mathschallenge.net

Problem

Consider the equation,

For a given x, determine the number of solutions.

Solution

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

Let k = xz, so z = zk, giving, y = x(xk)/k.

Therefore, y = x²/kx. 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 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 = xk 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).