20221115, 07:30  #12 
Sep 2002
Database er0rr
1001011001011_{2} Posts 
(Lucas) Q=3 seems the natural base to study as it is the discriminant of z^2+z+1.
In practice, avoidance of z^2z+1 seems sufficient. This divides z^3+1 which divides z^61. We have z^2((3)^(2*r1)3+32)*z+1, where (3^(2*(r1)+1)==0 is to be avoided. That is we don't want 3^(4*(r1))==1. Hence I will run a program to check for gcd(r1,n1)==1, and try to figure out a proof before someone else might do so Last fiddled with by paulunderwood on 20221121 at 06:17 Reason: fixed variable names 
20221120, 05:34  #13 
Sep 2002
Database er0rr
17×283 Posts 
Here is a short paper about this Lucas(n,3^r,3) test.
Now I have to get into coding up a verification program written with GMP+primesieve. Pari/GP is too slow for the purpose. As I say in the paper, it is me against the broader mathematical community on this topic. I shall be reaching out beyond mersenneforum on this. Please feel free to comment Last fiddled with by paulunderwood on 20221121 at 06:29 Reason: Fixed mistakes in paper 
20221120, 07:11  #14 
If I May
"Chris Halsall"
Sep 2002
Barbados
2^{2}×5×571 Posts 

20221121, 06:51  #15 
Sep 2002
Database er0rr
17·283 Posts 
I fixed a few things in the paper that were mistakes.
A GMP+primesieve program is now running here. It is much faster tha the Pari/GP one. I posted on Math.StackExchange but the thread was closed by a moderator with the reason given that the question was not focused. I suppose it is like asking the computer what the final digit if Pi is: My endeavour for a proof will continue! Last fiddled with by paulunderwood on 20221121 at 06:58 
20221121, 20:16  #16 
"Chereztynnoguzakidai"
Oct 2022
Ukraine, near Kyiv.
103_{8} Posts 
This bring to my memory the true story about Charles Babbage's Magnum Opus  The Infinity Engine from the Book of Never Told)

20221121, 20:34  #17  
If I May
"Chris Halsall"
Sep 2002
Barbados
2^{2}·5·571 Posts 
Quote:
Cryptonomicon is just the beginning. Deal With Now. 

20221126, 10:17  #18 
Sep 2002
Database er0rr
17×283 Posts 
partial results
The GCDless test for "12" was verrified using Par/GP my forumite mart_t and me up to 1.2*10^12.
The test for "3" has now reached 10^12 using GMP+primesieve. It will take a couple of Celeron CPU months to reach 10^13. For those interested please see the paper in post #13. I am still working on a proof but making little headway. Last fiddled with by paulunderwood on 20221126 at 10:18 
20221126, 17:05  #19 
Sep 2002
Database er0rr
12CB_{16} Posts 
Partial proof
For a basis we know prime \(p\) that \(x^{p+1} \equiv 3 \pmod{n, x^23^rx3}\) where the Jacobi symbol of the discriminant is \(1\).
Now suppose \(n=pq\) where \(q>1\) not necessarily prime. That we assume \(x^{pq+1} \equiv 3\) which must also be true mod \(p\). Thus we can write \(x^{pq} \equiv \frac{3}{x} \pmod{p}\). Since \(x^p\equiv \frac{3}{x} \mod{p}\), we can write \(\frac{3^q}{x^q}\equiv \frac{3}{x} \pmod{p}\). Rationalising the equation we get \(3^{q1}\equiv x^{q1} \pmod{p}\). Thus dividing the LHS by \(3^{q1}\) and the RHS by \(x^{(q1)(p+1)}\) to get \(3^{q1q1}\equiv x^{q1(p+1)(q1)} \pmod{p}\). X That is \(3^{2}\equiv x^{q1pqq+p1} \pmod{p}\). I.e. \(3^{2}\equiv x^{(pq+1)+p1} \pmod{p}\). Thus \(3^{2}\equiv x^{(pq+1)}x^{p1} \pmod{p}\). So \(3^{2}\equiv 3^{1}.x^{p1} \pmod{p}\). Cancelling gives \(x^{p1}\equiv 3^{1} \pmod{p}\) But \(x^{p+1}\equiv 3 \pmod{p}\). So \(x^2\equiv 9 \pmod{p}\). It must be that \(x\equiv 3\pmod{p}\) or \(x\equiv 3 \pmod{p}\) which is contradiction since then \(x\in\mathbb{Z}\), Where is the flaw? Continuing, \(3^{r+1} \equiv \pm 6 \pmod{p}\) or \(3^r \equiv \pm 2 \pmod{p}\). Our original equation becomes \(x^2\pm 2x 3\equiv 0 \pmod{p}\) which can be factored \((x\pm 1)(x\mp 3)\) How this works for mod \(n\) I am not sure. Last fiddled with by paulunderwood on 20221127 at 01:07 
20221126, 19:20  #20 
"Chereztynnoguzakidai"
Oct 2022
Ukraine, near Kyiv.
67 Posts 
You know, when i'm not shure, i'm take my favorite chainsaw, Stihl MS361 and go to forgotten realms of Desna river for a cut down the couple of dry trees, relax and do muscle some work) Its easy task in modern time to be above on any top mathematician in it; at the time of Plato this is has been very hard)))
MORE clearly:Just do Your best! Do not matter cause this eyebeelding to Taolevelmathematician or not))) And I'm personally appreciate you for Pari/GP. Former Fortran + IMSL is a solid, as old rock but Pari /GP is a breeze... Last fiddled with by RMLabrador on 20221126 at 19:39 
20221127, 01:04  #21  
Sep 2002
Database er0rr
4811_{10} Posts 
Quote:
Last fiddled with by paulunderwood on 20221127 at 01:15 

20221127, 09:43  #22  
If I May
"Chris Halsall"
Sep 2002
Barbados
2^{2}×5×571 Posts 
Quote:
Seriously... Mistakes must be allowed. Otherwise, nothing moves forward; everyone is afraid of making mistakes. I see this in so many of my spaces at the moment. It can be a bit depressing. 

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