20080201, 23:55  #12 
(loop (#_fork))
Feb 2006
Cambridge, England
2·3,191 Posts 
I'm pretty sure that your conjecture for the j=3 case is true  you don't get that many oneinthree chances happening by chance  though I'm not quite sure how to prove that you're always hitting cubic residues: I suspect the i(i+1) is disguising a norm in the Eisenstein integers. I think proving the conjecture would be an interesting problem to set in an undergraduate numbertheory problem sheet.
It took me a little work to decipher the conjecture for j=3  you really need to write this sort of thing without forwards references, so: Let n and i be integers with 3 not dividing n (this condition on n is required for your p to be even integral) Then, if p=(2n^2 + 1)/3 + (18i^2 + 18i + 4) is prime, and d=6p+1 is also prime and d is congruent to 7 mod 8 then d divides 2^p1 and I can't work out what your conjecture is trying to say for any case other than j=3. Last fiddled with by fivemack on 20080201 at 23:58 
20080202, 12:47  #13  
Jan 2008
2·11 Posts 
Mersenne conjecture
For a fixed j, for M(p)=2^p1 and p prime:
let define the set of all potential divisors like the numbers that verify:
Now, let define the set of "allowed divisors" (from table 4 page 5): set of "allowed divisors" = set of "all potential divisors"  set of "forbidden divisor" So, for the particular case j=3, the set of "allowed divisors" correspond to the set of "all divisors". for j>3 (and of course mod(j,3)=0) the set of "allowed divisors" contain not only the set of "all divisors" but also numbers that are potentialy but not divisors. This set of "allowed divisors" is of course always smaller that the set of "all potential divisors". Now that you have understood the case of j=3, you can made test for
But for j=5 there is no new restrictions with diophantine equation of order 2, so it was really a quest. I was convinced that there was an ultimated and unified theory for every j and for every base (C+=b^p+1) and I have finally obtain conjecture 1 (page 4). Quote:


20080203, 04:54  #14 
"Jason Goatcher"
Mar 2005
5·701 Posts 
I know nothing about number theory, so I'm going to chime in about "semantics."
I think a problem that may show up a lot on these forums is that the people who show up with "conjectures" aren't necessarily wrong, but simply have a looser definition of what a conjecture is. Obviously, some or all of them are badly wrong. But do people deserve to be called cranks simply because they're proud that they discovered some particular quality of the mathematics they're involved in and want to "show off?" A particular property of a number or equation may be obvious to some people, but not to everybody else. For example, I recently discovered that (xi)^2=(ix)^2 for every combination I've tried. This may be basic theory for imaginary numbers, it might even have a name for itself in one of the textbooks I have. But that doesn't change the fact that I discovered something fascinating without anyone giving me hints. If an aborigine has never seen a wheel, but invents a simple cart, does he deserve to be ridiculed and scorned for not already being aware of it's existence? 
20080203, 11:23  #15  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
3×3,529 Posts 
Quote:
Paul 

20080203, 20:28  #16 
Feb 2007
1B0_{16} Posts 

20080203, 20:44  #17 
Feb 2007
1B0_{16} Posts 
Olivier:
in eq. (1) you /define/ the d_j as divisors of C, so the statement "they are divisors of C" is true so this statement is equivalent to anything which is true thus "they are divisors of C" if and only if [anything which is true]. This is your conjecture 1, where you put for [anything which is true] some conditions which are verified for these divisors (afaics). Thus, your conjecture is correct, but (afaics) does not give any new information (because it is subordinate to the condition that d_j IS a divisor of C). PS: I don't claim that there is nothing of interest in the paper, but having slept about 1 hour per day on the average during the last week, I am not in condition to find it. Last fiddled with by m_f_h on 20080203 at 20:59 
20080206, 11:42  #18  
Jan 2008
10110_{2} Posts 
Hi Fivemack!
Have you made some new verification other than j=3 (conjecture 2)? Have you try to verify conjecture 1 (the ultimate and unified formulation)? Quote:


20080207, 09:32  #19 
Jan 2008
2·11 Posts 
Call
Hi everyone!
Until now, only conjecture no 2 has been checked by Fivemack. It would be of course very interesting that someone try to verify the two other conjectures of my paper. An "extraordinary" claim need always to be verified by other people. This work could have big implications for instance:
Conjecture 2 like Fivemack says, seems to him "correct" but he had no demonstration to give. Pr H.W. Lenstra think also in this direction: "... This is, partly, due to one of the difficulties that you refer to yourself already, namely that it is exceedingly hard to come up with rigorous proofs of observed properties of Mersenne numbers. " Best Regards, Olivier 
20080212, 09:20  #20 
Jan 2008
2×11 Posts 
Numerical exemple (conjecture no 2)
Hi everyone!
At the request of Tony Reix, you will find in attachment a pdf file that contain a numerical exemple for conjecture 2, for the special case j=3. All the potential divisors are examined up to p=2000000. I recall conjecture 2 (for j=3 and M): d=2*p*j+1 divide M(p)=2^p1 if and only if
Olivier 
20080213, 05:09  #21  
Feb 2007
2^{4}·3^{3} Posts 
Quote:
this is supposed to be true only for j=3 ?! Since e.g. 2^201 is a counter example (d=2*20*1+1 divides, but p=20 is not prime) So it should be rewritten: d=6p+1 divides M(p)=2^p1 if and only if
p=21 is not prime, but 6p+1=127 divides Mp=2097151 p=72 is not prime, but 6p+1=433 divides Mp=4722366482869645213695 p=72 != 1 (mod 4), but 6p+1=433 divides Mp=4722366482869645213695 p=76 is not prime, but 6p+1=457 divides Mp=75557863725914323419135 p=76 != 1 (mod 4), but 6p+1=457 divides Mp=75557863725914323419135 ??? 

20080213, 06:34  #22 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3^{5}·5^{2} Posts 
m_f_h: I thought p was meant to be prime? So your examples where p != prime would seem to be not relevant?

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