Baltic Way 2020 · Problem 4

Answer: $f(x)=0$ for all $x$.

We first notice that if there exists a number $\alpha$ so that $f(\alpha)=0$, then $f(\alpha+y)=$ $f(f(\alpha)+\alpha+y)=f(\alpha+y)+y f(y)$ for all real $y$. Hence $y f(y)=0$ for all $y$, meaning that $f(y)=0$ for all $y \neq 0$. We are therefore done if we can show that $f(0)=0$, as then $f(x)=0$ for all $x$, which is a solution.

Substituting $y=0$ in the equation yields that:

$$f(f(x)+x)=f(x) \quad \forall x$$

Substituting $y=f(x)$ in the equation yields that:

$$f(f(x)+x+f(x))=f(x+f(x))+f(x) f(f(x))$$

Let $z=x+f(x)$. Then:

$$\begin{aligned} f(x) & =f(x+f(x))=f(z)=f(f(z)+z) \\ & =f(f(x+f(x))+x+f(x)) \\ & =f(f(x)+x+f(x)) \\ & =f(x+f(x))+f(x) f(f(x)) \\ & =f(x)+f(x) f(f(x)) \end{aligned}$$

Hence $f(x) f(f(x))=0$ for all $x$. Letting $x=0$ in (1), we get that $f(f(0))=f(0)$, which means that $f(0)^{2}=f(0) f(f(0))=0$. But then we must have $f(0)=0$.

Substitute $x=0$ and $y=-1$. We obtain $f(f(0)-1)=f(-1)+(-1) \cdot f(-1)=0$.

Substitute $x=f(0)-1$. Then $f(x)=0$ and therefore $f(f(x)+x+y)$ and $f(x+y)$ cancel out. We obtain $0=y f(y)$ for all $y$. It follows that if $y \neq 0$ then $f(y)=0$.

Now, substitute $x=y=0$. We obtain $f(f(0))=f(0)$. Substituting $y=f(0)$ to $0=y f(y)$ yields $f(0) f(f(0))=0$, which means $f(0)^{2}=0$, and finally $f(0)=0$.

Therefore $f(x)=0$ for all $x$, which clearly satisfies the equation.