Find all positive $x,y\in\Bbb{Q}$ such that $x+\frac{1}{y}$ and $y+\frac{1}{x}$ are natural numbers. What I did is, the product is also a natural number.
$xy+\frac{1}{xy}=m$ is also a natural number. Assuming $xy=z$, we get the equation $z^2-mz+1=0$
Hence, $z=\frac{m\pm \sqrt{m^2-4}}{2}$. Since z is rational $\sqrt{m^2-4}$ has to be rational. It is rational only when $m=2$ (not sure). So $z=xy=1$. So, $y=\frac{1}{x}$
Hence $2x$ and $2y$ are natural numbers.
I can't proceed further. Is my way correct? Any help please.

Hint If $\sqrt{m^2-4}=\frac{p}{q}$ is rational with $\mbox{gcd}(p,q)=1$, then $\frac{p^2}{q^2} \in \mathbb Z$. Use $\mbox{gcd}(p,q)=1$ to conclude that $q=1$.
This gives that $\sqrt{m^2-4}=n$ for some $n \in \mathbb N$ and hence
$$
4=m^2-n^2=(m-n)(m+n) 
$$
Look at all possible factorizations to conclude that $m=2$.
To complete the problem, observe that $2x=n, 2y=k$ and $xy=1$ gives
$$
kn=4
$$
There are only 3 possibilities (or two by symmetry), just check them.
