# 2018 USAMO Problems/Problem 3

## Problem 3

For a given integer $n\ge 2,$ let $\{a_1,a_2,…,a_m\}$ be the set of positive integers less than $n$ that are relatively prime to $n.$ Prove that if every prime that divides $m$ also divides $n,$ then $a_1^k+a_2^k + \dots + a_m^k$ is divisible by $m$ for every positive integer $k.$

## Solution (NO FORMATTING)

https://maa.org/sites/default/files/pdf/AMC/usamo/2018/2018USAMO.pdf The integer m in the statement of the problem is ϕ(n), where ϕ is the Euler totient function. Throughout our proof we write p s || m, if s is the greatest power of p that divides m. We begin with the following lemma: Lemma 1. If p is a prime and p s divides n for some positive integer s, then 1k + 2k + · · · + n k is divisible by p s−1 for any integer k ≥ 1. 2018 USAMO – Solutions 4 Proof. Let {a1, a2, . . . , am} be a complete reduced residue set modulo p s and m = p s−1 (p − 1). First we prove by induction on s that for any positive integer k, a k 1 + a k 2 + · · · + a k m is divisible by p s−1 . The base case s = 1 is true. Suppose the statement holds for some value of s. Consider the statement for s + 1. Note that {a1, . . . , am, ps + a1, . . . , ps + am, . . . , ps (p − 1) + a1, . . . , ps (p − 1) + am} is a complete reduced residue set modulo p s+1. Therefore, the desired sum of k-th powers is equal to a k 1 + · · · + a k m + · · · + (p s (p − 1) + a1) k + · · · + (p s (p − 1) + am) k ≡ p(a k 1 + · · · + a k m) ≡ 0 (mod p s ), where we have used the induction hypothesis for the second congruence. This gives the induction step. Now we are ready to prove the lemma. Because numbers from 1 to n can be split into blocks of consecutive numbers of length p s , it is enough to show that 1k + 2k +· · · + (p s ) k is divisible by p s−1 for any positive integer k. We use induction on s. The statement is true for s = 1. Assume the statement is true for s − 1. The sum 1 k + 2k + · · · + (p s ) k = a k 1 + a k 2 + · · · + a k m + p k � 1 k + 2k + · · · + (p s−1 ) k � is divisible by p s−1 , because p s−1 | a k 1 + · · · + a k m and by the induction hypothesis p s−2 | 1 k + · · · + (p s−1 ) k . Now we proceed to prove a second lemma, from which the statement of the problem will immediately follow: Lemma 2. Suppose p is a prime dividing n. Let {a1, . . . , am} be a complete reduced residue set mod n, and define s by p s || m. Then p s divides a k 1 + · · · + a k m for any integer k ≥ 1. Proof. We fix p, and use induction on the number of prime factors of n (counted by multiplicity) that are different from p. If there are no prime factors other than p, then n = p s+1 , m = p s (p − 1), and we proved in Lemma 1 that a k 1 +· · ·+a k m is divisible by p s . Now suppose the statement is true for n. We show that it is true for nq, where q is a prime not equal to p. Case 1. q divides n. We have p s || ϕ(n) and p s || ϕ(nq), because ϕ(nq) = qϕ(n). If {a1, a2, . . . , am} is a complete reduced residue set modulo n, then {a1, . . . , am, n + a1, . . . , n + am, . . . , n(q − 1) + a1, . . . , n(q − 1) + am} is a complete reduced residue set modulo nq. The new sum of k-th powers is equal to a k 1 + · · · + a k m + · · · + (n(q − 1) + a1) k + · · · + (n(q − 1) + am) k = mnk � 1 k + · · · + (q − 1)k � + 2018 USAMO – Solutions 5 � k 1 � n k−1 � 1 k−1 + · · · + (q − 1)k−1 � (a1 + · · · + am) + · · · + q(a k 1 + · · · + a k m). This sum is divisible by p s because p s || m and p s | a j 1 + a j 2 + · · · + a j m for any positive integer j. Case 2. q doesn’t divide n. Suppose p b || q − 1, where b ≥ 0. Note that ϕ(nq) = ϕ(n)(q − 1), so p s || ϕ(n) and p s+b || ϕ(nq). Let {a1, . . . , am} be a complete reduced residue set modulo n. The complete reduced residue set modulo nq consists of the mq numbers {a1, . . . , am, n + a1, . . . , n + am, . . . , n(q − 1) + a1, . . . , n(q − 1) + am} with the m elements {qa1, qa2, . . . , qam} removed. The new sum of k-th powers is equal to a k 1 + · · · + a k m + · · · + (n(q − 1) + a1) k + · · · + (n(q − 1) + am) k − q k (a k 1 + · · · + a k m) = mnk � 1 k + · · · + (q − 1)k � + � k 1 � n k−1 � 1 k−1 + · · · + (q − 1)k−1 � (a1 + · · · + am) + · · · · · · + � k k − 1 � n (1 + · · · + (q − 1)) (a k−1 1 + · · · + a k−1 m ) + q(a k 1 + · · · + a k m) − q k (a k 1 + · · · + a k m). Each term � k j � n k−j � 1 k−j + · · · + (q − 1)k−j � (a j 1 + · · · + a j m), for 0 ≤ j ≤ k − 1, is divisible by p s+b because p | n k−j , p s | a j 1 + · · · + a j m, and p b−1 | 1 k−j + · · · + (q − 1)k−j by Lemma 1. Also (q k − q)(a k 1 + · · · + a k m) is divisible by p s+b because p b | q − 1 | q k − q and p s | a k 1 + · · · + a k m. Thus p s+b divides our sum and our proof is complete. Remark. In fact, one can also show the converse statement: if {a1, a2, . . . , am} is as defined in the problem and a k 1 + a k 2 + · · · + a k m is divisible by m for every positive integer k, then every prime that divides m also divides n. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

 2018 USAMO (Problems • Resources) Preceded byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 All USAMO Problems and Solutions