in Pascal’s triangle yield the Fibonacci numbers. (See
Figure 30.) It is common for a combinatorial enumeration problem to admit both a
bijective and a recursive proof; see 1996B5 for another example.

1
1 1
1 2 1 1
1 3 3 1 1
1 4 6 4 1 2
3
1 5 10 10 5 1 5
8

FIGURE 30.
Fibonacci in Pascal.
224 The William Lowell Putnam Mathematical Competition

B2. (85, 13, 6, 0, 0, 0, 0, 0, 15, 7, 28, 52)

Show that for every positive integer n,

 2n−1
 2n+1
2n − 1 2
2n + 1 2
< 1 · 3 · 5 · · · (2n − 1) < .
e e

Solution 1. By estimating the area under the graph of ln x using upper and lower
rectangles of width 2 (see Figure 31), we obtain
 2n−1  2n+1
ln x dx < 2(ln(3) + · · · + ln(2n − 1)) < ln x dx.
1 3

Since ln x dx = x ln x − x + C, exponentiating and taking square roots yields the
middle two inequalities in

 2n−1
2n − 1 2 2n−1
< (2n − 1) 2 e−n+1
e
≤ 1 · 3 · · · (2n − 1)
−n+1

 2n+1
2
2n+1 e 2n + 1
≤ (2n + 1) 2 3/2
< ,
3 e
and the inequalities at the ends follow from 1 < e < 3. 

x
1 3 5 7 … 2n – 1 2n + 1

FIGURE 31.


Estimating ln x dx.

Solution 2. We use induction on n. The n = 1 case follows from 1 < e < 3. For
the inductive step, we need to prove
 2n+1  2n+1
2
 2n+3  2n+3
2
e
 2n−1  2n−1 < 2n + 1 <  e  2n+1 ,
2 2n+1 2
e e
which is equivalent to

 2n−1
2

 2n+3
2
2n + 1 2n + 3
<e< .
2n − 1 2n + 1
Solutions: The Fifty-Seventh Competition (1996) 225

The left side is equal to (1 + u)1/u for u = 2/(2n − 1), and the right side is equal to
(1 + u)1/u+1 for u = 2/(2n + 1), so it suffices to prove the inequalities
(1 + u)1/u < e < (1 + u)1/u+1 for u > 0.
Apply ln and rearrange to rewrite this as
u
< ln(1 + u) < u,
1+u
which follows by integrating
1 1
< <1
(1 + t)2 1+t
from t = 0 to t = u. 
Related question. Use the method of Solution 1 to prove that
e(n/e)n < n! < en(n/e)n .
(This is a weak version of Stirling’s approximation. The standard form of Stirling’s
approximation is stated in the following remark.)
Remark. Better estimates for n! can be obtained using the Euler-Maclaurin
summation formula [At, Section 5.4]: For any fixed k > 0,
 b
B2i " (2i−1) #
b k
f (a) + f (b)
f (j) = f (t) dt + + f (b) − f (2i−1) (a) + Rk (a, b),
j=a a 2 i=1
(2i)!

where the Bernoulli numbers B2i are given by the power series
∞
x B2i 2i
= −x/2 + x ,
ex − 1 i=1
(2i)!

(see [HW, Section 17.2]) and the error term Rk (a, b) is given by
 b
−1
Rk (a, b) = B2k+2 (t − t)f (2k+2) (t) dt.
(2k + 2)! a
(See 1993B3 for another appearance of Bernoulli numbers.) Specifically, this formula
can be used to obtain Stirling’s approximation to n! [Ap2, Theorem 15.19],
√ " n #n
n! ∼ 2πn ,
e
where the tilde indicates that the ratio of the two sides tends to 1 as n → ∞. See
Solution 1 to 1995A6 for an application of Stirling’s approximation.
The Euler-Maclaurin formula has additional applications in numerical analysis, as
well as in combinatorics; see the remark following 1996B3 for an example.

B3. (20, 19, 8, 0, 0, 0, 0, 0, 14, 36, 56, 53)

Given that {x1 , x2 , . . . , xn } = {1, 2, . . . , n}, find, with proof, the largest
possible value, as a function of n (with n ≥ 2), of
x1 x2 + x2 x3 + · · · + xn−1 xn + xn x1 .
226 The William Lowell Putnam Mathematical Competition

Answer. The maximum is (2n3 + 3n2 − 11n + 18)/6, and the unique arrangement
(up to rotation and reflection) achieving the maximum is
. . . , n − 4, n − 2, n, n − 1, n − 3, . . . .

Solution 1. Since there are finitely many arrangements, at least one is optimal.
In an optimal arrangement, if a, b, c, d occur around the circle in that order, with a
adjacent to b, c adjacent to d and a > d, then b > c. Otherwise, reversing the segment
from b to c would increase the sum by (a − d)(c − b).
This implies first that in an optimal arrangement, n is adjacent to n − 1, or else
we could reverse the segment from n − 1 to either neighbor of n. Likewise, n must be
adjacent to n − 2, or else we could reverse the segment from n − 2 to the neighbor of
n other than n − 1.
Now n − 1 is adjacent to n but not to n − 2; we must have that n − 1 is also adjacent
to n − 3, or else we could reverse the segment from n − 3 to the neighbor of n − 1 other
than n. Likewise n − 2 must be adjacent to n − 4, and so on.
In this arrangement, the value of the function is
n(n − 1) + n(n − 2) + (n − 1)(n − 3) + · · · + 4 · 2 + 3 · 1 + 2 · 1.
n
Using the fact that k=1 k 2 = n(n + 1)(2n + 1)/6, we can rewrite the sum as
n−2 n−2
n(n − 1) + 2 + k(k + 2) = n2 − n + 2 + (k + 1)2 − 1
k=1 k=1
(n − 1)n(2n − 1)
=n −n+2+
2
− 1 − (n − 2)
6
2n3 + 3n2 − 11n + 18
= . 
6
Solution 2. We prove that the given arrangement is optimal by induction on
n. Suppose that the analogous arrangement of n − 1 numbers is optimal. Given
any arrangement of n − 1 numbers, inserting n between a and b changes the sum
by na + nb − ab = n2 − (n − a)(n − b) ≤ n2 − 2. Thus the maximum sum for n
numbers is at most n2 − 2 plus the maximum sum for n − 1 numbers, and this value is
achieved by inserting n between n − 1 and n − 2 in the optimal arrangement of n − 1
numbers. Thus the result is the optimal arrangement of n numbers; since the only
possible arrangement for n = 3 has value 6 + 3 + 2 = 11, the value for n numbers is
n(n + 1)(2n + 1)
11 + (42 − 2) + · · · + (n2 − 2) = 11 + − 2(n − 3) − (12 + 22 + 32 )
6
2n3 + 3n2 − 11n + 18
= . 
6
Related question. Can you determine the arrangement that minimizes the function?
Hint: the minimum value is (n3 + 3n2 + 5n − 6)/6 for n even and (n3 + 3n2 + 5n − 3)/6
for n odd.
n 2
Remark. The fact that k=1 k = n(n + 1)(2n + 1)/6, used above, can be
generalized as follows: for any polynomial P (x) of degree m, there exists a polynomial
n
Q(x) of degree m + 1 such that k=1 P (k) = Q(n) for all positive integers n. This
Solutions: The Fifty-Seventh Competition (1996) 227

follows from the Euler-Maclaurin summation formula (see the remark in 1996B2),
which can also be used to compute Q from P .

B4. (40, 4, 7, 0, 0, 0, 0, 0, 17, 4, 48, 86)

For any square matrix A, we can define sin A by the usual power series:
∞
(−1)n
sin A = A2n+1 .
n=0
(2n + 1)!
Prove or disprove: there exists a 2 × 2 matrix A with real entries such that


1 1996
sin A = .
0 1

Answer. There does not exist such a matrix A.

Solution 1. Over the complex numbers, if A has distinct eigenvalues, it is
diagonalizable. Since sin A is a convergent power series in A, eigenvectors of A are
also eigenvectors of sin A,

so A having
 distinct eigenvalues would imply that sin A
1 1996
is diagonalizable. Since is not diagonalizable, it can be sin A only for a
0 1


x y
matrix A with equal eigenvalues. This matrix can be conjugated into the form
0 x
for some x and y. Using the power series for sin, we compute

 ∞
2k+1
x y (−1)k x y
sin =
0 x (2k + 1)! 0 x
k=0
∞
2k+1
(−1)k x2k+1 1 y/x
=
(2k + 1)! 0 1
k=0
∞

(−1)k x2k+1 1 (2k + 1)y/x
=
(2k + 1)! 0 1

k=0
∞ (−1)k x2k+1 ∞ (−1)k x2k 
y k=0 (2k)!
= ∞ (−1)k x2k+1 
k=0 (2k+1)!

0 k=0 (2k+1)!


sin x y cos x
= .
0 sin x


x y
Thus if sin x = 1, then cos x = 0 and sin is the identity matrix. In other
0 x
words, sin A cannot equal a matrix whose eigenvalues are 1 but which is not the
identity matrix. 



x y x 0
Remark. The computation of sin can be simplified by taking A =
0 x 0 x


0 y
and B = in the identity
0 0
sin(A + B) = sin A cos B + cos A sin B,
which holds when A and B commute.
228 The William Lowell Putnam Mathematical Competition

∞
Solution 2. Put cos A = n=0 (−1)n A2n /n!. The identity sin2 x + cos2 x = 1
implies the identity
4∞ 52 4 ∞ 52
(−1)n 2n+1 (−1)n 2n
x + x =1
n=0
(2n + 1)! n=0
(2n)!

of formal power series. The series converge absolutely if we substitute x = A, so

sin2 A + cos2 A equals the identity matrix I. But

2

1 1996 0 3992
I− =
0 1 0 0
cannot be the square of a matrix; such a matrix would have to be nilpotent, and
the kth power of a k × k nilpotent matrix is always zero, by the Cayley-Hamilton
Theorem. 
Stronger result. We will show that the image of sin : M2 (R) → M2 (R) is the set of
matrices B ∈ M2 (R) such that at least one of the following holds:


λ 0
(1) B is conjugate to with λ, λ ∈ [−1, 1];
0 λ


λ 1
(2) B is conjugate to with λ ∈ (−1, 1);
0 λ
(3a) B = λI for some λ ∈ R; or
(3b) B has nonreal eigenvalues.
Recall that if A, B ∈ Mn (R) are conjugate over C, then they are also conjugate over
R. The entire function sin : C → C is surjective, since it equals the composition of
−1
the surjection eiz : C → C∗ followed by the surjection z−z2i : C∗ → C.
Every A ∈ M2 (R) is conjugate over C to a matrix of one of the following types:


r 0
(1 ) with r, r ∈ R;
0 r


 x 1
(2 ) with x ∈ R; or
0 x


z 0
(3 ) with z ∈ C.
0 z̄
By definition, sin(CAC −1 ) = C(sin A)C −1 for any invertible C, so B ∈ M2 (R) is in
the image of sin : M2 (R) → M2 (R) if and only if it is conjugate over C to the sine of
a matrix of one of types (1 ), (2 ), (3 ).
If A is of type (1 ), then sin A is of type (1), and conversely all type (1) matrices

are

conjugate to a matrix arising in this way. If A is of type (2 ), then sin A =
sin x cos x
is of type (1) or (2), according as cos x = 0 or not, and conversely, any
0 sin x


sin x cos x
matrix of type (2) is conjugate to for some x ∈ (−π/2, π/2). If A is
0 sin x


sin z 0
of type (3 ), then sin A = , which is of type (3a) if sin z ∈ R, and of
0 sin z
type (3b) otherwise. Conversely, if B = λI if of type (3a), then B = sin(zI) where
Solutions: The Fifty-Seventh Competition (1996) 229

z ∈ C satisfies sin z = λ. And if B is of type (3b) with eigenvalues λ, λ̄, choose z ∈ C

with sin z = λ, so that B is conjugate to sin A where A ∈ M2 (R) is of type (3b) with
eigenvalues z and z̄.

B5. (34, 4, 7, 0, 0, 0, 0, 0, 3, 2, 58, 98)

Given a finite string S of symbols X and O, we write ∆(S) for the number
of X’s in S minus the number of O’s. For example, ∆(XOOXOOX) = −1.
We call a string S balanced if every substring T of (consecutive symbols of )
S has −2 ≤ ∆(T ) ≤ 2. Thus, XOOXOOX is not balanced, since it contains
the substring OOXOO. Find, with proof, the number of balanced strings
of length n.
Answer. The number of balanced strings of length n is 3 · 2n/2 − 2 if n is even, and
2 − 2 if n is odd.
(n+1)/2

Solution 1. We give an explicit counting argument. Consider a 1×n checkerboard,

in which we write an n-letter string, one letter per square. If the string is balanced, we
can cover each pair of adjacent squares containing the same letter with a 1×2 domino,
and these will not overlap (because no three in a row can be the same). Moreover,
any domino is separated from the next by an even number of squares, since they must
cover opposite letters, and the sequence must alternate in between.
Conversely, any arrangement of dominoes where adjacent dominoes are separated
by an even number of squares corresponds to a unique balanced string, once we choose
whether the string starts with X or O. In other words, the number of balanced strings
is twice the number of acceptable domino arrangements.
We count these arrangements by numbering the squares 0, 1, . . . , n − 1 and distin-
guishing whether the dominoes start on even or odd numbers. Once this is decided,
one simply chooses whether or not to put a domino in each eligible position. Thus
we have 2 n/2 arrangements in the first case and 2 (n−1)/2 in the second, but the
case of no dominoes has been
 n/2  counted twice. Hence the number of balanced strings is
2 2 +2 (n−1)/2
− 1 , which equals the answer given above. 

Solution 2. Let bn denote the number of balanced strings of length n. We

establish the recursion bn+2 = 2bn + 2 for n ≥ 1. This recursion and the initial
conditions b1 = 2, b2 = 4 yield the desired result by induction.
Let xn , yn , zn denote the number of balanced strings of length n ending with XX,
the number that end in XO and whose last doubled letter is X, and the number that
end in XO and whose last doubled letter is O, respectively. (To keep things consistent,
we count a purely alternating string as if its last doubled letter were both X and O.)
The counts remain the same if X and O are interchanged. Thus bn = 2xn +2yn +2zn −2
for n ≥ 1 (we subtract 2 to avoid double-counting the purely alternating strings).
If a balanced string of length n + 2 ends in XX, the remainder must end in OO, or
end in XO and have last doubled letter O. Thus xn+2 = xn + zn . If a balanced string
of length n + 2 ends in XO and has last doubled letter X, the remainder (the first
n symbols) must end in XO and have last doubled letter X, or end in OX and have
last doubled letter O; conversely, appending XX to the end of such a balanced string
results in a balanced string of length n + 2. Thus yn+2 = 2yn . Finally, if a balanced
230 The William Lowell Putnam Mathematical Competition

string of length n + 2 ends in XO and has last doubled letter O, the remainder must
end in OO, or end in XO and have last doubled letter O. Thus zn+2 = xn + zn as
well. Putting this together,

bn+2 = 2xn+2 + 2yn+2 + 2zn+2 − 2

= 2(xn + zn ) + 2(2yn ) + 2(xn + zn ) − 2
= 2(bn + 2) − 2,

the desired recursion. 

B6. (0, 1, 9, 0, 0, 0, 0, 0, 2, 0, 23, 171)

Let (a1 , b1 ), (a2 , b2 ), . . . , (an , bn ) be the vertices of a convex polygon which
contains the origin in its interior. Prove that there exist positive real
numbers x and y such that

(a1 , b1 )xa1 y b1 + (a2 , b2 )xa2 y b2 + · · · + (an , bn )xan y bn = (0, 0).

 (ai ,bi )·:v

Solution 1. Let f (-v ) = ie . If ∇f (x0 , y0 ) = -0, then (ex0 , ey0 ) is a
solution of the original vector equation. Hence it suffices to show that f achieves a
global minimum somewhere.
For 0 ≤ θ ≤ 2π, define

g(θ) = max{(ai , bi ) · (cos θ, sin θ)}.

i

Then g(θ) is always positive, because the origin lies in the interior of the convex hull
of the points (ai , bi ). Let c > 0 be the minimum value of g(θ) on the interval [0, 2π];
then
f (-v ) ≥ max{e(ai ,bi )·:v } ≥ ec|:v|
i

for all -v . In particular, f is greater than ecr outside of a circle of radius r. Choose
r > 0 such that ecr > f ((0, 0)). Then the infimum of f on the entire plane equals
its infimum within the disc of radius r centered at the origin. Since the closed disc is
compact, the infimum of f there is the value of f at some point, which is the desired
global minimum. 
Solution 2. We retain the notation of the first solution. To prove that ∇f
vanishes somewhere, it suffices to exhibit a simple closed contour over which ∇f has
a nonzero winding number. In fact, any sufficiently large counterclockwise circle has
this property. As in the first solution, there exists c such that maxi {(ai , bi ) · -v } ≥ c|-v |,
so
n
-v · ∇f (-v ) = e(ai ,bi )·:v ((ai , bi ) · -v )
i=1
≥ c|-v |ec|:v| + (n − 1) inf xex
x
= c|-v |ec|:v| − (n − 1)/e,

since xex is minimized over x ∈ R at x = −1.

Solutions: The Fifty-Seventh Competition (1996) 231

Thus if -v runs over a large enough circle, we will have -v · ∇f (-v ) > 0 everywhere on
the circle, implying that the vector field ∇f (-v ) has the same winding number as -v on
the circle, namely 1. (In particular, the number of solutions -v to ∇f (-v ) = 0, counted
with multiplicity, is odd.) 
Remark. The notion of winding number also occurs in complex analysis, for
instance in the proof of Rouché’s Theorem, which is stated in Solution 3 to 1989A3.
Remark. More generally, the following useful fact is true [Fu, p. 83]:
Let u1 , . . . , ur be points in Rn , not contained in any affine hyperplane, and
let K be their convex hull. Let ε1 , . . . , εr be any positive real numbers, and
define H : Rn → Rn by
r
1
H(x) = εk e(uk ,x) uk ,
f (x)
k=1

where f (x) = ε1 e + · · · + εr e
(u1 ,x)
, and (·, ·) is the usual inner product
(ur ,x)

on Rn . Then H defines a real analytic isomorphism of Rn onto the interior of

K.
(We are grateful to Bernd Sturmfels for pointing this out.) In algebraic geometry, this
can be used to show that “the moment map is surjective” for toric varieties.
232 The William Lowell Putnam Mathematical Competition

The Fifty-Eighth William Lowell Putnam Mathematical Competition

December 6, 1997

A1. (138, 3, 8, 0, 0, 0, 0, 0, 1, 2, 21, 32)

A rectangle, HOM F , has sides HO = 11 and OM = 5. A triangle ABC has
H as the intersection of the altitudes, O the center of the circumscribed
circle, M the midpoint of BC, and F the foot of the altitude from A. What
is the length of BC?

H O

F 11 M

Answer. The length of BC is 28.

Solution 1 (Dave Rusin). (See Figure 32.) Introduce coordinates for which

O = (0, 0), H = (−11, 0), F = (−11, −5), and M = (0, −5).

Since B and C are both equidistant from M and O, the perpendicular bisector of BC
is the y-axis. Since BC contains F , we have B = (−x, −5) and C = (x, −5) for some
x. Also, A = (−11, y) for some y.
The altitude through B passes through H, so its slope is 5/(x − 11). It is
perpendicular to the line AC, whose slope is −(y + 5)/(x + 11), so these two slopes
have product −1. That is, 5(y + 5) = (x − 11)(x + 11).

A
◗◗
 ◗
 ◗
◗
 ◗
◗
 ◗
 ◗
◗
 ◗
 ◗
◗
 H O ◗
 ◗
◗
 ◗
◗
 ◗
 ◗
◗
 ◗
B F M C

FIGURE 32.
Solutions: The Fifty-Eighth Competition (1997) 233

On the other hand, A and B are also equidistant from O, so y 2 + 112 = x2 + 52 .

Comparing this equality with the previous one, we find y 2 − 5y − 50 = 0, so y = 10 or
y = −5. The latter is impossible, so y = 10 and x = 14, yielding BC = 2x = 28. 
Solution 2. The centroid G of the triangle is collinear with H and O (Euler line),
and G lies two-thirds of the way from A to M . Therefore H is two-thirds of the way
from A to F , so AF = 3OM = 15. Triangles BF H and AF C are similar, since they
are right triangles and
∠HBF = ∠HBC = π/2 − ∠C = ∠CAF.
Hence BF/F H = AF/F C, and BF ·F C = F H ·AF = 75. Now BC 2 = (BF +F C)2 =
(F C − BF )2 + 4BF · F C, but
F C − BF = (F M + M C) − (BM − F M ) = 2F M = 2HO = 22,
so
√ √
BC = 222 + 4 · 75 = 784 = 28. 

Remark. For more on the Euler line and related topics, see [CG, Section 1.7].
Solution 3. Introduce a coordinate system with origin at O, and for a point
- denote the vector from O to X. Then H
X, let X - = A -+B - +C - (see remark
-
below) and M - -
√ = (B + C)/2 √ = (H − A)/2; we now compute AH = 2OM = 10,
- -
OC = OA = AH 2 + OH 2 = 221, and
√ √
BC = 2M C = 2 OC 2 − OM 2 = 2 221 − 25 = 28. 

Remark. The vector equality H - = A - +B - +C - holds for any triangle with

circumcenter O at the origin, and can be proved either using the properties of the
-  = A+
Euler line mentioned in the second solution, or directly as follows. Let H - B- + C;
-
then
-  − A)
(H - · (B
- − C)
- = (B - · (B
- + C) - − C) - ·B
- =B - −C
- ·C
- = OB 2 − OC 2 = 0.

