Czech-Polish-Slovak Mathematics Competition

We rewrite the given equation into the equivalent form
$$f(x + f(y)) = (x + f(y))^4 - x^4 + f(x). \quad (1)$$
Setting $x = -f(z)$, $y = z$ in (1) we obtain
$$f(0) = -(f(z))^4 + f(-f(z)) \quad \text{for all } z \in \mathbb{R}. \quad (2)$$
Now, setting $x = -f(z)$ in (1) and using (2) we get
$$f(f(y) - f(z)) = (f(y) - f(z))^4 - (f(z))^4 + f(-f(z)) = (f(y) - f(z))^4 + f(0)$$
for all $y, z \in \mathbb{R}$. This means that if a number $t$ can be expressed as the difference of two values of $f$, that is $t = f(y) - f(z)$, then $f(t) = t^4 + f(0)$. We show that if $f$ takes any nonzero value then every number is a difference of two values of $f$.
Let $f(a) = b \ne 0$. Putting $y = a$ in the original equation we have
$$f(x + b) - f(x) = (x + b)^4 - x^4.$$
Since $b \ne 0$, the expression on the right hand side is a polynomial of degree 3, and therefore takes every real number as its value when $x$ run over the entire real axis. Hence, the left hand side, which is the difference of two values of $f$, can take any real value. Together with the previous observation we get $f(t) = t^4 + f(0)$ for all $t \in \mathbb{R}$. Finally, one can easily check that all functions of the form $f(x) = x^4 + k$ satisfy the given functional equation.
The zero function is obviously a solution as well.