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-   -   A property of prime Mersenne numbers under LLT (https://www.mersenneforum.org/showthread.php?t=4650)

 T.Rex 2005-09-10 20:59

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

 T.Rex 2005-09-10 21:18

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 .

 cyrix 2005-09-10 22:54

Hi T.Rex!

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

 T.Rex 2005-09-11 09:30

A clearer conjecture

[QUOTE=cyrix]Your Conjecture is false![/QUOTE] 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. [/QUOTE] 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! [/QUOTE] 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

 cyrix 2005-09-11 10:47

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

 T.Rex 2005-09-11 13:21

I do not understand the link

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

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

 cyrix 2005-09-11 15:00

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

[QUOTE=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 $$M_q=4*n+3$$).
[/QUOTE]

$$M_q \equiv 3 \pmod 4$$, 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

 T.Rex 2005-09-11 16:56

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.

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

Tony

 cyrix 2005-09-11 17:30

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

 cyrix 2005-09-11 18:51

[QUOTE=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[/QUOTE]

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

 T.Rex 2005-09-11 19:36

LLT and Pépin tests are for Mersenne and Fermat numbers ! (I think)

[QUOTE=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[/QUOTE] 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

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