All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Factorial and a Square (Posted on 2008-02-23)
If a and p are positive integers, then solve for (a,p):
a2=p!

 See The Solution Submitted by Praneeth Rating: 2.0000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 No Subject | Comment 1 of 9

We can equate a to the product of its prime factors:
a = (f1 * f2 * ... * fn)
And a2 would then be the product of the squares of its prime factors:
a2 = (f12 * f22 * ... * fn2)

The factorial of the positive integer p is the product of all positive integers less than or equal to p:
p! = (1 * 2 *  .... * p-1 * p)

For a2  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 a2=p! is, then, (1,1).

 Posted by Dej Mar on 2008-02-23 13:14:56

 Search: Search body:
Forums (0)