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Re: [PrimeNumbers] new Mersenne Divisors conj. ???

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  • Bill Bouris
    Theorem:  if p= 1 (mod 4) is prime, then 2*p +1 is also prime iff 2*p +1 divides Mp +2; a complimentary proof to that of Euler s result for Mersenne
    Message 1 of 4 , Jan 16, 2010
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      Theorem:  if p= 1 (mod 4) is prime, then 2*p +1 is also prime iff
      2*p +1 divides Mp +2; a 'complimentary' proof to that of Euler's
      result for Mersenne Divisors [shown by modifying Lagrange's proof
      of the original if p= 3 (mod 4),...].
      proof(forward): Suppose that q= 2*p +1 is prime; q= 3 (mod 8) and
      2 is a quadratic residue modulo q, and there exists an integer 'n'
      such that n^2= 4 (mod q). Thus, 2^p = 2^((q-1)/2) = n^(q-1) = -1
      (mod q), and adding 2 to each portion of this equation shows that
      (2*p +1) | (Mp +2).
      proof(backward): Let 2*p +1 be a composite factor of Mp +2 and let
      'q' be the least prime factor; 2^p = 1 (mod q) and the order of 2
      divides both p and q-1; so, the conclusion is the same as the orig-
      inal Lagrange proof, (2*p +1) +1 > q^2 > p^2 contradicts that p> 2.
       
      I think that proves it. Cheers, Bill



      --- On Tue, 1/12/10, Bill Bouris <leavemsg1@...> wrote:


      From: Bill Bouris <leavemsg1@...>
      Subject: [PrimeNumbers] new Mersenne Divisors conj. ???
      To: "pgroup" <primenumbers@yahoogroups.com>
      Date: Tuesday, January 12, 2010, 9:10 AM

      I think that I found a new Mersenne Divisors conjecture ???; I couldn't
      locate it on the web, but that doesn't mean anything.

      Let p == 1(mod 4) be prime. 2p +1 is also prime iff 2p +1 divides Mp +2.
      p = 5, 29, 41, 53, 89, etc. satisfy the conjecture.

      Does anyone like it ??? Bill Mp for p= 5 & 89 happens to be prime.
    • djbroadhurst
      ... http://primes.utm.edu/notes/proofs/MerDiv2.html proves that if p = 3 mod 4 and p is prime, then 2p+1 is prime iff 2p+1|2^p-1 Bill claims that if p = 1 mod
      Message 2 of 4 , Jan 16, 2010
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        --- In primenumbers@yahoogroups.com,
        Bill Bouris <leavemsg1@...> wrote:

        > I think that proves it

        http://primes.utm.edu/notes/proofs/MerDiv2.html proves that
        if p = 3 mod 4 and p is prime,
        then 2p+1 is prime iff 2p+1|2^p-1

        Bill claims that
        if p = 1 mod 4 and p is prime,
        then 2p+1 is prime iff 2p+1|2^p+1

        and his facsimile proof looks good to me.

        If Chris agrees, than he might like to add
        Bill's case to the Euler-Lagrange page, above.

        David
      • djbroadhurst
        ... Henri has remarked (off list) that he had already given essentially the same proof as Bill, in 1998: http://www.primenumbers.net/Henri/us/NouvTh1us.htm
        Message 3 of 4 , Jan 16, 2010
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          --- In primenumbers@yahoogroups.com,
          "djbroadhurst" <d.broadhurst@...> wrote:

          > http://primes.utm.edu/notes/proofs/MerDiv2.html proves that
          > if p = 3 mod 4 and p is prime,
          > then 2p+1 is prime iff 2p+1|2^p-1
          >
          > Bill claims that
          > if p = 1 mod 4 and p is prime,
          > then 2p+1 is prime iff 2p+1|2^p+1
          >
          > and his facsimile proof looks good to me.

          Henri has remarked (off list) that he had already given
          essentially the same proof as Bill, in 1998:

          http://www.primenumbers.net/Henri/us/NouvTh1us.htm

          It's always good to see clear minds agreeing.

          David
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