mersenneforum.org  

Go Back   mersenneforum.org > Great Internet Mersenne Prime Search > Math

Reply
 
Thread Tools
Old 2005-09-10, 20:59   #1
T.Rex
 
T.Rex's Avatar
 
Feb 2004
France

2×33×17 Posts
Default A property of prime Mersenne numbers under LLT

Hi,
I've found (by computation) a property that looks interesting.

This property applies only to Mersenne numbers M_q such that q \equiv 1 \pmod{4}.
This properties is true for q=5,13,17 and false for q=29 .
(Why so few examples ? Because it takes hours or days to find these numbers !)

For q=5,13,17, there is a number R that has the following properties:

R^2-2 \equiv -(R+1) \pmod{M_q} (1)

(R+1)^2-2 \equiv R \pmod{M_q} (2)

R*(R+1) \equiv 1 \pmod{M_q} (3)


q=5 -> R=12
q=13 -> R=394
q=17 -> R=41127


For q=29 there are 4 numbers R such that R^2-2 \equiv -T \pmod{M_q} and T^2-2 \equiv R \pmod{M_q} :
874680 , 37882537 , 137237467 , 199174227 .
But T is not equal to R+1 , and R*T (mod M_q) is not equal to 1.


So, I have the following conjecture:

For q=1 (mod 4) , if there exists 1 and only 1 number that verifies properties (1), (2) and (3) , then M_q is prime .


Very nice ! Isn't it ?


The only very small problem is: HOW CAN WE FIND THESE NUMBERS R ?
(I have NO idea yet !)

If one can find the formula that generates R, then we have a VERY fast test for Mersenne numbers ! (but I guess the cost of finding R is comparable to the LLT ... so this would be of no real use.)


Any help is welcome !

Regards,
Tony

Last fiddled with by T.Rex on 2005-09-10 at 21:05
T.Rex is offline   Reply With Quote
Old 2005-09-10, 21:18   #2
T.Rex
 
T.Rex's Avatar
 
Feb 2004
France

2×33×17 Posts
Default PARI/gp code

A simple PARI/gp code for finding R is :

q=13;M=2^q-1
for(i=1,(M-1)/2, j=(i^2-2)%M; if((M-j)==i+1, print(i)))

For a great M, one should start i with a number such that i^2> M .
T.Rex is offline   Reply With Quote
Old 2005-09-10, 22:54   #3
cyrix
 
Jul 2003
Thuringia; Germany

2·29 Posts
Default

Hi T.Rex!

Your Conjecture is false!

All conditions are equal: R^2+R-1=0 mod M, this means R_{1/2}= -1/2 +- \sqrt{1/4+1}= (-1+-sqrt(5))/2.

For q=5 we have M=31 and sqrt(5)= +-6, thus we have R_1=(-1-6)/2=12 AND R_2=(-1+6)/2=18. So For M=31 R_2=18 is also a solution!


Cyrix
cyrix is offline   Reply With Quote
Old 2005-09-11, 09:30   #4
T.Rex
 
T.Rex's Avatar
 
Feb 2004
France

2×33×17 Posts
Default A clearer conjecture

Quote:
Originally Posted by cyrix
Your Conjecture is false!
Not really. It was unclear on some points and not reduced. Thanks for your help !

Quote:
All conditions are equal: R^2+R-1=0 mod M, this means R_{1/2}= -1/2 +- \sqrt{1/4+1}= (-1+-sqrt(5))/2.
You are perfectly right !
I found this too after I switched off my PC and read my post quietly.
That means properties (1), (2) and (3) are the same:

\ \ \ R^2+R-1 \equiv 0 \ \ \pmod{M_q} (P) .

Quote:
For q=5 we have M=31 and sqrt(5)= +-6, thus we have R_1=(-1-6)/2=12 AND R_2=(-1+6)/2=18. So For M=31 R_2=18 is also a solution!
Not really.
Since 18 \equiv -13 = -(12+1) \ \ \pmod{M_q} 18 is the "same" solution than 12.
In fact, if you replace R by -(R+1) in property (P), you have: (-(R+1))^2+(-(R+1))-1 = R^2+2R+1-R-2=R^2+R-1 .
So, if R is a solution of (P), then -(R+1) is also a solution.

