Iranian Mathematical Olympiad

Let $P(x,y)$ denote the assertion given in the statement of the problem. If there exists $a$ such that $f(a)=0$, $P(a,0)$ results in $a=0$. Now $$P(x,x-y)\to f(2x-y)f(x^2-xy+y^2)=(2x-y)(x^2-xy+y^2)$$ By dividing this equation by $P(x,y)$ (where $x+y\ne 0$), it is obtained that $$\frac{f(2x-y)}{f(x+y)}=\frac{2x-y}{x+y}(1)$$ In equation (1) put $y=1-x$. This will result in $f(3x-1)=(3x-1)f(1)$ and because $3x-1$ is surjective over real numbers, we conclude that $f(x)=xf(1)$. By putting this equality in the original equality, it is deduced that $f(1)=\pm 1$ so functions $f(x)=x$ and $f(x)=-x$ are the only answers. ■