USA(J)MO 2025 draft 2025-03-22

Evan Chen

March 22, 2025

Problem 1 (JMO 2025/1). Prove that if f : Z → Z is any function, then there are
infinitely many integers c such that the function g(x) = f(x) + cx is not a bijection.

Assume for contradiction that there exists a finite “bad” set S such that f(x) + cx is
non-bijective for all c ∈
/ S.
The first observation is basically that given just f(0) and f(1), or any two consecutive
f-values, we can already find bad values of c.

Claim — The numbers f(0) − f(1), f(1) − f(2), f(2) − f(3), . . . are all in S. That
is, consecutive differences of f only take finitely many values.

Proof. To see that f(0) − f(1) ∈ S, just note that

x 7→ f(x) + (f(0) − f(1)) · x

is obviously not a bijective function, since it takes the same values at x = 0 and x = 1,
namely f(0). The same is true for other elements.

In particular, suppose M > maxs∈S |s|. Then the function

g(x) = f(x) + (M + 100)x

is supposed to be a bijection (because M + 100 ∈

/ S), and yet g(x + 1) − g(x) > 100 for
all x, which is a contradiction.

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Problem 2 (JMO 2025/4). Let n be a positive integer, and let a0 ≥ a1 ≥ · · · ≥ an ≥ 0

be integers. Prove that
n    
X ai 1 a0 + a1 + · · · + an
i ≤ .
2 2 2
i=0

Link: https://aops.com/community/p34335897

For n = 0 (which we permit) there is nothing to prove. Hence to prove by induction

on n, it would be sufficient to verify
     
an a0 + a1 + · · · + an a0 + a1 + · · · + an−1
2n ≤ − .
2 2 2

Rearranging the terms around, that’s equivalent to proving

⇐⇒ 2n(a2n − an ) ≤ a2n + an · (2(a0 + · · · + an−1 ) − 1)

⇐⇒ 0 ≤ 2an (a0 + · · · + an−1 − nan ) + an (an + 2n − 1).

However, the last line is obvious because min(a0 , . . . , an−1 ) ≥ an , and an ≥ 0.

Remark. The only equality case is when a0 ∈ {0, 1} and ai = 0 for i ≥ 1.

The bound in the problem is extremely loose and pretty much anything will work.

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Problem 3 (JMO 2025/6). Let S be a set of integers with the following properties:
• {1, 2, . . . , 2025} ⊆ S.
• If a, b ∈ S and gcd(a, b) = 1, then ab ∈ S.
• If for some s ∈ S, s + 1 is composite, then all positive divisors of s + 1 are in S.
Prove that S contains all positive integers.
Link: https://aops.com/community/p3532099

We prove by induction on N that S contains {1, . . . , N} with the base cases being
N = 1, . . . , 2025 already given.
For the inductive step, to show N + 1 ∈ S:
• If N + 1 is composite we’re already done from the third bullet.
• Otherwise, assume N + 1 = p ≥ 2025 is an (odd) prime number. We say a number
is good if the prime powers in its prime factorization are all less than p. Hence by
the second bullet (repeatedly), good numbers are in S. Now our proof is split into
three cases:
Case 1. Suppose neither p − 1 nor p + 1 is a power of 2 (but both are still even).
We claim that the number
s := p2 − 1 = (p − 1)(p + 1)
is good. Indeed, one of the numbers has only a single factor of 2, and the
other by hypothesis is not a power of 2 (but still even). So the largest power
of 2 dividing p2 − 2 is certainly less than p. And every other prime power
divides at most one of p − 1 and p + 1.
Hence s := p2 − 1 is good. As s + 1 = p2 , Case 1 is done.
Case 2. Suppose p + 1 is a power of 2; that is p = 2q − 1. Since p > 2025, we
assume q ≥ 11 is odd. First we contend that the number
  
