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Diophantine reciprocals I

Problem 108

Published on 04 November 2005 at 06:00 pm [Server Time]

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.


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