Let's find the positive integer solutions of the equation

$$
\frac{x}{y}=\frac{\left(x^{2}-y^{2}\right)^{\frac{y}{x}}+1}{\left(x^{2}-y^{2}\right)^{\frac{y}{x}}-1}
$$
I. Solution. In expression (1), neither of the denominators can be equal to 0, so $x^{2}-y^{2} \neq 1$. Let's perform the following equivalent transformations:

$$
\begin{gathered}
\frac{x}{y}=\frac{\left(x^{2}-y^{2}\right)^{\frac{y}{x}}+1}{\left(x^{2}-y^{2}\right)^{\frac{y}{x}}-1} \\
x\left(x^{2}-y^{2}\right)^{\frac{y}{x}}-x=y\left(x^{2}-y^{2}\right)^{\frac{y}{x}}+y \\
(x-y)\left(x^{2}-y^{2}\right)^{\frac{y}{x}}=x+y \\
(x-y)^{x}\left(x^{2}-y^{2}\right)^{y}=(x+y)^{x} \\
(x-y)^{x}(x-y)^{y}(x+y)^{y}=(x+y)^{x} \\
(x-y)^{x+y}=(x+y)^{x-y}
\end{gathered}
$$

Let's introduce the notations $a=x+y$ and $b=x-y$. Then $b^{a}=a^{b}$, where $a>b$, since $y$ is positive. Along with $a$, $b$ is also positive, because $a=x+y>0$, and if $b$ were negative, then the absolute value of the left side would be greater than 1, while the absolute value of the right side would be less. If $b=0$, then $x=y$ and $\frac{x}{y} \neq \frac{\left(x^{2}-y^{2}\right)^{\frac{y}{x}}+1}{\left(x^{2}-y^{2}\right)^{\frac{y}{x}}-1}$.

Let $(a, b)=d$. Then there exist positive integers $a_{1}, b_{1}$ such that $a=a_{1} d, b=b_{1} d$ and $\left(a_{1}, b_{1}\right)=1$. Thus $\left(d b_{1}\right)^{d a_{1}}=\left(d a_{1}\right)^{d b_{1}} \cdot$

Since $a_{1}, b_{1}, d_{1}$ are positive integers: $d^{a_{1}} \cdot b_{1}^{a_{1}}=d^{b_{1}} \cdot a_{1}^{b_{1}} \cdot a>b$ implies $a_{1}>b_{1}$. Therefore

$$
d^{a_{1}-b_{1}} b_{1}^{a_{1}}=a_{1}^{b_{1}}
$$

Both sides are positive integers and $\left(a_{1}, b_{1}\right)=1$, so the equality can only hold if $b_{1}=1$. Thus $d^{a_{1}-1}=a_{1}$, where $d \neq 1$, otherwise $a=b=1$, but we know $a>b$. Therefore, it can be assumed that $d \geq 2$. Use the following estimate:

$$
d^{a_{1}-b} \geq 2^{a_{1}-1} \geq a_{1}
$$

From this, $d=2$ immediately follows. $2^{a_{1}-1}=a_{1}$ can be easily verified to hold if $a_{1}$ is 1 or 2. If $a_{1}=1$, then $a=b=2$, which is not a solution to the problem. If $a_{1}=2$, then $a=4$ and $b=2$, from which $x=3$ and $y=1$ follow. These satisfy the conditions of the problem. Therefore, the only solution is $x=3$ and $y=1$.

Based on the work of Zoltán Léka (Makó, József Attila Gymn., 12th grade)

II. Solution. Just like in the previous solution, we reduce the problem to the equation $a^{b}=b^{a}$. From this:

$$
\sqrt[a]{a}=\sqrt[b]{b}
$$

Thus, the question can be rephrased as: Does the function $n \rightarrow \sqrt[n]{n}$, defined on positive integers, have two values that are equal to each other? If $n \geq 5$, then $\sqrt[n]{n}>1$ and the function is strictly decreasing, since

$$
n \geq 4>e=2.71 \cdots>\left(1+\frac{1}{n}\right)^{n} \quad \text { hence } \quad n>\left(1+\frac{1}{n}\right)^{n}=\left(\frac{n+1}{n}\right)^{n}
$$

Thus $n^{n+1}>(n+1)^{n}$, so $\sqrt[n]{n}>\sqrt[n+1]{n+1}$.

Accordingly, if $n \geq 5$, all values of the function are different, and

$$
1<\sqrt[n]{n} \leq \sqrt[5]{5}<\min \{\sqrt[n]{n}: n \in\{2,3,4\}\}
$$

For $1 \leq n \leq 4$, $\sqrt[4]{4}=\sqrt[2]{2}$ follows, from which the only solution is $x=3$ and $y=1$.

Based on the work of Tamás Zombori (Budapest, Toldy F. Gymn., 12th grade)