Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy

$$
f(x+y)=f(x-y)+4 x y
$$

for all real numbers $x$ and $y$.
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.