Diophantine reciprocals I
November 04, 2005
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.
Written by gamwe6 who lives and works in San Francisco building useful things. You should follow him on Twitter