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!
 A: 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 $x,y>0$. So we have $q=1$ and so $x=y$.
A: We have $$\frac{x}{y} + \frac{y}{x} = \frac{(x+y)^2}{xy} - 2 \in \mathbb{N} \iff \frac{(x+y)^2}{xy} \in \mathbb{N}.$$  Now, let $p$ be a prime dividing $x$, then $p | (x+y) \implies p \mid y$.  By symmetry, $x$ and $y$ share the same prime factors.  Now prove by contradiction that the primes occur to the same powers in $x$ and $y$...
A: Below is a solution by what is called the Vieta Jumping .
Assume $x>y$ and that $(x,y)$ is the pair such that $x+y$ is minimal. Then,  $x = yr+s, 0\leq s <y$ and 
$$\frac xy+\frac yx = r+\frac sy +\frac{y}{yr+s} <r+1+1=r+2.$$ 
So it must follow that $\dfrac{s}{y}+\dfrac{y}{yr+s} = 1.\implies s^2+y^2+yrs = y^2r+ys\implies r = \dfrac{s^2-ys+y^2}{y^2-ys}=1+\dfrac{s^2}{y^2-ys}.$
Let $y=s+t.\implies ty|(y-t)^2\implies ty|y^2+t^2\implies \dfrac{t}{y}+\dfrac{y}{t}$ is integer and $t+y<2y<x+y$, a contradiction.  
A: There's lots of ways to prove it, but your approach works if you continue it.

If $x$ is a multiple of $y$

That's one case.  The other case you next have to check if $x$ is not a multiple of $y$.
Without loss of generality we can assume $x > y$ so $x/y = n + f$ for some natural number $n$ and fraction $0 < f < 1$.
$$n + f + \frac1{n + f} \in \mathbb N \iff f + \frac1{n + f} \in \mathbb N$$
Since both terms are less than $1$, it is to find:
$$f + \frac1{n + f} = 1$$
$$f = \frac{\pm \sqrt{n^2 + 2n - 3} - (n - 1)}{2}$$
Nothing that $\sqrt{\mathbb N}$ must either be a natural number or irrational
$$\sqrt{n^2 + 2n - 3} \in \mathbb N$$
but $n^2 < n^2 + 2n - 3 < (n+1)^2$ for $n > \frac 32$, so $n=1, f =0$ is the only solution.
A: $$\frac xy +\frac yx \to \frac{x^2+y^2}{xy}$$ which is only an integer when $x=y$
