Find all functions $f:R\rightarrow R$ which satisfy $f(x+f(y))=f(x-f(y))+4xf(y)$ $\forall x,y \in R$.
I strongly suspect $0$ and $x^2+C$ to be the only solutions but, as is almost the case with functional equations, finding the set the solutions is easy; it is proving that a certain set of solutions represents ALL solutions is what is difficult.
It is trivial that if $f$ is bounded $f\equiv0$.
Ok I finally got it: the equation $f(x + f(y)) = f(x - f(y)) + 4xf(y)$ can be rewritten as $$f(x + f(y)) - f(x - f(y)) = (x + f(y))^2 - (x - f(y))^2$$ Replacing $x$ by $x + f(y)$ and rearranging terms we get $$f(x + 2f(y)) - (x + 2f(y))^2 = f(x) - x^2$$ Thus $f(x) - x^2$ has period $2f(y)$ for all $y$. In particular, it has period $2f(a + f(b))$ as well as $2f(a - f(b))$ for all $a$ and $b$. Since the difference of two periods is a period, it also has period $2f(a + f(b)) - 2f(a - f(b))$ for any $a$ and $b$, which is $8af(b)$ by the conditions given. Assuming $f(x)$ is not the zero function, we may pick $b$ so that $f(b)$ is not zero. Since $a$ can be anything, we conclude $f(x) - x^2$ has every real number as a period; in other words it is constant. So if $f(x)$ is not the zero function, then $f(x) = x^2 + C$ for some constant $C$.