Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f(x+f(y))-f(x)=(x+f(y))^{4}-x^{4}
$$

for all $x, y \in \mathbb{R}$.
The solutions are $f(x)=0$ and $f(x)=x^{4}+k$ for any real constant $k$. These clearly satisfy the equation.

Assume $f$ is not the 0 function and pick any $y$ such that $f(y) \neq 0$. Then $(x+f(y))^{4}-x^{4}$ is a cubic polynomial in $x$. In particular, it attains every real value. So every real number has the form $f(x+f(y))-f(x)$ and in particular the form $f(a)-f(b)$ for some $a$ and $b$.

Substituting $(-f(b), a)$ and $(-f(b), b)$ into the equation,

$$
\begin{aligned}
f(f(a)-f(b))-f(-f(b)) & =(f(a)-f(b))^{4}-f(b)^{4} \\
f(0)-f(-f(b)) & =-f(b)^{4}
\end{aligned}
$$

and subtracting,

$$
f(f(a)-f(b))-f(0)=(f(a)-f(b))^{4} .
$$

But $f(a)-f(b)$ can equal an arbitrary real number $x$, so $f(x)=x^{4}+f(0)$ as desired.