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Re: (n!+1)^(1/2)



  >For any number n, if the square root of (n!)+1 is an integer, it is also
  >prime.  (This is interesting, but rather useless in practice)

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.


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