Re: (n!+1)^(1/2) Oops! I'm wrong.

[collins@newton.apple.com (Scott Collins)]

    >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 :-)


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