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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 $yz$+$xz$ = $xy$, $xz$ = $xy$minus$yz$ = $y$($x$minusz), therefore, $y$ = $xz$/($x$minus$z$), which means $z$ less than $x$.

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

Therefore, $y$ = $x$2/$k$minus$x$. But if $k$ = $x$ then $y$ = 0, so in order to have $y$ greater than 0 we must have $k$ less than $x$; 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 = $x$minus$k$ 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

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