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.

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Graduate student

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