Determine all possible pair(s) (x, y) of nonnegative integers such that (x!*y! - x! - y!) is a perfect square.

(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