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。