Find positive integers $x$ and $y$ such that:$$\frac{x}{y}=\frac{(x^2-y^2)^\frac{y}{x}+1}{(x^2-y^2)^\frac{y}{x}-1}.$$
I tried to proceed with this problem as follows: $$\frac{y}{x}=\frac{(x^2-y^2)^\frac{y}{x}-1}{(x^2-y^2)^\frac{y}{x}+1}=1-\frac{2}{x^2(1-\frac{y^2}{x^2})^\frac{y}{x}+1}$$ but I was stuck there and I coulnd't find the relation between $x$ and $y$.
Please help me solve the problem.
Recall $$v = \frac{u+1}{u-1}\quad\iff\quad u = \frac{v+1}{v-1}$$ Substitute $v$ by $\frac{x}{y}$ and $u$ by $(x^2-y^2)^{y/x}$, one get
$$(x^2-y^2)^{y/x} = \frac{\frac{x}{y}+1}{\frac{x}{y}-1} = \frac{x+y}{x-y}$$ Raising to power $x$ on both sides, we obtain $$\begin{align} & (x+y)^y (x-y)^y = (x+y)^x (x-y)^{-x}\\ \iff & (x-y)^{x+y} = (x+y)^{x-y}\\ \iff & (x-y)^{1/(x-y)} = (x+y)^{1/(x+y)} \end{align} \tag{*1} $$ Notice for $t \in [1,\infty)$, the map $t \mapsto t^{1/t}$ is strictly increasing on $[1,e]$ and strictly decreasing on $[e,\infty)$, In order to satisfy last identity of $(*1)$, we need $x - y \in [1,e]$ and $x + y \in [e,\infty)$
Since $x - y \in \mathbb{Z}_{+}$, there are only two possible choices for $x - y$, i.e. $1$ or $2$. However, $x-y$ cannot equal to $1$ because for $t \in [e,\infty)$, $t^{1/t} > 1$ and there are no $x+y$ to satisfy last identity. As a result,
$$x - y = 2\quad\stackrel{\because (*1)}{\implies}\quad x + y = 4 \quad\implies\quad (x,y) = (3,1)$$