12th Czech-Polish-Slovak Mathematics Competition

We rewrite the given equation into the equivalent form $$f(x+f(y))=(x+f(y){)}^4-x^4+f(x).(1)$$ Setting $x=-f(z)$, $y=z$ in (1) we obtain $$f(0)=-(f(z){)}^4+f(-f(z))\text{for all }z\in \mathbb{R}.(2)$$ 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.