# 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 $(2n \pm 1)^2 = 4n^2\pm4n+1 \equiv 1 \pmod{4n}$. 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 mathmaster2012)

## Solution 2

Let $x$ be odd where $x>1$. We have $x^2-1=(x-1)(x+1),$ so $x^2-1 \equiv 0 \pmod{2x+2}.$ 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 $x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},$ as desired.

## Solution 3

Because problems such as this usually are related to expressions along the lines of $x\pm1$, it's tempting to try these. After a few cases, we see that $\left(a,b\right)=\left(2x-1,2x+1\right)$ is convenient due to the repeated occurrence of $4x$ when squared and added. We rewrite the given expressions as: $$\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.$$ After repeatedly factoring the initial equation,we can get: $$\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).$$ Expanding each of the squares, we can compute each product independently then sum them: $$\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},$$ $$\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.$$ Now we place the values back into the expression: $$\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.$$ Plugging any positive integer value for $x$ into $\left(a,b\right)=\left(2x-1,2x+1\right)$ yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs $\left(a,b\right)$. $\square$

-fatant

## Solution 4

Let $a = 2x + 1$ and $b = 2x-1$, where $x$ leaves a remainder of $1$ when divided by $4$.We seek to show that $(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x$ because that will show 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 $a+b$.

Claim 1: $(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4$. We have that the remainder when $2x+1$ is divided by $4$ is $3$ and the remainder when $2x-1$ is divided by $4$ is always $1$. Therefore, the remainder when $(2x+1)^{2x-1} + (2x-1)^{2x+1}$ is divided by $4$ is always going to be $(-1)^{2x-1} + 1^{2x+1} = 0$.

Claim 2: $(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod x$ We know that $(2x+1) \mod x \equiv 1$ and $(2x-1) \mod x \equiv 3$, so the remainder when $(2x+1)^{2x-1} + (2x-1)^{2x+1}$ is divided by $4$ is always going to be $(-1)^{2x-1} + 1^{2x+1} = 0$.

Claim 3: $(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x$ Trivial given claim $1,2$. $\boxed{}$

~AopsUser101

## Solution 5

I claim $(a,b) = (2n-1,2n+1)$, $n (\in \mathbb{N}) \geq 2$ always satisfies above conditions.

Note: We could have also substituted 2n with 2^n or 4n, 8n, ... any sequence of numbers such that they are all even. The proof will work the same.

Proof:


Since there are infinitely many integers larger than or equal to 2, there are infinitely many distinct pairs $(a,b)$.

We only need to prove:

$a^b+b^a \equiv 0 \pmod{a+b}$

We can expand $a^b + b^a = (2n-1)^{2n+1} + (2n+1)^{2n-1}$ using binomial theorem. However, since $a + b = 2n-1 + 2n+1 = 4n$, all the $2n$ terms (with more than 2 powers of) when evaluated modulo $4n$ equal to 0, and thus can be omitted. We are left with the terms: $(2n+1)(2n)^1-1+(2n-1)(2n)^1+1 = 4n \cdot 2n$, which is divisible by $4n$.

$(2n-1)^{2n+1} + (2n+1)^{2n-1} \equiv (2n+1)(2n)-1+(2n-1)(2n)+1 = 4n \cdot 2n \equiv 0 \pmod{4n}$

The proof is complete. $\blacksquare$


-AlexLikeMath