Thus H  A is perpendicular to BC, i.e., H  lies on the altitude through A. Similarly,

H  lies on the other two altitudes, so H  = H.

A2. (42, 42, 44, 0, 0, 0, 0, 0, 18, 7, 19, 33)

Players 1, 2, 3, . . . , n are seated around a table, and each has a single penny.
Player 1 passes a penny to Player 2, who then passes two pennies to Player
3. Player 3 then passes one penny to Player 4, who passes two pennies to
Player 5, and so on, players alternately passing one penny or two to the
next player who still has some pennies. A player who runs out of pennies
drops out of the game and leaves the table. Find an infinite set of numbers
n for which some player ends up with all n pennies.
Solution. We show that the game terminates with one player holding all of the
pennies if and only if n = 2m + 1 or n = 2m + 2 for some m; this assertion is clear
for n = 1 and n = 2, so we assume n ≥ 3. We begin by determining the state of the
game after each player has moved once. The first player passes one penny and drops
234 The William Lowell Putnam Mathematical Competition

out, and the second player passes two pennies and drops out. Then the third player
passes one penny and keeps two, the fourth player passes two pennies and drops out,
the fifth player passes one penny and keeps two, and so on. The net result is that
 n−1
2  players remain, each of whom has two pennies except for the player to move
next, who has 3 or 4 pennies.
Trying some small examples suggests the following induction argument. Suppose
that for some k ≥ 2, the game reaches a point where:
• Except for the player to move, each player has k pennies;
• The player to move has at least k pennies, and it is his turn to pass one penny.
We will show by induction that the game terminates if and only if the number of
players remaining is a power of 2. If the number of players is odd, then two complete
rounds leaves the situation unchanged (here we need k ≥ 2), so the game terminates
only if there is one player left to begin with. If the number of players is even, then
after k complete rounds, the player who made the first move and every second player
thereafter will have gained k pennies, and the other players will all have lost their k
pennies. Thus we are in a situation of the same form with half as many players, and
by induction the latter terminates with one player if and only if the number of players
is a power of 2.
Returning to the original game, we see that if the player to move is to pass one
penny, we are in the desired situation. If the player to move is to pass two pennies,
after that move we end up in the desired situation. Thus the game terminates if and
only if  n−1
2  is a power of 2, that is, if and only if n = 2 + 1 or n = 2 + 2 for some
m m

m. 

A3. (5, 28, 4, 0, 0, 0, 0, 0, 3, 36, 29, 100)

Evaluate
 ∞


x3 x5 x7 x2 x4 x6
x− + − + ··· 1 + 2 + 2 2 + 2 2 2 + · · · dx.
0 2 2·4 2·4·6 2 2 ·4 2 ·4 ·6

√
Answer. The value of the integral is e.
2
Solution 1. The series on the left is the Taylor series of xe−x /2 . Since the terms
of the second sum are nonnegative, we may interchange the sum and integral by the
Monotone Convergence Theorem [Ru, p. 319], so the expression becomes
∞  ∞
2 x2n
xe−x /2 2n dx.
n=0 0
2 (n!)2

To evaluate these integrals, first note that

 ∞ ∞
2 2
xe−x /2 dx = −e−x /2 = 1.
0 0

By integration by parts,
 ∞ " # ∞
 ∞
2 2 2
x2n xe−x /2 dx = −x2n e−x /2 + 2nx2n−1 e−x /2
dx
0 0 0
Solutions: The Fifty-Eighth Competition (1997) 235

since both integrals converge absolutely, and (for n ≥ 1) the first expression evaluates
to 0 at both endpoints. Thus by induction,
 ∞
2
x2n+1 e−x /2 dx = 2 × 4 × · · · × 2n.
0

Consequently, the desired integral is

∞
1 √
= e. 
n=0
2n n!

Remark. One can also first make the substitution u = x2 /2, in which case the
integral becomes
∞  ∞
1
n (n!)2
un e−u du.
n=0
2 0
 ∞ n −u
The reader may recognize 0 u e du as the integral defining the gamma function
Γ(n + 1) = n!.

Solution 2. Let J0 (x) denote the function

∞
x2n
J0 (x) = (−1)n ;
n=0
22n (n!)2

the series converges absolutely for all x ∈ C, by the Ratio Test. Applying the operator
d
x dx twice to J0 (x) yields −x2 J0 (x), so J0 (x) satisfies the differential equation

x2 y  + xy  + x2 y = 0. (1)

To justify certain calculations in the rest of the proof, we need some bounds on J0
2
and its derivatives. For any t ∈ C and β > 0, we have J0 (tx)/eβx → 0 as x → +∞:
this holds because for any C > 0, all but finitely many terms in the expansion of J0 (tx)
2
are bounded in absolute by C times the corresponding term in eβx ; those finitely many
2
terms also are negligible compared to eβx as x → ∞. A similar argument shows that
2 2
J0 (tx)/eβx → 0 and J0 (tx)/eβx → 0 as x → +∞, uniformly for t in any bounded
subset of C.
For t ∈ C, define
 ∞
2
F (t) = xe−x /2 J0 (tx) dx. (2)
0

The previous paragraph shows that the integral converges and justifies the claim that
integration by parts yields
 ∞
2
e−x /2 J0 (tx) dx = t−1 (F (t) − 1) (3)
0

for t = 0. The rest of this paragraph justifies the claim that differentiating (3) with
respect to t yields
 ∞
2
xe−x /2 J0 (tx) dx = −t−2 (F (t) − 1) + t−1 F  (t). (4)
0
236 The William Lowell Putnam Mathematical Competition

2
The nontrivial statement here is that if g(x, t) is the integrand e−x /2 
J0 (tx) on the
left of (3), then   ∞
d ∞ ∂g
g(x, t) dx = dx.
dt 0 0 ∂t
By definition of the derivative, the left side at t = t0 equals
∞ ∞  ∞
0
g(x, t) dx − 0 g(x, t0 ) dx g(x, t) − g(x, t0 )
lim = lim dx, (5)
t→t0 t − t0 t→t0 0 t − t0
so to obtain the right side, what we need is to interchange the limit and the integral on
the right of (5). The interchange is justified by theDominated Convergence Theorem
∞
[Ru, p. 321], provided that there exists G(x) with 0 G(x) dx < ∞ such that
g(x, t) − g(x, t0 )
≤ G(x)
t − t0
for all t in a punctured neighborhood of t0 and for all x ≥ 0. For fixed x, the difference
0)
quotient g(x,t)−g(x,t
t−t0 is the average value of ∂g/∂t over the interval [t0 , t], so we need
only prove ∂g/∂t ≤ G(x). The limits in the previous paragraph show
∂g 2 2
= x2 e−x /2 J0 (tx) ≤ C(t)e−x /4 ,
∂t
where C(t) is uniformly bounded in a neighborhood of t0 . Hence we may take
2
G(x) = Ce−x /4 for some C > 0 depending on t0 to complete the justification.
By (1), (2), (3), and (4),
 ∞  ∞
−2 −1  −1 −x2 /2  2
−t (F (t) − 1) + t F (t) = −t e J0 (tx) dx − xe−x /2 J0 (tx) dx
0 0
= −t−2 (F (t) − 1) − F (t).

Therefore F  (t) = −tF (t). Separating variables and integrating, we get F (t) =
2
Ce−t /2 . Here
 ∞  ∞
−x2 /2 2
C = F (0) = xe J0 (0) dx = xe−x /2 dx = 1,
0 0
2
so F (t) = e−t /2
. The integral of the problem is
 ∞

−x2 /2 x2 x4 √
xe 1 + 2 + 2 2 + · · · dx = F (i) = e. 
0 2 2 · 4

Remark. The above solution can ∞be reinterpreted in terms of the Laplace transform,
which takes f (x) to (Lf )(s) = 0 e−sx f (x) dx. See Chapter 6 of [BD] for further
applications of this technique.
Remark. The function J0 (x) is an example of a Bessel function. See [O, Section 2.9]
for the definitions and properties of these commonly occurring functions.
Remark. The trick of differentiating an integral with respect to an auxiliary
parameter was a favorite of 1939 Putnam Fellow (and Nobel Laureate in physics)
Richard Feynman, according to [Fe, p. 72]. A systematic treatment of the method can
be found in [AZ].
Solutions: The Fifty-Eighth Competition (1997) 237

A4. (100, 21, 1, 0, 0, 0, 0, 0, 3, 8, 24, 48)

Let G be a group with identity e and φ : G → G a function such that

φ(g1 )φ(g2 )φ(g3 ) = φ(h1 )φ(h2 )φ(h3 )

whenever g1 g2 g3 = e = h1 h2 h3 . Prove that there exists an element a ∈ G

such that ψ(x) = aφ(x) is a homomorphism (that is, ψ(xy) = ψ(x)ψ(y) for all
x, y ∈ G).

Solution. Homomorphisms map e to e, so we take a = φ(e)−1 and define

ψ(x) = aφ(x) in order to have ψ(e) = e. The hypothesis on φ implies

φ(g)φ(e)φ(g −1 ) = φ(e)φ(g)φ(g −1 ),

and cancelling φ(g −1 ) shows that φ(g) commutes with φ(e) for all g. The hypothesis
also implies

φ(x)φ(y)φ(y −1 x−1 ) = φ(e)φ(xy)φ(y −1 x−1 ).

Since φ(e) commutes with anything in the image of φ, φ(e)−1 does too, so we deduce

φ(e)−1 φ(x)φ(e)−1 φ(y) = φ(e)−1 φ(xy)

or equivalently ψ(xy) = ψ(x)ψ(y), as desired. 

Remark. It is not necessarily true that φ(e) commutes with all elements of the
group. For example, φ could be a constant map whose image is an element not in the
center.

A5. (22, 10, 15, 0, 0, 0, 0, 0, 24, 11, 37, 86)

Let Nn denote the number of ordered n-tuples of positive integers
(a1 , a2 , . . . , an ) such that 1/a1 + 1/a2 + · · · + 1/an = 1. Determine whether
N10 is even or odd.
Answer. The number N10 is odd.

Solution 1. Since we are only looking for the parity of the number of solutions, we
may discard solutions in pairs. For example, any solution with a1 = a2 may be paired
with the solution obtained from this one by interchanging a1 and a2 . The solutions
left unpaired are those with a1 = a2 , so the parity of N10 equals the parity of the
number of solutions with a1 = a2 .
Similarly, we may restrict attention to solutions with a3 = a4 , a5 = a6 , a7 = a8 ,
a9 = a10 . Such solutions must satisfy the equation

2/a1 + 2/a3 + 2/a5 + 2/a7 + 2/a9 = 1.

As above, we may restrict attention to solutions with a1 = a3 and a5 = a7 , so we

get 4/a1 + 4/a5 + 2/a9 = 1. We may restrict once more to solutions with a1 = a5 ,
which satisfy 8/a1 + 2/a9 = 1. This is equivalent to (a1 − 8)(a9 − 2) = 16, which has 5
solutions, corresponding to the factorizations of 16 as 2i × 24−i for i = 0, . . . , 4. Thus
N10 is odd. 
238 The William Lowell Putnam Mathematical Competition

Solution 2. Suppose we start with a solution in which the smallest ai occurs p1

times, the next smallest occurs p2 times, and so on. Then the number of ordered 10-
tuples which are rearrangements of this solution is given by the multinomial coefficient


10 10!
= .
p1 , p2 , . . . p1 ! p2 ! · · ·
The exponent of 2 in the prime factorization of this number is given by Kummer’s
Theorem.
Kummer’s Theorem. a1The exponent
 of the prime p in the factorization of the
+···+an
multinomial coefficient a1 ,...,an equals the number of carries that occur when
a1 , . . . , an are added in base p.
Proof. The exponent of p in the prime factorization of n! can be written as
∞ . /
n
.
i=1
pi

(This sum counts each multiple of p up to n once, each multiple of p2 a second time,
and so on.) Hence the exponent of p in aa1 1+···+a
,...,an
n
is
∞
. / . / . /
a 1 + · · · + an a1 an
− i − ··· − ,
i=1
pi p pi

and the summand is precisely the number of carries into the pi column when a1 , . . . , an
are added in base p. 
 
In particular, p1 ,p102 ,... is even unless p1 = 10 or {p1 , p2 } = {2, 8}. Up to
rearrangement, there is one solution of the former shape, namely (10, . . . , 10), and
four of the latter shape. Hence N10 is odd. 
Reinterpretation. The symmetric group S10 on ten symbols acts on the set of
ordered tuples. The number of elements in the orbit of a given tuple is 10!/|G|, where
|G| is the order of the stabilizer G of the tuple. This orbit size is odd if and only
if G contains a 2-Sylow subgroup. All 2-Sylow subgroups are conjugate, and one
2-Sylow subgroup contains the product of a disjoint 8-cycle and 2-cycle, so all 2-Sylow
subgroups must contain such a product. But G is a product of symmetric groups, so G
contains a 2-Sylow subgroup if and only if G contains a subgroup isomorphic to S8 ×S2 .
As seen above there are five solutions stabilized by S8 × S2 , up to rearrangement.

A6. (0, 1, 1, 0, 0, 0, 0, 0, 8, 8, 35, 152)

For a positive integer n and any real number c, define xk recursively by
x0 = 0, x1 = 1, and for k ≥ 0,
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.
 
Answer. For 1 ≤ k ≤ n, we have xk = n−1
k−1 .
Solutions: The Fifty-Eighth Competition (1997) 239

Solution 1. Introduce the generating function

p(t) = xk+1 tk .
k≥0

The condition xn+1 = 0 forces xm = 0 for all m ≥ n + 1, so p(t) is a polynomial of

degree at most n − 1.

Now p (t) = k
k≥0 (k + 1)xk+2 t , so the recursion implies that p(t) satisfies the
differential equation

p (t) = cp(t) − (n − 1)tp(t) + t2 p (t). (1)

Since p(0) = x1 = 1, p(t) is not identically zero. Rearranging (1), we get

p (t) c − (n − 1)t A B
(ln p(t)) = = = + (2)
p(t) 1 − t2 1+t 1−t

for some A and B depending on c and n. This implies that the only linear factors
of the polynomial p(t) over C are 1 − t and 1 + t. (See the discussion of logarithmic
differentiation in 1991A3.) Moreover, p(0) = 1, so

p(t) = (1 + t)r (1 − t)s

for some integers r, s ≥ 0. Then (2) implies A = r, B = −s, and c − (n − 1)t =

r(1 − t) − s(1 + t) as polynomials in t, so r = n − 1 − s and c = r − s = n − 1 − 2s.
Since s ≥ 0, c ≤ n − 1.
On the other hand, if we take c = n − 1, then reversing our steps shows that

the sequence xk defined by k≥0 xk+1 tk = p(t) = (1 + t)n−1 satisfies the original
recursion.
 Therefore the theoretical maximum c = n − 1 is attained, and in that case
xk = n−1k−1 . 

Solution 2. For fixed n, induction on k shows that xk is a polynomial in c of

degree exactly k − 1, so there are at most n values of c making xn+1 = 0. We will
prove by induction on n that the possible values of c are
−(n − 1), −(n − 3), . . . , n − 5, n − 3, n − 1
 
and that if c = n − 1 the sequence is xk = n−1 k−1 for k ≥ 1.
If n = 1, then the recursion gives x"
2 = c.# Thus the only possible value of c is 0. In
this case, the recursion says xk+2 = k−1 k+1 xk for k ≥ 0. Starting from x0 = 0 and
 0 
x1 = 1, this implies xk = 0 for all k ≥ 2. In other words, xk = k−1 for all k ≥ 1, as
claimed.
Now suppose that n ≥ 1, and that the claim has been verified for n. Suppose that
c is one of
−(n − 1), −(n − 3), . . . , n − 5, n − 3, n − 1
and that {yk } satisfies the hypotheses of the problem for n and c. Thus yn+1 = 0 and

(k + 1)yk+2 + (n − k)yk = cyk+1 for k ≥ 0. (3)

240 The William Lowell Putnam Mathematical Competition

The recursion holds for k = −1 as well, if we define y−1 = 0. Let xk = yk + yk−1 for
k ≥ 0. Then for k ≥ 0,

(k + 1)xk+2 + (n + 1 − k)xk = (k + 1)(yk+2 + yk+1 ) + (n + 1 − k)(yk + yk−1 )

= ((k + 1)yk+2 + (n − k)yk )
+ (kyk+1 + (n − k + 1)yk−1 ) + yk + yk+1
= cyk+1 + cyk + yk + yk+1 (by the y recursion (3))
= (c + 1)(yk+1 + yk )
= (c + 1)xk+1 ,

so {xk } satisfies the recursion for n + 1 and c + 1. Taking k = n in the y recursion (3)
shows that yn+1 = 0 implies yn+2 = 0, so xn+2 = yn+2 + yn+1 = 0. Thus {xk } is a
solution sequence for n + 1 and c + 1. Hence adding 1 to each value of c for n gives a
possible value of c for n + 1, so the inductive hypothesis implies that
−(n − 2), −(n − 4), . . . , n − 4, n − 2, n
are possible values of c for n + 1. If {xk } is the sequence for c = n, then {(−1)k−1 xk }
is a solution sequence for c = −n, so
−n, −(n − 2), −(n − 4), . . . , n − 4, n − 2, n
are n + 1 possible values of c for n + 1. By the first sentence of this solution, there
cannot be any others.  
k−1 , k ≥ 0, is a solution for n
Finally, the inductive hypothesis implies that yk = n−1
n−1
and c = n − 1 (we interpret −1 as 0), so




n−1 n−1 n
xk = yk + yk−1 = + = for k ≥ 1,
k−1 k−2 k−1
gives the solution for n + 1 and c = n. This completes the inductive step. 
Remark. Both of the previous solutions can be used to show that the sequences
for which xn+1 = 0 are the sequences of coefficients of t(1 + t)j (1 − t)n−1−j for some
j ∈ {0, 1, . . . , n − 1}.

Solution 3 (Byron Walden). Rewrite the recursion as

(k + 1)xk+2 − (k − 1)xk = cxk+1 − (n − 1)xk .

If m ≥ 0, then summing from k = 0 to m yields

(m + 1)xm+2 + mxm+1 = cxm+1 + (c − (n − 1))(x1 + x2 + · · · + xm ),

since x0 = 0. Thus, for m ≥ 0,

c−m c − (n − 1)
xm+2 = xm+1 + (x1 + x2 + · · · + xm ). (4)
m+1 m+1
If c > n − 1, then applying (4) for m = 0, 1, . . . , n − 1 in turn shows that x2 , x3 ,
. . . , xn+1 are all positive, since at each step, all terms on the right-hand side of (4)
are positive. Hence the assumption xn+1 = 0 forces c ≤ n − 1.
Solutions: The Fifty-Eighth Competition (1997) 241

Suppose we take c = n − 1. Then (4) becomes xm+2 = n−m−1

m+1 xm+1 for m ≥ 0. In
 
particular, if xm+1 = n−1
m , then


n−m−1 (n − 1)! (n − 1)! n−1
xm+2 = = = .
m + 1 m! (n − m − 1)! (m + 1)! (n − m − 2)! m+1
   
Hence by induction, starting from x1 = n−1 , we can prove that xk = n−1 for all
n−1 0 k−1
k ≥ 1. In particular, xn+1 = n = 0. Thus n − 1 is a possible value for c, and by
the previous paragraph it is the largest possible value. 
Solution 4 (Greg Kuperberg). The condition xn+1 = 0 states that (x1 , . . . , xn )
is an eigenvector of the n × n matrix



i if j = i + 1
Aij = n − j if j = i − 1


0 otherwise,
with eigenvalue c. In particular, A has nonnegative entries. By the Perron-Frobenius
Theorem (see remark below), A has an eigenvector with positive entries, unique up
to a positive scalar multiple, and the corresponding eigenvalue has multiplicity one
and
n−1has absolute value
 greater
n−1than
 or equal
n−1to that of any other eigenvalue. Using
n−k−1 n−1 k
k+1 = k+1 k and k−1 = n−k k (which follow from the definition of
 
binomial coefficients in terms of factorials), one shows that the vector with xk = n−1 k−1
is an eigenvector with eigenvalue n − 1, and it has positive entries, so c = n − 1 is the
desired maximum. 
Remark (the Perron-Frobenius Theorem). In its simplest form, this theorem states
that any matrix with positive entries has a unique eigenvector with positive entries,
and that the corresponding eigenvalue has multiplicity one and has absolute value
strictly greater than that of any other eigenvalue. In the study of random walks, the
theorem implies that any random walk on finitely many states, in which there is a
positive transition probability from any state to any other state, has a unique steady
state. One proof of the existence of the eigenvector, due to G. Birkhoff [Bi], is that a
matrix with positive entries acts on the set of lines passing through the first orthant
as a contraction mapping under the Hilbert metric:
" #
d((v1 , . . . , vn ), (w1 , . . . , wn )) = ln max{vi /wi }/ min{vi /wi } .
i i

In many applications to random walks, a more refined version of the Perron-

Frobenius Theorem is needed. Namely, let A be a matrix with nonnegative entries,
and assume that for some k > 0, Ak has strictly positive entries. The refined theorem
states that A then has a unique eigenvector with positive entries, whose eigenvalue
has multiplicity one and has absolute value strictly greater than that of any other
eigenvalue. This assertion can be deduced from the simpler form of the theorem as
follows. If v is a positive eigenvector of A, then it is also a positive eigenvector of
Ak , to which we may apply the simpler form; thus the positive eigenvector of A, if
it exists, is unique. To show existence, let w be a positive eigenvector of Ak with
eigenvalue λ > 0; then w + λ−1 Aw + · · · + λ−k+1 Ak−1 w is a positive eigenvector of
A with eigenvalue λ. Finally, if λ is the eigenvalue of the positive eigenvector of A,
242 The William Lowell Putnam Mathematical Competition

and ρ is any other eigenvalue, then |λk | > |ρk | by the simpler form of the theorem, so
|λ| > |ρ|.
There is also a formulation of the theorem that applies directly to matrices like
the one occurring in Solution 4, in which An may not have positive entries for any
n. Let A be a matrix with nonnegative entries. Let G be the directed graph with
vertices {1, . . . , n} and with an edge from i to j if and only if Aij > 0. Call A
irreducible if G is strongly connected (see 1990B4 for the definition). In this case, A
has a unique eigenvector with positive entries, whose eigenvalue has multiplicity one
and has absolute value greater than or equal to that of any other eigenvalue. This
can be deduced by applying the version in the previous paragraph to A + CI for each
C > 0.
In fact, the matrix in Solution 4 shows that “greater than or equal to” cannot be
replaced by “strictly greater than” in this final version: the matrix A there has 1 − n
as an eigenvalue in addition to the eigenvalue n − 1 of the positive eigenvector.
Related question. The following is an example of a problem that can be solved
by an appropriate application of the Perron-Frobenius Theorem. (We are grateful to
Mira Bernstein for passing it on to us.) It can also be solved by other means.
The Seven Dwarfs are sitting around the breakfast table; Snow White has just
poured them some milk. Before they drink, they perform a little ritual. First,
Dwarf #1 distributes all the milk in his mug equally among his brothers’ mugs
(leaving none for himself). Then Dwarf #2 does the same, then Dwarf #3,
#4, etc., finishing with Dwarf # 7. At the end of the process, the amount of
milk in each dwarf’s mug is the same as at the beginning! If the total amount
of milk is 42 ounces, how much milk did each of them originally have?
See 1995A5 for another application of the Perron-Frobenius Theorem, and see
1993B4 for an application of a continuous analogue of the theorem. For more
on matrices with nonnegative entries, including the Perron-Frobenius Theorem and
related results, see [HJ, Chapter 8].

