Problem 110

Diophantine Reciprocals II

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

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

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.

丢番图倒数（二）

在如下方程中， $x$ 、 $y$ 、 $n$ 均为正整数。

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

可以验证当 $n = 1260$ 时，恰好有 $113$ 种不同的解，这也是最小的使得不同的解的数目超过一百的 $n$ 。

最小的使得不同的解的数目超过四百万的 $n$ 是多少？

注意：这是第108题一个极其困难的版本，而且远远超过暴力解法的能力范围，因此需要更加聪明的手段。