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Problem 454

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Diophantine reciprocals III

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

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

For a limit L we define F(L) as the number of solutions which satisfy x < y ≤ L.

We can verify that F(15) = 4 and F(1000) = 1069.
Find F(10¹²).

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丢番图倒数III

在如下方程中，x、y、n均为正整数。

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

我们记上述方程满足x < y ≤ L的解的数目为F(L)。

可以验证F(15) = 4以及F(1000) = 1069。
求F(10¹²)。

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