x=-f(y) we have f(0)=f(x)-x^4 so we have f(x)=x^4+f(0).(f(0)=c)so f(x)=x^4+c.and if you test this answer you can see that it is true so the answer is f(x)=(x^4)+c which c is a rational number

Sayan wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying \[f(x+f(y))-f(x)=(x+f(y))^4-x^4\] for all $x,y \in \mathbb{R}$ $f(x)=0$ $\forall x$ is a solution. So let us from now look only for non all zero solutions. Let $P(x,y)$ be the assertion $f(x+f(y))-f(x)=(x+f(y))^4-x^4$ Let $a=f(0)$ Let $u$ such that $f(u)\ne 0$ and let then $v=f(u)\ne 0$ The equation $(x+v)^4-x^4=t$ is a polynomial of degree $3$ (since $v\ne 0$) and so always has at least one real solution. Then $P(x,u)$ $\implies$ $f(x+v)-f(x)=t$ and so any real $t$ may be written as $f(r)-f(s)$ for some other real numbers $r,s$ $P(-f(y),y)$ $\implies$ $f(-f(y))=f(y)^4+a$ $P(-f(y),x)$ $\implies$ $f(f(x)-f(y))-f(-f(y))=(f(x)-f(y))^4-f(y)^4$ Adding, we get $f(f(x)-f(y))=(f(x)-f(y))^4+a$ An since any real may be written as $f(x)-f(y)$ for some real numbers $x,y$, we got $f(x)=x^4+a$ $\forall x\in\mathbb R$ which indeed is a solution, whatever is $a$ Hence the solutions : $f(x)=0$ $\forall x$ $f(x)=x^4+a$ $\forall x$ whatever is $a\in\mathbb R$

thanks patrick iwas wrong we can't given that f(y)=-x when f(x) is constant so we must first given that f(x)=a which a is constant and test it then we can given that f(y)=-x and we must prove that this is surjective.

msop wrote: ...and we must prove that this is surjective. Which unfortunately is impossible since none of the solutions we found is surjective ...

Obviously $\boxed{f=0}$ is a solution. Assume that $f$ is not identically 0. Setting $x=0$ gives $f(f(y))=f(0)+f(y)^4$. Replacing $x$ with $f(x)-f(y)$, yields $f(f(x)-f(y))=(f(x)-f(y))^4+f(0)$. Finally, taking $y=y_0$ such that $f(y_0)\neq 0$, we get that $\text{Im} f-\text{Im} f=\mathbb{R}$. Therefore, $f(x)-f(y)$ spans reals, so $\boxed{f=x^4+c}$ is the other solution. btw this was Czech-Polish-Slovak 2012