Another variation on Wilsonm's Theorem:

Both n! and (n+1)! = -nmodn^2 iff n is prime

Jon Perry

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-----Original Message-----

From: CRESGE - Hervé LELEU [mailto:

h.leleu@...]

Sent: 16 April 2002 16:57

To:

primenumbers@yahoogroups.com
Subject: RE : [PrimeNumbers] variation on Wilson theorem

Does it mean that n is prime iff (a-1)!*(n-a)! = -1 (mod n) ?

where a=1 for Wilson theorem.

And also that: n is prime iff ((n-1)/2)! ^ 2 = (+/-) 1 (mod n) where

(+/-) 1 depends if (n-1)/2 is odd or even.

> -----Message d'origine-----

> De : Paul Jobling [mailto:Paul.Jobling@...]

> Envoyé : mardi 16 avril 2002 17:07

> À : 'CRESGE - Hervé LELEU'; primenumbers@yahoogroups.com

> Objet : RE: [PrimeNumbers] variation on Wilson theorem

>

>

> > Let n>2.

> > n is prime if and only if 2(n-3)! = -1 mod n

> > Is this result well-known?

> > Thanks for answers.

>

> Wilson's theorem states that n is prime iff (n-1)! = -1 (mod

> n). This is the same as your result: you have explicitly

> multiplied n-1 and n-2 together and removed them from the

> factorial. (n-1)*(n-2) (mod n) = -1*-2 (mod n) = 2 (mod n).

>

> Regards,

>

> Paul.

>

>

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