We can equate a to the product of its prime factors:
a = (f_{1} * f_{2} * ... * f_{n})
And a^{2} would then be the product of the squares of its prime factors:
a^{2} = (f_{1}^{2} * f_{2}^{2} * ... * f_{n}^{2})
The factorial of the positive integer p is the product of all positive integers less than or equal to p:
p! = (1 * 2 * .... * p1 * p)
For a^{2} to equal p!, then, the number of each distinct prime factor of p must be divisible by 2. As a prime is, by definition, a number that has no factor other than itself and 1, p would need be a composite number greater than its last prime that included an odd multiple of its last prime as one of its factors. This would mean, in the least, the existence of a prime gap equal or greater than the number's last prime. No such prime gap has been found, and the probable existence of one is, or almost is, nil.
The only solution for a^{2}=p! is, then, (1,1).

Posted by Dej Mar
on 20080223 13:14:56 