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Re: (n!+1)^(1/2)
[email protected] (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.)