, 2015

Notice that the function $f$, defined by $f(x)=x^{2}$ for all $x \in \mathbb{R}$, is a solution of the equation.

Assume that there exists $a \in \mathbb{R}$ such that $f(a) \neq a^{2}$ and let $b=a^{2}-f(a) \neq 0$. Then for all $x \in \mathbb{R}$ we have $f(x+b)=f(x)$. Therefore, for all $x, y \in \mathbb{R}$, let $z=\frac{y-x-b^{2}}{2b}$ we have
$$\begin{aligned} f(x) & =f\left(x+(z+b)^{2}-f(z+b)\right)=f\left(x+2bz+b^{2}+z^{2}-f(z)\right) \\ & =f\left(x+2bz+b^{2}\right)=f(y) \end{aligned}$$
We deduce that $f$ is constant. Conversely, any constant function satisfies the given functional equation.