Let $\mathbb{R}$ be the set of all real numbers. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying for all $x, y \in \mathbb{R}$ the equation

$$
f(x)+f(y)=f(f(x) f(y)) .
$$
Answer: $f(x) \equiv 0$ is the only such function.

Choose an arbitrary real number $x_{0}$ and denote $f\left(x_{0}\right)=c$. Setting $x=y=x_{0}$ in the equation we obtain $f\left(c^{2}\right)=2 c$. For $x=y=c^{2}$ the equation now gives $f\left(4 c^{2}\right)=4 c$. On the other hand, substituting $x=x_{0}$ and $y=4 c^{2}$ we obtain $f\left(4 c^{2}\right)=5 c$. Hence $4 c=5 c$, implying $c=0$. As $x_{0}$ was chosen arbitrarily, we have $f(x)=0$ for all real numbers $x$.

Obviously, the function $f(x) \equiv 0$ satisfies the equation. So it is the only solution.