
03.4. Let $\mathbb{R}^{*}=\mathbb{R} \backslash\{0\}$ be the set of non-zero real numbers. Find all functions $f: \mathbb{R}^{*} \rightarrow \mathbb{R}^{*}$ satisfying

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

for $x, y \in \mathbb{R}^{*}$ and $x+y \neq 0$.


Solution. If $x \neq y$, then

$$
f(y)+f(x-y)=f(y(x-y) f(x))
$$

Because $f(y) \neq 0$, we cannot have $f(x-y)=f(y(x-y) f(x))$ or $x-y=y(x-y) f(x)$. So for all $x \neq y, y f(x) \neq 1$. The only remaining possibility is $f(x)=\frac{1}{x}$. - One easily checks that $f, f(x)=\frac{1}{x}$, indeed satisfies the original functional equation.
