## Task 5 - 231245

Determine all functions $f$ that are defined for all non-zero real numbers $x$ and satisfy the following conditions:

(1) For all $x_{1}, x_{2}$ with $x_{1} \neq 0, x_{2} \neq 0, x_{1}+x_{2} \neq 0$, $f\left(\frac{1}{x_{1}+x_{2}}\right)=f\left(\frac{1}{x_{1}}\right)+f\left(\frac{1}{x_{2}}\right)$.

(2) For all $x_{1}, x_{2}$ with $x_{1} \neq 0, x_{2} \neq 0, x_{1}+x_{2} \neq 0$, $\left(x_{1}+x_{2}\right) \cdot f\left(x_{1}+x_{2}\right)=x_{1} x_{2} \cdot f\left(x_{1}\right) \cdot f\left(x_{2}\right)$.

(3) $f(1)=1$.
Let $r \neq 0$ be any non-zero real number. Setting $x_{1}=x_{2}=\frac{1}{r}$, we have that $x_{1}$ and $x_{2}$ are well-defined and (as well as $x_{1}+x_{2}$) are non-zero, so from (1) we obtain the relation

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

On the other hand, setting $x_{1}=x_{2}=\frac{r}{2}$, the conditions for (2) are again satisfied, so we get

$$
r \cdot f(r)=\left(\frac{r}{2}+\frac{r}{2}\right) \cdot f\left(\frac{r}{2}+\frac{r}{2}\right)=\frac{r}{2} \cdot \frac{r}{2} \cdot f\left(\frac{r}{2}\right) \cdot\left(\frac{r}{2}\right)=\frac{r^{2}}{4} \cdot(2 f(r))^{2}=(r \cdot f(r))^{2}
$$

which implies $r \cdot f(r) \in\{0 ; 1\}$, since these are the only real numbers that are equal to their square.

If for some $r \neq 0$ the product $r \cdot f(r)$ were 0, then $f(r)$ would also be 0. However, $r$ cannot be 1 due to (3), so we can choose $x_{1}=r$ and $x_{2}=1-r$ and substitute them into (2). Since $f\left(x_{1}\right)=f(r)=0$, it follows that

$$
f(1)=f\left(x_{1}+x_{2}\right)=x_{1} x_{2} \cdot f\left(x_{1}\right) \cdot f\left(x_{2}\right)=0
$$

which contradicts (3).

Therefore, for all $r \neq 0$, the equation $r \cdot f(r)=1$ or $f(r)=\frac{1}{r}$ must hold. Indeed, this function satisfies all three required properties, as can be easily verified by substitution.