s0 := 2q+1 − 1 = 2(q+1)/2 − 1 2(q+1)/2 + 1

is good. Indeed, this follows from the two factors being coprime and both less
than p. Hence s0 + 1 = 2q+1 is in S.
Thus, we again have
s := p2 − 1 = (p − 1)(p + 1) ∈ S
as we did in the previous case, because the largest power of 2 dividing p2 − 1
will be exactly 2q+1 which is known to be in S. And since s + 1 = p2 , Case 2
is done.
Case 3. Finally suppose p − 1 is a power of 2; that is p = 22 + 1 is a Fermat
e

prime. Then in particular, p ≡ 2 (mod 3). Now observe that

s := 2p − 1 ≡ 0 (mod 3)
and moreover 2p − 1 is not a power of 3 (it would imply 22 +1 + 1 = 3k , which
e

is impossible for k ≥ 3 by Zsigmondy/Mihailescu/etc.). So s is good, and

since s = 3p, Case 3 is done.
Having finished all the cases, we conclude p ∈ S and the induction is done.

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Remark. In fact just 2025 ∈ S is sufficient as a base case; however this requires a bit more
work to check. Here is how:
• From 2025 ∈ S we get 2026 = 2 · 1013, so 2, 1013 ∈ S.
• From 1013 + 1 = 1014 = 2 · 3 · 132 we get 3, 13 ∈ S.
• From 3 + 1 = 4 we get 4 ∈ S.
• 3 · 13 + 1 = 40 = 23 · 5 we get 5 ∈ S.
• Once {1, 2, . . . , 5} ⊆ S, the induction above actually works fine; that is, N ≤ 6 are
sufficient as base cases for the earlier cases to finish the rest of the problem. (Case 2
works once q ≥ 3, and Case 3 works once e ≥ 2.)
However, {1, 2, 3, 4} ⊆ S is not sufficient; for example S = {1, 2, 3, 4, 6, 12} satisfies all the
problem conditions.

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Problem 4 (USAMO 2025/1). Fix positive integers k and d. Prove that for all suffi-
ciently large odd positive integers n, the digits of the base-2n representation of nk are
all greater than d.

The problem actually doesn’t have much to do with digits: the idea is to pick any
length ` ≤ k, and look at the rightmost ` digits of nk ; that is, the remainder upon
division by (2n)` . We compute it exactly:

Claim — Let n ≥ 1 be an odd integer, and k ≥ ` ≥ 1 integers. Then

nk mod (2n)i = c(k, `) · ni

for some odd integer 1 ≤ c(k, `) ≤ 2` − 1.

Proof. This follows directly by the Chinese remainder theorem, with c(k, `) being the
residue class of n−k (mod 2` ) (which makes sense because n was odd).

In particular, for the `th digit from the right to be greater than d, it would be enough
that
c(k, `) · n` ≥ (d + 1) · (2n)`−1 .
But this inequality holds whenever n ≥ (d + 1) · 2`−1 .
Putting this together by varying `, we find that for all odd

n ≥ (d + 1) · 2k−1

we have that

• nk has k digits in base-2n; and

• for each ` = 1, . . . , k, the `th digit from the right is at least d + 1

so the problem is solved.

Remark. Note it doesn’t really matter that c(k, i) is odd per se; we only need that c(k, i) ≥
1.

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Problem 5 (USAMO 2025/2). Let n > k ≥ 1 be integers. Let P(x) ∈ R[x] be a

polynomial of degree n with no repeated roots and P(0) 6= 0. Suppose that for any real
numbers a0 , . . . , ak such that the polynomial ak xk + · · · + a1 x + a0 divides P(x), the
product a0 a1 . . . ak is zero. Prove that P(x) has a nonreal root.

