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Problem 108

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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.

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丢番图倒数I

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3个不同的解：

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

使得不同的解的数目超过1000的最小n值是多少？

注意：这个问题是第110题的简单版本；强烈推荐先解决这一题。

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