## Task 3A - 191233A

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

(1) For all pairs $\left(x_{1} ; x_{2}\right)$ of real numbers, $f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right)+f\left(x_{2}\right)$.

(2) It holds that $f(1)=1$.

(3) For all real numbers $x \neq 0$, $f\left(\frac{1}{x}\right)=\frac{1}{x^{2}} f(x)$.
From (1) it follows that $f(0+0)=f(0)+f(0)$, hence $f(0)=2 f(0)$ and thus, after subtracting $f(0)$ from both sides, $f(0)=0$. Furthermore, $f(x-x)=f(x)+f(-x)$ for all $x \in \mathbb{R}$, i.e., $0=f(x)+f(-x)$ or $f(-x)=-f(x)$. Given $f(1)=1$, we particularly have $f(-1)=-1$ and $f(x+1)=f(x)+1$.

For all $x \notin\{0,-1\}$, we now have

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

and

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

which combined then gives

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

thus

$$
2 \frac{f(x)}{x^{2}}=\frac{2}{x}
$$

and therefore, together with $f(0)=0$ and $f(-1)=-1$ (from the preliminary considerations or (1) and (2)), we have $f(x)=x$ for all $x$.