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).