2017 USAJMO Problems/Problem 2

Problem:

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

Solution

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.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

2017 USAJMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
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2017 USAJMO Problems/Problem 2

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Solution

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