So I propose to reformulate the conjecture:

For \ q \equiv 1\ \ \pmod{M_q} , if there exists one and only one number R that verifies the property (P), then M_q is prime.
-(R+1) is called the dual solution of (P).


Do you agree with this new conjecture ?


Now, we have 2 problems:
- provide a proof !
- find a way to build this mysterious number R !

Help is welcome !


Also, finding R for other q would be great !
But the next one is: 89 . It may take months or years of computation before finding R_89 ...

Regards,
Tony

Last fiddled with by T.Rex on 2005-09-11 at 09:40
T.Rex is offline   Reply With Quote
Old 2005-09-11, 10:47   #5
cyrix
 
Jul 2003
Thuringia; Germany

2·29 Posts
Default

Hi T.Rex!

For prime M_q with  q \equiv 1 \pmod 4 (and q>1) your conjecture is true: Because of the quadratic reciprocity law (Gauß) 5 is a quadratic residue of  M_q , iff  M_q is quadratic residue of 5 (because both are of the from 4n+1). But  M_q=2^q-1=2^{4n+1}-1 \equiv 2^1-1=1^2 \pmod 5 so 5 is a quadratic residue of  M_q and there existists two numbers X and Y, which have the properties  X \equiv -Y \pmod {M_q} and  x^2 \equiv y^2 \equiv 5 \pmod {M_q}


Yours,
Cyrix
cyrix is offline   Reply With Quote
Old 2005-09-11, 13:21   #6
T.Rex
 
T.Rex's Avatar
 
Feb 2004
France

2×33×17 Posts
Default I do not understand the link

Hi Cyrix,
Sorry, I do not see the link between (5/M_q) and the conjecture.

I know the quadratic reciprocity and I understand your explanations.
But, are you saying that M_q is of the form 4n+1 ?

What is the link between (5/M_q) and what I said ?

Tony
T.Rex is offline   Reply With Quote
Old 2005-09-11, 15:00   #7
cyrix
 
Jul 2003
Thuringia; Germany

2·29 Posts
Default

sorry Tony!
I messed something up. Now in a better form:

Quote:
Originally Posted by cyrix
Hi T.Rex!

For prime M_q with  q \equiv 1 \pmod 4 (and q>1) your conjecture is true: Because of the quadratic reciprocity law (Gauß) 5 is a quadratic residue of  M_q , iff  M_q is quadratic NON residue of 5 (because 5=4*1+1 and  M_q=4*n+3 ).
 M_q \equiv 3 \pmod 4 , but since  5 \equiv 1 \pmod 4 the reciprocity law works in the same way, So you can find a solution X with  X^2 \equiv 5 \pmod {M_q} . This means, you can find the two solutions R_1 and R_2 with  R^2+R-1 \equiv 0 \pmod {M_q} because of the formel in my first post. (with the second solution  Y^2 \equiv 5 \pmod {M_q} you get the same solutions R_2, R_1)

Your conjecture reduces to: 5 is a quadratic residue of  M_q with q a prime and  q \equiv 1 \pmod 4 , iff  M_q is prime itself.

Yours,
Cyrix

Last fiddled with by cyrix on 2005-09-11 at 15:01
cyrix is offline   Reply With Quote
Old 2005-09-11, 16:56   #8
T.Rex
 
T.Rex's Avatar
 
Feb 2004
France

2·33·17 Posts
Default Clear now

OK. I understand your points now.
So, seems we have a conjecture for a Pépin-like test for Mersenne numbers ?!
Is it something new ? I've searched in my books and found nothing.

q \equiv 1 \ \pmod{4} , \ M_q\text{ is prime }  \Longleftrightarrow  \ \ 5^{\frac{M_q-1}{2}} \ \equiv \ 1 \  \ \pmod{M_q}

Your opinion ?

