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Old 2008-02-01, 23:55   #12
fivemack
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I'm pretty sure that your conjecture for the j=3 case is true - you don't get that many one-in-three 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 number-theory 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^p-1

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 2008-02-01 at 23:58
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Old 2008-02-02, 12:47   #13
olivier_latinne
 
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Default Mersenne conjecture

For a fixed j, for M(p)=2^p-1 and p prime:
let define the set of all potential divisors like the numbers that verify:
  • mod(2*p*j+1,8)=7
  • 2*p*j+1 prime
These are the "classical" restrictions that we know for 3 century.
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
  • M+=2^p+1, try also the case j=3, you only need to change the condition: mod(2*p*j+1,8)=3 and you will get all the divisors.
  • For mod(j,3)=0 and j>3
  • Try the case j=4 and mod(j,4)=0 (table 6 page 7).
  • You can check that there is no other diophantine equation of order 2 that will lead to new restrictions.
  • You can extend that (conjecture 2 and 3) for unspecified p.
  • You can extend that (conjecture 2 and 3) also for Cunningham numbers.
So, with conjecture 2 and 3 we are at the point that we understand better where does come from the Mersenne divisors for j=1 (Sophie Germain prime), j=2 is forbidden (for M+-), for j=3 (conjecture 2) and for j=4 (conjecture 3).
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:
Originally Posted by fivemack View Post
I'm pretty sure that your conjecture for the j=3 case is true - you don't get that many one-in-three 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 number-theory 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^p-1

and I can't work out what your conjecture is trying to say for any case other than j=3.
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Old 2008-02-03, 04:54   #14
jasong
 
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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 (x-i)^2=(i-x)^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?
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Old 2008-02-03, 11:23   #15
xilman
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Quote:
Originally Posted by jasong View Post
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 (x-i)^2=(i-x)^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?
Note that (x-i) = -(i-x) and that for all y, y^2 = (-y)^2, thus proving your conjecture.


Paul
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Old 2008-02-03, 20:28   #16
m_f_h
 
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Quote:
Originally Posted by xilman View Post
Note that (x-i) = -(i-x) and that for all y, y^2 = (-y)^2, thus proving your conjecture.
Paul
The straightforward expansion, using only distributivity, shows that (a-b)^2=(b-a)^2 for any two elements a,b in any (not necessarily associative) algebra.
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Old 2008-02-03, 20:44   #17
m_f_h
 
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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 2008-02-03 at 20:59
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Old 2008-02-06, 11:42   #18
olivier_latinne
 
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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:
Originally Posted by fivemack View Post
I'm pretty sure that your conjecture for the j=3 case is true - you don't get that many one-in-three 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 number-theory 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^p-1

and I can't work out what your conjecture is trying to say for any case other than j=3.
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Old 2008-02-07, 09:32   #19
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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:
  • to explain or demonstrate the square free conjecture.
  • to maybe speed up the search of factors of Cunningham numbers, includind Fermat numbers.
  • If we want to know if C5=M(M(7)) (Catalan serie) is prime or not, I don't think it would be ever with the Lucas Lehmer test.
  • ...
I think it would be also a chalenge only to demonstate the three conjectures.
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
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Old 2008-02-12, 09:20   #20
olivier_latinne
 
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Default 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^p-1 if and only if
  • d=2*p*j+1 is prime
  • mod(d,8)=7
  • p prime
  • and there exists integer n and i such that: d=4*n^2 + 3*(3+6*i)^2
Best Regards,

Olivier
Attached Files
File Type: pdf conjecture2.pdf (920.6 KB, 239 views)
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Old 2008-02-13, 05:09   #21
m_f_h
 
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Quote:
Originally Posted by olivier_latinne View Post
d=2*p*j+1 divide M(p)=2^p-1 if and only if
  • d=2*p*j+1 is prime
  • mod(d,8)=7
  • p prime
  • and there exists integer n and i such that: d=4*n^2 + 3*(3+6*i)^2
I don't understand why you leave j here :
this is supposed to be true only for j=3 ?!
Since e.g. 2^20-1 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^p-1 if and only if
  • d=6p+1 is prime
  • mod(d,8)=7 [or, a bit simpler (subtract 1 and divide by 2):
    3p=3 (mod 4), i.e.: p=1 (mod 4)]
  • p prime
  • and there exist integers n and i such that: d=4*n^2 + 3*(3+6*i)^2
But this is clearly wrong:

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

???
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Old 2008-02-13, 06:34   #22
retina
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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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