PRIMALITY PROOF for Wagstaff numbers! - Page 3

R.D. Silverman · 2008-10-06, 12:07

Quote:

Originally Posted by T.Rex

Seems better now !
I'll look when back home.

However, did you see the comments of David Broadhurst and Phil Carmody on [openpfgw] mailing list ?

Comments of David Broadhurst:

And the comments of Phil Carmody:

Sure that their comments are based on more solid Maths than mines...
But I hope you'll be able to answer their comments and fix the proof.

I wish you to succeed !

Bob did not say a word... Maybe he will provide us with another proof soon !

Regards,

Tony

It is Monday, AM, EST. I am just now seeing the weekend comments.
I don't need to look at the claimed proof, as others have found flaws.

I am unlikely to provide a proof, because I don't think the
conjecture is correct. As I have said, it seems to be working in the
*wrong subgroup" of GF(p^2).

AntonVrba · 2008-10-07, 16:35

Quote:

Originally Posted by R.D. Silverman

I am unlikely to provide a proof, because I don't think the
conjecture is correct. As I have said, it seems to be working in the
*wrong subgroup" of GF(p^2).

Dear Bob, It gives me pleasure to prove you wrong.

I thank all contributions received pointing out my mistakes in the previous papers.

I am a quick learner, and I knew I was on the right track but did not see the obvious solution. All my previous horrific garbling, my previous method has been totally discarded.

The clue is in the paper of Shallit&Vasiga Theorem 12 showing the elegant way to express $S_2 = S_p$ in terms of $\alpha$ which then is factorised to give $\alpha^m=1$ . This alone does not define the order of $\alpha$ but after doing the same on $S_1 = -S_{p-1}$ the second equality was derived for $\alpha^n=-1$ .

This result was then plugged into the method of Bruce to complete the proof. There is no real innovative thinking other than knowing that a second identity had to be found - the result was always there but nobody picked up the penny. My only claim to fame is $S_1 = -S_p-1$ and to use it and find the elusive proof -the mathematical brains are by others.

The proof is now oh so simple, but it gave me a big headache for a number of days till the obvious was found.

best regards
Anton

AntonVrba · 2008-10-07, 17:50

Sorry I just noticed a small typo which I have now corrected, please use this File

Any ideas where I can safe files to link to so that I can make small changes transparrently.

I now have concluded that the proof can also be used for the numbers of the for 2^q+p.
the small doubt I express in concluding part of the paper I have cleared in my mind.

But now it is time for me to sleep - in a day or two I will give the explanation.

Mini-Geek · 2008-10-07, 18:10

Quote:

Originally Posted by AntonVrba

Any ideas where I can safe files to link to so that I can make small changes transparrently.

http://sites.google.com/
I use Google Pages, but they're not accepting new sign ups there any more to change over to Google Sites.

uigrad · 2008-10-07, 18:28

Wow, I'm pulling for you too Anton. It looks like a really big story, if everything checks out!

Once the conjecture is proved (if indeed it is correct), I expect we'll start a Wagstaff equivalent to GIMPS. I'd like to nominate a name for it:
Wagstaff
Internet
Massive
Prime
Search

Ok, so maybe that wouldn't be best. Well, I'm jumping the gun anyway.

henryzz · 2008-10-07, 19:10

Quote:

Originally Posted by uigrad

Wow, I'm pulling for you too Anton. It looks like a really big story, if everything checks out!

Once the conjecture is proved (if indeed it is correct), I expect we'll start a Wagstaff equivalent to GIMPS. I'd like to nominate a name for it:
Wagstaff
Internet
Massive
Prime
Search

Ok, so maybe that wouldn't be best. Well, I'm jumping the gun anyway.

if this conjecture is proved true then i am sure a project will spring up to test them as i cant see why testing times should be any different to lucas lehmer tests
they will not be searched anywhere near as high as mersenne primes though unless we come up with a special form for their factors
there are 40 wagstaff primes up to n=1mil and only 33 mersenne primes up til there

the wikipedia entry for wagstaff primes could do with some work

T.Rex · 2008-10-07, 20:28

Hi Anton,
I'm a little lost with the notation... (I'm used with the notation of Ribenboim or Williams).

However, in (1.1), you say : $h^p(6) -h^n(2) = 0$

And then later you use $h^n(a)=\alpha^n+\beta^n \text{ where } \alpha+\beta=a$ and where $\alpha$ and $\beta$ are the roots of $X^2-aX+1=0$ .

