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 $a = 2n-1$ and $b = 2n+1$. We see that $(2n \pm 1)^2 = 4n^2-4n+1 \equiv 1 \pmod{4n}$. Therefore, we have $(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}$, as desired.

(Credits to laegolas)

Solution 2

Let $x$ be any odd number above 1. We have $x^2-1=(x-1)(x+1).$ Since $x-1$ is even, $x^2-1 \equiv 0 \pmod{2x+2}.$ This means that $x^{x+2}-x^x \equiv 0 \pmod{2x+2},$ and since x is odd, $x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},$ or $x^{x+2}+x+2^x \equiv 0 \pmod{2x+2}.$ This means for any odd x, the ordered triple $(x,x+2)$ satisfies the condition. Since there are infinitely many values of $x$ possible, there are infinitely many ordered triples, as desired.

(Credits to kingofgeedorah, 2016 MATHCOUNTS Champion)

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

See also

 2017 USAJMO (Problems • Resources) First Problem Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 All USAJMO Problems and Solutions
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