B1. (171, 6, 8, 0, 0, 0, 0, 0, 1, 8, 10, 1)

Let {x} denote the distance between the real number x and the nearest
integer. For each positive integer n, evaluate
6n−1 "& m ' & m '#
Sn = min , .
m=1
6n 3n

(Here min(a, b) denotes the minimum of a and b.)

Answer. We have Sn = n.

Solution 1. It is trivial to check that 6n m

= { 6n
m
} ≤ { 3n
m
} for 1 ≤ m ≤ 2n, that
1 − 3n = { 3n } ≤ { 6n } for 2n ≤ m ≤ 3n, that 3n − 1 = { 3n } ≤ { 6n
m m m m m m
} for 3n ≤ m ≤ 4n,
and that 2 − 6n = { 6n } ≤ { 3n } for 4n ≤ m ≤ 6n. (See Figure 33 for a graph.)
m m m

Therefore
3n−1 "
m#
4n−1 " # 6n−1 " m#
2n−1
m m
Sn = + 1− + −1 + 2− .
m=1
6n m=2n 3n m=3n
3n m=4n
6n
Solutions: The Fifty-Eighth Competition (1997) 243

m
2n 3n 4n 6n

FIGURE 33.
 m   m 
Graph of y = min 6n
, 3n
.

Each of the four sums is an arithmetic progression, which can be evaluated as the
number of terms times the average of the first and last terms. This yields





2n n+1 n−1 2n + 1
Sn = (2n − 1) +n +n + 2n = n. 
12n 6n 6n 12n

Remark. Solution 1 can be simplified by exploiting symmetry: since {1 − x} = {x}

and {2 − x} = {x}, the mth term in the series Sn equals the (6n − m)th term.
Solution 2. The series Sn is the approximation given by the Trapezoid Rule to
the area A under the graph of the function
"& x ' & x '#
f (x) = min ,
6n 3n
for 0 ≤ x ≤ 6n, using the sampling points 0, 1, . . . , 6n. But f (x) is piecewise linear and
the break points x = 2n, 3n, 4n are also sample points, so Sn = A. The area A consists
of the triangle with vertices (0, 0), (3n, 0), (2n, 1/3), and the congruent triangle with
vertices (6n, 0), (3n, 0), (4n, 1/3). Thus A = n/2 + n/2 = n, and Sn = n. 

B2. (28, 1, 1, 0, 0, 0, 0, 0, 1, 3, 38, 133)

Let f be a twice-differentiable real-valued function satisfying
f (x) + f  (x) = −xg(x)f  (x),
where g(x) ≥ 0 for all real x. Prove that |f (x)| is bounded.
Solution. Multiplying both sides of the given differential equation by 2f  (x), we
have
2f (x)f  (x) + 2f  (x)f  (x) = −2xg(x)f  (x)2 .
The left side of the equation is the derivative of f (x)2 + f  (x)2 , whereas the right side
is nonnegative for x < 0 and nonpositive for x > 0. Thus f (x)2 + f  (x)2 increases to
its maximum value at x = 0 and decreases thereafter. In particular, it is bounded, so
f (x) and f  (x) are bounded. 
Reinterpretation. This problem has a physical interpretation: f (x) is the amplitude
of an oscillator with time-dependent damping xg(x). Since the damping is negative
for x < 0 and positive for x > 0, the oscillator gains energy before time 0 and loses
energy thereafter.
244 The William Lowell Putnam Mathematical Competition

B3. (1, 4, 7, 0, 0, 0, 0, 0, 22, 20, 89, 62) n

For each positive integer n, write the sum m=1 m 1
in the form pqnn , where
pn and qn are relatively prime positive integers. Determine all n such that
5 does not divide qn .
Answer. The only such n are the 19 integers in the ranges 1–4, 20–24, 100–104, and
120–124; i.e., the set of such n is

{1, 2, 3, 4, 20, 21, 22, 23, 24, 100, 101, 102, 103, 104, 120, 121, 122, 123, 124}.

Solution. To simplify the discussion, we introduce some terminology. For s a

positive integer, we say the rational number a/b, in lowest terms, is s-integral if b is
coprime to s, and divisible by s if a is divisible by s.
n
Let Hn = m=1 m−1 . (By convention, the empty sum H0 is 0.) Let Sn be the sum
of m−1 over those m ∈ {1, . . . , n} not divisible by 5. Then
1
Hn = Sn + H n/5 . (1)
5
By design, Sn is always 5-integral, so Hn is 5-integral if and only if H n/5 is divisible
by 5.
In particular, if Hn is 5-integral, then H n/5 must have been 5-integral. Thus the
nonnegative integers n such that Hn is 5-integral are precisely the integers in the set
S after carrying out the following algorithm.
(a) Start with the set S = {0} and the integer n = 0.
(b) If Hn is divisible by 5, add 5n, 5n + 1, 5n + 2, 5n + 3, 5n + 4 to S.
(c) If n is the largest element of S, stop; otherwise, increase n to the smallest element
of S greater than the current n, and go to (b).
To carry out the algorithm, we first take n = 0 and add 1, 2, 3, 4 to S. Now
H1 = 1, H2 = 3/2, H3 = 11/6 are not divisible by 5, but H4 = 25/12 is. Thus we add
20, 21, 22, 23, 24 to S and keep going.
We next must determine which of H20 , . . . , H24 are divisible by 5. From (1), we
have H20+i = S20+i + 15 H4 for i = 0, . . . , 4. Before proceeding further, we verify that
1 1 1
+ + ··· + (2)
5k + 1 5k + 2 5k + j
is not divisible by 5 if j = 1, 2, 3 but is divisible by 25 if j = 4. For j = 1, 2, 3, the
sum (2) differs from Hj by a rational number divisible by 5 (since this already holds
termwise), and Hj is not divisible by 5. For j = 4, we combine terms to rewrite (2) as
(10k + 5)(50k 2 + 50k + 10)
;
(5k + 1)(5k + 2)(5k + 3)(5k + 4)
in this expression, the numerator is divisible by 25 while the denominator is coprime
to 5.
Summing (2) for k = 0, 1, 2, 3 and j = 4, we find that S20 is divisible by 25. Adding
to this (2) for k = 4 and appropriate j, we find that S21 , S22 , and S23 are not divisible
by 5, but S24 is. Thus we add 100, 101, 102, 103, 104, 120, 121, 122, 123, 124 to S.
Solutions: The Fifty-Eighth Competition (1997) 245

To complete the proof, we must check that Hn is not divisible by 5 for any of the ten
integers n just added to S. Suppose n = 100 + j with j = 0, . . . , 4. Applying (1) twice
gives Hn = S100+j + 15 S20 + 25
1
H4 . We know 15 S20 is divisible by 5, and H4 = 25/12,
so it suffices to show that S100+j + 1/12 is not divisible by 5. For j = 0, . . . , 4,

 j

1 1 1 1
S100+j + = S100 + Hj + + − .
12 12 i=1
100 + i i

Here S100 is a sum of expressions of the form (2), hence divisible by 5, and 100+i 1
− 1i =
−100
i(100+i) is divisible by 5, but one can check (for j = 0, . . . , 4 individually) that Hj +1/12
is not divisible by 5. Thus S100+j is not divisible by 5, and neither is H100+j . By a
similar argument, if n = 120 + j with j = 0, . . . , 4, then Hn is not divisible by 5.
Thus the algorithm terminates, and the list of integers n such that Hn is 5-integral
is as given in the answer above. 
Reinterpretation. Arguments of this sort are often more easily expressed in terms of
the p-adic valuation on the rational numbers, for p a prime. Given r ∈ Q nonzero, let
vp (r) be the largest integer m (positive, negative or zero) such that p−m r is p-integral.
Then the problem is to determine when v5 (Hn ) ≥ 0, and the first step is to note that
this is equivalent to v5 (H n/5 ) ≥ 1. The rest of the solution amounts to computing
some low order terms in the base 5 expansions of 1/m for some small m.
Remark. The fact that (2) is divisible by 25 for j = 4 is a special case of the fact
that for a prime p > 3 and an integer x,
1 1 1
+ + ··· + ≡0 (mod p2 ).
px + 1 px + 2 px + (p − 1)
The case x = 0 is known as Wolstenholme’s Theorem, which arises in a different
context in 1991B4. To prove the general case, first combine opposite terms in the sum
to rewrite it as
(p−1)/2
1
(2px + p) .
i=1
(px + i)(px + (p − i))

Hence it suffices to show that

(p−1)/2
1
≡0 (mod p).
i=1
(px + i)(px + (p − i))
(p−1)/2 −2
This sum is congruent to −S where S = i=1 i . If a2 is a perfect square not
congruent to 0 or 1 modulo p (e.g. a = 4), then the terms of a2 S are congruent
2

modulo p to a permutation of the terms of S, and we may solve a2 S ≡ S (mod p) to

obtain S ≡ 0 (mod p), as desired.
By a similar argument, one can prove that for a prime p > 3, positive integer n,
and integer x,
p−1 pn−1 −1
1
≡0 (mod p2n ).
i=1 j=0
pn x + pj + i

See the remarks following 1991B4 for an application of this generalization.

246 The William Lowell Putnam Mathematical Competition

B4. (23, 6, 7, 0, 0, 0, 0, 0, 5, 4, 35, 125)

Let am,n denote the coefficient of xn in the expansion of (1 + x + x2 )m .
Prove that for all integers k ≥ 0,
2k
3

0≤ (−1)i ak−i,i ≤ 1.
i=0

 i
Solution 1. Let sk = i (−1) ak−i,i be the given sum. (Note that ak−i,i is
nonzero precisely for i = 0, . . . ,  3 .) Since
2k

am+1,n = am,n + am,n−1 + am,n−2 ,

we have

sk − sk+1 + sk+2 = (−1)i (ak−i,i + ak−i,i+1 + ak−i,i+2 )

i

= (−1)i ak−i+1,i+2 = sk+3 .

i

After computing s0 = 1, s1 = 1, s2 = 0, we may prove by induction that s4j = s4j+1 =

1 and s4j+2 = s4j+3 = 0 for all j ≥ 0. 
Reinterpretation. The characteristic polynomial of the linear recursion sk+3 −
sk+2 + sk+1 − sk is x3 − x2 + x − 1, which divides x4 − 1. Thus the sequence also
satisfies sk+4 −sk = 0, which explains the periodicity. See the remark following 1988A5
for more on linear recursive sequences.

Solution 2. We use a generating function. Define sk as in Solution 1. Then

∞
sk xk = (−1)i ak−i,i xk
k=0 i,k

= xj aj,i (−x)i (change of variable: k = i + j)

i,j

= xj (1 − x + x2 )j (since aj,i z i = (1 + z + z 2 )j )
j i
1
= (geometric series)
1 − x + x2 − x3
1+x
=
1 − x4
= (1 + x) + (x4 + x5 ) + (x8 + x9 ) + · · · ,

so 0 ≤ sk ≤ 1 for all k ≥ 0. 
Remark. If one does not notice that
1 1+x
= ,
1 − x + x2 − x3 1 − x4
one can instead use partial fractions to recover its coefficients.
Solutions: The Fifty-Eighth Competition (1997) 247

B5. (9, 4, 3, 0, 0, 0, 0, 0, 17, 24, 44, 104)

Prove that for n ≥ 2,
2
' 2
'
·· ··
2· n 2· n−1
2 ≡2 (mod n).

Solution. Define a sequence by x0 = 1 and xm = 2xm−1 for m > 0. We are asked

to prove xn ≡ xn−1 (mod n) for all n ≥ 2. We will use strong induction on n to prove
the stronger result xn−1 ≡ xn ≡ xn+1 ≡ · · · (mod n) for n ≥ 1. The n = 1 case is
obvious.
Now suppose n ≥ 2. Write n = 2a b, where b is odd. It suffices to show that xn−1 ≡
xn ≡ · · · modulo 2a and modulo b. For the former, we only need xn−1 ≥ a, which
holds, since xn−1 ≥ n by induction on n. For the latter, note that xm ≡ xm+1 ≡ · · ·
(mod b) as long as xm−1 ≡ xm ≡ · · · (mod φ(b)), where φ(n) is the Euler φ-function.
By the inductive hypothesis, the latter holds for m = φ(b); but m = φ(b) ≤ n − 1, so
xn−1 ≡ xn ≡ · · · (mod b), as desired. 
Remark. This solution also yields a solution to Problem 3 on the 1991 USA
Mathematical Olympiad [USAMO]:
Show that, for any fixed integer n ≥ 1, the sequence
2 22
2, 22 , 22 , 22 , . . . (mod n)

is eventually constant.
See 1985A4 for a similar problem and for the properties of φ(n).
Remark. For n ≥ 1, let f (n) be the smallest integer k ≥ 0 such that xk ≡ xk+1 ≡
xk+2 ≡ · · · (mod n). The above solution shows that f (n) ≤ n − 1, but the actual
value of f (n) is much smaller.
We now prove f (n) ≤ log2 n by strong induction on n. The base case n = 1
is trivial. If n = 2a for some a ≥ 1, then 2a divides xa , xa+1 , . . . (since xa−1 ≥ a
was proved in the solution above), so f (n) ≤ a. If n = 2a b for some a ≥ 1 and odd
b > 1, then f (n) ≤ max{f (2a ), f (b)}, so we apply the inductive hypothesis at 2a and
b. Finally, if n > 1 is odd, write φ(n) = 2c d with d odd, and observe that

f (n) ≤ 1 + f (φ(n)) (as in the solution above)

≤ 1 + max{f (2 ), f (d)}c

≤ 1 + max{c, log2 d} (by the inductive hypothesis)

≤ log2 n,

since c ≤ log2 φ(n) < log2 n and d ≤ φ(n)/2 < n/2. This completes the inductive
step.
When n is a power of 3, the bound just proved is best possible up to a constant
factor: we now show that f (3m ) = m + 1 for m ≥ 1 by induction on m. First,
xk+1 ≡ (−1)xk = 1 (mod 3), if and only if k ≥ 1, so f (3) = 2. Now suppose
m > 1. Since the image of 2 generates the cyclic group (Z/3m Z)∗ (i.e., 2 is a primitive
root of 3m [NZM, Theorem 2.40]), we have xk ≡ xk+1 ≡ · · · (mod 3m ) if and only
248 The William Lowell Putnam Mathematical Competition

if xk−1 ≡ xk ≡ · · · (mod 2 · 3m−1 ), but all the xk except x0 are even, so this is
equivalent to xk−1 ≡ xk ≡ · · · (mod 3m−1 ) if k ≥ 2. Hence f (3m ) = f (3m−1 ) + 1, so
f (3m ) = m + 1 by induction.
It is not true in general that f (n) is bounded below by a constant multiple of ln n,
because there are some values of n, such as large powers of 2, for which f (n) is much,
much less than ln n. We can at least say that f (n) → ∞ as n → ∞, though: we have
xf (n)+1 ≥ xf (n)+1 − xf (n) ≥ n, since n divides xf (n)+1 − xf (n) .

B6. (5, 4, 5, 0, 0, 0, 0, 0, 3, 0, 51, 137)

The dissection of the 3–4–5 triangle shown below has diameter 5/2.

❙
❙
❙
❙ 5
4 ❙

 ❙
 ❙
 ❙
 ❙
3

Find the least diameter of a dissection of this triangle into four parts. (The
diameter of a dissection is the least upper bound of the distances between
pairs of points belonging to the same part.)
Answer. The minimum diameter is 25/13.
Solution. (See Figure 34.) Place the triangle on the cartesian plane so that its
vertices are A = (4, 0), B = (0, 0), C = (0, 3). The five points A, B, C, D = (27/13, 0),
E = (20/13, 24/13) lie at distance at least 25/13 apart from each other (note that
AD = DE = EC = 25/13). By the Pigeonhole Principle, any dissection of the
triangle into four parts has a part containing at least two of these points, and hence
has diameter at least 25/13.

C




 E

✑
✑ 
✑  H
G ✭✭✭✑I
L ✂
L ✂ 
L ✂ 
L ✂ 
L ✂ 
B F D A

FIGURE 34.
Solutions: The Fifty-Eighth Competition (1997) 249

On the other hand, we now construct a dissection into four parts in which each part
has diameter 25/13. Let H = (32/13, 15/13), so that AH = 25/13. Next, construct a
point I inside quadrilateral BDEC so that BI, CI, DI, EI, HI have length less than
25/13. For example, we may take I = (7/13, 15/13) such that DECI and ADIH are
parallelograms. Take F = (1, 0) on AB so that EF = 25/13, and take G = (0, 14/13)
on BC so that CG = 25/13.
Our parts are the triangle ADH, the pentagon DHEIF , the quadrilateral CEIG,
and the quadrilateral BF IG. Verifying that each has diameter 25/13 entails checking
that the distance between any two vertices of a polygon is at most 25/13. For starters,
AD = AH = CE = CG = CI = DE = DI = EF = HI = 25/13.
Next, we see without any calculations that
BF, BG < F G F I, F D < DI DH < F H GI < CI.
Finally, we compute the remaining distances:
274 500 225
BI 2 = , EG2 = , EH 2 = ,
169 169 169
260 365 586
EI 2 = , F G2 = , F H2 = . 
169 169 169
Remark. Once the “skeleton” is in place, some variations in the placements of the
points are possible.
250 The William Lowell Putnam Mathematical Competition

The Fifty-Ninth William Lowell Putnam Mathematical Competition

December 5, 1998

A1. (156, 23, 4, 0, 0, 0, 0, 0, 0, 0, 16, 0)

A right circular cone has base of radius 1 and height 3. 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 side-length of the cube?
√
Answer. The side length of the cube is (9 2 − 6)/7.
Solution. Consider a plane cross-section through a vertex of the cube and the
axis of the cone as shown in Figure 35.

A
❙
 ❙
 ❙
 ❙
 F ❙
D ❙E
 ❙
 ❙
 ❙
 ❙
 ❙
B G C

FIGURE 35.

Let s be the side length√of the cube. Segment DE is a diagonal of the top of
the cube, so its length is s √ 2. From the similar triangles ADE √ ∼ ABC, we have
DE
BC = AF
AG , or equivalently s 2
2 = 3−s
3 . Solving for s, we get s = (9 2 − 6)/7. 

A2. (103, 35, 26, 0, 0, 0, 0, 0, 9, 8, 12, 6)

Let s be any arc of the unit circle lying entirely in the first quadrant. Let
A be the area of the region lying below s and above the x-axis and let B be
the area of the region lying to the right of the y-axis and to the left of s.
Prove that A + B depends only on the arc length, and not on the position,
of s.
Solution 1. Let O be the center of the circle, K and H the endpoints of the
arc (with K above H), C and F the projections of K onto the x-axis and y-axis,
respectively, D and E the projections of K onto the x-axis and y-axis, respectively,
and G the intersection of KC and HE, as in Figure 36.
Denote by [XY Z] the area of the region with vertices X, Y , Z. In this notation,
A + B = [EF GK] + [CDHG] + 2[GHK]
= 2([OGK] + [OGH] + [GHK])
= 2[OHK] = θ,
where θ is the length of the arc s. 
Solutions: The Fifty-Ninth Competition (1998) 251

F K

G
E H

O C D

FIGURE 36.

Solution 2. We obtain A and B by integrating −y dx and x dy, respectively,

counterclockwise along the arc. Thus A + B is the integral of x dy − y dx over the arc.
If we parameterize the arc by arc length, that is, by setting x = cos θ and y = sin θ,
the integrand becomes cos2 θ +sin2 θ = 1. Thus the integral is precisely the arc length,
and so does not depend on the position of the arc. 
Remark. For a straightforward but less elegant solution, one can also compute
the integrals separately. Alternatively, for a fixed arc length, one can express the sum
of areas as a function F (θ) of the argument θ at one endpoint of s, and show that
F  (θ) = 0.

