# Difference between revisions of "2019 USAJMO Problems/Problem 2"

## Problem

Let $\mathbb Z$ be the set of all integers. Find all pairs of integers $(a,b)$ for which there exist functions $f:\mathbb Z\rightarrow\mathbb Z$ and $g:\mathbb Z\rightarrow\mathbb Z$ satisfying $$f(g(x))=x+a\quad\text{and}\quad g(f(x))=x+b$$ for all integers $x$.

## Solution

$f$ and $g$ are surjective because $x+a$ and $x+b$ can take on any integral value, and by evaluating the parentheses in different order, we find $f(g(f(x)))=f(x+b)=f(x)+a$ and $g(f(g(x)))=g(x+a)=g(x)+b$. We see that if $a=0$ then $g(x)=g(x)+b$ to $b=0$ as well, so similarly if $b=0$ then $a=0$, so now assume $a, b\ne 0$.

We see that if $x=|b|n$ then $f(x)\equiv f(0) \pmod{|a|}$, if $x=|b|n+1$ then $f(x)\equiv f(1)\pmod{|a|}$, if $x=|b|n+2$ then $f(x)\equiv f(2)\pmod{|a|}$... if $x=|b|(n+1)-1$ then $f(x)\equiv f(|b|-1)\pmod{|a|}$. This means that the $b$-element collection $\{f(0), f(1), f(2), ... ,f(|b|-1)\}$ contains all $|a|$ residues mod $|a|$ since $f$ is surjective, so $|b|\ge |a|$. Doing the same to $g$ yields that $|a|\ge |b|$, so this means that only $|a|=|b|$ can work.

For $a=b$ let $f(x)=g(x)=x+\frac{a}{2}$ and for $a=-b$ let $f(x)=-x+\frac{a}{2}$ and $g(x)=-x-\frac{a}{s}$, so $|a|=|b|$ does work.

-Stormersyle