On the diophantine equation $xy=n(x+y)$ with $\gcd(x,y,n)=1$

Question

Prove that if $xy=n(x+y)$ with $\gcd(x,y,n)=1$ has a solution in natural numbers $x,y,n\in \mathbb{N}$, then it follows necessarily that:

1. $n = ab, \gcd(a,b)=1$
2. $x = n+a^2$
3. $y = n+b^2$

I have a proof for this, but it would be interesting to see other solutions.

Answer 1

We get $\frac{1}{n} = \frac{1}{x} + \frac{1}{y}$ so $n \lt x$ and $n \lt y$

Now rewrite this as

$$(x-n)(y-n) = n^2$$

Now if a prime $p$ divides $x-n$ and $y-n$ and $n$, then the gcd won't be $1$.

Thus both $x-n$ and $y-n$ are both perfect squares, with no common factor.