A3. (82, 34, 2, 0, 0, 0, 0, 0, 5, 0, 39, 37)

Let f be a real function on the real line with continuous third derivative.
Prove that there exists a point a such that
f (a) · f  (a) · f  (a) · f  (a) ≥ 0.

Solution. If at least one of f (a), f  (a), f  (a), or f  (a) vanishes at some point a,
then we are done. Otherwise by the Intermediate Value Theorem, each of f (x), f  (x),
f  (x), and f  (x) is either strictly positive or strictly negative on the real line, and we
only need to show that their product is positive for a single value of x. By replacing
f (x) by −f (x) if necessary, we may assume f  (x) > 0; by replacing f (x) by f (−x) if
necessary, we may assume f  (x) > 0. Notice that these substitutions do not change
the sign of f (x)f  (x)f  (x)f  (x).
Now f  (x) > 0 implies that f  (x) is increasing. Thus for a > 0,
 a
f  (a) = f  (0) + f  (t) dt ≥ f  (0) + af  (0).
0

In particular, f (a) > 0 for large a. Similarly, since f  (x) > 0, f (a) is positive for


large a. Therefore f (x)f  (x)f  (x)f  (x) > 0 for sufficiently large x. 
Reinterpretation. More succinctly, f cannot both be positive and strictly concave-
down everywhere, nor negative and strictly concave-up everywhere. So f (x)f  (x)
must be positive for some range of x, as must be f  (x)f  (x) by the same reasoning
applied to f  instead of f .
252 The William Lowell Putnam Mathematical Competition

A4. (39, 27, 52, 0, 0, 0, 0, 0, 49, 7, 14, 11)

Let A1 = 0 and A2 = 1. For n > 2, the number An is defined by
concatenating the decimal expansions of An−1 and An−2 from left to right.
For example A3 = A2 A1 = 10, A4 = A3 A2 = 101, A5 = A4 A3 = 10110, and so
forth. Determine all n such that 11 divides An .
Answer. The number 11 divides An if and only if n ≡ 1 (mod 6).

Solution. The number of digits in the decimal expansion of An is the nth

Fibonacci number Fn . It follows that the sequence {An } modulo 11 satisfies a
recursion:

An = 10Fn−2 An−1 + An−2

≡ (−1)Fn−2 An−1 + An−2 (mod 11).

By induction, Fn is even if and only if 3 divides n; hence (−1)Fn−2 is periodic with

period 3.
Computing An modulo 11 for small n using the recursion, we find

A1 , . . . , A8 ≡ 0, 1, −1, 2, 1, 1, 0, 1 (mod 11).

By induction, we deduce that An+6 ≡ An (mod 11) for all n, and so An is divisible
by 11 if and only if n ≡ 1 (mod 6). 
Remark. See 1988A5 for more on linear recursions and Fibonacci numbers.

A5. (85, 24, 15, 0, 0, 0, 0, 0, 5, 2, 8, 60)

Let F be a finite collection of open discs in R2 whose union contains a
set E ⊆ R2 . Show that there is a pairwise disjoint subcollection D1 , . . . , Dn
in F such that
(
n
E⊆ 3Dj .
j=1

Here, if D is the disc of radius r and center P , then 3D is the disc of radius
3r and center P .

Solution. Define the sequence Di by the following greedy algorithm: let D1 be

the disc of largest radius (breaking ties arbitrarily), let D2 be the disc of largest radius
not meeting D1 , let D3 be the disc of largest radius not meeting D1 or D2 , and so
on, up to some final disc Dn (since F is finite). To see that E ⊆ ∪nj=1 3Dj , consider
a point P in E; if P lies in one of the Di , we are done. Otherwise, P lies in a disc
D ∈ F, say of radius r, which meets one of the Di having radius s ≥ r. (This is the
only reason a disc can be skipped in our algorithm.) Choose Q in D ∩ Di . If O and
Oi are the centers of D and Di , respectively, then by the triangle inequality

P Oi ≤ P O + OQ + QOi < r + r + s ≤ 3s.

Thus P ∈ 3Di . 
Solutions: The Fifty-Ninth Competition (1998) 253

A6. (12, 1, 3, 0, 0, 0, 0, 0, 3, 9, 92, 79)

Let A, B, C denote distinct points with integer coordinates 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.
Solution 1. Given that A, B, C have integer coordinates, it follows that
|AB|2 , |BC|2 , |CA|2 are integers, and that 2[ABC] is an integer (see the first remark
in 1990A3). These quantities are related via the AM-GM Inequality (see the end of
1985A2) in the form |AB|2 + |BC|2 ≥ 2|AB||BC|, and the area formula
2[ABC] = |AB||BC| sin ∠ABC ≤ |AB||BC|.
The quantity |AB|2 + |BC|2 + 4[ABC] is thus an integer which, by the above
inequalities and the given condition, “sandwiches” 8[ABC] as follows:
|AB|2 + |BC|2 + 4[ABC] ≤ (|AB| + |BC|)2
< 8[ABC] + 1
≤ |AB|2 + |BC|2 + 4[ABC] + 1,
We may thus conclude that |AB|2 + |BC|2 + 4[ABC] = 8[ABC], since these quantities
are integers and their difference is at least 0 but less than 1. In particular, we have
equality both in AM-GM and the area formula, so AB = BC and ∠ABC is a right
angle, as desired. 
Solution 2. Set up a new coordinate system with the same unit length as the
original system, but with B = (0, 0) and A = (s, 0) for some s > 0. We may assume
that C is in the upper half plane in these coordinates, and put C = (x, y + s). Let C 
be the image of A under 90◦ counterclockwise rotation around B, so C  = (0, s). The
given inequality is equivalent to all of the following:
"  #2
s + x2 + (y + s)2 < 4s(y + s) + 1

s2 + x2 + (y + s)2 + 2s x2 + (y + s)2 < 4s(y + s) + 1
" #
2s x2 + (y + s)2 − (y + s) < 1 − x2 − y 2 .

The left-hand side in the last inequality is nonnegative, so x2 + y 2 < 1. But

x2 + y 2 = |CC  |2 , and C, C  have integer coordinates in the old coordinate system, so
C = C  . Thus A, B, C are three consecutive vertices of a square. 

B1. (112, 30, 30, 0, 0, 0, 0, 0, 3, 3, 12, 9)

Find the minimum value of
(x + 1/x)6 − (x6 + 1/x6 ) − 2
(x + 1/x)3 + (x3 + 1/x3 )
for x > 0.
Answer. The minimum value is 6.
254 The William Lowell Putnam Mathematical Competition

Solution. If we put a = (x + 1/x)3 and b = x3 + 1/x3 , the given expression can

be rewritten as
a2 − b2
= a − b = 3(x + 1/x).
a+b
The minimum of x + 1/x for x > 0 is 2 by the AM-GM inequality (see the end of
√
1985A2), by calculus, or by the identity x + 1/x = ( x − 1/x)2 + 2. Thus the
minimum value of the original expression is 6, attained when x = 1. 

B2. (82, 10, 9, 0, 0, 0, 0, 0, 3, 11, 38, 46)

Given a point (a, b) with 0 < b < a, determine the minimum perimeter of
a triangle with one vertex at (a, b), one on the x-axis, and one on the line
y = x. You may assume that a triangle of minimum perimeter exists.
√
Answer. The minimum perimeter of such a triangle is 2a2 + 2b2 .

Solution. Consider a triangle as described by the problem. Label its vertices

A, B, C so that A = (a, b), B lies on the x-axis, and C lies on the line y = x; see
Figure 37. Further let D = (a, −b) be 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 AC = CE,
so the perimeter of ABC is
 
DB + BC + CE ≥ DE = (a − b)2 + (a + b)2 = 2a2 + 2b2 .

This lower bound can be achieved: set B (resp., C) to be the intersection between the
segment√DE and the x-axis (resp., the line x = y). Thus the minimum perimeter is
in fact 2a2 + 2b2 . 
Remark. The reflection trick comes up in various geometric problems where the
sum of certain distances is to be minimized. For many applications of this principle,
see [CG, Sections 4.4–4.6] and [Pó, Section IX].

A
C

FIGURE 37.
The reflection trick: AB + BC + CA = DB + BC + CD ≥ DE.
Solutions: The Fifty-Ninth Competition (1998) 255

Remark. Our proof did not use the assumption that a triangle of minimum
perimeter exists. This assumption might be slightly useful in a “brute force” approach
using calculus to find the minimum of a two-variable function. Such an approach is
painful to execute, however.

B3. (44, 4, 3, 0, 0, 0, 0, 0, 7, 2, 62, 77)

Let H be the unit hemisphere { (x, y, z) : x2 + y 2 + z 2 = 1, z ≥ 0 }, C the
unit circle { (x, y, 0) : x2 + y 2 = 1 }, and P the regular pentagon inscribed in
C. Determine the surface area of that portion of H lying over the planar
region inside P , and write your answer in the form A sin α + B cos β, where
A, B, α, β are real numbers.
Answer. The surface area is 5π cos π5 − 3π. In particular, we may take

A = −3π, α = π/2, B = 5π, and β = π/5.

Solution. The surface area of the spherical cap

{ (x, y, z) : x2 + y 2 + z 2 = 1, z ≥ z0 }

can be computed in spherical coordinates as

 cos−1 z0  2π
sin φ dθ dφ = 2π(1 − z0 ).
φ=0 θ=0

The desired surface area is the area of a hemisphere minus the surface areas of five
identical halves of spherical caps; these caps, up to isometry, correspond to z0 being
the distance from the center of the pentagon to any of its sides, i.e., z0 = cos π5 . Since
the area of a hemisphere is 2π (take z0 = 0 above), the desired area is
5" " π ## π
2π − 2π 1 − cos = 5π cos − 3π. 
2 5 5

Remark. The surface area of a spherical cap can also be computed by noting that
the cap is the surface of revolution
√obtained by rotating part of the graph of y = f (x)
around the x-axis, where f (x) = 1 − x2 . This gives
 1   1 
2
2πy dx + dy = 2 2πf (x) 1 + f  (x)2 dx
x=z0 z0
 2
1  x2
= 2π 1 − x2 1+ dx
z0 1 − x2
 2
1  1
= 2π 1− x2 dx
z0 1 − x2
= 2π(1 − z0 ).

Notice an amazing consequence of the formula: the area of a slice of a unit sphere
depends only on the width of the slice, and not on its position on the sphere! This
was known already to Archimedes, and forms the basis of the Lambert equal-area
cylindrical projection and the Gall-Peters equal-area projection used in some maps.
256 The William Lowell Putnam Mathematical Competition

This fact can be used to give a beautiful solution to the following problem [She]:
Consider a disk of diameter d drawn on a plane and a number of very long
white paper strips of different widths. Show that the disk can be covered by
the strips if and only if the sum of the widths is at least d.

B4. (42, 9, 22, 0, 0, 0, 0, 0, 21, 28, 24, 53)

Find necessary and sufficient conditions on positive integers m and n so
that
mn−1
i/m + i/n
(−1) = 0.
i=0

m n
Answer. We have S(m, n) = 0 if and only if gcd(m,n) and gcd(m,n) are not both odd.
Solution. For convenience, define fm,n (i) = +  mi   ni ,
so that the given sum is
mn−1 fm,n (i)
S(m, n) = i=0 (−1) . If m and n are both odd, then S(m, n) is the sum of
an odd number of ±1’s, and thus cannot be zero. If m and n are of opposite parity,
. / . / . / . /
i i mn − 1 − i mn − 1 − i
fm,n (i) + fm,n (mn − 1 − i) = + + +
m n m n
= (n − 1) + (m − 1),
since  ac  + b − c+1
a  = b − 1 for all integers a, b, c with a > 0. Thus the terms of
S(m, n) cancel in pairs and so the sum is zero.
Now suppose that m = dk and n = dl for some d. For i = 0, . . . , d − 1, we have
 dj+i
dk  =  k  and  dl  =  l , so fm,n (dj + i) = fk,l (j). Hence
j dj+i j

dkl−1
S(dk, dl) = d (−1)fk,l (j)
j=0
kl−1
=d (−1)fk,l (j) + (−1)fk,l (j+kl) + · · · + (−1)fk,l (j+(d−1)kl)
j=0
kl−1 d−1
=d (−1)fk,l (j) (−1)i(k+l)
j=0 i=0

(since fk,l (j + ikl) = i(k + l) + fk,l (j))

d−1
= dS(k, l) (−1)i(k+l) .
i=0

Thus we have S(dk, dl) = d2 S(k, l) if k + l is even; this equality also holds if k + l
is odd, since then S(k, l) = 0. Thus S(dk, dl) vanishes if and only if S(k, l) vanishes.
Piecing together the various cases gives the desired result. 
Remark. There are many equivalent ways to state the answer. Here are some other
possibilities, mostly from student papers:
• The highest powers of 2 dividing m and n are different.
• There exist integers a, b, k ≥ 0 such that m = 2k a, n = 2k b, and a and b have
opposite parity.
Solutions: The Fifty-Ninth Competition (1998) 257

• lcm(m, n) is an even integer times m, or an even integer times n, but not both.
• Exactly one of m
gcd(m,n) and n
gcd(m,n) is even.
• m+n
gcd(m,n) is odd.
• mn
gcd(m,n)2 is even.
lcm(m,n)
• gcd(m,n) is even.
• When expressed in lowest terms, m/n has even numerator or even denominator.
• The 2-adic valuation of m/n is nonzero. (See the remark following 1997B3 for a
definition of the p-adic valuation for any prime p.)
Remark. If S(m, n) is nonzero, it equals gcd(m, n)2 . To show this, we use the
equation S(dk, dl) = S(k, l) from the above solution to reduce to the case where m, n
are coprime and odd. In this case,
fm,n (i) ≡ mi/m + ni/n ≡ i − mi/m + i − ni/n (mod 2).
Now i − mi/m is the remainder upon dividing i by m. By the Chinese Remainder
Theorem, the pairs (i − mi/m, i − ni/n) run through all pairs (j, k) of integers
with 0 ≤ j ≤ m − 1, 0 ≤ k ≤ n − 1 exactly once. Hence
m−1 n−1
S(m, n) = (−1)j (−1)k = 1.
j=0 k=0

B5. (55, 2, 29, 0, 0, 0, 0, 0, 8, 0, 29, 76)

Let N be the positive integer with 1998 decimal digits, all of them 1;
that is,
N = 1111 · · · 11.
√
Find the thousandth digit after the decimal point of N.
Answer. The 1000th digit is 1.
Solution 1. Write N = (101998 − 1)/9. Then
√ 10999  10999 1
N= 1 − 10−1998 = (1 − 10−1998 + r),
3 3 2
where r can be bounded using Taylor’s Formula with remainder: see Solution 2 to
2
1992A4. Recall that Taylor’s Formula says that f (x) = f (0) + xf  (0) + x2 f  (c) for
√
some c between 0 and x. For f (x) = 1 + x, f  (x) = − 14 (1 + x)−3/2 ; therefore for
x = −10−1998 , |r| < 10−3996 /8.
The digits after the decimal point of 10999 /3 are given by .3333 . . . , while the digits
1 −999
after the decimal
√ point of 6 10 are given by .00000 . . . 1666666 . . . . Thus the first
1000 digits of N after the decimal point are given by .33333 . . . 3331; in particular,
the thousandth digit is 1. 
√
Solution 2 (Hoeteck Wee). The 1000th digit after the decimal point of N is
the unit digit of
√ √
√ 101998 − 1 103998 − 102000
101000 N = 10001000 = .
3 3
258 The William Lowell Putnam Mathematical Competition

Now
(101999 − 7)2 < 103998 − 102000 < (101999 − 4)2 ,

so
101999 − 7 √ 101999 − 4
< 101000 N < .
3 3
Hence
√
33 . . . 31 < 101000 N < 33 . . . 32,

and the answer is 1. 

Related question. What are the next thousand digits? (Hint: extend Solution 1.)

B6. (25, 8, 8, 0, 0, 0, 0, 0, 4, 5, 31, 118)

Prove
√ that, for any integers a, b, c, there exists a positive integer n such
that n3 + an2 + bn + c is not an integer.

Solution 1. Recall that all perfect squares are congruent to 0 or 1 modulo 4.

Suppose P (n) = n3 + an2 + bn + c is a square for n = 1, 2, 3, 4. Since P (2) and P (4)
are perfect squares of the same parity, their difference 56 + 12a + 2b must be a multiple
of 4; that is, b must be even. On the other hand, since P (1) and P (3) are also perfect
squares of the same parity, their difference 26 + 8a + 2b must be a multiple of 4; that
is, b must be odd. This is a contradiction. 
2
Solution 2. If 4b − a2 = 0 and c = 0, then n3 + an2 + bn + c = n (n + a/2) ,
which is not a perfect square if n is not a perfect square and n = −a/2.
So suppose that 4b − a2 and c are not both zero. Take n = 4m2 . Then

n3 + an2 + bn + c = (8m3 + am)2 + (4b − a2 )m2 + c.

If 4b−a2 > 0, or 4b−a2 = 0 and c > 0, then for m sufficiently large, (4b−a2 )m2 +c > 0,
and
|(4b − a2 )m2 + c| < 2(8m3 + am) − 1,

so n3 + an2 + bn + c lies between (8m3 + am)2 and (8m3 + am + 1)2 and is hence not a
perfect square. Similarly, if 4b−a2 < 0, or 4b−a2 = 0 and c < 0, then n3 +an2 +bn+c
lies between (8m3 + am − 1)2 and (8m3 + am)2 for sufficiently large m. 

Solution 3. We will use the theory of height functions on elliptic curves (see
Chapter 8 of [Sil] for this theory). Suppose first that P (n) = n3 + an2 + bn + c is not
squarefree as a polynomial in Q[n]. Then by Gauss’s Lemma (see the remark below),
P (n) factors as (n − d)2 (n − e) for some d, e ∈ Z. Choose a positive integer n = d
such that n − e is not a square. Then P (n) is not a square.
Now suppose instead that P (n) is squarefree. Then y 2 = P (x) is an elliptic curve
E over Q. Let N (B) be the number of rational numbers x = u/v in lowest terms with
|u|, |v| ≤ B such that P (x) is the square of a rational number. The theory of height
functions implies that N (B) = O((ln B)r/2 ) as B → ∞, where r is the rank of E(Q).
In particular, if B is a sufficiently large integer, then N (B) < B, so not all integers 1,
2, . . . , B can be x-values making P (x) a square. 
Solutions: The Fifty-Ninth Competition (1998) 259

Remark (Gauss’s Lemma). Define the content c(f ) of a polynomial f ∈ Z[x] as

the greatest common divisor of its coefficients. “Gauss’s Lemma” refers to any of the
three following statements:
1. For any f, g ∈ Z[x], we have c(f g) = c(f )c(g).
2. If a monic polynomial f with integer coefficients factors into monic polynoimals
with rational coefficients, say f = gh with g, h ∈ Q[x], then g and h have integer
coefficients [NZM, Theorem 9.7].
3. If a polynomial f ∈ Z[x] factors nontrivially into polynomials with rational
coefficients, then it also factors nontrivially into (possibly different) polynomials
with integer coefficients [Lar1, Problem 4.2.16].
The first statement immediately implies the other two. All three statements can be
generalized to the case where Z is replaced by an arbitrary unique factorization domain
(UFD). For the generalization of the first statement, see [Lan1, p. 181].
Stronger result. For any polynomial P with integer coefficients that is not the
square of a polynomial with integer coefficients, there are arbitrarily large n such that
P (n) is not a perfect square. We present three proofs, using quite different techniques.
Proof 1 (Paul Cohen). Suppose
 that P (n) is the square of a nonnegative integer for
all n ≥ n0 . Express f (n) = P (n) as an infinite Puiseux series (see the first remark
after this solution)
f (n) = c0 nk/2 + c1 nk/2−1 + · · ·
with c0 > 0, convergent for n sufficiently large. Define the difference operator ∆ taking
the function g to the new function (∆g)(n) = g(n + 1) − g(n). For i ≥ 1, the function
(i) (i)
(∆i f )(n) is represented by the series c0 nk/2−i + · · · for some coefficients cj , and this
series converges for n sufficiently large. In particular, if i > k/2, then |(∆i f )(n)| < 1
for sufficiently large n, but (∆i f )(n) is also an integer, so it is zero. By the lemma
in the Lagrange interpolation remark below, f (n) agrees with a polynomial F (n) for
sufficiently large integers n. Then P (n) − F (n)2 is a polynomial with infinitely many
zeros, so P (n) = F (n)2 as polynomials. Since F (n) is an integer for sufficiently large
integers n, Lagrange interpolation (described in the second remark below) implies
that F has rational coefficients. Finally, Gauss’s Lemma (given in the remark after
Solution 3) implies that P , being the square of a polynomial with rational coefficients,
is also the square of a polynomial with integer coefficients. 
Remark (Puiseux series). A Puiseux series in the variable x over a field k is a series
∞
of the form j=M aj xj/r for some M ∈ Z (possibly negative), some integer r ≥ 1,
and some elements aj ∈ k. In other words, it is a Laurentseries in the parameter x1/r
for some integer r ≥ 1. (In Proof 1 above, we expressed f (n) as a Puiseux series in
x = n−1 .) The set of all Puiseux series over k forms a field.
If k is an algebraically closed field of characteristic zero, such as C, then the field of
Puiseux series is an algebraic closure of the field k((x)) of Laurent series; in particular,
under these hypotheses, if f ∈ k(x)[t] is a one-variable polynomial with coefficients in
the field of rational functions k(x), then the zeros of f can be √ expressed as Puiseux
series. Such zeros are called algebraic functions; for instance 3 7 + x2 is an algebraic
function. The fact that algebraic functions can be expanded in Puiseux series was
260 The William Lowell Putnam Mathematical Competition

observed by Newton; hence the term Newton-Puiseux series is also sometimes used.
The statements in this paragraph are false when k is an algebraically closed field of
characteristic p > 0, but see [Ke] for an analogue.

