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 $(a,b) = (2n-1,2n+1)$ always satisfies the conditions, n is integer, $n \geq 2$ ,

Proof:

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

Since there are infinitely many integers larger than or equal to 2, the proof is complete . $\blacksquare$

-AlexLikeMath

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

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