# 2007 BMO Problems/Problem 2

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## Problem

(Bulgaria) Find all functions $\displaystyle f : \mathbb{R} \mapsto \mathbb{R}$ such that

$\displaystyle f(f(x) + y) = f(f(x) - y) + 4f(x)y$.

## Solution

We first note that $\displaystyle f(x) = 0$ is a solution to the equation. Henceforth we shall consider other solutions to the equations, i.e., functions $\displaystyle f$ such that for some $\displaystyle c$, $\displaystyle f(c) \neq 0$.

Setting $\displaystyle y = f(x)$ gives us $\displaystyle f(2f(x)) = f(0) + [f(x)]^2$.

We note that for any $\displaystyle x$, $\displaystyle f \left[ f(c) + \frac{x}{4f(c)} \right] - f \left[ f(c) - \frac{x}{4f(c)} \right] = f(c)\cdot \frac{x}{4f(c)} = x$, i.e., as $\displaystyle a$ and $\displaystyle b$ assume all real values, $\displaystyle f(b) - f(a)$ assume all real values.

Now, setting $\displaystyle x=b$ and $\displaystyle y = 2f(a) - f(b)$, we obtain

$\displaystyle f[2f(a)] = f\{ 2[f(b)-f(a)]\} + 4f(b)[2f(a)-f(b)]$,

or

$\displaystyle f\{ 2[f(b) - f(a)]\} = 4[f(a)]^2 +f(0) - 8f(a)f(b) + 4[f(b)]^2 + f(0) = \{ 2[f(b) - f(a)] \}^2 + f(0)$.

Since $\displaystyle 2[ f(b) - f(a) ]$ takes on all real values, it follows that for all $\displaystyle x$, $\displaystyle f(x) = x^2 + f(0)$. It is easy to see that any value of $\displaystyle f(0)$ will satisfy the desired condition. Thus the only solutions to the functional equation are $\displaystyle f(x) = 0$ and $\displaystyle f(x) = x^2 + c$, $\displaystyle c$ an arbitrary constant.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.