Imo small question about the functional equation

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SUMMARY

The discussion revolves around the functional equation from the IMO 2012, specifically the equation f: Z -> Z defined by a + b + c = 0 and f²(a) + f²(b) + f²(c) = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a). The user successfully deduces that f(0) = 0 and that f(a) = f(-a), indicating that f is an even function. However, confusion arises regarding the conclusion that not all even functions satisfy the original functional equation, highlighting a common misconception in functional equations.

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Homework Statement


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 can't we just conclude that all even functions are the solution ,well it doesn't work for all even functions which is weird since we got f(a)=f(-a) can someone please tell me what I am doing wrong here?

Homework Equations


The Attempt at a Solution

 
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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?
 

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