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How to prove that this conjecture about a new primality test for Mersenne numbers is true ?

Definition: Let $M_{q}=2^{q}-1 , S_{0} = 3^{2} + 1/3^{2} , \ and: \ S_{i+1} = S_{i}^{2}-2 \pmod{M_{q}}$

Conjecture: $$M_{q}\text{ is a prime iff: } \ S_{q-1} \equiv S_{0} \pmod{M_{q}}$$ $$\text{ and iff: } \prod_{0}^{q-2} S_i \equiv 1 \pmod{M_{q}}$$ $$\text{ and iff: } S_i \ncong S_0 \pmod{M_{q}} , 0 < i < q-1$$

The easy part part (if $M_{q} \text{ is prime then } S_{q-1}\equiv S_{0} \pmod{M_{q}}$), has already been proven. See: http://tony.reix.free.fr/Mersenne/ConjectureLLTCyclesMersenne.pdf .

Now, a proof of the converse is needed.

The "classic" LLT by Lucas and Lehmer for Mersenne numbers is based on the binary tree of the digraph (under $x^2-2 \pmod{M_q}$). This new conjectured test makes use of a cycle of the same digraph.

The goal of this question is to validate (or not) the method (use a cycle of the digraph) for Wagstaff numbers ($\dfrac{2^q+1}{3}$ , q prime) for which we also lack a proof for the converse. See: https://trex58.files.wordpress.com/2009/01/wagstaffandfermat.pdf for a proof of the easy part, by Robert Gerbicz .

I and a team (Vincent and Paul) had found a very large Wagstaff PRP with this algorithm. See: http://www.primenumbers.net/prptop/prptop.php (2^4031399+1)/3 , rank 9 for now.

Pang
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Tony Reix
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    This is likely better asked at mathoverflow.net, which is more of a site for research questions. – Thomas Andrews Aug 31 '15 at 19:40
  • Hummm I saw this similar question: http://math.stackexchange.com/questions/1394160/conjectured-compositeness-tests-for-n-k-cdot-2n-pm-c (conjecture 2 with k=1 and c=1), and I thought I could ask mine too. However, my question is more difficult I think. So I'll have a look at mathoverflow.net . However, I'm not a professional. I'm just an amateur. – Tony Reix Sep 01 '15 at 20:46
  • I didn't say it was wrong here, just that you might be more likely to get appropriate eyeballs on the question if you posted there. Note: no answers here so far. Not even an indication that anybody but me tried to read it. – Thomas Andrews Sep 01 '15 at 20:59
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    I have posted on mathoverflow too. About the proof needed, I know that it would be very difficult to provide a proof, since H.C. Williams said to me that it would be a great result, and since Mr Wagstaff had a student working on this who did not find a proof for the converse. I think that, if the conjecture is in fact a theorem, it requires brand new technics for building a proof. Or maybe there are pseudoprimes and it is only a PRP algorithm. I'd like to know ! – Tony Reix Sep 02 '15 at 15:57
  • I'm not clear about your statement. Do you need a "partial converse" assuming all three points, or do you want to prove the exact converse of your (if, ..., then ...) in parentheses? Also, will we get credited in a paper if we do prove it? – Daniel Donnelly Feb 12 '23 at 05:24
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    @DLeftAdjointtoU I offered a bounty not a paper. Maybe ping TonyReix – mick Feb 12 '23 at 23:14
  • This deserves more attention imo hence my bounty ! – mick Feb 15 '23 at 22:40
  • This needs the converse of the "easy part". In short: if Property P is true, then it implies that the number is prime.

    The seed (here 3^2+1/3^2) can be anything else, under the condition that it is a Universal Seed (works for all q prime).

    The second condition (the product of Si) is not relevant. I've found non-prime Mersenne and Wagstaff numbers which have a cycle with the expected q-1 length and verifying that the product of all Si is 1 Mod Mq. However, such a cycle of length q-1 of a non-prime Mq does not start with a well-known Universal Seed and thus not with 3^2+1/3^2.

     – Tony Reix Feb 17 '23 at 09:04
  • If someone provides a proof (for Mersenne or Wagstaff), he'll have to talk about my work (during more than 15 years). – Tony Reix Feb 17 '23 at 09:04

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