By considering any k + 1 of the roots of P, we may as well assume WLOG that

n = k + 1. Suppose that P(x) = (x + r1 ) . . . (x + rn ) ∈ R[x] has P(0) 6= 0. Then the
problem hypothesis is that each of the n polynomials (of degree n − 1) given by

P1 (x) = (x + r2 )(x + r3 )(x + r4 ) . . . (x + rn )

P2 (x) = (x + r1 )(x + r3 )(x + r4 ) . . . (x + rn )
P3 (x) = (x + r1 )(x + r2 )(x + r4 ) . . . (x + rn )
..
.
Pn (x) = (x + r1 )(x + r2 )(x + r3 ) . . . (x + rn+1 )

has at least one coefficient equal to zero. We’ll prove that at least one ri is not real.
Obviously the leading and constant coefficients of each Pi are nonzero, and there are
n − 2 other coefficients to choose between. So by pigeonhole principle, we may assume,
say, that P1 and P2 share the position of a zero coefficient, say the xk one, for some
1 ≤ k < n − 1.

Claim — If P1 and P2 both have xk coefficient equal to zero, then the polynomial

Q(x) = (x + r3 )(x + r4 ) . . . (x + rn )

has two consecutive zero coefficients, namely bk = bk−1 = 0.

Proof. Invoking Vieta formulas, suppose that

Q(x) = xn−2 + bn−3 xn−3 + · · · + b0 .

(And let bn−2 = 1.) Then the fact that the xk coefficient of P1 and P2 are both zero
means
r1 bk + bk−1 = r2 bk + bk−1 = 0
and hence that bk = bk−1 = 0 (since the r − i are nonzero).

To solve the problem, we finish by noting:

Lemma
If F(x) ∈ R[x] is a polynomial with two consecutive zero coefficients, it cannot have
all distinct real roots.

Here are two possible proofs of the lemma I know (there are more).

First proof using Rolle’s theorem. Say xt and xt+1 coefficients of F are both zero.
Assume for contradiction all the roots of F are real and distinct. Then by Rolle’s
theorem, every higher-order derivative of F should have this property too. However, the
tth order derivative of F has a double root of 0, contradiction.

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Evan Chen USA(J)MO 2025 draft 2025-03-22

Second proof using Descartes rule of signs. The number of (nonzero) roots of F is bounded
above by the number of sign changes of F(x) (for the positive roots) and the number of
sign changes of F(−x) (for the negative roots). Now consider each pair of consecutive
nonzero coefficients in F, say ?xi and ?xj for i > j.

• If i − j = 1, then this sign change will only count for one of F(x) or F(−x)

• If i − j ≥ 2, then the sign change could count towards both F(x) or F(−x) (i.e.
counted twice), but also there is at least one zero coefficient between them.

Hence if b is the number of nonzero coefficients of F, and z is the number of consecutive

runs of zero coefficients of F, then the number of real roots is bounded above by

1 · (b − 1 − z) + 2 · z = b − 1 + z ≤ deg F .

However, if F has two consecutive zero coefficients, then the inequality is strict.

Remark. The final claim has appeared before apparently in the HUST Team Selection
Test for the Vietnamese Math Society’s undergraduate olympiad; see https://aops.com/
community/p33893374 for citation.
The claim also follows by Descartes rule of signs. If F

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Problem 6 (USAMO 2025/3). Alice the architect and Bob the builder play a game.
First, Alice chooses two points P and Q in the plane and a subset S of the plane, which
are announced to Bob. Next, Bob marks infinitely many points in the plane, designating
each a city. He may not place two cities within distance at most one unit of each other,
and no three cities he places may be collinear. Finally, roads are constructed between
the cities as follows: each pair A, B of cities is connected with a road along the line
segment AB if and only if the following condition holds:

For every city C distinct from A and B, there exists R ∈ S such that 4P QR
is directly similar to either 4ABC or 4BAC.

Alice wins the game if (i) the resulting roads allow for travel between any pair of cities via
a finite sequence of roads and (ii) no two roads cross. Otherwise, Bob wins. Determine,
with proof, which player has a winning strategy.

