mersenneforum.org Amazing pattern with -12 (was -3)
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 2022-11-15, 07:30 #12 paulunderwood     Sep 2002 Database er0rr 7·23·29 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^2-z+1 seems sufficient. This divides z^3+1 which divides z^6-1. We have z^2-((-3)^(2*r-1)-3+3-2)*z+1, where (3^(2*(r-1)+1)==0 is to be avoided. That is we don't want 3^(4*(r-1))==1. Hence I will run a program to check for gcd(r-1,n-1)==1, and try to figure out a proof before someone else might do so Last fiddled with by paulunderwood on 2022-11-21 at 06:17 Reason: fixed variable names
2022-11-20, 05:34   #13
paulunderwood

Sep 2002
Database er0rr

7×23×29 Posts

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

Attached Files
 Amazing_Q_is_Negative_3_Test.pdf (50.2 KB, 53 views)

Last fiddled with by paulunderwood on 2022-11-21 at 06:29 Reason: Fixed mistakes in paper

2022-11-20, 07:11   #14
chalsall
If I May

"Chris Halsall"
Sep 2002

260378 Posts

Quote:
 Originally Posted by paulunderwood Please feel free to comment
Ummm... [jaw drops] Wow!

There is nothing quite like a challenge to keep one motivated, is there? 8^)

 2022-11-21, 06:51 #15 paulunderwood     Sep 2002 Database er0rr 110758 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 2022-11-21 at 06:58
 2022-11-21, 20:16 #16 RMLabrador   "Chereztynnoguzakidai" Oct 2022 Ukraine, near Kyiv. 3×19 Posts This bring to my memory the true story about Charles Babbage's Magnum Opus - The Infinity Engine from the Book of Never Told)
2022-11-21, 20:34   #17
chalsall
If I May

"Chris Halsall"
Sep 2002

32×5×251 Posts

Quote:
 Originally Posted by RMLabrador This bring to my memory the true story about Charles Babbage's Magnum Opus - The Infinity Engine from the Book of Never Told)
Have you read any Neal Stephenson?

Cryptonomicon is just the beginning.

Deal With Now.

 2022-11-26, 10:17 #18 paulunderwood     Sep 2002 Database er0rr 7·23·29 Posts partial results The GCD-less 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 2022-11-26 at 10:18
 2022-11-26, 17:05 #19 paulunderwood     Sep 2002 Database er0rr 7×23×29 Posts Partial proof For a basis we know prime $$p$$ that $$x^{p+1} \equiv -3 \pmod{n, x^2-3^rx-3}$$ 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 $$3^{q-1}\equiv x^{q-1} \pmod{p}$$. Thus dividing the LHS by $$3^{q-1}$$ and the RHS by $$x^{(q-1)(p+1)}$$ to get $$3^{q-1-q-1}\equiv x^{q-1-(p+1)(q-1)} \pmod{p}$$. X That is $$3^{-2}\equiv x^{q-1-pq-q+p-1} \pmod{p}$$. I.e. $$3^{-2}\equiv x^{-(pq+1)+p-1} \pmod{p}$$. Thus $$3^{-2}\equiv x^{-(pq+1)}x^{p-1} \pmod{p}$$. So $$3^{-2}\equiv -3^{-1}.x^{p-1} \pmod{p}$$. Cancelling gives $$x^{p-1}\equiv -3^{-1} \pmod{p}$$ 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 2022-11-27 at 01:07
 2022-11-26, 19:20 #20 RMLabrador   "Chereztynnoguzakidai" Oct 2022 Ukraine, near Kyiv. 3×19 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 Tao-level-mathematician 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 2022-11-26 at 19:39
2022-11-27, 01:04   #21
paulunderwood

Sep 2002
Database er0rr

7·23·29 Posts

Quote:
 Originally Posted by RMLabrador 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 Tao-level-mathematician 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...
Hah, I made a complete blunder. The conclusion is 1=1. I have marked it with a red X.

Last fiddled with by paulunderwood on 2022-11-27 at 01:15

2022-11-27, 09:43   #22
chalsall
If I May

"Chris Halsall"
Sep 2002

32·5·251 Posts

Quote:
 Originally Posted by paulunderwood Hah, I made a complete blunder. The conclusion is 1=1. I have marked it with a red X.
Mistakes happen from time to time. Just try to not make the same mistakes too many times...

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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