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 $(x \neq y)$

Question

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 $(x \neq y)$.

Possibly related to Project Euler problems 108/110: https://projecteuler.net/problem=108 — Alex Wertheim, Jul 25 '13 at 18:08
@labbhattacharjee i didn't knew that it has been asked before. — , Jul 25 '13 at 18:14
@SamSharma, it's not easy to find duplicate here. I could recount because I answered it:) — lab bhattacharjee, Jul 25 '13 at 18:18
I am going to edit the title of the duplicate question, because it would not be immediately obvious to someone approaching this for the first time that they were the same. — Mark Bennet, Jul 25 '13 at 20:59

score 4 · Accepted Answer · answered Jul 25 '13 at 18:09

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 $n^2.$

score 4 · Answer 2 · answered Jul 25 '13 at 18:16

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.

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 $(x \neq y)$

2 Answers2