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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 $f\equiv0$.

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What is the motivation behind the equation? It looks pretty random... – Mariano Suárez-Alvarez Jan 26 '11 at 23:28
Looks like $x^2 + 1$ is also a solution. Of course, I might have screwed up the algebra. I am guessing $x^2 + C$ will also be a solution... – Aryabhata Jan 26 '11 at 23:34
@Moron You are correct. – Benji Jan 26 '11 at 23:42
@Mariano Suárez-Alvarez: I was given this problem by a friend; it may very well be contrived. – Benji Jan 26 '11 at 23:43
@bobobinks: One observation is if $z = f(y)$, then $f(2z) = 4z^2 + f(0)$. This can be seen by plugging $x= f(y)$ in the functional equation. – user17762 Jan 27 '11 at 0:04
up vote 7 down vote accepted

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 $C$.

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