Amazing pattern with -12 (was -3)

[paulunderwood]

(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

[paulunderwood]

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

[chalsall]

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^)

[paulunderwood]

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!

[RMLabrador]

This bring to my memory the true story about Charles Babbage's Magnum Opus - The Infinity Engine from the Book of Never Told)

[chalsall]

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.

[paulunderwood]

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.

[paulunderwood]

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.

[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...

[paulunderwood]

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.

[chalsall]

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.