# Difference between revisions of "2017 USAJMO Problems/Problem 1"

## Problem

Prove that there are infinitely many distinct pairs $(a,b)$ of relatively prime integers $a>1$ and $b>1$ such that $a^b+b^a$ is divisible by $a+b$.

## Solution 1

Let $a = 2^n - 1$ and $b = 2^n + 1$. We see that $a$ and $b$ are relatively prime (they are consecutive positive odd integers).

Lemma: $(2^k + 1)^{-1} \equiv 2^k + 1 \pmod{2^{k+1}}$.

Since every number has a unique modular inverse, the lemma is equivalent to proving that $(2^k+1)^2 \equiv 1 \pmod{2^{k+1}}$. Expanding, we have the result.

Substituting for $a$ and $b$, we have $$(2^k+1)^{2^k-1} + (2^k-1)^{2^k+1} \equiv 2^k - 1 + 2^k + 1 \equiv 0 \pmod{2^{k+1}},$$ where we use our lemma and the Euler totient theorem: $a^{\phi{n}} \equiv 1 \pmod{n}$ when $a$ and $n$ are relatively prime.

## Solution 2

Let $x$ be any odd number above 1 The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.