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 2008-09-18, 07:17 #1 T.Rex     Feb 2004 France 16268 Posts A Euro Reward for a proof Hello, I will give a 100Euro reward for the author of the first (mathematical !) proof of one of the conjectures I will provide in this thread. I invite other GIMPS addicts to participate and to also offer an amount of reward. I'm also looking for reviewers. I invite Bob Silverman to participate. It is required that he candidate proofs are written in LaTeX (or in any other Math-enabled tool) and that they will be provided as a post in this thread with a link to a .pdf paper. Or they can be sent to me by private email, so that I put them on my site. For sure, the winner will not be rich ! However, Maths students may be interested, for fame and glory. So, I ask Math teachers/searchers in this forum for talking of this to their students and to any Math journal or web-site. As you may know, these conjectures deal with LLT and cycles in a digraph under x^2-2 modulo a prime. Some of these conjectures are mine, or are Anton Vrba and mine, based on my work. My hope/goal is that, at the end, a new proof technique will emerge and will enable to build a faster proof, based on LLT cycles (instead of the tree) under x^2-2 modulo a Mersenne prime and thus in (q-1)/n iterations, for saying that a Mersenne number is composite. Which will speed up the GIMPS project ! Look at this (not perfect, I know) paper for some information about the underlying facts I've found for Fermat and mainly Mersenne numbers. I perfectly know that, though all our experiments have shown that the conjectures say yes for known primes and no for composites, either there exist exceptions or there is no way to prove them (Gödel !). Now, go and work ! Regards, Tony Last fiddled with by T.Rex on 2008-09-18 at 08:13
 2008-09-18, 07:33 #2 T.Rex     Feb 2004 France 2·33·17 Posts Conjecture N°1 : A big q-1 Cycle for Mersennes Here is the first conjecture to prove: A LLT-like test for Mersenne numbers, based on cycles of the Digraph under x^2-2 modulo a Mersenne prime. $\large \text{Let } q \text{ be a prime integer } > \ 3 \text{ , and } M_q = 2^q-1 \ .$ $\large M_q \text{ is prime } \Longleftrightarrow \ S_{q-1} \equiv S_0 \ \pmod{M_q} \text{ , where: } S_0=3^2+1/3^2 , \ S_{i+1}=S_i^2-2 \ \pmod{M_q} \ .$ Happy guys, I've already done half the work ! You only have to prove the converse ! (which is the most difficult part, though...) . Look at my paper: First part of the proof. Tony Last fiddled with by T.Rex on 2008-09-18 at 08:08
 2008-09-18, 08:06 #3 T.Rex     Feb 2004 France 2×33×17 Posts Conjecture N°2 : A Big q-1 Cycle for Wagstaffs Here is the second conjecture (by Anton Vrba and my-self) to prove: A LLT-like test for Wagstaff numbers, based on cycles of the Digraph under x^2-2 modulo a Wagstaff prime. $\large \text{Let } q \text{ be a prime integer } > \ 3 \text{ , } N_q = 2^q+1 \text{ and } W_q = \frac{N_q}{3} \ .$ $W_q \text{ is prime } \Longleftrightarrow \ S_{q-1} \equiv S_0 \ \pmod{W_q} \text{ , where: } S_0=3/2 \text{ (or }1/4 \text{ ) , and } \ S_{i+1}=S_i^2-2 \ \pmod{N_q} \ .$ This "Vrba-Reix PRP" conjecture for Wagstaff numbers has been tested (private communication) very high by Jean Penné by means of his new version 3.7.2 of his famous LLR tool based on Prime95 library v24, which now implements this "Vrba-Reix PRP" test. Tony Last fiddled with by T.Rex on 2008-09-18 at 08:08
 2008-09-18, 10:07 #4 T.Rex     Feb 2004 France 39616 Posts Conjecture N°3: A Big 2^n-1 Cycle for Fermats Here is the third conjecture (by Anton Vrba and my-self) to prove: A LLT-like test for Fermat numbers, based on cycles of the Digraph under x^2-2 modulo a Fermat prime. $\large \text{Let } n \text{ be an integer } > \ 1 \text{ , and } F_n = 2^{2^n}+\ 1 \ .$ $F_n \text{ is prime } \ \Longleftrightarrow \ \ S_{2^n-1} \equiv S_0 \ \pmod{F_n} \text{ , where: } S_0=-3/2 \text{ , and } \ S_{i+1}=S_i^2-2 \ \pmod{F_n} \ .$ It is perfectly clear that such a test will not speed up proving F33 is prime or not (and programs like Mlucas or Prime95 are already ready to test it) ! But it could be a first step for speeding up a proof that F33 is not prime (I cannot wait for about 2025 !). Note that I already provided a proof that the LLT can be used for Fermat numbers and that the proof technic used for proving this theorem also proves the Pépin's test (see bottom of page 4 of this paper). Edouard Lucas already provided such a proof of LLT for Fermat numbers but, as usual, it was not a clear and complete proof. Maybe other people have built such a proof in the past, by I have found no paper about that. Tony Last fiddled with by akruppa on 2008-09-21 at 14:23 Reason: fixed link
 2008-09-18, 10:23 #5 T.Rex     Feb 2004 France 91810 Posts Some PARI/gp code for experimenting For those wanting to see the conjectures at work, here is some simple PARI/gp code that exercises the 3 conjectures: Code for conjecture about Mersennes: Code: q=5;M=2^q-1;S=Mod(3^2+1/3^2,M);print(S) for(i=1,q-1,S=Mod(S^2-2,M);print(S)) Mod(3, 31) Mod(7, 31) Mod(16, 31) Mod(6, 31) Mod(3, 31) Code for conjecture about Wagstaffs: Code: q=5;N=2^q+1;W=N/3;S=Mod(3/2,N);print(Mod(S,W)) for(i=1,q-1,S=Mod(S^2-2,N);print(Mod(S,W))) Mod(7, 11) Mod(3, 11) Mod(7, 11) Mod(3, 11) Mod(7, 11) Code for conjecture about Fermats: Code: n=2;F=2^2^n+1;S=Mod(-3/2,F);print(S) for(i=1,2^n-1,S=Mod(S^2-2,F);print(S)) Mod(7, 17) Mod(13, 17) Mod(14, 17) Mod(7, 17) Have fun ! Tony
 2008-09-18, 13:58 #6 T.Rex     Feb 2004 France 2·33·17 Posts A common property Note that we have the following properties: Mersennes: $\prod_{i=1}^{q-1} S_i \equiv 1 \ \pmod{M_q} \ .$ Wagstaffs: $\prod_{i=1}^{q-1} S_i \equiv 1 \ \pmod{W_q} \ .$ Fermats: $\prod_{i=1}^{2^n-1} S_i \equiv -1 \ \pmod{F_n} \ .$ Tony
2008-09-18, 14:05   #7
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by T.Rex Note that we have the following properties: Mersennes: $\prod_{i=1}^{q-1} S_i \equiv 1 \ \pmod{M_q} \ .$ Wagstaffs: $\prod_{i=1}^{q-1} S_i \equiv 1 \ \pmod{W_q} \ .$ Fermats: $\prod_{i=1}^{2^n-1} S_i \equiv -1 \ \pmod{F_n} \ .$ Tony