The answer is that Alice wins. Let’s define a Bob-set V to be a set of points in the
plane with no three collinear and with all distances at least 1. The point of the problem
is to prove the following fact.

Claim — Given a Bob-set V ⊆ R2 , consider the Bob-graph with vertex set V defined
as follows: draw edge ab if and only if the disk with diameter ab contains no other
points of V on or inside it. Then the Bob-graph is (i) connected, and (ii) planar.

Proving this claim shows that Alice wins since Alice can specify S to be the set of points
outside the disk of diameter P Q.

Proof that every Bob-graph is connected. Assume for contradiction the graph is discon-
nected. Let p and q be two points in different connected components. Since pq is not an
edge, there exists a third point r inside the disk with diameter pq.
Hence, r is in a different connected component from at least one of p or q — let’s
say point p. Then we repeat the same argument on the disk with diameter pr to find
a new point s, non-adjacent to either p or r. See the figure below, where the X’ed out
dashed edges indicate points which are not only non-adjacent but in different connected
components.

δ3 r
s

δ2
δ1
p q

In this way we generate an infinite sequence of distances δ1 , δ2 , δ3 , . . . among the

non-edges in the picture above. By the “Pythagorean theorem” (or really the inequality

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Evan Chen USA(J)MO 2025 draft 2025-03-22

for it), we have

δi2 ≤ δi−1
2
−1
and this eventually generates a contradiction for large i, since we get 0 ≤ δi2 ≤ δ12 − (i −
1).

Proof that every Bob-graph is planar. Assume for contradiction edges ac and bd meet,
meaning abcd is a convex quadrilateral. WLOG assume ∠bad ≥ 90◦ (each quadrilateral
has an angle at least 90◦ ). Then the disk with diameter bd contains a, contradiction.

Remark. In real life, the Bob-graph is actually called the Gabriel graph. Note that we
never require the Bob-set to be infinite; the solution works unchanged for finite Bob-sets.
However, there are approaches that work for finite Bob-sets that don’t work for infinite
sets, such as the relative neighbor graph, in which one joins a and b iff there is no c such
that d(a, b) ≤ max{d(a, c), d(b, c)}. In other words, edges are blocked by triangles where
ab is the longest edge (rather than by triangles where ab is the longest edge of a right or
obtuse triangle as in the Gabriel graph).
The relative neighbor graph has fewer edges than the Gabriel graph, so it is planar
too. When the Bob-set is finite, the relative distance graph is still connected. The same
argument above works where the distances now satisfy

δ1 > δ 2 > . . .

instead, and since there are finitely many distances one arrives at a contradiction.
However for infinite Bob-sets the descending condition is insufficient, and connectedness
actually fails altogether. A counterexample (communicated
√ to me by Carl Schildkraut) is
to start by taking An ≈ (2n, 0) and Bn ≈ (2n + 1, 3) for all n ≥ 1, then perturb all the
points slightly so that

B1 A1 > A1 A2 > A2 B1 > B1 B2 > B2 A2

> A2 A3 > A3 B2 > B2 B3 > B3 A3
> ··· .

A cartoon of the graph is shown below.

B1 2.06 B2 2.02 B3
...
9

1
2.0

2.0

2.0
2.0

2.0
7

...
A1 2.08 A2 2.04 A3

In that case, {An } and {Bn } will be disconnected from each other: none of the edges
An Bn or Bn An+1 are formed. In this case the relative neighbor graph consists of the edges
A1 A2 A3 A4 · · · and B1 B2 B3 B4 · · · . That’s why for the present problem, the inequality

δi2 ≤ δi−1
2
−1

plays such an important role, because it causes the (squared) distances to decrease appre-
ciably enough to give the final contradiction.

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Problem 7 (USAMO 2025/4). Let H be the orthocenter of an acute triangle ABC, let
F be the foot of the altitude from C to AB, and let P be the reflection of H across BC.
Suppose that the circumcircle of triangle AF P intersects line BC at two distinct points
X and Y. Prove that CX = CY.

