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# 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.

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

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