# Imo small question about the functional equation

1. Mar 31, 2013

### Andrax

1. The problem statement, all variables and given/known data
as some of you might've done it this is the functional eqUATION FROM THE IMO 2012 / a + b + c = 0 f2(a)+f2(b)+f2(c)=2f(a)f(b)+2f(b)f(c)+2f(c)f(a). f:Z->Z http://www.cut-the-knot.org/arithmetic/algebra/2012IMO-4.shtml <- link of the problem and its SOLUTION
now i worked with this equatio nyesterday and i just need someone to correct me
setting 0 0 0 just like the solution is doing we get f(0) = 0 now setting a -a 0 gives us (f(a)-f(-a))^2=0 thus f(a)=f(-a) this is the part where i get reall confused , why cant we just conclude that all even functions are the solution ,well it dosen't work for all even functions which is weird since we got f(a)=f(-a) can someone please tell me what im doing wrong here?

2. Relevant equations

3. The attempt at a solution

2. Apr 2, 2013

### haruspex

Function f has property P. You deduce from this that it must be an even function. How does it follow that all even functions have property P?