Tony
T.Rex is offline   Reply With Quote
Old 2005-09-11, 17:30   #9
cyrix
 
Jul 2003
Thuringia; Germany

3A16 Posts
Default In the End: Your (second) Conjecture is false, sorry

Hi Tony!

For q=53 we have M_q=6,361 \cdot 69,431 \cdot 20,394,401 , for all of these primefactors 5 is a quadratic residue, so 5 is also a quadratic residue of M_q=M_{53} , so this is a counter example for the conjecture, that there exists exactly two solutions of  R^2+R-1 \equiv 0 \pmod {M_q} for prime q>1 and q=4n+1.

EDIT: But your last statement holds for q=53:
 5^{\left(\left(2^{53}-1\right)-1\right)/2} \equiv 6,364,152,535,243,836 \pmod {2^{53}-1} , means: M_{53} is not a prime.
Cyrix

Last fiddled with by cyrix on 2005-09-11 at 17:45
cyrix is offline   Reply With Quote
Old 2005-09-11, 18:51   #10
cyrix
 
Jul 2003
Thuringia; Germany

2×29 Posts
Default

Quote:
Originally Posted by T.Rex
q \equiv 1 \ \pmod{4} , \ M_q\text{ is prime }  \Longleftrightarrow  \ \ 5^{\frac{M_q-1}{2}} \ \equiv \ 1 \  \ \pmod{M_q}
Tony
We proved the "\Rightarrow", the "\Leftarrow" is true for all prime q=4n+1<3000 (I tested it with Maple).

But even when this is a real test (and the equvialence is true), it would cost as much time as a LL-Test would do...

Cyrix
cyrix is offline   Reply With Quote
Old 2005-09-11, 19:36   #11
T.Rex
 
T.Rex's Avatar
 
Feb 2004
France

2×33×17 Posts
Default LLT and Pépin tests are for Mersenne and Fermat numbers ! (I think)

Quote:
Originally Posted by cyrix
But even when this is a real test (and the equivalence is true), it would cost as much time as a LL-Test would do... Cyrix
Yes, I know about the cost. But it seems interesting to be able to say that a Pépin's like test applies to Mersenne numbers. I (and Lucas before) provided LLT-like tests for Fermat numbers.
Many people think that LLT is for Mersenne numbers (N-1) and that Pépin's test is for Fermat numbers (N+1). I think it is interesting to be able to say that these 2 tests can apply both to N-1 or N+1 numbers. (done for LLT)

I've studied the LLT function llt(x)=x^2-2 \ \pmod{N} with Mersenne numbers and other numbers and found interesting properties. But I must write down this since 1 or 2 years ...
About Mersenne numbers, some of these properties have been described before by Shanks. But, since he said "prove it if you can", I'm not sure he had a proof !
As an example of these properties, if you start the LLT with L_0=3 and do the test q-2 times, then L_{q-2}=3 if M_q is prime (q=1 (mod 4) .
If you start with L_0=2^{\frac{q+1}{2}}-1 then you get L_{q-2} has the same value as L_0 if M_q is prime, for any prime q.

About q=53 I don't understand how 1 statement is true (5^...) though the other one is false (R^2+R-1 ...) since they are related. Do you ?

I really appreciate our discussion !

Tony
T.Rex is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Property of pseudoprime numbers by base 2 AND / OR 3 allasc And now for something completely different 1 2017-05-17 15:00
new property of prime numbers discovered? ixfd64 Math 1 2016-03-14 21:53
A conjecture on a new property of Mersenne primes Thiele Math 18 2010-05-23 05:35
Curious property of Mersenne numbers. arithmeticae Lounge 5 2008-10-27 06:15
A property of Fermat numbers. Already known ? T.Rex Math 6 2006-09-17 22:11

All times are UTC. The time now is 22:04.


Thu Oct 28 22:04:10 UTC 2021 up 97 days, 16:33, 0 users, load averages: 1.23, 1.21, 1.27

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.