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February 29th, 2008, 04:27 PM   #1
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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^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 )
Regards,
Olivier Latinne
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March 1st, 2008, 10:10 AM   #2
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Re: New Mersenne numbers conjecture

I don't understand what you mean by "demonstration".

Quote:
Originally Posted by o.latinne
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
p = 21, d = 127: 127 | M(21), but 21 is not prime. There are many similar examples.

But if I change the conjecture to

Given a prime p and d = 6p + 1, then d | M(p) if and only if
  • d ≡ 7 (mod 8)[/*:m:2ywqvqyb]
  • d is prime[/*:m:2ywqvqyb]
  • There exist nonnegative integers n and i with d = 4n^2 + 3(3 + 6i)^2[/*:m:2ywqvqyb]
then the conjecture seems to hold for p up to 10 million. (I verified this in 70 seconds with Pari/GP.)
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March 2nd, 2008, 07:10 AM   #3
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I see that Dr. Sun solved it on the NMBRTHRY list.
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March 2nd, 2008, 12:27 PM   #4
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Quote:
Originally Posted by CRGreathouse
I see that Dr. Sun solved it on the NMBRTHRY list.
Dr. Sun has solved conjecture #2 and #3 but conjecture #1 remain unsolved:

For any prime p:
d=2*p*j+1 divide M(p)= 2^p-1 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^p-1
and modifying adequately the modularities of d

Olivier Latinne
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March 2nd, 2008, 07:58 PM   #5
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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:
Originally Posted by CRGreathouse
I see that Dr. Sun solved it on the NMBRTHRY list.
Link?
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March 2nd, 2008, 08:23 PM   #6
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Ah, no link, where are my manners. Here you go:
http://listserv.nodak.edu/cgi-bin/wa.ex ... &T=0&P=163

Quote:
Originally Posted by skipjack
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.
Yes, I noticed his similar posts on three other forums.
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March 2nd, 2008, 09:29 PM   #7
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Quote:
Originally Posted by skipjack
Link?
http://listserv.nodak.edu/cgi-bin/wa.ex ... &T=0&P=163

Cheers.
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March 12th, 2008, 12:40 AM   #8
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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^{(d-1)/(2j)}=1 (mod d)."

Best Regards,

Olivier

Quote:
Dear number theorists,

If (j,m)=1, then 1=aj+bm for some integers a and b. If j is even and
(j,m)=1, then we also have (2j,m)=1. Thus the latest conjecture of
Olivier Latinne can be restated in the following simple form that I will
prove.

THEOREM. Let p be an odd prime and let d=2pj+1 (0<j<2p+2).

(1) When j is odd, d divides 2^p-1 if and only if d is a prime with d=7
(mod 8 ) and x^j=2 (mod d) is solvable.

(2) When j is even, d divides 2^p-1 if and only if d is a prime with d=1
(mod 8 ) and x^{2j}=2 (mod d) is solvable.

Proof. Suppose that d|2^p-1. If d is composite, then d has a prime
divisor q not exceeding sqrt(2pj+1)<2p+1 but any prime divisor of 2^p-1
has the form 2pk+1. So d must be a prime. As 2^p=1 (mod d) we have
1=(2/d)^p=(2/d). So d=1 or 7 (mod 8 ). If j is odd then d=3 (mod 4) and
hence d=7 (mod 8 ). If j is even, then d=1 (mod 4) and hence d=1 (mod 8 ).

Now assume that d is a prime with d=(-1)^j (mod 8 ). Observe that
x^{2j}=2 (mod d) is solvable if and only if 2^p=2^{(d-1)/(2j)}=1 (mod
d). So (2) holds. Note that x^j=2 (mod d) is solvable if and only if
2^{2p}=2^{(d-1)/j}=1 (mod d), i.e., d|(2^p-1)(2^p+1). If j is odd, then
d=7 (mod 8 ) and (2/d)^p=1 is different from (-1/d)=-1, and thus d does
not divide 2^p+1. So, when j is odd, x^j=2 (mod d) is solvable if and
only if d|2^p-1.

Zhi-Wei Sun

http://math.nju.edu.cn/~zwsun
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