If $x, y \in \mathbb{N^{*}}$, such that $\frac{x}{y}+\frac{y}{x} \in \mathbb{N}$, show that $x=y$. If $x, y \in \mathbb{N^{*}}$, such that $\frac{x}{y}+\frac{y}{x} \in \mathbb{N}$, show that $x=y$.

My try: If $x$ is a multiple of $y$, such that $x \neq y$, $\frac{x}{y}$ would be greater than $1$, and natural, while $\frac{y}{x}$ would be less than $1$, so not a natural number. Same if we swap $x$ with $y$. So, in those two cases, the given sum can't be natural.

I don't know how to continue it from here. Can you help me, please? Thanks!
Let $q=\frac{x}{y}$, we want $q+\frac{1}{q}=n$ for some natural number $n$, or in other words $q$ is a rational root of $p(x)=x^2-nx+1$. Now by Rational root theorem, only possible values are $q=\pm 1$. Now it cannot be $-1$ since originally $x,y&gt;0$. So we have $q=1$ and so $x=y$.
$$\frac xy +\frac yx \to \frac{x^2+y^2}{xy}$$ which is only an integer when $x=y$