--- In primenumbers@yahoogroups.com, Bill Bouris <leavemsg1@...> wrote:

>

> I only have two strikes against the theorem; 2^2-1 could be the problem,

> since if R = 3, then 2^((R-1)/2) = 2 mod 3 == -1, and violates the theory.

> I hope that's correct--- gcd(b, 3) = 1 should be the hangup! Bill

>

> From: Bill Bouris <leavemsg1@...>

> To: Bernardo Boncompagni <redgolpe@...>; pgroup <primenumbers@yahoogroups.com>

> Sent: Wednesday, September 14, 2011 4:45 PM

> Subject: Re: [PrimeNumbers] Riesel's number theorem

>

>

> only trying to further my theorem; maybe it has to do with gcd(b, 3) = 1,

> and n >2; just digging; that might need to be the restriction ??? Bill

> thank you for investigating it, thus far.

>

> From: Bernardo Boncompagni <redgolpe@...>

> To: primenumbers@yahoogroups.com

> Cc: leavemsg1@...

> Sent: Wednesday, September 14, 2011 3:23 PM

> Subject: Re: [PrimeNumbers] woodall or riesel number theorem

>

>

>

>

>

> It was failing ONLY because the gcd(R -1, b) = 1 needs to be a part of it.

> >gcd(174, 51) = 3; otherwise, it holds, not because b <= 2*ln(R). it's a nice

> >theorem!

> >

>

> n=5, k=21, R=671=11*61, b=9<2*ln(671), gcd(670,9)=1

>

> b^335==1 (mod 671)

>

> [Non-text portions of this message have been removed]

>