## Diophantine reciprocals I

Published on Friday, 4th November 2005, 06:00 pm; Solved by 11397;
Difficulty rating: 30%

### Problem 108

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.