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New Mersenne numbers conjecture

  1. Feb 29, 2008 #1
    Hi everyone!

    Is anyone able to find the demonstration of the following Mersenne conjecture?

    for j=3, d=2*p*j+1=6*p+1 divide M(p)=2^p-1 if and only if

    d is prime
    and mod(d,8)=7
    and p prime
    and there exists integer n and i such that: d=4*n^2 + 3*(3+6*i)^2

    This conjecture has been numericaly tested for p up to 10^11 and is a particular case of one of three new Mersenne and Cunningham conjectures that I have introduced in the Math Mersenne numbers forum four weeks ago (http://mersenneforum.org/showthread.php?t=9945)
    But unfortunately up to now, no one of the three conjectures has been demonstrated. On this forum you will also find one numerical example (pdf file) for each of the three conjectures (see thread #20, #25 and #38)
    Best Regards,
    Olivier Latinne
     
  2. jcsd
  3. Mar 1, 2008 #2
    Counter-example: p=13, d=79. d divides 2^p-1, but there are no integers n,i that make d.
     
  4. Mar 1, 2008 #3
    for p=13, d=79 but mod(2^13-1)%79=54 so d=79 is not a divisor of 2^13-1
    So, it is not a counter example !
     
  5. Mar 4, 2008 #4
    Hot off the press in the forum The poster linked
    Zhi-Wei SUN to NMBRTHRY - Mar 1, 2008

    Dear number theorist,

    Olivier Latinne conjectured that
    d=6p+1 divides 2^p-1 if and only if
    p is a prime with 6p+1=7 (mod 8) (i.e., p=1(mod 4))
    and d=6p+1 is a prime in the form x^2+27y^2.
    (since d=3 (mod 4), x cannot be odd).

    Below I prove this conjecture for any prime p.

    Proof of the "if" part. Let p=1 (mod 4) be a prime such that d=6p+1 is a
    prime in the form x^2+27y^2. By Corollary 9.6.2 of K. Ireland and M.
    Rosen's book "A Classical Introduction to Modern Number Theory" [GTM
    84, Springer, 1990], 2 is a cubic residue mod d, and hence d divides
    2^{(d-1)/3}-1=(2^p-1)(2^p+1). If 2^p=-1 (mod d), then
    1^p=(2/d)^p=(-1/d)=-1, a contradiction! So we have 2^p=1 (mod d).

    Proof of the "only if" part. Suppose that p is a prime and d=6p+1
    divides 2^p-1. If d=6p+1 is composite then it has a prime factor q not
    exceeding sqrt(6p+1)<2p+1. But any prime divisor of 2^p-1 is of the form
    2pk+1, so d=6p+1 must be a prime.
    As d=1 (mod 3) and 2^{(d-1)/3}=2^{2p}=1 (mod d), again by Corollary
    9.6.2 of Ireland and Rosen's book, d can be written in the form
    x^2+27y^2. Since d=3 (mod 4), x must be even.

    Zhi-Wei Sun
     
  6. Mar 5, 2008 #5
    a demonstration for the particular case j=4 of conjecture #3 has also been found

     
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