20221115, 07:30  #12 
Sep 2002
Database er0rr
10617_{8} 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
5·29·31 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
3·5·739 Posts 

20221121, 06:51  #15 
Sep 2002
Database er0rr
5·29·31 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.
53 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
3×5×739 Posts 
Quote:
Cryptonomicon is just the beginning. Deal With Now. 

20221126, 10:17  #18 
Sep 2002
Database er0rr
5·29·31 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
5·29·31 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.
53_{10} 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
1000110001111_{2} Posts 
Quote:
Last fiddled with by paulunderwood on 20221127 at 01:15 

20221127, 09:43  #22  
If I May
"Chris Halsall"
Sep 2002
Barbados
3×5×739 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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