Let Q be the antipode of B.

Claim — AHQC is a parallelogram, and AP CQ is an isosceles trapezoid.

Proof. As AH ⊥ BC ⊥ CQ and CF ⊥ AB ⊥ AQ.

Q
N

F
M
H O

B C

Let M be the midpoint of QC.

Claim — Point M is the circumcenter of 4AF P.

Proof. It’s clear that M A = M P from the isosceles trapezoid, so we show M A = M F.

Let N denote the midpoint of AH. Then M N is a midline of the parallelogram. Thus,
M N ⊥ AF.
However, we also know N A = N F from since 4AF H is right. So M N is the perpen-
dicular bisector of AF, as needed.

Since CM ⊥ BC and M is the center of (AF P), it follows CX = CY.

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Problem 8 (USAMO 2025/5). Find all positive integers k such that: for every positive
integer n, the sum
 k  k  k
n n n
+ + ··· +
0 1 n
is divisible by n + 1.

Link: https://aops.com/community/p34335836

The answer is all even k.

Let’s abbreviate S(n) := n k n k
for the sum in the problem.



0 + ··· + n

¶ Proof that even k is necessary. Choose n = 2. We need 3 | S(2) = 2 + 2k , which

requires k to be even.

Remark. It’s actually not much more difficult to just use n = p − 1 for prime p, since
p−1
(mod p). Hence S(p − 1) ≡ 1 + (−1)k + 1 + (−1)k + · · · + 1 (mod p), and

i
i ≡ (−1)
this also requires k to be even. This special case is instructive in figuring out the proof to
follow.

¶ Proof that k is sufficient. From now on we treat k as fixed, and we let pe be a prime
fully dividing n + 1. The basic idea is to reduce from n + 1 to (n + 1)/p by an induction.

Remark. Here is a concrete illustration that makes it clear what’s going on. Let p = 5.
When n = p − 1 = 4, we have

S(4) = 1k + 4k + 6k + 4k + 1k ≡ 1 + 1 + 1 + 1 + 1 ≡ 0 (mod 5).

When n = p2 − 1 = 24, the 25 terms of S(24) in order are, modulo 25,

S(24) ≡ 1k + 1k + 1k + 1k + 1k
+ 4k + 4k + 4k + 4k + 4k
+ 6k + 6k + 6k + 6k + 6k
+ 4k + 4k + 4k + 4k + 4k
+ 1k + 1k + 1k + 1k + 1k
= 5(1k + 4k + 6k + 4k + 1k ).

The point is that S(24) has five copies of S(4), modulo 25.

To make the pattern in the remark explicit, we prove the following lemma on each
individual binomial coefficient.

Lemma 9
Suppose pe is a prime power which fully divides n + 1. Then
 n+1
−1
  
n
≡± p (mod pe ).
i bi/pc

Proof of lemma. It’s easiest to understand the proof by looking at the cases bi/pc ∈
{0, 1, 2} first.

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Evan Chen USA(J)MO 2025 draft 2025-03-22

• For 0 ≤ i < p, since n ≡ −1 mod pe , we have

 
n n(n − 1) . . . (n − i + 1) (−1)(−2) . . . (−i)
= ≡ ≡ ±1 (mod pe ).
i 1 · 2 · ··· · i 1 · 2 · ··· · i

• For p ≤ i < 2p we have

 
n n − p + 1 (n − p)(n − p − 1) . . . (n − i + 1)
≡ ±1 · ·
i p (p + 1)(p + 2) . . . i
n−p+1
p
≡ ±1 · · ±1
1
 n+1
−1

≡± p (mod pe ).
1

• For 2p ≤ i < 3p the analogous reasoning gives

 
n n−p+1 n − 2p + 1
≡ ±1 · · ±1 · · ±1
i p 2p
  
n+1 n+1
p − 1 p − 2
≡±
1·2
 n+1
−1

≡± p (mod pe ).
2

. . .And so on. The point is that in general, if we write

  Y n − (j − 1)
n
=
i j
0≤j≤i

then the fractions for p - j are all ±1 (mod pe ). So only considers those j with p | j; in
n+1
−1

that case one obtains the claimed bi/pcp
exactly (even without having to take modulo
pe ).

