## Reciprocal Sum

#### Problem

Consider the equation,

1 x | + | 1 y | = | 1 z | where x, y, z are positive integers |

For a given `x`, determine the number of solutions.

#### Solution

From 1/`x` + 1/`y` = 1/`z`, we get `yz`+`xz` = `xy`, `xz` = `xy``yz` = `y`(`x`z), therefore, `y` = `xz`/(`x``z`), which means `z` `x`.

Let `k` = `x``z`, so `z` = `z``k`, giving, `y` = `x`(`x``k`)/`k`.

Therefore, `y` = `x`^{2}/`k``x`. But if `k` = `x` then `y` = 0, so in order to have `y` 0 we must have `k` `x`; in addition, `y` will only be integer if `x`^{2} divides by `k`.

That is, the number of solutions for a given `x` is the same as the number of divisors of `x`^{2} less than `x`.

For example, when `x` = 12, `x`^{2} = 144, `k` = {1, 2, 3, 4, 6, 8, 9}.

As z = `x``k` and `y` = `xz`/`k`, we get the seven solutions: (12,132,11) (12,60,10) (12,36,9) (12,24,8) (12,12,6), (12,6,4) and (12,4,3).

Problem ID: 169 (Apr 2004) Difficulty: 4 Star