n is+ve integer, how many solutions $(x,y)$ exist for $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ with $x$, $y$ being positive integers and $(x \neq y)$ I wanted to know, how can i solve this.
For a given positive integer n, how many solutions $(x,y)$ exist for $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ with $x$ and $y$ being positive integers and $(x \neq y)$.

If $(x,y)$ is a solution of $\frac{1}{x}+\frac{1}{y}$ = $\frac{1}{n}$ then $x>n$ and $y>n$.
We can write the equation as 
$n(x+y)=xy$ = $(x-n)(y-n) = n^2$ 
and for any $k$ positive $k|n^2$, therefore the solutions are given by $x-n=k$ and $y-n=\frac{n^2}{k}$ or $x=n+k$ and $y=n+\frac{n^2}{k}$.
The only case for $x=y$ is when $n=k$, therefore the number of solutions for $(x,y)$ with $(x\neq y)$ is $d(n^2)-1$ where $d(n^2)$ is the number of divisors of $n^2.$
