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# Problem 454

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).

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