Remark (Lagrange interpolation). Suppose one is given n + 1 pairs of numbers

(ai , bi ) such that a1 , . . . , an+1 are distinct. The theorem of Lagrange interpolation
states that there exists a unique polynomial f (x) of degree less than or equal to n
such that f (ai ) = bi for 1 ≤ i ≤ n + 1. Uniqueness follows from the observation that
the difference of two such polynomials f is a polynomial of degree at most n, having
at least n + 1 zeros (the ai ): such a difference must be identically zero. Existence
follows from the explicit formula

n+1
(x − a1 ) · · · (x − ai ) · · · (x − an+1 )
f (x) = bi ,
i=1 (ai − a1 ) · · · (ai − ai ) · · · (ai − an+1 )

where 9· indicates that the corresponding term is omitted from the product. (It is easy
to check that this f (x) is a polynomial of degree less than or equal to n such that
f (ai ) = bi for all i.) As an immediate consequence, if ai , bi ∈ Q, then f (x) has rational
coefficients.
One application of Lagrange interpolation is the lemma below, used in Proof 1
above. For further applications, see [Lar1, Section 4.3].
Lemma. If n0 ∈ Z, and g(n) is a function defined on integers n ≥ n0 such that
(∆k g)(n) = 0 for all n ≥ n0 , then there is a polynomial G(x) of degree at most k − 1
such that g(n) = G(n) for all n ≥ n0 .
Proof of Lemma. First, if h(x) is a polynomial of degree m ≥ 1, then (∆h)(x) is a
polynomial of degree m − 1. It follows that (∆k h)(n) = 0 for all k > m.
Use Lagrange interpolation to find the polynomial G(x) of degree at most k − 1
such that G(n) = g(n) for n = n0 , n0 + 1, . . . , n0 + k − 1. Let H(n) = G(n) − g(n) for
integers n ≥ n0 . Then (∆k G)(n) = 0 by the previous paragraph, and (∆k g)(n) = 0
for integers n ≥ n0 is given, so (∆k H)(n) = 0 for integers n ≥ n0 .
It remains to prove that if H is a function such that (∆k H)(n) = 0 for integers
n ≥ n0 and H(n0 ) = H(n0 + 1) = · · · = H(n0 + k − 1) = 0, then H(n) = 0 for all
n ≥ n0 . We use induction on k. The case k = 0 is trivial, since ∆0 H is just H. For
k ≥ 1, applying the inductive hypothesis to ∆H, which satisfies the assumptions with
k − 1 instead of k, shows that (∆H)(n) = 0 for integers n ≥ n0 . But H(n0 ) = 0 too,
so induction on n proves H(n) = 0 for all integers n ≥ n0 . 

Proof 2. We may reduce to the case that P has no repeated factors by dividing by
squares of factors, so that the discriminant D of P is nonzero. We construct a prime
p not dividing D and an integer n such that p divides P (n). Pick an integer m such
that P (m) = 0, and for each prime pi dividing D, let ei be the exponent to which
3 ei +1
pi divides P (m). As n varies over integers congruent to m modulo pi , P (n)
assumes arbitrarily large values (since P is nonconstant), and none of these values is
divisible by piei +1 , since P (m) is not. Thus there exist arbitrarily large integers n for
which P (n) is divisible by a prime p not dividing D.
Solutions: The Fifty-Ninth Competition (1998) 261

If p2 does not divide P (n), then P (n) is not a square. Otherwise,

P (n + p) ≡ P (n) + pP  (n) ≡ pP  (n) (mod p2 ),
and p does not divide P  (n) since p does not divide the discriminant of P . Thus
P (n + p) is divisible by p but not p2 , so P (n + p) is not a square. 

Remark. The end of the argument is related to Hensel’s Lemma; a similar idea
arises in 1986B3.
Proof 3. We show that for any sufficiently large prime p, there exist arbitrarily
large integers n such that P (n) is not a square modulo p. Again we reduce to the case
where P is squarefree and nonconstant, say of degree d. Then the discriminant ∆ of
P is a nonzero integer; let p be any odd prime not dividing ∆. Suppose that P (n)
is a square modulo p for all sufficiently large n. Then for each x in the field Fp of p
elements, the equation y 2 = P (x) has two solutions y, unless x is a zero of P modulo
p in which case there is only one solution y. The number of zeros of P modulo p is
at most d, so y 2 = P (x) has at least 2p − d solutions (x, y) ∈ Fp × Fp . On the other
hand, the Weil Conjectures (see the end of 1991B5) for the curve y 2 = P (x) imply
√
that the number of solutions is at most p + c p, where the constant c depends only
on d. This yields a contradiction for sufficiently large p. 

Remark. A fourth proof of the stronger result can be given using the Thue-Siegel
Theorem [HiS, Theorem D.8.3], which asserts that if P is a polynomial of degree at
least 3 with no repeated factors, then P (n) is a perfect square for only finitely many
integers n. For an effective bound on the size of such n, see Section 4.3 of [Bak2].
(The Thue-Siegel Theorem can fail if P has degree 2. For example, 2n2 + 1 is a square
for infinitely many integers n. See Solution 2 to 2000A2.)
Related question. A similar problem is 1971A6 [PutnamII, p. 15]:
Let c be a real number such that nc is an integer for every positive integer n.
Show that c is a nonnegative integer.
262 The William Lowell Putnam Mathematical Competition

The Sixtieth William Lowell Putnam Mathematical Competition

December 4, 1999

A1. (124, 17, 34, 0, 0, 0, 0, 0, 10, 4, 11, 5)

Find polynomials f (x), g(x), and h(x), if they exist, such that, for all x,


−1
 if x < −1
|f (x)| − |g(x)| + h(x) = 3x + 2 if −1 ≤ x ≤ 0


−2x + 2 if x > 0.

Answer. Take
3x + 3 5x 1
f (x) = , g(x) = , and h(x) = −x + .
2 2 2

Solution. Let


−1
 if x < −1
F (x) = 3x + 2 if −1 ≤ x ≤ 0


−2x + 2 if x > 0.

The function G(x) = max{−1, 3x + 2} agrees with F (x) for x ≤ 0, but

F (x) − G(x) = −5x for x > 0.
Thus
F (x) = max{−1, 3x + 2} − max{0, 5x}
= (3x + 1 + | − 3x − 3|)/2 − (5x + |5x|)/2
(by the identity max{r, s} = (r + s + |r − s|)/2)
1
= |(3x + 3)/2| − |5x/2| − x + ,
2
so we may take
f (x) = (3x + 3)/2, g(x) = 5x/2, h(x) = −x + 1/2. 

Remark. Alternatively, based on Figure 38, one may guess that f, g, h are linear,
and that f changes sign at −1 and g changes sign at 0. We may assume that f and g
have positive leading coefficients. Then one has the linear equations
−1 = −f (x) + g(x) + h(x)
3x + 2 = f (x) + g(x) + h(x)
−2x + 2 = f (x) − g(x) + h(x)
from which one solves for f, g, h as given above.
In fact, both guesses can be justified, and we can prove that the solution is unique
up to the signs of f and g. Of the four polynomials ±f ± g + h, three must be equal
to −1, 3x + 2, −2x + 2 in some order. But of any three of these functions, two sum
to 2h, and the difference between some two is 2f or 2g. Thus f, g, h are each linear.
As for the sign changes, assume that the graphs of f and g have positive slope. The
Solutions: The Sixtieth Competition (1999) 263

y = 3x + 2

y = –2x + 2

y = –1

FIGURE 38.
Graph of F (x) (in bold)

slope of the graph of |f (x)| − |g(x)| + h(x) jumps up at the zero of f and jumps down
at the zero of g, so these points must be x = −1 and x = 0, respectively.

A2. (61, 16, 28, 0, 0, 0, 0, 0, 2, 4, 52, 42)

Let p(x) be a polynomial that is nonnegative for all real x. Prove that
for some k, there are polynomials f1 (x), . . . , fk (x) such that
k
p(x) = (fj (x))2 .
j=1

Solution 1. If p(x) is identically zero, we are done. Otherwise factor p(x)

into linear and quadratic factors over the real numbers. Each linear factor must
occur with even multiplicity, or else p(x) would have a sign change at the zero
of the linear factor. Thus p(x) can be factored into squares of linear factors and
irreducible quadratic factors. The latter can be written as sums of squares (namely,
x2 + ax + b = (x + a/2)2 + (b − a2 /4)), so p(x) is the product of polynomials which
can be written as sums of squares, so is itself a sum of squares. 
Solution 2. We proceed by induction on the degree of p, with base case where p
has degree 0. As noted in Solution 1, real zeros of p occur with even multiplicity, so
we may divide off the linear factors having such zeros to reduce to the case where p
has no real zeros (and is nonconstant, or else we are already done). Then p(x) > 0 for
all real x, and since p(x) → +∞ as x → ±∞, p attains a minimum value c > 0. Now
p(x)−c has real zeros, so as above, we deduce that p(x)−c is a sum of squares. Adding
√
one more square, namely ( c)2 , to p(x) − c expresses p(x) as a sum of squares. 
Stronger result. In fact, only two polynomials are needed. This can be deduced
from Solution 1, provided that one knows the identity (1) in Solution 3 to 2000A2,
which says that a sum of two squares times a sum of two squares can be expressed as
a sum of two squares.
264 The William Lowell Putnam Mathematical Competition

Here is another proof that k = 2 suffices. Factor p(x) = q(x)r(x), where q has
all real zeros and r has all nonreal zeros and r is monic. Each zero of q has even
multiplicity, since otherwise p would have a sign change at that zero. Also q has
positive leading coefficient, since p must. Thus q(x) has a square root s(x) with real
3k
coefficients. Now write r(x) = j=1 (x − aj )(x − aj ) (possible because r has zeros in
3k
complex conjugate pairs). Write j=1 (x − aj ) = t(x) + iu(x) with t and u having real
coefficients. Then for x real,
p(x) = q(x)r(x)
= s(x)2 (t(x) + iu(x)) (t(x) + iu(x))
= (s(x)t(x))2 + (s(x)u(x))2 .

Literature note. This problem appeared as [Lar1, Problem 4.2.21], and is credited
there to [MathS].
Remark. A polynomial in more than one variable with real coefficients taking
nonnegative values need not be a sum of squares. (The reader may check that
1 − x2 y 2 (1 − x2 − y 2 ) is a counterexample.) Hilbert’s Seventeenth Problem asked
whether such a polynomial can always be written as the sum of squares of rational
functions; this was answered affirmatively by E. Artin and O. Schreier. See [J,
Section 11.4] for a proof and for generalizations.

A3. (103, 10, 2, 0, 0, 0, 0, 0, 1, 18, 27, 44)

Consider the power series expansion
∞
1
= an xn .
1 − 2x − x2 n=0

Prove that, for each integer n ≥ 0, there is an integer m such that

a2n + a2n+1 = am .

Solution 1. Note that

4 √ √ 5
1 1 1+ 2 1− 2
= √ √ − √
1 − 2x − x2 2 2 1 − (1 + 2)x 1 − (1 − 2)x
and
∞
1 √
√ = (1 ± 2)n xn ,
1 − (1 ± 2)x n=0
so
1 " √ √ #
an = √ (1 + 2)n+1 − (1 − 2)n+1 .
2 2
A simple computation now shows that a2n + a2n+1 = a2n+2 . 
n n
Solution 2. As in Solution 1, we see that an = Cα + Dβ for some constants
C, D, α, β with αβ = −1 (since α, β are the zeros of x2 − 2x − 1). Thus
a2n = Eα2n + F (−1)n + Gβ 2n
Solutions: The Sixtieth Competition (1999) 265

for some constants E, F, G, and

a2n + a2n+1 = Hα2n + Iβ 2n

for some constants H, I. But a2n+2 has the same form, so a2n + a2n+1 and a2n+2 satisfy
the same second-order linear recursion. (See the remarks in 1988A5 for more on linear
recursions.) Hence we can prove a2n + a2n+1 = a2n+2 for all n ≥ 0 by checking it for
n = 0 and n = 1, which is easy. 
Remark. From the recursion an+1 = 2an + an−1 , one can also find the recursion
satisfied by a2n+2 , then directly prove that a2n + a2n+1 satisfies the same recursion.


0 1
Solution 3 (Richard Stanley). Let A be the matrix . By induction,
1 2
the recursion an+1 = 2an + an−1 implies


an an+1
An+2 = .
an+1 an+2
The desired result follows from equating the top left entries in the matrix equality
An+2 An+2 = A2n+4 . 
Related question. This sequence, translated by one, also appeared in the following
proposal by Bulgaria for the 1988 International Mathematical Olympiad [IMO88,
p. 62].
An integer sequence is defined by an = 2an−1 + an−2 (n > 1), a0 = 0, a1 = 1.
Prove that 2k divides an if and only if 2k divides n.

A4. (33, 27, 3, 0, 0, 0, 0, 0, 3, 2, 14, 123)

Sum the series
∞ ∞
m2 n
.
m=1 n=1
3m (n3m + m3n )

Answer. The series converges to 9/32.

Solution. Denote the series by S, and let an = 3n /n. Note that
∞ ∞ ∞ ∞
1 1
S= = ,
a (a + an )
m=1 n=1 m m
a (a
m=1 n=1 n m
+ an )
where the second equality follows by interchanging m and n. Thus


1 1
2S = +
m n
am (am + an ) an (am + an )
4∞ 52
1 n
= = .
m n
am a n n=1
3n
Finally, if
∞
n
A= , (1)
n=1
3n
266 The William Lowell Putnam Mathematical Competition

then
∞ ∞
n n+1
3A = = , (2)
n=1
3n−1 n=0
3n

so subtracting (1) from (2) gives

∞
1 3
2A = 1 + n
= .
n=1
3 2
Hence
3
A= , (3)
4
so S = 9/32. 
Remark. Equation (3) also follows by differentiating both sides of
∞
xn 3
= ,
n=0
3 n 3 − x
and then evaluating at x = 1. Either method can be generalized to evaluate arbitrary
∞
sums of the form n=0 P (n)αn where P (n) is a polynomial, and |α| < 1.
Remark. The rearrangement of terms in various parts of the solution is justified,
since all terms are nonnegative.

A5. (5, 1, 4, 0, 0, 0, 0, 0, 3, 10, 63, 119)

Prove that there is a constant C such that, if p(x) is a polynomial of
degree 1999, then
 1
|p(0)| ≤ C |p(x)| dx.
−1

Solution 1. Let P denote the set of polynomials of degree at most 1999. Identify
1999
P with R2000 by identifying i=0 ai xi with (a0 , a1 , . . . , a1999 ). Let S be the set of
1999
polynomials i=0 ai xi such that max{|ai |} = 1. Then S is a closed and bounded
subset of P ≈ R2000 , so S is compact. The function P × R → R mapping (p, x) to
|p(x)| is the absolute value of a polynomial in all 2001 coordinates,
1 so it is continuous.
Therefore the function g : P → R defined by g(p) = −1 |p(x)| dx is continuous, as
is its restriction to S. Similarly the function f : S → R defined by f (p) = |p(0)| is
continuous. Since g(p) = 0 for p ∈ S, the quotient f /g : S → R is continuous. By
the Extreme Value Theorem, there exists a constant C such that f (p)/g(p) ≤ C for
all p ∈ S. An arbitrary p ∈ P can be written as cq for some c ∈ R and q ∈ S; then
f (p) = |c|f (q) ≤ |c|Cg(q) = Cg(p), as desired. 
Remark. The same method proves the standard result that any two norms on a
finite-dimensional vector space are equivalent [Hof, p. 249]. In fact, another way to
solve the problem would be to apply this result to the two norms
 1
sup |p(x)| and |p(x)| dx
x∈[−1,1] −1

on P .
Solutions: The Sixtieth Competition (1999) 267

Solution 2 (Reid Barton). We exhibit an explicit constant C, by showing that

|p(x)| must be large compared to |p(0)| on some subinterval of [−1, 1] whose length is
bounded below. Assume p(0) = 1 without loss of generality.
31999
Then p(x) = i=1 (1 − x/ri ), where r1 , . . . , r1999 are the complex zeros of p, listed
with multiplicity. Fix C < 1/3998, and draw closed discs of radius C centered at the
ri in the complex plane. These discs intersect (−1/2, 1/2) in at most 1999 intervals
of total length at most 3998C, so their complement consists of at most 2000 intervals
of total length at least 1 − 3998C. By the Pigeonhole Principle, one of these intervals,
say (c, d), has length at least δ = (1 − 3998C)/2000 > 0. For x ∈ (c, d), if |ri | ≤ 1 then
|1 − x/ri | ≥ |x − ri | > C, whereas if |r1 | ≥ 1 then |1 − x/ri | ≥ 1 − |x/ri | ≥ 1/2 > C.
Hence  1  d 1999
+
|p(x)| dx ≥ |1 − x/ri | dx ≥ δC1999 ,
−1 c i=1
so  1
|p(0)| ≤ C |p(x)| dx
−1

with C = 1/(δC1999 ). Taking C = 1/4000 yields the explicit value C = 21999 20002001 . 
Related question. Can you improve the constant in Solution 2? (We do not know
what the smallest possible C is.)

A6. (31, 11, 9, 0, 0, 0, 0, 0, 1, 5, 28, 120)

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.
Solution. Rearranging the given equation yields the much more tractable equation
an an−1 an−2
=6 −8 .
an−1 an−2 an−3
Let bn = an /an−1 . With the initial conditions b2 = 2, b3 = 12, one obtains bn =
2n−1 (2n−1 − 1), by induction or by the theory of linear recursive sequences: see the
remark in 1988A5. Thus
+
n−1
an = a1 b2 b3 . . . bn = 2n(n−1)/2 (2i − 1). (1)
i=1

If n = 1, then n divides an . Otherwise factor n as 2k m, with m odd. Then

k ≤ n − 1 ≤ n(n − 1)/2, and there exists i ≤ n − 1 such that m divides 2i − 1, namely
i = φ(m). (Here φ denotes the Euler φ-function: see 1985A4.) Hence n divides an for
all n ≥ 1. 
Remark. Alternatively, the result for n ≥ 3 can be proved from the following two
facts:
(a) The right side of the formula (1) for an equals 2n−1 #GLn−1 (F2 ). (See page xi for
the definition of GLn−1 (F2 ).)
(b) If n ≥ 3, GLn−1 (F2 ) contains an element of exact order n.
268 The William Lowell Putnam Mathematical Competition

These suffice because of Lagrange’s Theorem, which states that the order of an element
of a finite group G divides the order of G.
Let us prove (a). Matrices in GLn−1 (F2 ) can be constructed one row at a time: the
first row may be any nonzero vector, and then each successive row may be any vector
not in the span of the previous rows, which by construction are independent. Hence
the number of possibilities for the jth row, given the previous ones, is 2n−1 − 2j−1 .
Thus
+
n−1 +
n−1
#GLn−1 (F2 ) = (2n−1 − 2j−1 ) = 2(n−1)(n−2)/2 (2n−j − 1).
j=1 j=1

Setting i = n − j and comparing with (1) proves (a).


It remains to prove (b). Let V = { (x1 , . . . , xn ) ∈ (F2 )n : xi = 0 }. Since
dimF2 V = n − 1, the group of automorphisms of the vector space V is isomor-
phic to GLn−1 (F2 ). Let T : V → V be the automorphism (x1 , x2 , . . . , xn ) !→
(xn , x1 , . . . , xn−1 ). Then T n is the identity. If m < n and n ≥ 3, then there exists
v = (v1 , . . . , vn ) ∈ V with v1 = 1 and vm+1 = 0: make just one other vi equal to 1,
to make the sum zero. Then T m v = v, so T m is not the identity. Hence T has exact
order n.
(The matrix of T with respect to the basis

C1 = (1, 1, 0, 0, . . . , 0), C2 = (0, 1, 1, 0, . . . , 0), . . . , Cn−1 = (0, 0, . . . , 0, 1, 1)

of V is  
0 0 0 ··· 0 1
1 0 0 ··· 0 1
 
 
0 1 0 ··· 0 1
. .. .. 
. .. .. . .  ∈ GLn−1 (F2 ).
. . . . . .
 
0 0 0 ··· 0 1
0 0 0 ··· 1 1
This also equals the companion matrix of the polynomial

f (x) = xn−1 + xn−2 + · · · + x + 1.)

B1. (126, 17, 17, 0, 0, 0, 0, 0, 6, 0, 23, 16)

Right triangle ABC has right angle at C and ∠BAC = θ; the point D is
chosen on AB so that |AC| = |AD| = 1; the point E is chosen on BC so
that ∠CDE = θ. The perpendicular to BC at E meets AB at F . Evaluate
limθ→0 |EF |. [Here |P Q| denotes the length of the line segment P Q.]†
Answer. The limit of |EF | as θ → 0 is 1/3.

Solution. (See Figure 39.) The triangles BEF and BCA are similar. Since
AC = 1, we have EF = BE/BC. Also, since 4ACD is isosceles, ∠ACD =
∠ADC = π/2 − θ/2. We then compute ∠BCD = π/2 − ∠DCA = θ/2 and
∠BDE = π − ∠CDE − ∠CDA = π/2 − θ/2.

† The figure is omitted here, since it is contained in Figure 39 used in the solution.
Solutions: The Sixtieth Competition (1999) 269

q/2
E

q
q p/2 – q
A F D B

3q/2

E¢

FIGURE 39.

