# IMO 2019, Problem 1

The International Math Olympiad 2019 had the following question:

Find all functions $f:\Bbb{Z}\to \Bbb{Z}$ such that $f(2a)+2f(b)=f(f(a+b))$.

The reason that I decided to record this is because I thought I’d made an interesting observation that allowed me to solve the problem in only a couple of steps. However, I later realized that at least one other person has solved the problem the same way.

The right hand side is symmetric in $a,b$. Clearly, $f(f(a+b))=f(f(b+a))$. Hence, symmetrizing the left side as well, we get $f(2a)+2f(b)=f(2b)+2f(a)$. This implies that $f(2a)-f(2b)=2(f(a)-f(b))$. Assuming $b=0$, we get $f(2a)=2f(a)-f(0)$.

Now use $a=x+y$ and $b=0$ to show that $f(x)-f(0)$ is linear. This shows us that $f(x)=2x-f(0)$ or $f(x)=0$ are the only solutions to this question.