Berkeley Math Circle

Solution:
The answer is the functions $f(x)=x^{2}+c$, where $c$ is a constant. It is easy to check that all such functions work, since
$$(x+y)^{2}+c=(x-y)^{2}+c+4 x y$$
is trivially true.

Now, we prove these are the only functions. Put $x=y=\frac{1}{2} a$ to obtain
$$f(a)=f(0)+a^{2}$$
for all real numbers $a$, which implies the conclusion.