Project Euler Lab - Problem 108

#108 - Diophantine Reciprocals I

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In the following equation \(x\), \(y\), and \(n\) are positive integers.

\[\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}\]

For \(n = 4\) there are exactly three distinct solutions:

\[\begin{align} \dfrac{1}{5} + \dfrac{1}{20} &= \dfrac{1}{4}\\ \dfrac{1}{6} + \dfrac{1}{12} &= \dfrac{1}{4}\\ \dfrac{1}{8} + \dfrac{1}{8} &= \dfrac{1}{4} \end{align} \]

What is the least value of \(n\) for which the number of distinct solutions exceeds one-thousand?

NOTE: This problem is an easier version of Problem 110; it is strongly advised that you solve this one first.

This problem is taken from Project Euler, Problem 108.
Problem text © Project Euler, licensed under CC BY-NC-SA 4.0. Original: projecteuler.net/problem=108. Published Friday, 4th November 2005, 06:00 pm. Solved by 14,555 members at time of mirroring.

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