
4. Find all functions $f: \mathbb{R} \backslash\{0\} \rightarrow \mathbb{R}$ such that for all non-zero numbers $x, y$,

$$
x \cdot f(x y)+f(-y)=x \cdot f(x) .
$$

(Pavel Calábek)


Solution. Substituting $x=1$ gives

$$
f(y)+f(-y)=f(1) .
$$

Denoting $f(1)=a$, we have $f(-y)=a-f(y)$. Substituting $y=-1$, we get

$$
x \cdot f(-x)+f(1)=x \cdot f(x),
$$

i. e.,

$$
x(a-f(x))+a=x \cdot f(x),
$$

hence

$$
f(x)=\frac{a(x+1)}{2 x}=\frac{a}{2}\left(1+\frac{1}{x}\right) .
$$

Finally, we check that for any real number $c$, the function $f(x)=c(1+1 / x)$ satisfies the conditions:

$$
\begin{aligned}
x \cdot f(x y)+f(-y) & =x \cdot c\left(1+\frac{1}{x y}\right)+c\left(1+\frac{1}{-y}\right)=c\left(x+\frac{1}{y}+1-\frac{1}{y}\right)= \\
& =c(x+1)=c x\left(1+\frac{1}{x}\right)=x \cdot f(x) .
\end{aligned}
$$

Other Solution. Let us set $f(1)=a$. Substituting $y=-1$ into the given equation yields

$$
x f(-x)+a=x f(x),
$$

i. e.,

$$
f(-x)=f(x)-\frac{a}{x} \text {. }
$$

The given equation can be rearranged into the form

$$
f(x y)+\frac{1}{x} \cdot f(-y)=f(x)
$$

whence, using (1), we have

$$
\begin{gathered}
f(x y)+\frac{1}{x}\left(f(y)-\frac{a}{y}\right)=f(x), \\
f(x y)+\frac{f(y)}{x}-\frac{a}{x y}=f(x) .
\end{gathered}
$$

Interchanging $x$ and $y$ gives

$$
f(y x)+\frac{f(x)}{y}-\frac{a}{y x}=f(y)
$$

so, combining the last two equations, we get

$$
\frac{f(x)}{y}-\frac{f(y)}{x}=f(y)-f(x)
$$

and substituting $y=1$ now gives

$$
2 f(x)=a\left(1+\frac{1}{x}\right)
$$

Again, we can easily verify that every function $f(x)=c(1+1 / x)$ is a solution of the given functional equation.
