Problem 108

Diophantine Reciprocals I

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

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

For $n = 4$ there are exactly three distinct solutions:

$\frac{1}{5} + \frac{1}{20} = \frac{1}{4}$ $\frac{1}{6} + \frac{1}{12} = \frac{1}{4}$ $\frac{1}{8} + \frac{1}{8} = \frac{1}{4}$

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.

在如下方程中， $x$ 、 $y$ 、 $n$ 均为正整数：

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

对于 $n = 4$ ，上述方程恰好有三个不同的解：

$\frac{1}{5} + \frac{1}{20} = \frac{1}{4}$ $\frac{1}{6} + \frac{1}{12} = \frac{1}{4}$ $\frac{1}{8} + \frac{1}{8} = \frac{1}{4}$

最小的使得不同的解的数目超过 $1000$ 的 $n$ 是多少？

注意：这个问题是第110题的简单版本；强烈建议读者先解决这一题。