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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 = xyminusyz = y(xminusz), therefore, y = xz/(xminusz), which means z less than x.

Let k = xminusz, so z = zminusk, giving, y = x(xminusk)/k.

Therefore, y = x2/kminusx. 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 x2 divides by k.

That is, the number of solutions for a given x is the same as the number of divisors of x2 less than x.

For example, when x = 12, x2 = 144, k = {1, 2, 3, 4, 6, 8, 9}.

As z = xminusk 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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