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Factorial Difference = Perfect Square (Posted on 2011-03-28) Difficulty: 3 of 5
Determine all possible pair(s) (x, y) of nonnegative integers such that (x!*y! - x! - y!) is a perfect square.

No Solution Yet Submitted by K Sengupta    
Rating: 2.5000 (2 votes)

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possible pairs | Comment 1 of 3

(2,2), (2,3), (3,2)

Some define 0 to be a perfect square, thus (2,2) is included as a possible pair.

2!*2! - 2! - 2! = 2*2 - 2 - 2 = 0 = 02
2!*3! - 2! - 3! = 2*6 - 2 - 6 = 4 = 22
3!*2! - 3! - 2! = 6*2 - 6 - 2 = 4 = 22


  Posted by Dej Mar on 2011-03-28 11:39:42
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