The reflection E  of E across AB lies on CD, so by the Law of Sines,

BE BE  sin ∠BCD
= = .
BC BC sin ∠CE  B
Since
π π
∠CBD = − ∠BAC = − θ
2 2
and
π 3θ
∠CE  B = ∠DEB = − ∠CBD − ∠BDE = ,
2 2
we obtain
BE sin(θ/2)
EF = = .
BC sin(3θ/2)
By L’Hôpital’s Rule,
sin(θ/2) cos(θ/2) 1
lim = lim = . 
θ→0 sin(3θ/2) θ→0 3 cos(3θ/2) 3

Remark. One can avoid using the reflection by writing

BE BE BD
= ·
BC BD BC
and using the Law of Sines in triangles BDE and BCD.

B2. (38, 27, 9, 0, 0, 0, 0, 0, 9, 0, 74, 48)

Let P (x) be a polynomial of degree n such that P (x) = Q(x)P  (x), where
Q(x) is a quadratic polynomial and P  (x) is the second derivative of P (x).
Show that if P (x) has at least two distinct roots then it must have n distinct
roots. [The roots may be either real or complex.]
270 The William Lowell Putnam Mathematical Competition

Solution. Suppose that P does not have n distinct zeros; then it has a zero
of multiplicity k ≥ 2, which we may assume without loss of generality is x = 0.
Differentiating P term by term shows that the highest power of x dividing P  (x) is
xk−2 . But P (x) = Q(x)P  (x), so x2 divides Q(x). Since Q is quadratic, Q(x) = Cx2
for some constant C. Comparing the leading coefficients of P (x) and Q(x)P  (x) yields
1
C = n(n−1) .
n
Write P (x) = j=0 aj xj ; equating coefficients in P (x) = Cx2 P  (x) implies that
aj = Cj(j − 1)aj for all j. Hence aj = 0 for j ≤ n − 1, and P (x) = an xn , which has
all identical zeros. 
Remark. If Q has distinct zeros, then after replacing x by ax+b for suitable a, b ∈ C
we may assume that Q = c(1 − x2 ) for some c ∈ C. Then P  (x) is a scalar multiple
of Rn−2 (x), where Rn (x) is the unique monic polynomial of degree n satisfying the
differential equation

d2
[(1 − x2 )Rn (x)] = −(n + 1)(n + 2)Rn (x).
dx2

(This equation implies a recursion which uniquely determines the coefficients of Rn .)

Integration by parts yields
 1  1
1
(1 − x2 )Rm (x)Rn (x) dx = [(1 − x2 )Rm (x)] [(1 − x2 )Rn (x)] dx;
−1 (n + 1)(n + 2) −1

interchanging m and n and comparing, we find

 1
(1 − x2 )Rm (x)Rn (x) dx = 0
−1

for m = n. That is, the Rm (x) form a family of orthogonal polynomials with respect
to the measure (1 − x2 ) dx. (In other terminology, the Rm are eigenvectors of the
d2
operator f !→ dx 2 [(1 − x )f ]. This operator is self-adjoint with respect to the measure
2

(1 − x2 ) dx, so the eigenvectors are orthogonal with respect to that measure, because
eigenspaces corresponding to different eigenvalues are orthogonal.)
The solution above implies that Rn has distinct zeros. A result from the theory
of orthogonal polynomials implies that the zeros of Rn are real numbers in [−1, 1].
This can also be deduced by applying Lucas’ Theorem (1991A3) twice. Namely, if the
convex hull of the zeros of the polynomial P (x) = (1 − x2 )Rn (x) has a vertex z other
than ±1, then Lucas’ Theorem implies that z lies outside of the convex hull of the
zeros of P  (because z is not a multiple zero of P ) and then implies that z lies outside
of the convex hull of the zeros of P  (x) = Rn (x), a contradiction.
Noam Elkies points out that according to [GR], Rn (x) is a scalar multiple of the
nth Gegenbauer polynomial with parameter λ = 3/2, also called the nth ultraspherical
polynomial. Also, if we put Pn (x) = (1 − x2 )Rn (x), then Pn (x) is (a scalar multiple)
of the (n + 1)st Legendre polynomial.
Related question. Problem 6 from the final round of the 1978 Swedish Mathematical
Olympiad [SMO] is related:
Solutions: The Sixtieth Competition (1999) 271

The polynomials
P (x) = cxn + an−1 xn−1 + · · · + a1 x + a0
Q(x) = cxm + bm−1 xm−1 + · · · + b1 x + b0

with c = 0 satisfy the identity

P (x)2 = (x2 − 1)Q(x)2 + 1.
Show that P  (x) = nQ(x).

B3. (44, 2, 2, 0, 0, 0, 0, 0, 13, 14, 56, 74)

Let A = { (x, y) : 0 ≤ x, y < 1 }. For (x, y) ∈ A, let

S(x, y) = xm y n ,
1 m
2 ≤ n ≤2

where the sum ranges over all pairs (m, n) of positive integers satisfying the
indicated inequalities. Evaluate

lim (1 − xy 2 )(1 − x2 y)S(x, y).

(x,y)→(1,1)
(x,y)∈A

Answer. The limit is equal to 3.

Solution 1. For (x, y) ∈ A,

xy
xm y n = .
m,n>0
(1 − x)(1 − y)

Subtracting S from this gives two sums, one of which is

x2n+1 x3 y
xm y n = yn =
n
1−x (1 − x)(1 − x2 y)
m≥2n+1

and the other of which analogously sums to

xy 3
.
(1 − y)(1 − xy 2 )
Therefore
xy x3 y xy 3
S(x, y) = − −
(1 − x)(1 − y) (1 − x)(1 − x2 y) (1 − y)(1 − xy 2 )
xy(1 + x + y + xy − x2 y 2 )
=
(1 − x2 y)(1 − xy 2 )
and the desired limit is lim(x,y)→(1,1) xy(1 + x + y + xy − x2 y 2 ) = 3. 

Solution 2. A pair (m, n) of positive integers satisfies 1/2 ≤ m/n ≤ 2 if

and only if (m, n) = a(1, 2) + b(2, 1) for some nonnegative rational numbers a, b.
The pairs (a, b) with 0 ≤ a, b < 1 which yield integral values of m and n are
(0, 0), (1/3, 1/3), (2/3, 2/3), whose corresponding pairs are (0, 0), (1, 1), (2, 2). Thus
every pair (m, n) with 1/2 ≤ m/n ≤ 2 can be written as one of these three pairs plus
272 The William Lowell Putnam Mathematical Competition

✁ ❜ ❜
✁ ✟ ✟
✁ ✟✟ ✁ ✁
✁✟
❜✟ ✁ ❜ ❜
✁ ✁ ✟✟✟✁
✁ ✁ ✟ ✁
✁ ❜ ✟❜✟ ✁✟ ✁ ❜
✁ ✟✟ ✁ ✁ ✟
✁ ✟✟ ✁ ✁ ✟✟
✁❜✟ ✁ ❜ ✟ ❜✁ ✟
✁ ✟ ✟
✁ ✁ ✟
✁ ❜ ❜✁ ✟
✟
✁ ✟✟
✁✟✟
✁❜✟

FIGURE 40.
Points of the form a(1, 2) + b(1, 2), where a and b are nonnegative integers; two translates of
this semigroup are also indicated.

an integral linear combination of (1, 2) and (2, 1). (See Figure 40 for those points of
the form a(1, 2) + b(1, 2) where a and b are nonnegative integers.) In particular,
∞ ∞
S(x, y) + 1 = (1 + xy + x2 y 2 ) xa y 2a x2b y b
a=0 b=0
1 + xy + x2 y 2
=
(1 − xy 2 )(1 − x2 y)
and the desired limit is 3. 
Stronger result. Let p/q and r/s be positive rational numbers with p/q < r/s, and

define T (x, y) = (m,n) xm y n , the sum taken over all pairs (m, n) of positive integers
with p/q ≤ m/n ≤ r/s. Then

lim (1 − xp y q )(1 − xr y s )T (x, y) = qr − ps.

(x,y)→(1,1)
(x,y)∈A

This can be proved by either of the above methods.

B4. (0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 31, 173)

Let f be a real function with a continuous third derivative such that f (x),
f  (x), f  (x), f  (x) are positive for all x. Suppose that f  (x) ≤ f (x) for all
x. Show that f  (x) < 2f (x) for all x.
Remark. The total score of the top 205 writers in 1999 on this problem was 2.
The same is true of the next problem, 1999B5. By this measure, these two are the
hardest Putnam problems in the years covered in this volume. Nevertheless, we feel
that there are other problems in this volume that are at least as difficult as these two.
At the other extreme, the problems having the highest average score of the top 200 or
so participants were 1988A1 and then 1988B1.
Solution 1. For simplicity, we will only show f  (0) < 2f (0). Applying this result
to f (x + c) shows that f  (c) < 2f (c) for all c.
Solutions: The Sixtieth Competition (1999) 273

Since f  is positive, f is an increasing function. Thus for x ≤ 0, f  (x) ≤ f (x) ≤

f (0). Integrating f  (x) ≤ f (0) from x to 0 gives f  (x) ≥ f  (0)+f (0)x for x ≤ 0, and
integrating again gives the second inequality in 0 < f  (x) ≤ f  (0) + f  (0)x+ f (0)x2 /2.
Thus the polynomial f  (0) + f  (0)x + f (0)x2 /2 has no negative zeros. Since its
coefficients are positive, it also has no nonnegative zeros. Therefore its discriminant
f  (0)2 − 2f (0)f  (0) must be negative.
In a similar vein, since f  is positive, f  is increasing. Thus for x ≤ 0, f  (x) ≤
f (0), so f  (x) ≥ f  (0) + f  (0)x, and 0 < f (x) ≤ f (0) + f  (0)x + f  (0)x2 /2. Again,


the discriminant of the quadratic must be negative: f  (0)2 − 2f (0)f  (0) < 0.
Combining the conclusions of the last two paragraphs, we obtain

f  (0)4 < 4f (0)2 f  (0)2 < 8f (0)3 f  (0),

which implies f  (0) < 2f (0). 

Solution 2. We prove that f  (x) < (9/2)1/3 f (x) < 1.651f (x). Since f (x) is
bounded below (by 0) and increasing, it has an infimum m, and limx→−∞ f (x) = m.
Then f  (x) tends to 0 as x → −∞ since it is positive and its integral from −∞ to 0
converges. Similarly f  (x) and f  (x) tend to 0 as x → −∞.
By Taylor’s Formula with remainder (or integration by parts), for any function g,
 s
g(x) − g(x − s) = g  (x − t) dt,
0
 s
g(x) − g(x − s) = sg  (x − s) + tg  (x − t) dt, (2)
0
 s
1 1 2 
g(x) − g(x − s) = sg  (x − s) + s2 g  (x − s) + t g (x − t) dt, (3)
2 0 2
as long as the specified derivatives exist and are continuous. For g = f , as s → +∞
for fixed x, the left side of (2) is bounded above, the term sf  (x − s) is nonnegative
and the integrand tf  (x − t) is everywhere nonnegative; consequently, the integrand
tends to 0 as t → ∞. Thus in (2) with g = f  , letting s tend to ∞ yields
 ∞
f  (x) = tf  (x − t) dt.
0

We cannot a priori take limits in any of the equations with g = f , but from (3) we
have
 s
1 2  1
f (x) − t f (x − t) dt = f (x − s) + sf  (x − s) + s2 f  (x − s) ≥ 0
0 2 2
for all s, so
 ∞
1 2 
f (x) ≥ t f (x − t) dt.
0 2
Thus for c ≥ 0, we have
 ∞

 1 2
cf (x) − f (x) ≥ ct − t f  (x − t) dt.
0 2
274 The William Lowell Putnam Mathematical Competition

The integrand is positive precisely when t > 2/c. Moreover, f  (x − t) ≤ f (x − t) by

hypothesis and f (x − t) ≤ f (x) for t ≥ 0, since f  is positive. We conclude
 2/c

1 2 2
cf (x) − f  (x) > ct − t f (x) dt = − 2 f (x),
0 2 3c
or equivalently


2
f  (x) < f (x) c + c−2 .
3
The minimum value of the expression in parentheses is (9/2)1/3 , achieved when
c = (4/3)1/3 : this can be proved by calculus or by the n = 3 case of the AM-GM
Inequality applied to c/2, c/2, and 23 c−2 . (The AM-GM Inequality is given at the end
of 1985A2.) 

Solution 3. We prove that f  (x) < 21/6 f (x) < 1.123f (x). As in the
previous solution, f (x) → 0 as x → −∞ for i = 1, 2, 3. For similar reasons,
(i)

limx→−∞ f (x)f  (x) = 0.

First notice that

2f (x)f  (x) + 2f  (x)2 − 2f  (x)f  (x) > 0

because the first term plus the third term is nonnegative and the second term is
positive. Consequently,

2f (x)f  (x) − f  (x)2 > 0

since this expression has positive derivative and tends to 0 as x → −∞.

Combining this with f  (x) ≤ f (x) yields

2f (x)2 f  (x) − f  (x)2 f  (x) > 0,

from which we obtain

2f (x)3 − f  (x)3 > 2m3 ≥ 0

by similar reasoning, where m = limx→−∞ f (x). This inequality implies

21/3 f (x)f  (x) − f  (x)f  (x) > 0,

and integrating yields

21/3 f (x)2 − f  (x)2 > 21/3 m2 ≥ 0,

which implies the desired result. 

Remark. Let C be the infimum of the set of γ > 0 such that the conditions of
the problem imply f  (x) ≤ γf (x) for all x. Solution 3 shows that C ≤ 21/6 , and the
example f (x) = ex shows that C ≥ 1. One might guess from this that C = 1, but we
now give an example to show C > 1. (We do not know the exact value of C.)
We first construct an example where f  has a discontinuity at 0, then show that
f can be modified slightly while retaining the property that f  (x) > f (x) at some

Solutions: The Sixtieth Competition (1999) 275

point. Our strategy is to consider

,
ex , if x ≤ 0
H3 (x) =
e − cg(x),
x
if x > 0,
where
x g(x) is a solution of the differential equation g  = g, to set Hj (x) =
H (t) dt for j = 2, 1, 0, then to take f (x) = H0 (x). For x > 0, we have
−∞ j+1

H3 (x) = ex − cg  (x)

H2 (x) = ex − cg  (x) + cg  (0)
H1 (x) = ex − cg  (x) + c[g  (0) + g  (0)x]
7 8
1
H0 (x) = ex − cg(x) + c g(0) + g  (0)x + g  (0)x2 .
2
All solutions of g  = g are O(ex ) as x → ∞, so for c sufficiently small, H3 (x) > 0 for
all x > 0. Hence Hi (x) > 0 for x > 0, i = 2, 1, 0 as well. Now


 1 
H0 (x) − H3 (x) = c g(0) + g (0)x + g (0)x . 2
2
We make the right-hand side positive by taking g to be the solution to g  = g whose
Taylor polynomial of degree 2 equals (x − d)2 for some d > 0, For sufficiently small
c, the function f (x) = H0 (x) now satisfies the conditions of the problem, except for
continuity of the third derivative.
Also,
f  (d) − f (d) = H1 (d) − H0 (d) = c[g(d) − g  (d)].
If there exists d > 0 such that the corresponding g satisfies g(d) > g  (d), we have
f  (d) > f (d) as desired. A numerical computation shows that this occurs for d = 8.
To modify this example so that f  is continuous, choose δ > 0 and define h3 to
be the function that is equal  x to H3 outside (0, δ) and is linear between 0 and δ: see
Figure 41. Define hi (x) = −∞ hi+1 (t) dt for i = 2, 1, 0 and take f (x) = h0 (x). Since

x
y = e – cg(x)

x
y=e

FIGURE 41.
The graph of y = h3 (x).
276 The William Lowell Putnam Mathematical Competition

h3 (x) ≥ H3 (x) everywhere (for δ sufficiently small), we also have f (i) (x) = hi (x) ≥
Hi (x) > 0 for i = 2, 1, 0. For x > δ, we have f (x) ≥ H0 (x) ≥ H3 (x) = f  (x); for
0 < x < δ, we have f (x) ≥ f (0) = 1 ≥ h3 (x) = f  (x), provided that δ is small enough
that H3 (δ) < 1. Thus we have f  (x) ≤ f (x) for all x.

B5. (0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 43, 161)

For an integer n ≥ 3, let θ = 2π/n. Evaluate the determinant of the n × n
matrix I + A, where I is the n × n identity matrix and A = (ajk ) has entries
ajk = cos(jθ + kθ) for all j, k.
Remark. The total score of the top 205 writers in 1999 on this problem was 2. See
the comment at the beginning of 1999B4.
Answer. The determinant of I + A is 1 − n2 /4.

Solution 1. We compute the determinant of I + A by computing its eigenvalues.

The eigenvalues of I + A are obtained by adding 1 to each of the eigenvalues of A, so
it suffices to compute the latter.
We claim that the eigenvalues of A are n/2, −n/2, 0, . . . , 0, where 0 occurs with
multiplicity n − 2. To prove this claim, define vectors v (m) , 0 ≤ m ≤ n − 1,
componentwise by (v (m) )k = eikmθ (where θ = 2π/n). If we form a matrix from the
v (m) , its determinant is a Vandermonde product and hence is nonzero. (See 1986A6
for a short explanation of the Vandermonde determinant.) Thus the v (m) form a basis
for Cn . Since cos z = (eiz + e−iz )/2 for any z,
n
(m)
(Av )j = cos(jθ + kθ)eikmθ
k=1
n n
eijθ e−ijθ
= eik(m+1)θ + eik(m−1)θ .
2 2
k=1 k=1
n
Since k=1 eik!θ = 0 for integer @ unless n | @, we conclude that Av (m) = 0 for m = 0
and for 2 ≤ m ≤ n − 1. In addition, we find that (Av (1) )j = n2 e−ijθ = n2 (v (n−1) )j
and (Av (n−1) )j = n2 eijθ = n2 (v (1) )j , so A(v (1) ± v (n−1) ) = ± n2 (v (1) ± v (n−1) ). Thus
{v (0) , v (2) , v (3) , . . . , v (n−2) , v (1) + v (n−1) , v (1) − v (n−1) } is a basis for Cn of eigenvectors
of A with the claimed eigenvalues.
Finally, the determinant of I + A is the product of (1 + λ) over all eigenvalues λ of
A, namely,
det(I + A) = (1 + n/2)(1 − n/2) = 1 − n2 /4. 

Motivation. An n × n matrix A with the property that Ajk depends only on j − k

(mod n) is called a circulant matrix. Such a matrix always has v (0) , . . . , v (n−1) as
eigenvectors. In this problem, A is not circulant but A2 is. See Solution 3 to 1988B5
for more on circulant matrices.

Solution 2. As in Solution 1, to compute det(I + A), it suffices to compute the

eigenvalues of A. Let C be the vector with components (cos θ, cos 2θ, . . . , cos nθ), and
let D be the vector with components (sin θ, sin 2θ, . . . , sin nθ). If we identify C and D
Solutions: The Sixtieth Competition (1999) 277

with the corresponding n × 1 matrices, then the addition formula

cos(j + k)θ = cos jθ cos kθ − sin jθ sin kθ
implies the matrix identity A = CC T − DDT . Since CC T and DDT are matrices of
rank 1, A has rank at most 2. In fact, the image of A (as a linear transformation on
column vectors) is spanned by C and D. Thus the zero eigenspace of A has dimension
at least n − 2, and the remaining eigenspaces lie in the span of C and D.
Now
n
 2 
n= cos jθ + sin2 jθ = C · C + D · D
j=1

and
n
 2
0= e2πij n
j=1
n
 
= cos2 jθ − sin2 jθ + 2i cos jθ sin jθ
j=1

= C · C − D · D + 2iC · D.
Therefore, C · C = D · D = n/2 and C · D = 0. We now compute
 
AC = CC T − DDT C = C(C · C) − D(D · C) = (n/2)C
 
AD = CC T − DDT D = C(C · D) − D(D · D) = −(n/2)D
and conclude that the two remaining eigenvalues of A are n/2 and −n/2. 

B6. (3, 5, 5, 0, 0, 0, 0, 0, 3, 0, 80, 109)

Let S be a finite set of integers, each greater than 1. Suppose that for
each integer n there is some s ∈ S such that gcd(s, n) = 1 or gcd(s, n) = s.
Show that there exist s, t ∈ S such that gcd(s, t) is prime.
Solution. Let n be the smallest positive integer such that gcd(s, n) > 1 for all s
in n; note that n has no repeated prime factors. By the condition on S, there exists
s ∈ S which divides n.
On the other hand, if p is a prime divisor of s, then by the minimality of n, n/p is
relatively prime to some element t of S. Since n cannot be relatively prime to t, t is
divisible by p, but not by any other prime divisor of s (any such prime divides n/p).
Thus gcd(s, t) = p, as desired. 
Remark. The problem fails if S is allowed to be infinite. For example, let p1 , p2 , . . .
be distinct primes and let S = {p1 p2 , p3 p4 , p5 p6 , . . . }.
Remark. One can rephrase the problem in combinatorial terms, by replacing each
integer n with the sequence (n2 , n3 , n5 , . . . ), where np is the exponent of p in the
factorization of n.
278 The William Lowell Putnam Mathematical Competition

The Sixty-First William Lowell Putnam Mathematical Competition

December 2, 2000

A1. (61, 9, 5, 0, 0, 0, 0, 0, 14, 54, 41, 11) ∞

Let A be a positive real number. What are the possible values of x2j ,
∞ j=0
given that x0 , x1 , . . . are positive numbers for which j=0 xj = A?
Answer. The possible values comprise the interval (0, A2 ).
Solution 1. Since all terms in the series are positive, we may rearrange terms to
deduce " #2
0< x2i < x2i + 2xi xj = xi = A2 .
i<j
 2
Thus it remains to show that each number in (0, A2 ) is a possible value of x .
 i
We use geometric series. Given 0 < r < 1, there is a geometric series xi with
common ratio r and sum A: it has x0 /(1−r) = A so x0 = (1−r)A and xj = rj (1−r)A.
Then ∞
x20 1−r 2
x2j = = A .
j=0
1−r 2 1+r

To make this equal a given number B ∈ (0, A2 ), take r = (A2 − B)/(A2 + B). 
 2
Solution 2. As in Solution 1, we prove 0 < x < A2 , and it remains to show
2
 2i
that each number in (0, A ) is a possible value of xi . There exists a series of positive

numbers xi with sum A. Then x0 /2, x0 /2, x1 /2, x1 /2, x2 /2, . . . also sums to A
but its squares sum to half the previous sum of squares. Iterating shows that the sum
of squares can be arbitrarily small.

