Let $\alpha \in \mathbb{Q}^{+}$. Determine all functions $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ such that for all $x, y \in \mathbb{Q}^{+}$

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

holds.

Here, $\mathbb{Q}^{+}$ denotes the set of positive rational numbers.

(Walther Janous)
We set $y=x$ and obtain

$$
f(x+1)=1+f(x)+\alpha x .
$$

Furthermore, we set $y=1$ and obtain

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

Equating (1) and (2) yields

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

Since the function $f$ cannot be constant due to (1), it must be that $f(1)=1$. Using induction, we then obtain from equation (1)

$$
f(x)=\frac{\alpha}{2} x(x-1)+x \quad \text { for all } x \in \mathbb{Z}^{+} \text {. }
$$

This particularly gives $f(2)=\alpha+2$ and $f(4)=6 \alpha+4$. We now set $x=4$ and $y=2$ in the original equation and obtain, after simplification,

$$
\alpha^{2}-2 \alpha=0 .
$$

Thus, $\alpha=2$ must hold for there to be solutions. From now on, we only consider this case.

From (3), it follows that $f(x)=x^{2}$ for $x \in \mathbb{Z}^{+}$. Using induction and from (1), we infer that if $f(x+n)=(x+n)^{2}$ for some $x \in \mathbb{Q}^{+}$ and $n \in \mathbb{Z}^{+}$, then $f(x)=x^{2}$ follows.

Let $\frac{a}{b} \in \mathbb{Q}^{+}$ with $a, b \in \mathbb{Z}^{+}$. We set $x=a$ and $y=b$ and obtain

$$
f\left(\frac{a}{b}+b\right)=\frac{a^{2}}{b^{2}}+b^{2}+2 a=\left(\frac{a}{b}+b\right)^{2} .
$$

Thus, according to the above remark, $f\left(\frac{a}{b}\right)=\left(\frac{a}{b}\right)^{2}$ also holds. Checking shows that $f(x)=x^{2}$ is indeed a solution.

In summary, there is no solution for $\alpha \neq 2$ and the solution $f(x)=x^{2}$ for $\alpha=2$.

(Theresia Eisenkölbl)