17 Dec
2003
17 Dec
'03
11:17 p.m.
collins@newton.apple.com (Scott Collins):
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.
Well, it was quoted from memory, so it's possible that I made an error, but it seems to work as stated... For example : (4!+1)^(1/2)=5 (5!+1)^(1/2)=11 (7!+1)^(1/2)=71 I can't find a value which produces a result that is a non-prime integer. (Of course that doesn't prove that there isn't one.)