Let $\alpha \neq 0$ be a real number.

Determine all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that

$$
f(f(x)+y)=\alpha x+\frac{1}{f\left(\frac{1}{y}\right)}
$$

for all $x, y \in \mathbb{R}_{>0}$.

(Walther Janous)
Answer: For $\alpha=1$ the only solution is $f(x)=x$, for other values of $\alpha$ there is no solution.

Of course, $\alpha>0$ must hold, otherwise the right side becomes negative for large $x$. By considering $x$ in the original equation, we see that $f$ is injective $\left(f\left(x_{1}\right)=f\left(x_{2}\right)\right.$ implies $\alpha x_{1}=\alpha x_{2}$ ). Similarly, by the arbitrary choice of $x$ on the right side, the function is surjective onto an interval $(a, \infty)$ with a suitable $a$ (e.g., $a=\frac{1}{f(1)}$ ) and thus, by choosing a small $x$ and a large function value $f\left(\frac{1}{y}\right)$ on the right side, it is generally surjective.

We now replace $y$ with $f(y)$ and obtain

$$
f(f(x)+f(y))=\alpha x+\frac{1}{f\left(\frac{1}{f(y)}\right)}
$$

The left side is symmetric in $x$ and $y$, so it also holds that

$$
\alpha x+\frac{1}{f\left(\frac{1}{f(y)}\right)}=\alpha y+\frac{1}{f\left(\frac{1}{f(x)}\right)}
$$

Choosing any fixed value for $y$, we get

$$
\frac{1}{f\left(\frac{1}{f(x)}\right)}=\alpha x+C
$$

for a suitable $C$. Clearly, $C \geq 0$ must hold because the function only takes positive values. We substitute the obtained identity into equation (1) and get

$$
f(f(x)+f(y))=\alpha x+\alpha y+C .
$$

Due to injectivity, it follows that

$$
f(x)+f(y)=f(z)+f(w) \text {, if } x+y=z+w \text {. }
$$

In particular, for $x, y \geq 0$, it holds that

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

With $g(x)=f(x+1)$, it follows for the function $g: \mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}$ that

$$
g(x)+g(y)=g(x+y)+g(0) .
$$

For $h(x)=g(x)-g(0)$, it holds that $h(x) \geq-g(0)$ and the Cauchy functional equation

$$
h(x)+h(y)=h(x+y) .
$$

If a function value $h(t)1$, then $f(x)=h(x-1)+g(0)=c x+d$ for suitable constants $c$ and $d$. This also holds for smaller $x$, as can be seen from equation (2) for example with the values $y=3$, $z=2$ and $w=x+1$.

Since $f$ is surjective and $c$ must be positive, the constant term must of course be zero, otherwise small values would not be assumed or negative values would be assumed. Substituting $f(x)=c x$ into the original equation yields (by comparing coefficients) the conditions $c^{2}=\alpha$ and $c^{2}=1$. Since $c$ must be positive, we obtain the only solution $f(x)=x$, if $\alpha=1$.

(Theresia Eisenkölbl)