A5. Determine all functions $f:(0, \infty) \rightarrow \mathbb{R}$ satisfying
$$
\left(x+\frac{1}{x}\right) f(y)=f(x y)+f\left(\frac{y}{x}\right)
$$
for all $x, y>0$.
Answer: $f(x)=C_{1} x+\frac{C_{2}}{x}$ with arbitrary constants $C_{1}$ and $C_{2}$.
Solution 1. Fix a real number $a>1$, and take a new variable $t$. For the values $f(t), f\left(t^{2}\right)$, $f(a t)$ and $f\left(a^{2} t^{2}\right)$, the relation (1) provides a system of linear equations:
$$
\begin{array}{lll}
x=y=t: & \left(t+\frac{1}{t}\right) f(t) & =f\left(t^{2}\right)+f(1) \\
x=\frac{t}{a}, y=a t: & \left(\frac{t}{a}+\frac{a}{t}\right) f(a t) & =f\left(t^{2}\right)+f\left(a^{2}\right) \\
x=a^{2} t, y=t: & \left(a^{2} t+\frac{1}{a^{2} t}\right) f(t)=f\left(a^{2} t^{2}\right)+f\left(\frac{1}{a^{2}}\right) \\
x=y=a t: & \left(a t+\frac{1}{a t}\right) f(a t)=f\left(a^{2} t^{2}\right)+f(1)
\end{array}
$$

In order to eliminate $f\left(t^{2}\right)$, take the difference of (2a) and (2b); from (2c) and (2d) eliminate $f\left(a^{2} t^{2}\right)$; then by taking a linear combination, eliminate $f(a t)$ as well:
$$
\begin{array}{c}
\left(t+\frac{1}{t}\right) f(t)-\left(\frac{t}{a}+\frac{a}{t}\right) f(a t)=f(1)-f\left(a^{2}\right) \text { and } \\
\left(a^{2} t+\frac{1}{a^{2} t}\right) f(t)-\left(a t+\frac{1}{a t}\right) f(a t)=f\left(1 / a^{2}\right)-f(1), \text { so } \\
\left(\left(a t+\frac{1}{a t}\right)\left(t+\frac{1}{t}\right)-\left(\frac{t}{a}+\frac{a}{t}\right)\left(a^{2} t+\frac{1}{a^{2} t}\right)\right) f(t) \\
=\left(a t+\frac{1}{a t}\right)\left(f(1)-f\left(a^{2}\right)\right)-\left(\frac{t}{a}+\frac{a}{t}\right)\left(f\left(1 / a^{2}\right)-f(1)\right) .
\end{array}
$$

Notice that on the left-hand side, the coefficient of $f(t)$ is nonzero and does not depend on $t$ :
$$
\left(a t+\frac{1}{a t}\right)\left(t+\frac{1}{t}\right)-\left(\frac{t}{a}+\frac{a}{t}\right)\left(a^{2} t+\frac{1}{a^{2} t}\right)=a+\frac{1}{a}-\left(a^{3}+\frac{1}{a^{3}}\right)b>1$ arbitrarily and substitute $x=(a / b)^{n}$ and $y=(a b)^{n}$ in (1):
$$
\begin{aligned}
\left(\frac{a^{n}}{b^{n}}+\frac{b^{n}}{a^{n}}\right) f_{0}\left((a b)^{n}\right) & =f_{0}\left(a^{2 n}\right)+f_{0}\left(b^{2 n}\right), \quad \text { so } \\
\left(\frac{a^{n}}{b^{n}}+\frac{b^{n}}{a^{n}}\right) K(a b)\left((a b)^{n}-\frac{1}{(a b)^{n}}\right) & =K(a)\left(a^{2 n}-\frac{1}{a^{2 n}}\right)+K(b)\left(b^{2 n}-\frac{1}{b^{2 n}}\right), \quad \text { or equivalently } \\
K(a b)\left(a^{2 n}-\frac{1}{a^{2 n}}+b^{2 n}-\frac{1}{b^{2 n}}\right) & =K(a)\left(a^{2 n}-\frac{1}{a^{2 n}}\right)+K(b)\left(b^{2 n}-\frac{1}{b^{2 n}}\right) .
\end{aligned}
$$

By dividing (5) by $a^{2 n}$ and then taking limit with $n \rightarrow+\infty$ we get $K(a b)=K(a)$. Then (5) reduces to $K(a)=K(b)$. Hence, $K(a)=K(b)$ for all $a>b>1$.

Fix $a>1$. For every $x>0$ there is some $b$ and an integer $n$ such that $1<b<a$ and $x=b^{n}$. Then
$$
f_{0}(x)=f_{0}\left(b^{n}\right)=K(b)\left(b^{n}-\frac{1}{b^{n}}\right)=K(a)\left(x-\frac{1}{x}\right) .
$$

Hence, we have $f(x)=f_{0}(x)+f(1) x=C_{1} x+\frac{C_{2}}{x}$ with $C_{1}=K(a)+f(1)$ and $C_{2}=-K(a)$.
Comment. After establishing (5), there are several variants of finishing the solution. For example, instead of taking a limit, we can obtain a system of linear equations for $K(a), K(b)$ and $K(a b)$ by substituting two positive integers $n$ in (5), say $n=1$ and $n=2$. This approach leads to a similar ending as in the first solution.

Optionally, we define another function $f_{1}(x)=f_{0}(x)-C\left(x-\frac{1}{x}\right)$ and prescribe $K(c)=0$ for another fixed $c$. Then we can choose $a b=c$ and decrease the number of terms in (5).