(n!+1)^(1/2)
Scott Collins
collins at newton.apple.com
Mon Apr 11 12:21:30 PDT 1994
>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.
Scott Collins | "That's not fair!" -- Sarah
| "You say that so often. I wonder what your basis
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