20050910, 20:59  #1 
Feb 2004
France
2·3^{3}·17 Posts 
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 such that . 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: (1) (2) (3) q=5 > R=12 q=13 > R=394 q=17 > R=41127 For q=29 there are 4 numbers R such that and : 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 20050910 at 21:05 
20050910, 21:18  #2 
Feb 2004
France
2·3^{3}·17 Posts 
PARI/gp code
A simple PARI/gp code for finding R is :
q=13;M=2^q1 for(i=1,(M1)/2, j=(i^22)%M; if((Mj)==i+1, print(i))) For a great M, one should start i with a number such that i^2> M . 
20050910, 22:54  #3 
Jul 2003
Thuringia; Germany
58_{10} Posts 
Hi T.Rex!
Your Conjecture is false! All conditions are equal: R^2+R1=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=(16)/2=12 AND R_2=(1+6)/2=18. So For M=31 R_2=18 is also a solution! Cyrix 
20050911, 09:30  #4  
Feb 2004
France
2×3^{3}×17 Posts 
A clearer conjecture
Quote:
Quote:
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: (P) . Quote:
Since 18 is the "same" solution than 12. In fact, if you replace R by (R+1) in property (P), you have: . So, if R is a solution of (P), then (R+1) is also a solution. So I propose to reformulate the conjecture: For , if there exists one and only one number R that verifies the property (P), then 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 20050911 at 09:40 

20050911, 10:47  #5 
Jul 2003
Thuringia; Germany
2·29 Posts 
Hi T.Rex!
For prime with (and q>1) your conjecture is true: Because of the quadratic reciprocity law (Gauß) 5 is a quadratic residue of , iff is quadratic residue of 5 (because both are of the from 4n+1). But so 5 is a quadratic residue of and there existists two numbers X and Y, which have the properties and Yours, Cyrix 
20050911, 13:21  #6 
Feb 2004
France
918_{10} Posts 
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 
20050911, 15:00  #7  
Jul 2003
Thuringia; Germany
2·29 Posts 
sorry Tony!
I messed something up. Now in a better form: Quote:
Your conjecture reduces to: 5 is a quadratic residue of with q a prime and , iff is prime itself. Yours, Cyrix Last fiddled with by cyrix on 20050911 at 15:01 

20050911, 16:56  #8 
Feb 2004
France
2×3^{3}×17 Posts 
Clear now
OK. I understand your points now.
So, seems we have a conjecture for a Pépinlike test for Mersenne numbers ?! Is it something new ? I've searched in my books and found nothing. Your opinion ? Tony 
20050911, 17:30  #9 
Jul 2003
Thuringia; Germany
2·29 Posts 
In the End: Your (second) Conjecture is false, sorry
Hi Tony!
For q=53 we have , for all of these primefactors 5 is a quadratic residue, so 5 is also a quadratic residue of , so this is a counter example for the conjecture, that there exists exactly two solutions of for prime q>1 and q=4n+1. EDIT: But your last statement holds for q=53: , means: is not a prime. Cyrix Last fiddled with by cyrix on 20050911 at 17:45 
20050911, 18:51  #10  
Jul 2003
Thuringia; Germany
2·29 Posts 
Quote:
But even when this is a real test (and the equvialence is true), it would cost as much time as a LLTest would do... Cyrix 

20050911, 19:36  #11  
Feb 2004
France
1110010110_{2} Posts 
LLT and Pépin tests are for Mersenne and Fermat numbers ! (I think)
Quote:
Many people think that LLT is for Mersenne numbers (N1) 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 N1 or N+1 numbers. (done for LLT) I've studied the LLT function 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 and do the test q2 times, then if M_q is prime (q=1 (mod 4) . If you start with then you get 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+R1 ...) since they are related. Do you ? I really appreciate our discussion ! Tony 

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