Given any series xi of positive terms with sum A, form a weighted average of it
with the series A + 0 + 0 + · · · ; in other words, choose t ∈ (0, 1) and define a new series

of positive terms yi by setting y0 = tx0 + (1 − t)A and yi = txi for i ≥ 1. Then

yi = A, and
yi2 = t2 x2i + 2t(1 − t)x0 A + (1 − t)2 A2 .
As t runs from 0 to 1, the Intermediate Value Theorem shows that this quadratic
 2  2
polynomial in t takes on all values strictly between xi and A2 . Since xi can be
 2
2
made arbitarily small, any number in (0, A ) can occur as yi . 

A2. (150, 1, 0, 0, 0, 0, 0, 0, 1, 0, 23, 20)

Prove that there exist infinitely many integers n such that n, n + 1, n + 2
are each the sum of two squares of integers. [Example: 0 = 02 +02 , 1 = 02 +12 ,
and 2 = 12 + 12 .]

In all of the following solutions, we take n = x2 − 1. Then n + 1 = x2 + 02 ,

n + 2 = x2 + 12 and it suffices to exhibit infinitely many x so that x2 − 1 is the sum
of two squares.
Solution 1. Let a be an even integer such that a2 + 1 is not prime. (For example,
choose a = 10k + 2 for some integer k ≥ 1, so that a2 + 1 = 100k 2 + 40k + 5 is divisible
by 5.) Then we can write a2 + 1 as a difference of squares x2 − b2 , by factoring a2 + 1
Solutions: The Sixty-First Competition (2000) 279

as rs with r ≥ s > 1, and setting x = (r + s)/2, b = (r − s)/2. (These are integers

because r and s must both be odd.) It follows that x2 − 1 is the sum of two squares
a2 + b2 , as desired. 
Solution 2. The equation u2 − 2v 2 = 1 is an example of Pell’s equation [NZM,
Section 7.8], so it has infinitely many solutions, and we can take x = u. 
Remark.
√ The√positive integer solutions to u2 −2v 2 = 1 are the pairs (u, v) satisfying
u + v 2 = (1 + 2)2n for some n ≥ 1. Incidentally, the positive possibilities for v are
the same as the odd terms in the sequence (an )n≥0 of Problem 1999A3.
Solution 3. Take x = 2a2 + 1; then x2 − 1 = (2a)2 + (2a2 )2 . 
Solution 4 (Abhinav Kumar). If x2 − 1 is the sum of two squares, then so is
x − 1, using
4

x4 − 1 = (x2 − 1)(x2 + 1)
and the identity
(a2 + b2 )(c2 + d2 ) = (ac − bd)2 + (ad + bc)2 (1)
(obtained by computing the norm of (a + bi)(c + di) in two ways). Hence by induction
n
on n, if x2 − 1 is the sum of two squares (for instance, if x = 3) then so is (x2 )2 − 1
for all nonnegative integers n. 
Related question. Let S = { a2 + b2 : a, b ∈ Z }. Show that S does not contain four
consecutive integers. But show that S does contain a (nonconstant) 4-term arithmetic
progression.
Related question. Does S contain an arithmetic progression of length k for every
integer k ≥ 1? This is currently an unsolved problem! A positive answer would follow
from either of the following conjectures:
• Dickson’s Conjecture. Given linear polynomials a1 n + b1 , . . . , ak n + bk in n, with
ai , bi ∈ Z and ai > 0 for all i, such that no prime p divides (a1 n + b1 ) · · · (ak n + bk )
for all n ∈ Z, there exist infinitely many integers n ≥ 1 such that the values
a1 n+b1 , . . . , ak n+bk are simultaneously prime. (The hypothesis that p not divide
(a1 n + b1 ) · · · (ak n + bk ) is automatic if p > k and p  gcd(a1 , b1 ) · · · gcd(ak , bk ), so
checking it for all p is a finite computation.)
The special case a1 = · · · = ak = 1 of Dickson’s Conjecture is the qualitative
form of the Hardy-Littlewood Prime k-tuple Conjecture, which itself is a gener-
alization of the Twin Prime Conjecture, which is the statement that there exist
infinitely many n ≥ 1 such that n and n + 2 are both prime. On the other hand,
Dickson’s Conjecture is a special case of “Hypothesis H” of Schinzel and Sierpiński,
in which the linear polynomials are replaced by distinct irreducible polynomials
f1 (n), . . . , fk (n) with positive leading coefficients. Moreover, for each of these
conjectures, there is a heuristic that predicts that the number of n less than or
equal to x satisfying the conclusion is (c + o(1))x/(ln x)k as x → ∞, where c is a
constant given by an explicit formula in terms of the k polynomials. (See 1988A3
for the definition of o(1).) The quantitative form of Hypothesis H is known as the
Bateman-Horn Conjecture. For more on all of these conjectures, see Chapter 6
of [Ri], especially pages 372, 391, and 409.
280 The William Lowell Putnam Mathematical Competition


• A conjecture of P. Erdős. If T is a set of positive integers such that n∈T 1/n
diverges, then T contains arbitrarily long finite arithmetic progressions. Erdős,
who frequently offered cash rewards for the solution to problems, offered his
highest-valued reward of $3000 for a proof or disproof of this statement. See [Gr1,
p. 24] or [Guy, p. 16].
Let us explain why either of the two conjectures above would imply that our set
S contains arithmetic progressions of arbitrary length. We will use the theorem
mentioned in 1991B5, that any prime congruent to 1 modulo 4 is in S.
Fix k ≥ 4, and let @i (n) = 4n + 1 + i(k!) for i = 1, 2, . . . , k. Let P (n) =
@1 (n)@2 (n) . . . @k (n). If p is prime and p ≤ k, then p does not divide P (0). If p is
prime and p > k, then for each i, the set { n ∈ Z : p | @i (n) } is a residue class
modulo p, so there remains at least one residue class modulo p consisting of n such
that p  P (n). Hence Dickson’s Conjecture predicts that there are infinitely many n
such that @1 (n), . . . , @k (n) are simultaneously prime. Each of these k primes would
be congruent to 1 modulo 4, and hence would be in S.
Now we show that the conjecture of Erdős implies that S contains arithmetic
progressions of arbitrary length, and even better, that the subset T of primes congruent

to 1 mod 4 contains such progressions. It suffices to prove that p∈T 1/p diverges,

or equivalently that lims→1+ p∈T 1/ps = ∞. The proof of Dirichlet’s Theorem on
primes in arithmetic progressions gives a precise form of this, namely:
 


1 : 1 1
lim  ln = .
s→1+ ps s−1 2
p∈T

1
Since lims→1+ ln s−1 = ∞, this implies lims→1+ p∈T 1/ps = ∞.
More generally, one says that a subset P of the set of prime numbers has Dirichlet
density α if  
1 :
1

lim   ln = α.
s→1+ ps s−1
p∈P

Dirichlet’s Theorem states that if a and m are relatively prime positive integers,
then the Dirichlet density of the set of primes congruent to a modulo m equals
1/φ(m), where φ(m) is the Euler φ-function defined in 1985A4. See Theorem 2 in
Chapter VI, §4 of [Se1] for details.

A3. (64, 20, 20, 0, 0, 0, 0, 0, 23, 9, 27, 32)

The octagon P1 P2 P3 P4 P5 P6 P7 P8 is inscribed in a circle, with the vertices
around the circumference in the given order. Given that the polygon
P1 P3 P5 P7 is a square of area 5 and the polygon P2 P4 P6 P8 is a rectangle of
area 4, find the maximum possible area of the octagon.
√
Answer. The maximum possible area is 3 5.

Solution.  (See Figure 42.) We deduce from the area of P1 P3 P5 P7 that the radius
of the circle is 5/2. If s > t are the sides of the rectangle P2 P4 P6 P√8 , then s2 +t2 =√10
and st = 4, so (s + t)2 = 18 and (s − t)2 = 2. Therefore s + t = 3 2 and s − t = 2,
Solutions: The Sixty-First Competition (2000) 281

P3

P2 P4

P5
P1

P6

P7

FIGURE 42.

√ √ √
2 2 and t = 2. Without loss of generality, assume that P2 P4 = 2 2
yielding s = √
and P4 P6 = 2.
For notational ease, let [Q1 Q2 · · · Qn ] denote the area of the polygon Q1 Q2 · · · Qn .
By symmetry, the area of the octagon can be expressed as
[P2 P4 P6 P8 ] + 2[P2 P3 P4 ] + 2[P4 P5 P6 ].
√
Note that [P2 P3 P4 ] is 2 times the distance from P3 to P2 P4 , which
√ is maximized when
P3 lies on the midpoint of arc P2 P4 ; similarly, [P4 P5 P6 ] is 2/2 times the distance
from P5 to P4 P6 , which is maximized when P5 lies on the midpoint of arc P4 P6 . Thus
the area of the octagon is maximized when P3 is the midpoint of arc P2 P4 and P5
is the midpoint of arc P4 P6 (in which case P1 P3 P5 P7 is indeed a square). In this
case, the distance
√ from P3 to P2 P4 equals the radius
√ of the circle minus half of P4 P6 ,
√2 P3 P4 ] = 5 − 1. Similarly [P4 P5 P6 ] = 5/2 − 1, so the area of the octagon
so [P
is 3 5. 

A4. (10, 2, 2, 0, 0, 0, 0, 0, 6, 2, 43, 130)

Show that the improper integral
 B
lim sin(x) sin(x2 ) dx
B→∞ 0
converges.
Solution 1. We may shift the lower limit to 1 without affecting convergence.
That done, we use integration by parts:
 B  B
sin x
sin x sin x2 dx = sin x2 (2x dx)
1 1 2x
B  B

sin x cos x sin x
=− cos x 2
+ − cos x2 dx.
2x 1 1 2x 2x2
282 The William Lowell Putnam Mathematical Competition

2x cos x tends to 0 as x → ∞, and the integral of 2x2 cos x converges

Now sin x 2 sin x 2

absolutely as B → ∞ by comparison to 1/x . It remains to consider

2

 B  B
cos x cos x
cos x2 dx = cos x2 (2x dx)
1 2x 1 4x2
 B
cos x B −2 cos x − x sin x
= 2
sin x 2
− sin x2 dx.
4x 0 1 4x3

4B 2 sin B → 0 as B → ∞, and the final integral converges absolutely as B → ∞

Now cos B 2

by comparison to the integral of 1/x2 . 

Solution 2. The addition formula for cosine implies that
1 
sin x sin x2 = cos(x2 − x) − cos(x2 + x) .
2
The
 ∞ substitution x = y + 1 transforms x2 − x into y 2 + y, so it suffices to show
 that
0
cos(x2
+ x) dx converges. Now substitute u = x2
+ x; then x = −1/2 + u + 1/4
and
 ∞  ∞
cos u
cos(x2 + x) dx =  du. (1)
0 0 2 u + 1/4
The latter integrand is bounded, so we may replace the lower limit of integration with
π/2 without affecting convergence. Moreover, the integrand tends to zero as u → ∞,
so the error introduced by replacing the upper limit of integration in
 B
cos u
 du
π/2 2 u + 1/4

by the nearest odd integer multiple of π/2 tends to zero as B → ∞. Therefore (1)
∞
converges if and only if n=1 an converges, where
 (n+ 12 )π  π/2
cos u cos t
an =  du = (−1) n
 dt.
(n− 2 )π 2 u + 1/4
1
−π/2 2 t + nπ + 1/4

The integrand in the last expression decreases to 0 uniformly as n → ∞. Therefore

an alternate in sign, tend to 0 as n → ∞ and satisfy |an | ≥ |an+1 |. By the alternating
∞
series test, n=1 an converges, so the original integral converges. 
Remark. One can also combine the first two solutions, by rewriting in terms of
cos(x2 + x) and then integrating by parts.
2
Solution 3. The integrand is the imaginary part of (sin x)eix , and sin x =
ix
e −e−ix
2i , so it suffices to show that
 ∞  ∞
ix2 +ix 2
e dx and eix −ix dx
0 0
ix2 ±ix
converge. The functions e are entire, so for any real B > 0, Cauchy’s
Theorem [Ah, p. 141] applied to the counterclockwise triangular path with vertices at
0, B, and B + Bi in the complex plane (Figure 43) yields
 B  B  B
ix2 ±ix i(t+ti)2 ±i(t+ti) 2
e dx = e (1 + i) dt − ei(B+ui) ±i(B+ui) i du. (2)
0 0 0
Solutions: The Sixty-First Competition (2000) 283

B + Bi

z = t (1 + i)
z = B + ti

z=t
0 B

FIGURE 43.

Since
2 2 2
±i(t+ti)
|ei(t+ti) | = e−2t ∓t
≤ e−t

for t ≥ 1, and
2
±i(B+ui)
|ei(B+ui) | = e−2Bu∓u ≤ e−u

once B ≥ 1, the two integrals on the right of (2) have finite limits as B → ∞, as
desired. 
Remark. By extending the previous analysis, we can evaluate
 ∞
2
I(t) = (sin tx)eix dx
0

for t ∈ R in terms of generalized hypergeometric functions, and hence we can evaluate

the integral of the problem, which is Im(I(1)). (A definition of the generalized
hypergeometric
∞ functions is given below; for more on these functions, see [O, p. 168].)
Since 0 e−2Bu+u du → 0 as B → ∞, the Cauchy’s Theorem argument implies that
the path of integration in  ∞
2
(sin x)eix dx
0

can be changed to the diagonal ray x = eiπ/4 y with y ranging from 0 to ∞, without
changing its value. More generally, this argument shows that for any t ∈ R,
 ∞
2
I(t) = sin(eiπ/4 ty)e−y eiπ/4 dy.
0

The substitution z = y 2 transforms this into


1 ∞ iπ/4 −1/2 √
e z sin(eiπ/4 t z)e−z dz.
2 0
We now expand in a power series in t and integrate term-by-term: this is justified
by the Dominated Convergence Theorem [Ru, p. 321], since the series of integrals of
absolute values of the terms,
∞  ∞ √ 2n+1 ∞
−1/2 |t z| n! |t|2n+1
z e−z dz = ,
n=0 0
(2n + 1)! n=0
(2n + 1)!
284 The William Lowell Putnam Mathematical Competition

converges for any real t, by the Ratio Test. (We used the identity
 ∞
z n e−z dz = Γ(n + 1) = n!.)
0

Thus
∞  ∞ ∞
1 ei(2n+2)π/4 (−1)n t2n+1 n −z 1 (−1)n in+1 n! t2n+1
I(t) = z e dz =
2 n=0 (2n + 1)! 0 2 n=0 (2n + 1)!
∞
t (−1)m (2m)! t4m
Im(I(t)) = .
2 m=0 (4m + 1)!

Pairing the factor j in (2m)! with the factor 2j of (4m + 1)!, and dividing numerator
and denominator by 2m factors of 4, which we distribute among the odd factors of
(4m + 1)! in the denominator, we obtain



t 3 5 t4
Im(I(t)) = F
1 2 1; , ; − ,
2 4 4 64
where the generalized hypergeometric function is defined by
∞
(a1 )m · · · (ap )m z m
p Fq (a1 , . . . , ap ; b1 , . . . , bq ; z) = · ,
m=0
(b1 )m · · · (bq )m m!

using the Pochhammer symbol

Γ(a + n)
(a)n = a(a + 1) · · · (a + n − 1) = .
Γ(a)
(One can similarly evaluate Re(I(t)).) In particular, the integral in the original
problem converges to



1 3 5 1
Im(I(1)) = 1F2 1; , ; − .
2 4 4 64

Remark. One can also show that

 ∞ 2
  π         
sin(bx) sin ax2 dx = cos b2 /4a C b2 /4a + sin b2 /4a S b2 /4a
0 2a
where the two Fresnel integrals are
 x
1 dt
C(x) = √ cos t √
2π 0 t
and  x
1 dt
S(x) = √ sin t √ .
2π 0 t
Similarly,
 ∞ 2
 2 π   2       
sin(bx) cos ax dx = sin b /4a C b2 /4a − cos b2 /4a S b2 /4a .
0 2a
Solutions: The Sixty-First Competition (2000) 285

If sin(bx) is replaced by cos(bx) in the above two integrals, then they can be evaluated
in terms of elementary functions:
 ∞ 2
1 π  
cos(bx) sin(ax2 ) dx = cos(b2 /4a) − sin(b2 /4a)
0 2 2a
 ∞ 2
2 1 π  
cos(bx) cos(ax ) dx = cos(b2 /4a) + sin(b2 /4a) .
0 2 2a
All of these integrals are originally due to Cauchy [Tal].
Remark. Oscillatory integrals like the one in this problem occur frequently in
analysis and mathematical physics. They can be bounded or evaluated by the methods
given here in some cases, and often can be estimated by a technique known as the
stationary phase approximation [CKP, Section 6.4]. This method was first used by
Stokes, to obtain an asymptotic representation for the function
 ∞
 
f (x) = cos x(ω 3 − ω) dω
0
valid for large positive real values of x.

A5. (30, 11, 4, 0, 0, 0, 0, 0, 2, 0, 19, 129)

Three distinct points with integer coordinates lie in the plane on a circle
of radius r > 0. Show that two of these points are separated by a distance
of at least r1/3 .
Solution 1. We will prove a stronger result, with r1/3 replaced by (4r)1/3 . Let A,
B, C be the three points. Examining
√ small triangles with integer coordinates shows
that either 4ABC has sides√1, 1, 2, or else some
√ side has length at least 2 and hence
r ≥ 1. In the first case, r = 2/2, so (4r)1/3 = 2 and we are done. So assume r ≥ 1.
If the minor arcs AB, BC, CA cover the circle, then one has measure at least 2π/3,
and the length of the corresponding chord is at least
√
2r sin(π/3) = r 3 > (4r)1/3 ,
where the last equality follows from r ≥ 1. Thus we may assume, without loss of
generality, that C lies in the interior of minor arc AB.
Since A, B, C have integer coordinates, the area K of 4ABC is half an integer (see
the first remark in 1990A3). Thus K ≥ 1/2. Let 2θ be the measure of minor arc AB.
Then the base AB of 4ABC is 2r sin θ, and the height is at most r − r cos θ, with
equality if and only if C is the midpoint of arc AB (see Figure 44). Thus
1
≤K
2
1
≤ (2r sin θ)(r − r cos θ)
2
r2 sin θ(1 − cos2 θ)
=
1 + cos θ
r sin3 θ
2
=
1 + cos θ
≤ r2 sin3 θ (since 0 < θ ≤ π/2),
286 The William Lowell Putnam Mathematical Competition

C
A B

r
q q

FIGURE 44.
The point C on arc AB maximizing the height of ABC in Solution 1 to 2000A5.

so sin θ ≥ 1/(2r2 )1/3 and

AB = 2r sin θ ≥ 2r/(2r2 )1/3 = (4r)1/3 . 

Related question. Examine Solution 1 to prove that if the longest side of√4ABC
has length equal to (4r)1/3 , it is an isosceles right triangle with sides 1, 1, 2. (We
will prove a stronger result soon.)
Solution 2. We will prove the result with r1/3 replaced by (2r)1/3 , using:
Lemma. If a, b, c are the lengths of sides of a triangle with area K and circumradius
r, then K = abc/(4r).
Proof. Let A, B, C be the vertices opposite the sides of length a, b, c, respectively.
If we view BC as base, the height is b sin C, so K = 12 ab sin C. Now use the Extended
Law of Sines,
sin A sin B sin C 1
= = = ,
a b c 2r
to replace sin C by c/(2r). 
Let a, b, c be the distances between the points. By the lemma, the area of the
triangle with the three points as vertices is abc/(4r). On the other hand, the area is
half an integer (see the first remark in 1990A3). Thus abc/(4r) ≥ 1/2, and
max{a, b, c} ≥ (abc)1/3 ≥ (2r)1/3 . 

Stronger result. In the notation of Solution 2, we prove that if r is sufficiently

large, then max{a, b, c} ≥ 2r1/3 .
If K ≥ 1, then the lower bound (4r)1/3 of Solution 1 is improved to 2r1/3 , and we
are done. Therefore assume K = 1/2. Without loss of generality, assume a ≤ b ≤ c.
Suppose that c < 2r1/3 . Then c3 < 8r = 8(abc/4K) = 4abc, so 4ab > c2 . If a ≤ 12 r1/3 ,
then abc < 12 r1/3 (2r1/3 )(2r1/3 ) = 2r, contradicting abc/(4r) = 1/2 (the lemma of
Solution 2). Thus 12 r1/3 < a ≤ b ≤ c < 2r1/3 .
Viewing AB as base of 4ABC, the height h equals 1/c, since K = 1/2. Let a
and b be the lengths of the projections of the sides of lengths a and b onto AB (see
Solutions: The Sixty-First Competition (2000) 287

C
✑PPP
✑ PP b
a ✑ PP
✑ h PP
✑ a b  PP
✑ A
B c

FIGURE 45.
Defining a and b , and constructing the parallelogram ACBD.

