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。