<s>[INST] 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 $(x \neq y)$.
 [/INST] It is clear that the two positive fractions on the left have to be less than $\frac 1n$, so we have $x,y \gt n$.
Let $x=n+a, y=n+b$ and clear fractions then $$n(n+b)+n(n+a)=(n+a)(n+b)$$ which simplifies to $$ab=n^2$$
Any such $a,b$ give a pair of fractions.
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