Figure 45). Since AB is the longest side of 4ABC, point C lies directly above the
segment AB, so a + b = c. By the Pythagorean Theorem,
%

 1

a = a −h = a 1− 2 2
2 2 2
a c




1 1
=a 1−O =a−O
a2 c2 ac2


1
=a−O 3 ,
c
and similarly b = b − O(1/c3 ).
If we form a parallelogram ACBD as in Figure 45, then D also has integer
coordinates, so

1 ≤ CD2 = (a − b )2 + (2h)2 = (a − b )2 + O(1/c2 ),

by the Pythagorean Theorem.

Now
  
4ab = 4 a + O(1/c3 ) b + O(1/c3 )
= 4a b + O(max{a , b }/c3 )
= 4a b + O(1/c2 )
= (a + b )2 − (a − b )2 + O(1/c2 )
 
≤ c2 − 1 − O(1/c2 ) + O(1/c2 )
= c2 − 1 + O(1/c2 ).

If r is sufficiently large, then c ≥ (4r)1/3 is also large, so this contradicts 4ab > c2 .

√ √we can prove that max{a, b, c} ≥ 2r

1/3
Stronger result. With a little
√ more work,
except when a, b, c are 1, 1, 2 or 1, 2, 5 (in some order). We continue with
the notation and results introduced in the previous proof. Let d = a − b , so
d2 + 4h2 = CD2 ≥ 1. Thus 1 − 4h2 ≤ d2 ≤ c2 . Also a = (c + d)/2 and b = (c − d)/2.
Recall that in a counterexample, K = 1/2 and 4ab > c2 . Thus
  
c4 < (4ab)2 = 16(a + h2 )(b + h2 ) = (c + d)2 + 4h2 (c − d)2 + 4h2 .
2 2
288 The William Lowell Putnam Mathematical Competition

For fixed c, the right-hand side is a monic quadratic in d2 so its maximum value for
d2 ∈ [1 − 4h2 , c2 ] is attained at an endpoint. In other words, the inequality remains
true either when d2 is replaced by 1−4h2 or when d2 is replaced by c2 . At d2 = 1−4h2 ,
the inequality becomes
c4 < (c2 + 2cd + 1)(c2 − 2cd + 1) = (c4 + 2c2 + 1) − 4c2 (1 − 4h2 ),
which, since h = 1/c, is equivalent to c2 < 17/2. At d2 = c2 , the inequality becomes
c4 < (4c2 + 4h2 )(4h2 ) = 16(1 + c−4 ),
which implies c4 < 17. Thus in either case, c2 < 17/2. But c2 is a sum of integer
squares, so c2 ∈ {1, 2, 4, 5, 8}. The only√lattice triangles
√ √ of area 1/2 with such a value
of c2 are those with side lengths 1, 1, 2 and 1, 2, 5.
Remark. The result just proved is nearly best possible. We now construct examples
with r → ∞ such that max{a, b, c} = 2r1/3 + O(r−1/3 ).
Let n be a large positive integer, and take the circle √passing  through (0, 0),
(n, 1), (2n + 1, 2). Then the three sides of the triangle are n2 + 1, (n + 1)2 + 1,

2
(2n + 1) + 4, and K = 1/2, so the lemma of Solution 2 implies
1 2
r= (n + 1)(n2 + 2n + 2)(4n2 + 4n + 5).
2
Thus
 1/6
2r1/3 = 2 (n2 + 1)(n2 + 2n + 2)(n2 + n + 5/4)
 1/6
= 2n (1 + n−2 )(1 + 2n−1 + 2n−2 )(1 + n−1 + (5/4)n−2 )
5
= 2n + 1 + n−1 + O(n−2 ),
6

and similarly

max{a, b, c} = (2n + 1)2 + 4 = 2n + 1 + n−1 + O(n−2 ),
so
1 1
max{a, b, c} = 2r1/3 + n−1 + O(n−2 ) = 2r1/3 + r−1/3 + O(r−2/3 ).
6 6
A6. (1, 2, 2, 0, 0, 0, 0, 0, 3, 19, 57, 111)
Let f (x) be a polynomial with integer coefficients. Define a sequence
a0 , a1 , . . . of integers such that a0 = 0 and an+1 = f (an ) for all n ≥ 0. Prove
that if there exists a positive integer m for which am = 0 then either a1 = 0
or a2 = 0.
Solution 1. Recall that if f (x) is a polynomial with integer coefficients, then m−n
divides f (m) − f (n) for any integers m and n. In particular, if we put bn = an+1 − an ,
then bn divides bn+1 for all n. On the other hand, we are given that a0 = am = 0,
which implies that a1 = am+1 and so b0 = bn . If b0 = 0, then a0 = a1 = · · · = am and
we are done. Otherwise, |b0 | = |b1 | = |b2 | = · · · , so bn = ±b0 for all n.
Now b0 + · · · + bm−1 = am − a0 = 0, so half of the integers b0 , . . . , bm−1 are positive
and half are negative. In particular, there exists an integer 0 < k < m such that
Solutions: The Sixty-First Competition (2000) 289

bk−1 = −bk , which is to say, ak−1 = ak+1 . From this it follows that an = an+2 for all
n ≥ k − 1; in particular, for n = m, we have
   
0 = am = am+2 = f f (am ) = f f (a0 ) = a2 . 

Solution 2. Choose m ≥ 1 minimal such that am = 0. If ai = aj for some

0 ≤ i < j ≤ m − 1, then m − j applications of f lead to am−(j−i) = am = 0,
contradicting the minimality of m. Hence, for arbitrary i and j, we have ai = aj if
and only if i − j is divisible by m.
If m = 1, we are done. Otherwise let ai > aj be the maximum and minimum
terms of a0 , . . . , am−1 . Then ai − aj divides f (ai ) − f (aj ) = ai+1 − aj+1 , which is
nonzero since (i + 1) − (j + 1) = i − j is not divisible by m. On the other hand,
|ai+1 − aj+1 | ≤ ai − aj because ai and aj are the maximum and minimum of all terms
in the sequence. Therefore equality must occur, which implies that ai+1 and aj+1
equal ai and aj in some order. If ai+1 = ai , then m = 1; otherwise ai+1 = aj implies
ai+2 = aj+1 , which with aj+1 = ai yields ai+2 = ai , so m divides 2. Hence m is 1 or
2. 
Remark. A special case of this problem (the fact that a3 = 0 implies a1 = 0 or
a2 = 0) was Problem 1 on the 1974 USA Mathematical Olympiad; the solution given
in [USAMO7286] is similar to Solution 1 above.
Literature note. By a shift of variable, it follows that for f (x) ∈ Z[x], if a ∈ Z is
such that a, f (a), f (f (a)), . . . is periodic, then the minimal period is at most 2. See
Chapter XII of [Na] for many other results of this type.
If one replaces Z everywhere with Q, then the period can be arbitrarily long: given
distinct a1 , . . . , an ∈ Q, Lagrange interpolation (see 1998B6) lets one construct a
polynomial f (x) ∈ Q[x] such that f (ai ) = ai+1 for 1 ≤ i ≤ n − 1 and f (an ) = a1 . On
the other hand, there might be a bound on the period in terms of the degree of the
polynomial f (x) ∈ Q[x]. Whether such a bound exists is unknown even in the case
of quadratic polynomials. For quadratic polynomials, Lagrange interpolation shows
that periods 1, 2, and 3 are possible; it is also true (but much harder to prove) that
periods 4 and 5 are not possible. See [Mort], [MS], [FPS], and [P2] for more details.

B1. (126, 14, 4, 0, 0, 0, 0, 0, 8, 14, 14, 15)

Let aj , bj , cj be integers for 1 ≤ j ≤ N . Assume, for each j, at least one of
aj , bj , cj is odd. Show that there exist integers r, s, t such that raj + sbj + tcj
is odd for at least 4N/7 values of j, 1 ≤ j ≤ N .

Solution. Consider the seven triples (r, s, t) with r, s, t ∈ {0, 1} not all zero. If
aj , bj , cj are not all even, then four of the sums raj + bsj + tcj with r, s, t ∈ {0, 1} are
even and four are odd. (For example, if aj is odd, then flipping r changes the sum
between even and odd, so half of the sums must be even and half odd.) Of course the
sum with r = s = t = 0 is even, so at least four of the seven triples with r, s, t not all
zero yield an odd sum. In other words, at least 4N of the quadruples (r, s, t, j) yield
odd sums. By the Pigeonhole Principle, there is a triple (r, s, t) for which at least
4N/7 of the sums are odd. 
290 The William Lowell Putnam Mathematical Competition

Reinterpretation. If (r, s, t) is chosen randomly from the seven triples with

(r, s, t) ∈ {0, 1} not all zero, then for each j, the probability that raj + sbj + tcj
is odd is 4/7, so the expected number of j for which this sum is odd equals 4N/7. For
some particular choice of (r, s, t), the number of such j must be greater than or equal
to the expectation.

B2. (114, 7, 2, 0, 0, 0, 0, 0, 2, 6, 35, 29)

Prove that the expression


gcd(m, n) n
n m
is an integer for all pairs of integers n ≥ m ≥ 1.
m n
Solution 1. Let a = and b = gcd(m,n)
gcd(m,n) . Then




a n m n n−1
= =
b m n m m−1
n n
is an integer, so b | a m . But gcd(a, b) = 1, so b | m . Hence



gcd(m, n) n 1 n
=
n m b m
is an integer. 
Solution 2. Since gcd(m, n) is an integer linear combination of m and n, it follows
that


gcd(m, n) n
n m
is an integer linear combination of the integers





m n n−1 n n n
= and =
n m m−1 n m m
and hence is itself an integer. 
Solution 3. To show that a nonzero rational number is an integer, it suffices to
check that the exponent of each prime in its factorization is nonnegative. So let p be
a prime, and suppose that the highest powers of p dividing n and m are pa and pb ,
respectively.
n If a ≤ b, then p has a nonnegative
n exponent in both gcd(m, n)/n and in
m . If a > b, it suffices to show that m is divisible by pa−b , but this follows from
Kummer’s Theorem, described in Solution 2 to 1997A5. 

B3. (5, 3, 3,1, 0, 0, 0, 16, 9, 26, 3, 129)

N
Let f (t) = j=1 aj sin(2πjt), where each aj is real and aN is not equal to 0.
dk f
Let Nk denote the number of zeros† (including multiplicities) of dtk
. Prove
that
N0 ≤ N1 ≤ N2 ≤ · · · and lim Nk = 2N.
k→∞

† The proposers intended for Nk to count only the zeros in the interval [0, 1).
Solutions: The Sixty-First Competition (2000) 291

df k
Solution. We first show Nk ≤ 2N for all k ≥ 0. Write f (k) (t) for dtk
. If we use
the identity sin x = (eix − e−ix )/(2i) and set z = e2πit , then
N
1
f (k) (t) = (2πij)k aj (z j − z −j )
2i j=1

is rewritten as z −N times a polynomial of degree 2N in z. Hence as a function of

z, it has at most 2N zeros. Therefore fk (t) has at most 2N zeros in [0, 1); that is,
Nk ≤ 2N . In particular, f (k) has at most finitely many zeros in [0, 1).
To prove Nk ≤ Nk+1 , we use Rolle’s Theorem, and the fact that at every zero of
f , f  has a zero of multiplicity one less. If 0 ≤ t1 < t2 < · · · < tr < 1 are the zeros
of f (k) in [0, 1), occurring with respective multiplicities m1 , . . . , mr , then f (k+1) has
at least one zero in each of the open intervals (t1 , t2 ), (t2 , t3 ), . . . , (tr−1 , tr ), (tr , t1 + 1);
we may translate the part [1, t1 + 1) of the last interval to [0, t1 ), on which f (k+1)
takes the same values. This gives r zeros of f (k+1) . Adding to these the multiplicities
m1 − 1, . . . , mr − 1 at t1 , . . . , tr , we find that f (k+1) has at least m1 + · · · + mr = Nk
zeros. Thus Nk+1 ≥ Nk .
To establish that Nk → 2N , it suffices to prove N4k ≥ 2N for sufficiently large k.
This we do by making precise the assertion that
N
f (4k) (t) = (2πj)4k aj sin(2πjt)
j=1

is dominated by the term with j = N at each point t = tm = (2m + 1)/(4N ) for

m = 0, 1, . . . , 2N − 1. At t = tm , the j = N term is
(2πN )4k aN sin(2πN tm ) = (−1)m (2πN )4k aN .
The absolute value of the sum of the other terms at t = tm , divided by the j = N
term, is bounded by

4k
4k
1 N −1
|a1 | + · · · + |aN −1 | ,
N N
which tends to 0 as k → ∞; in particular, this ratio is less than 1 for sufficiently large
k. Then f (4k) (tm ) has the same sign as (−1)m (2πN )4k aN . In particular, the sequence
f (4k) (t0 ), f (4k) (t1 ), . . . , f (4k) (t2N −1 )
alternates in sign, when k is sufficiently large. Between these points (again including
a final “wraparound” interval) we find 2N sign changes of f (4k) . By the Intermediate
Value Theorem , this implies N4k ≥ 2N for large k, and we are done. 
Remark. A similar argument was used in Solution 1 to 1989A3.
Remark. A more analytic proof that Nk → 2N involves the observation
that (2πN )−4k f (4k) (x) and its derivative converge uniformly to aN sin(2πN t) and
2πN aN cos(2πN t), respectively, on [0, 1]. That implies that for large k, we can divide
[0, 1] into intervals on which f (4k) (x) does not change sign, and intervals on which
f (4k+1) (x) does not change sign while f (4k) (x) has a sign change.
292 The William Lowell Putnam Mathematical Competition

B4. (14, 5, 1, 0, 0, 0, 0, 0, 3, 6, 106, 60)

Let f (x) be a continuous function such that f (2x2 − 1) = 2xf (x) for all x.
Show that f (x) = 0 for −1 ≤ x ≤ 1.

Solution 1. Since x = −1/2 and x = 1 satisfy 2x2 − 1 = x, the given equation

implies f (−1/2) = −f (−1/2) and f (1) = 2f (1), so f (−1/2) and f (1) both equal 0.
The former equation can be rewritten
 
f cos(2π/3 + 2πn) = 0

for any n ∈ Z. Taking x = cos θ in the functional equation shows that if f (cos 2θ) = 0
and cos θ = 0, then f (cos θ) = 0 as well. Therefore
"  #
f cos 2−k (2π/3 + 2πn) = 0

for all k, n ∈ Z with k ≥ 0. The numbers 2−k (2π/3+2πn) are dense in R, so continuity
of f (cos x) yields f (cos r) = 0 for all r ∈ R. Thus f is zero on [−1, 1]. 
Remark. We started with f (−1/2) = 0 instead of f (1) = 0 to avoid complications
arising from the condition cos θ = 0 in the iteration.

Solution 2. For t real and not a multiple of π, write g(t) = f (cos t)

sin t . Then
g(t + 2π) = g(t); furthermore, the given equation implies that for t not a multiple of
π/2,
f (2 cos2 t − 1) 2(cos t)f (cos t)
g(2t) = = = g(t).
2 sin t cos t 2 sin t cos t
In particular, for any integers n and k, we have
     
g 1 + nπ/2k−1 = g 2k + 2nπ = g 2k = g(1).

Since f is continuous, g is continuous where it is defined; but the set { 1 + nπ/2k−1 :

n, k ∈ Z } is dense in the reals (because π is irrational), and so g must be constant on
its domain. Since g(−t) = −g(t) for any t not a multiple of π, this constant must be
zero. Hence f (x) = 0 for x ∈ (−1, 1). Finally, setting x = 0 and x = 1 in the given
equation yields f (−1) = f (1) = 0 (which can also be deduced by continuity). 
Remark. A variation of the above solution starts by noting that for x = −1/2, we
have f (−1/2) = −f (−1/2) = 0, that is, f (cos(2π/3)) = 0. As above, we deduce that
f (cos θ) = 0 for θ = (2π/3 + 2n)/2k . These θ are dense in the interval [0, 2π].
Remark. We describe all functions f that satisfy the conditions of the problem: in
particular, we see that they need not be constant outside [−1, 1]. As shown above, f
must be zero on [−1, 1]. From the functional equation,

2xf (x) = f (2x2 − 1) = 2(−x)f (−x),

so f (x) = −f (−x); so it suffices to classify continuous functions f on [1, ∞) with

f (1) = 0 that satisfy the functional equation.
Given such a function, the function g(t) = f (coshsinh t
t)
on (0, ∞) is continuous and
satisfies g(2t) = g(t), so the function h(u) = g(eu ) is continuous on R and periodic with
period ln 2. Conversely, given h : R → R continuous of period ln 2, set g(t) = h(ln t) for
Solutions: The Sixty-First Competition (2000) 293
√
t > 0; then f (x) = g(cosh−1 x) x2 − 1 for x > 1 and f (1) = 0 define an f satisfying
the conditions of the problem. (Continuity of f at 1 follows because h is bounded.)

B5. (29, 16, 14, 0, 0, 0, 0, 0, 13, 1, 11, 111)

Let S0 be a finite set of positive integers. We define finite sets S1 , S2 , . . .
of positive integers as follows: the integer a is in Sn+1 if and only if exactly
one of a − 1 or a is in Sn . Show that there exist infinitely many integers N
for which SN = S0 ∪ { N + a : a ∈ S0 }.

Solution. We claim that all integers N of the form 2k , with k a positive integer
and N > max S0 , satisfy the desired condition.
It follows from the definition of Sn that

xj ≡ (1 + x) xj (mod 2).
j∈Sn j∈Sn−1
 
(When we write that two polynomials ai xi and bi xi in Z[x] are congruent modulo
2, we mean that ai ≡ bi (mod 2) for all i ≥ 0, or equivalently, that the two polynomials
are equal when considered in F2 [x]. Thus for instance, x2 ≡ x (mod 2) as polynomials,
even though m2 ≡ m (mod 2) holds for any particular integer m.) By induction on
n,

xj ≡ (1 + x)n xj (mod 2).

j∈Sn j∈S0

k
From the identity (x + y)2 ≡ x2 + y 2 (mod 2) and induction on n, we have (x + y)2 ≡
k k
x2 +y 2 (mod 2). (This also follows from Kummer’s Theorem, described in Solution 2
to 1997A5.) Hence if N = 2k for some k ≥ 1, then

xj ≡ (1 + xN ) xj .
j∈Sn j∈S0

If moreover N > max S0 , then SN = S0 ∪ { N + a : a ∈ S0 }, as desired. 

Remark. This solution can also be written in terms of binomial coefficients, without
reference to polynomials with mod 2 coefficients.
Remark. See 1989A5 for another problem involving generating functions modulo 2.
Related question. Problem 6 of the 1993 International Mathematical Olympiad
[IMO93] is related:
There are n lamps L0 , . . . , Ln−1 in a circle (n > 1), where we denote
Ln+k = Lk . (A lamp at all times is either on or off.) Perform steps s0 ,
s1 , . . . as follows: at step si , if Li−1 is lit, switch Li from on to off or vice
versa; otherwise do nothing. Initially, all lamps are on. Show that:
(a) There is a positive integer M (n) such that after M (n) steps all the lamps
are on again;
(b) If n = 2k , we can take M (n) = n2 − 1;
(c) If n = 2k + 1, we can take M (n) = n2 − n + 1.
294 The William Lowell Putnam Mathematical Competition

B6. (8, 1, 0, 0, 0, 0, 0, 0, 0, 0, 41, 145)

Let B be a set of more than 2n+1 /n distinct points with coordinates of
the form (±1, ±1, . . . , ±1) in n-dimensional space with n ≥ 3. Show that
there are three distinct points in B which are the vertices of an equilateral
triangle.
Solution. Let S be the set of points with all coordinates equal to ±1. For each
P ∈ B, let SP be the set of points in S which differ from P in exactly one coordinate.
Since there are more than 2n+1 /n points in B, and each SP has n elements, the
cardinalities of the sets SP sum to more than 2n+1 , which is to say, more than twice
the number of points in S. By the Pigeonhole Principle, there must be a point of S
in at least three of the sets, say in SP , SQ , SR . But then any two of P, Q,√
R differ in
exactly two coordinates, so P QR is an equilateral triangle of side length 2 2, by the
Pythagorean Theorem. 
Remark. Noam Elkies points out that when n = 2k , there exist sets of 2n+1 /n =
n−k+1
2 points having no subset of three points, each pair of which differ in two
coordinates. For example, for i = 1, . . . , k − 1, let Ti be the set of numbers in

{0, 1, . . . , n − 1} that, when expanded in base 2 as aj 2j , have ai = 1. Then take B
to be the set of points (e1 , . . . , en ) ∈ S with
+ + +
ej = ej = · · · = ej = 1.
j∈T1 j∈T2 j∈Tk−1

Now |B| = 2n−k+1 , since we can choose ej for all j except 2, 22 , . . . , 2k−1 , and then
3
the condition j∈Ti ej = 1 determines eti in terms of the ej already chosen. To see
that there cannot be three points in B, any pair of which differ in two coordinates,
suppose that such points exist; then there exist coordinates l, m, n at each of which
two of the three points are the same and the third is different. At least two of l, m, n
must have the same parity; suppose l and m have the same parity. Then there exists
i ∈ {1, . . . , k − 1} such that Ti contains one of l and m but not the other. Hence at the
3
two points that only differ in coordinates l and m, the values of j∈Ti ei are different.
This contradicts the definition of B, which says that both values should equal 1.
Remark. The construction of patterns such as in the previous remark is known as
design theory. This topic is the subject of [BJL].