(n!+1)^(1/2) Oops! I'm wrong.
Scott Collins
collins at newton.apple.com
Mon Apr 11 14:59:54 PDT 1994
>For any number a, 1<a<=n, n! mod a == 0; therefore, n!+1 mod a == 1. n!+1
>is prime. Prime numbers don't have integral square roots.
>For example :
>
>(4!+1)^(1/2)=5
>(5!+1)^(1/2)=11
>(7!+1)^(1/2)=71
I am completely wrong. I replied too hastily. Please accept my apologies.
In fact, n!+1 is relatively prime to any a, 1<a<=n, however plainly it is
much larger than n itself and when n>3, (n!+1)>(n^2) and may have factors
(including an integral square root) larger than n.
Oops :-)
Scott Collins | "That's not fair!" -- Sarah
| "You say that so often. I wonder what your basis
408.862.0540 | for comparison is." -- Goblin King
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