2014 USAMO Problems/Problem 2
Problem
Let be the set of integers. Find all functions such that for all with .
Solution
Note: This solution is kind of rough. I didn't want to put my 7-page solution all over again. It would be nice if someone could edit in the details of the expansions.
Lemma 1: . Proof: Assume the opposite for a contradiction. Plug in (because we assumed that ), . What you get eventually reduces to: which is a contradiction since the RHS is divisible by 2 but not 4.
Then plug in into the original equation and simplify by Lemma 1. We get: Then:
$\begin{align*} x^6f(x) &= x^4(-x)^2f(-(-x)) \\ &= x^4f(-x)^2 \\ &= f(x)^4 \end{align*}$ (Error compiling LaTeX. Unknown error_msg)
Therefore, must be 0 or .
Now either is for all or there exists such that . The first case gives a valid solution. In the second case, we let in the original equation and simplify to get: But we know that , so: Since is not 0, is 0 for all (including 0). Now either is 0 for all , or there exists some such that . Then must be odd. We can let in the original equation, and since is 0 for all , stuff cancels and we get: [b]for .[/b] Now, let and we get: Now, either both sides are 0 or both are equal to . If both are then: which simplifies to: Since and is odd, both cases are impossible, so we must have: Then we can let be anything except 0, and get is 0 for all except . Also since , we have , so is 0 for all except . So is 0 for all except . Since , . Squaring, and dividing by , . Since , , which is a contradiction, so our only solutions are and .