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

3 expanded posts ยท every known parent and child

NODE a0f3bd9eRe: (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.


Scott Collins   | "That's not fair!"                         -- Sarah
                | "You say that so often.  I wonder what your basis
   408.862.0540 |  for comparison is."                 -- Goblin King
................|....................................................
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NODE 811ee15aRe: (n!+1)^(1/2)
collins@newton.apple.com (Scott Collins) once said:

CO>   >For any number n, if the square root of (n!)+1 is an integer, it is also
CO>   >prime.  (This is interesting, but rather useless in practice)
CO> 
CO> For any number a, 1<a<=n, n! mod a == 0; therefore, n!+1 mod a == 1.  n!+1
CO> is prime.  Prime numbers don't have integral square roots.
CO> 
CO> 
CO> Scott Collins   | "That's not fair!"                         -- Sarah
CO>                 | "You say that so often.  I wonder what your basis
CO>    408.862.0540 |  for comparison is."                 -- Goblin King
CO> ................|....................................................
CO> BUSINESS.    fax:974.6094    R254(IL5-2N)    collins@newton.apple.com
CO> Apple Computer, Inc.  5 Infinite Loop, MS 305-2D  Cupertino, CA 95014
CO> .....................................................................
CO> PERSONAL.    408.257.1746       1024:669687       catalyst@netcom.com
CO> 
CO> 


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NODE d906c7c0Re: (n!+1)^(1/2)
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.)