From the lemma, it follows if pe is a prime power which fully divides n + 1, then
 
n+1
S(n) ≡ p · S −1 (mod pe )
p

by grouping the n + 1 terms (for 0 ≤ i ≤ n) into consecutive ranges of length p (by the
value of bi/pc).

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Problem 10 (USAMO 2025/6). Let m and n be positive integers with m ≥ n. There are
m cupcakes of different flavors arranged around a circle and n people who like cupcakes.
Each person assigns a nonnegative real number score to each cupcake, depending on how
much they like the cupcake. Suppose that for each person P, it is possible to partition
the circle of m cupcakes into n groups of consecutive cupcakes so that the sum of P’s
scores of the cupcakes in each group is at least 1. Prove that it is possible to distribute
the m cupcakes to the n people so that each person P receives cupcakes of total score
at least 1 with respect to P.

Link: https://aops.com/community/p34335840

Arbitrarily pick any one person — call her Pip — and her n arcs. The initial idea is
to try to apply Hall’s marriage lemma to match the n people with Pip’s arcs (such that
each such person is happy with their matched arc). To that end, construct the obvious
bipartite graph G between the people and the arcs for Pip.
We now consider the following algorithm, which takes several steps.

• If a perfect matching of G exists, we’re done!

• We’re probably not that lucky. Per Hall’s condition, this means there is a bad set
B1 of people, who are compatible with fewer than |B1 | of the arcs. Let’s imagine
deleting B1 and those neighbors of B1 , then try to find a matching on the remaining
graph.

• If a matching exists, terminate the algorithm. Otherwise, that means there’s an-
other bad set B2 for the remaining graph. We again delete B2 and the fewer than
B2 neighbors.

• Repeat until some perfect matching M is possible in the remaining graph, i.e. there
are no more bad sets (and then terminate once that occurs).
Since Pip is a universal vertex, it’s impossible to delete Pip, so the algorithm does
indeed terminate with nonempty M.

A cartoon of this picture is shown below.

People Arcs of Pip

Bad set B1

Bad set B2

Bad set B3

Final perfect matching M

Pip

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Evan Chen USA(J)MO 2025 draft 2025-03-22

We commit to assigning each of person in M their matched arc (in particular if there
are no bad sets at all, the problem is already solved). Now we finish the problem by
induction on n (for the remaining people) by simply deleting the arcs used up by M.
To see why this deletion-induction works, consider any particular person Q not in M.
By definition, Q is not happy with any of the arcs in M So when an arc A of M is
deleted, it had value less than 1 for Q, so in particular it couldn’t contain entirely any
of Q’s arcs. Hence exactly one endpoint among Q’s arcs was in the deleted arc A. This
causes two arcs of Q to merge, and the merged value is

(≥ 1) + (≥ 1) − (≤ 1) ≥ 1

meaning the induction is OK. See below for a cartoon of the deletion, where Pip’s arcs
are drawn in blue while Q’s arcs and scores are drawn in red (in this example n = 3).
Pip arc to delete

0.2 0.3

0.8 Q’s values 0.7 0.8 Q’s values 0.7

Pip

arc

Pip

arc
0.43 0.57 0.43 0.57
Pip

Pip
arc

arc

Remark. This deletion argument can be thought of in some special cases even before the
realization of Hall, in the case where M has only one person (Pip). This amounts to
saying that if one of Pip’s arcs isn’t liked by anybody, then that arc can be deleted and the
induction carries through.

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