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Diophantine reciprocals I

Problem 108

Published on Friday, 4th November 2005, 06:00 pm; Solved by 7317

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

1
x
+
1
y
=
1
n

For n = 4 there are exactly three distinct solutions:

1
5
+
1
20
=
1
4
1
6
+
1
12
=
1
4
1
8
+
1
8
=
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.