New Mersenne numbers conjecture

In summary, Olivier Latinne has introduced a Mersenne conjecture that states that for j=3, d=2*p*j+1=6*p+1 divide M(p)=2^p-1 if and only if d is prime, mod(d,8)=7, p prime, and there exists an integer n and i such that d=4*n^2 + 3*(3+6*i)^2. This conjecture has been tested numerically and is a particular case of one of three new Mersenne and Cunningham conjectures. However, none of the three conjectures have been demonstrated yet. Zhi-Wei Sun has proven a similar conjecture for the particular case j=4.
  • #1
o.latinne
3
0
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
 
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  • #2
Counter-example: p=13, d=79. d divides 2^p-1, but there are no integers n,i that make d.
 
  • #3
Dodo said:
Counter-example: p=13, d=79. d divides 2^p-1, but there are no integers n,i that make d.

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 !
 
  • #4
o.latinne said:
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
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
 
  • #5
a demonstration for the particular case j=4 of conjecture #3 has also been found

Zhi-Wei SUN to NMBRTHRY, me

Dear number theorists,
Here I give a simple proof of another conjecture of Olivier Latinne.
THEOREM. Let p be a prime, and let d=8p+1. Then d divides 2^p-1
if and only if d is a prime in the form x^2+64y^2 with y odd.
Proof. If d=8p+1 divides 2^p-1 then d cannot be composite since
sqrt(8p+1)<2p+1 and any prime divisor of 2^p-1 is =1 (mod 2p).
Observe that d=9 (mod 16). When d is a prime, by Corollary 7.5.8 in
the book [B.C. Berndt, R.J. Evans and K.S. Williams, Gauss and Jacobi
Sums, Wiley, 1998], d is in the form x^2+64y^2 with x,y odd if and only
if 2 is an octic residue mod d (i.e., 2^p=2^{(d-1)/8}=1 (mod d)). So the
desired result follows.

Zhi-Wei Sun
http://math.nju.edu.cn/~zwsun
 

1. What is the "New Mersenne numbers conjecture"?

The "New Mersenne numbers conjecture" is a mathematical conjecture that states that there are infinitely many Mersenne numbers that are also prime numbers. A Mersenne number is a number of the form 2n-1, where n is a positive integer.

2. Who came up with the "New Mersenne numbers conjecture"?

The "New Mersenne numbers conjecture" was proposed by mathematician Marin Mersenne in the 17th century. However, the conjecture was revised and expanded upon by mathematicians Ernst Kummer and Édouard Lucas in the 19th century.

3. Has the "New Mersenne numbers conjecture" been proven?

No, the "New Mersenne numbers conjecture" has not been proven. It remains an open mathematical problem and is considered one of the most difficult and important unsolved problems in mathematics.

4. What is the significance of the "New Mersenne numbers conjecture"?

The "New Mersenne numbers conjecture" has significant implications in number theory and prime number research. If proven to be true, it would provide a better understanding of the distribution of prime numbers and could potentially lead to new discoveries in the field.

5. Are there any known examples of Mersenne primes?

Yes, there are currently 51 known Mersenne primes. The largest known Mersenne prime is 282,589,933-1, which has over 24 million digits and was discovered in 2018. However, it is not known if there are infinitely many Mersenne primes, which is what the "New Mersenne numbers conjecture" aims to prove.

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