From: Matthew J Ghio <mg5n+@andrew.cmu.edu>
Date: Mon, 11 Apr 94 13:16:59 PDT
To: collins@newton.apple.com (Scott Collins)
Subject: Re: (n!+1)^(1/2)
In-Reply-To: <9404111823.AA19530@newton.apple.com>
Message-ID: <AheOyHC00WC0QH7kpb@andrew.cmu.edu>
MIME-Version: 1.0
Content-Type: text/plain


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.)