General hint: Think 'analogue of Wilson's Theorem". What does one
get when you multiply the elements of an Abelian group?

Now ask "what group are we working in"? It is either the full twisted group
[of GF(p^2)] or a subgroup of the same.

Proving the conjectures amounts to showing that the starting element,
S0 always has the correct order in this group.

2008-09-18, 14:22   #8
T.Rex

Feb 2004
France

2×33×17 Posts

Welcome to the game, Bob !

If there are proof proposals, would you mind contributing in reviewing them ?

Quote:
 Originally Posted by R.D. Silverman ... Abelian group? ... the full twisted group [of GF(p^2)] or a subgroup of the same. ... the correct order in this group.
Nuts ! Seems I have to refresh my memory or to buy new books...
Never mind, thanks to help !

Tony

2008-09-18, 14:25   #9
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by T.Rex Welcome to the game, Bob ! If there are proof proposals, would you mind contributing in reviewing them ? Nuts ! Seems I have to refresh my memory or to buy new books... Never mind, thanks to help ! Tony
I will try to find a proof. I am particularly intested in Conjecture 2.

2008-09-18, 15:29   #10
Primeinator

"Kyle"
Feb 2005
Somewhere near M52..

3×5×61 Posts

Quote:
 Happy guys, I've already done half the work ! You only have to prove the converse ! (which is the most difficult part, though...) . Look at my paper: First part of the proof.
I do not have enough math to attempt this proof, however, I did show Conjecture 1 to my math professor and he may have an idea. However, he is a busy individual (teaching, family, etc) but will attempt a proof in his spare time.

 2008-09-18, 19:11 #11 T.Rex     Feb 2004 France 11100101102 Posts Thanks for your help, Bob and Primeinator ! You're welcome ! Primeinator, my Number Theory Math level is... very low. However, I bought and read (and read again, and again, and again...) several books, like the one of Paulo Ribenboim. I have some (limited !) knowledge, now. With some work, you can make terrific progress ! Bob, would you mind recommending some books to Primeinator ? Tony

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