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February 29th, 2008, 04:27 PM  #1 
Newbie Joined: Feb 2008 From: Brussels Posts: 3 Thanks: 0  New Mersenne numbers conjecture
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^p1 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 ) Regards, Olivier Latinne 
March 1st, 2008, 10:10 AM  #2  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: New Mersenne numbers conjecture
I don't understand what you mean by "demonstration". Quote:
But if I change the conjecture to Given a prime p and d = 6p + 1, then d  M(p) if and only if
 
March 2nd, 2008, 07:10 AM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
I see that Dr. Sun solved it on the NMBRTHRY list.

March 2nd, 2008, 12:27 PM  #4  
Newbie Joined: Feb 2008 From: Brussels Posts: 3 Thanks: 0  Quote:
For any prime p: d=2*p*j+1 divide M(p)= 2^p1 if and only if there exists integers x, y, m and k, with x and y coprime, m and j coprime, mod(y,2) ≠ 0 and mod(m,p) ≠ 0, such that: for j odd : d= (2^m * x^j + y^j) / k is prime and mod(d,8 )=7 for j even: d= (2^m * x^(2*j)  y^(2*j) ) / k is prime and mod(d,8 )=1 This conjecture can be generalized to Cunningham numbers by replacing base 2 by base b of C= b^p1 and modifying adequately the modularities of d Olivier Latinne  
March 2nd, 2008, 07:58 PM  #5  
Global Moderator Joined: Dec 2006 Posts: 18,037 Thanks: 1394 
I find it odd that o.latinne is using the uncorrected form of his conjecture (both here and elsewhere) even though he seemed to agree 16 days ago in the forum he linked to that the correction given above is needed. Quote:
 
March 2nd, 2008, 08:23 PM  #6  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Ah, no link, where are my manners. Here you go: http://listserv.nodak.edu/cgibin/wa.ex ... &T=0&P=163 Quote:
 
March 2nd, 2008, 09:29 PM  #7  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Quote:
Cheers.  
March 12th, 2008, 12:40 AM  #8  
Newbie Joined: Feb 2008 From: Brussels Posts: 3 Thanks: 0  Last Mersenne conjecture solved?
Hi everyone! Pr Sun provide a demonstration for conjecture #1 of my paper. Can anyone explain me in details (it's maybe obvious, but not for me) the following transition of that demonstration: " ... Observe that x^{2j}=2 (mod d) is solvable if and only if 2^p=2^{(d1)/(2j)}=1 (mod d)." Best Regards, Olivier Quote:
 

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