(In reply to
No Subject by Dej Mar)
According to the Chebyshev's Theorem that's also known as Bertrand's Postulate, (there's another Chebyshev's Theorem that's not related) there is always a prime number between any number larger than 1 and its double (i.e., between n and 2n for n>=2).
So between the first appearance of a given prime in the formation of n! as 1x2x3x..., and its second appearance as a factor (as 2n), there is always some other prime, by its lonesome, in between, as 2n will come before 2p (where p is that other prime). So there will always be an unpaired prime in the construction of any factorial, and so the factorial can't be a perfect square so long as it's greater than 1.

Posted by Charlie
on 20080223 14:54:44 