## Diophantine reciprocals II

### Problem 110

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

$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}$$

It can be verified that when $n = 1260$ there are 113 distinct solutions and this is the least value of $n$ for which the total number of distinct solutions exceeds one hundred.

What is the least value of $n$ for which the number of distinct solutions exceeds four million?

NOTE: This problem is a much more difficult version of Problem 108 and as it is well beyond the limitations of a brute force approach it requires a clever implementation.