IMO 2019, Problem 1

-November 21, 2020November 21, 2020Uncategorized

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 $f(2a)=2f(a)-f(0)$ .