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woodall or riesel number theorem

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  • leavemsg1
    Hi, Group, ... Woodall s theorem (as it should be named): ... let R= k*2^n -1, n 1 is a natural number and k
    Message 1 of 8 , Sep 13, 2011
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      Hi, Group,
      ...
      Woodall's theorem (as it should be named):
      ...
      let R= k*2^n -1, n >1 is a natural number and k <= 2^n-1.
      If for some 'b', b^((R-1)/2) == +1 (mod R), then 'R' is prime;
      (similar to Proth's theorem.)
      ...
      proof:
      if 'm' is from the set of natural numbers, then every odd prime
      divisor 'r' of a^(2^m) +1 implies that q == +1(mod a^(m+1))
      [concluded from generalized Fermat-number 'proofs' by Proth,
      and with me examining Proth's theorem].
      ...
      now, if 'q' is any prime divisor of 'S', then b^((R-1)/2)= (b^k)^
      (2^(n-1)) == +1 (mod q) implies that q == +1 (mod 2^n).
      ...
      thus, if 'S' is composite, 'S' will be the product of at least two
      primes each of which may have a maximum value of (2^n -1), and it
      follows that...
      ...
      k*2^n +1 = (2^n)*(2^n) -1 = (2^n +1)*(2^n -1) <= (2^n)*(2^n) -2*(2^n)
      +1; implies that -2 <= -2*(2^n) and dividing by 2^n and multiplying
      by -1, we have.. 1 > 2^n, contradiction!
      ...
      hence, for some 'b', if k <= 2^n -1 and b^((R-1)/2) == +1 (mod R),
      then 'R' is prime.
      *QED
      ...

      Warm regards,

      Bill Bouris
      the proof can be found at...
      www.oddperfectnumbers.com
    • Bill Bouris
      oh yeah,... I forgot to mention that b
      Message 2 of 8 , Sep 14, 2011
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        oh yeah,... I forgot to mention that b <= 2*ln(R).
        it's further implications... when k=1, means that Mersenne
        numbers could be found prime with a single non-pseudo-
        prime test (eg. 2^15 mod 31 == 1; you're right; 'b' needs
        to be restricted similar to 'k'. Bill (happy hunting!)

        From: Bernardo Boncompagni <redgolpe@...>
        To: leavemsg1 <leavemsg1@...>
        Sent: Wednesday, September 14, 2011 2:08 AM
        Subject: Re: [PrimeNumbers] woodall or riesel number theorem





        On Wed, Sep 14, 2011 at 8:22 AM, leavemsg1 <leavemsg1@...> wrote:

         
        >let R= k*2^n -1, n >1 is a natural number and k <= 2^n-1.
        >If for some 'b', b^((R-1)/2) == +1 (mod R), then 'R' is prime;
        >(similar to Proth's theorem.)
        >

        n=4, k=11, R=175, b=51,
        b^87==1 (mod 175)

        [Non-text portions of this message have been removed]
      • leavemsg1
        duh? I thought about it in the car while running some errands; the remedy is that along with gcd(R -1, b)= 1, because gcd(174, 51) = 3. that is the ONLY thing
        Message 3 of 8 , Sep 14, 2011
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          duh? I thought about it in the car while running some
          errands; the remedy is that along with gcd(R -1, b)= 1,
          because gcd(174, 51) = 3. that is the ONLY thing keep-
          ing it from working in your example... enjoy!

          --- In primenumbers@yahoogroups.com, Bill Bouris <leavemsg1@...> wrote:
          >
          > oh yeah,... I forgot to mention that b <= 2*ln(R).
          > it's further implications... when k=1, means that Mersenne
          > numbers could be found prime with a single non-pseudo-
          > prime test (eg. 2^15 mod 31 == 1; you're right; 'b' needs
          > to be restricted similar to 'k'. Bill (happy hunting!)
          >
          > From: Bernardo Boncompagni <redgolpe@...>
          > To: leavemsg1 <leavemsg1@...>
          > Sent: Wednesday, September 14, 2011 2:08 AM
          > Subject: Re: [PrimeNumbers] woodall or riesel number theorem
          >
          >
          >
          >
          >
          > On Wed, Sep 14, 2011 at 8:22 AM, leavemsg1 <leavemsg1@...> wrote:
          >
          >  
          > >let R= k*2^n -1, n >1 is a natural number and k <= 2^n-1.
          > >If for some 'b', b^((R-1)/2) == +1 (mod R), then 'R' is prime;
          > >(similar to Proth's theorem.)
          > >
          >
          > n=4, k=11, R=175, b=51,
          > b^87==1 (mod 175)
          >
          > [Non-text portions of this message have been removed]
          >
        • Bill Bouris
          Bernard,   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
          Message 4 of 8 , Sep 14, 2011
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            Bernard,
             
            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!
             
            Rewards,
             
            Bill
             
            From: Bill Bouris <leavemsg1@...>
            To: Bernardo Boncompagni <redgolpe@...>; pgroup <primenumbers@yahoogroups.com>
            Sent: Wednesday, September 14, 2011 12:51 PM
            Subject: Re: [PrimeNumbers] woodall or riesel number theorem


            oh yeah,... I forgot to mention that b <= 2*ln(R).
            it's further implications... when k=1, means that Mersenne
            numbers could be found prime with a single non-pseudo-
            prime test (eg. 2^15 mod 31 == 1; you're right; 'b' needs
            to be restricted similar to 'k'. Bill (happy hunting!)

            From: Bernardo Boncompagni <redgolpe@...>
            To: leavemsg1 <leavemsg1@...>
            Sent: Wednesday, September 14, 2011 2:08 AM
            Subject: Re: [PrimeNumbers] woodall or riesel number theorem





            On Wed, Sep 14, 2011 at 8:22 AM, leavemsg1 <leavemsg1@...> wrote:

             
            >let R= k*2^n -1, n >1 is a natural number and k <= 2^n-1.
            >If for some 'b', b^((R-1)/2) == +1 (mod R), then 'R' is prime;
            >(similar to Proth's theorem.)
            >

            n=4, k=11, R=175, b=51,
            b^87==1 (mod 175)

            [Non-text portions of this message have been removed]
          • Bernardo Boncompagni
            It was failing ONLY because the gcd(R -1, b) = 1 needs to be a part of it. ... n=5, k=21, R=671=11*61, b=9
            Message 5 of 8 , Sep 14, 2011
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              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]
            • Bill Bouris
              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
              Message 6 of 8 , Sep 14, 2011
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                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]
              • Bill Bouris
                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
                Message 7 of 8 , Sep 14, 2011
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                  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]
                • leavemsg1
                  n 2 also...
                  Message 8 of 8 , Sep 14, 2011
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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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