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Re: Riesel's number theorem

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  • leavemsg1
    n 2 also...
    Message 1 of 8 , Sep 14 3:49 PM
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      n > 2 also...

      --- 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]
      >
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