Am I too tired, or should you not use the same names $\alpha$ and $\beta$ to the roots of $X^2-6X+1=0$ and $X^2-2X+1=0$ like you do on top of page 2 ?

Or maybe you wanted to write: (1.1) $h^p(6) - h^2(6) = 0$ ?

$a$ should'nt be the same (6) in all this part ?

(or maybe I too need a sleep...)

T.

T.Rex · 2008-10-07, 20:32

Quote:

Originally Posted by henryzz

...there are 40 wagstaff primes up to n=1mil...

Up to now, there are only 30 Wagstaff numbers proved prime and 10 PRP up to 1M.
If/when the conjecture is proved, and when some code implements it, then we'll have 40 primes. A number is prime only once there is a proof...
T.

ixfd64 · 2008-10-07, 20:40

You know, my gut feeling is telling me that this conjecture is correct. I've verified this conjecture for values of q up to 101 using a program I wrote on my TI-89.

Below is the program I used:

Code:

:wagstaff(n)
:Prgm
:ClrIO
:floor(n)→n
:If n<3 Then
:Disp "invalid exponent"
:Disp "must be 3 or greater"
:ElseIf n>1013 Then
:Disp "exponent too large"
:Disp "must be 1013 or less"
:ElseIf not isPrime(n) Then
:Disp "exponent must be prime"
:Disp string(n)&" factors to:"
:Disp factor(n)
:ElseIf n=3 Then
:Disp "W3 is prime"
:Else
:6→j
:For t,1,n
:mod(j^2-2,(2^n+1)/3)→j
:If mod(t,25)=0
:Disp "iteration: "&string(t)&"/"string(n)
:If t=2
:j→k
:EndFor
:ClrIO
:If j=k Then
:Disp "W"&string(n)&" is prime!"
:Else
:Disp "W"&string(n)&" is not prime"
:EndIf
:Disp "residue:"
:Disp mod(j,10^25)
:If j>=10^25
:Disp "(last 25 digits)"
:EndIf
:DelVar j,k,t
:EndPrgm

AntonVrba · 2008-10-07, 21:02

Quote:

Originally Posted by T.Rex

Hi Anton,
I'm a little lost with the notation... (I'm used with the notation of Ribenboim or Williams).

However, in (1.1), you say : $h^p(6) -h^n(2) = 0$

And then later you use $h^n(a)=\alpha^n+\beta^n \text{ where } \alpha+\beta=a$ and where $\alpha$ and $\beta$ are the roots of $X^2-aX+1=0$ .

Am I too tired, or should you not use the same names $\alpha$ and $\beta$ to the roots of $X^2-6X+1=0$ and $X^2-2X+1=0$ like you do on top of page 2 ?

Or maybe you wanted to write: (1.1) $h^p(6) - h^2(6) = 0$ ?

$a$ should'nt be the same (6) in all this part ?

(or maybe I too need a sleep...)

T.

No I need the sleep, you know after a while one reads what one wants to read and not what is on paper, or these days on the screen

Yes you are correct - another typo! the joys of Ctrl+c and Ctrl+v

obviously I want to write as you correctly point out
(1.1) $h^p(6) - h^2(6) = 0$ ?

You should be familiar with the notation it is lifted from Shallit&Vasiga.

regards

AntonVrba · 2008-10-07, 21:31

Quote:

Originally Posted by T.Rex

If/when the conjecture is proved, T.

I would prefer "when the proposed theory is verified", I cannot see how anybody can argue against it now! I unashamedly copied both Bruce and Shallit and given the dew credit - so the only verification remains is the simple schoolboy algebra

S_1 = -S_p_1 which is true as both s=s^2-2=(-s)^2-2 (a point that you queried earlier on) but I think I now explained it well enough for every one to follow.

Expressing S_2 = S_p and S_1 = -S_p-1 in terms of alpha and then factoring both equations, follows exactly what Shallit writes in his theorem 12 bottom half of page 14 - nothing new and accepted!

Thereafter I quote Bruce word for word, who will argue against that - here is a link to his full theory "A Really Trivial Proof of the Lucas-Lehmer Test" J. W. Bruce - now you may understand the word play in my title.

To those who believed it impossible - I say Chess Mate

Quote:

Originally Posted by T.Rex

Up to now, there are only 30 Wagstaff numbers proved prime and 10 PRP up to 1M.
T.

if quantity is a measure, there are 20 Mersenne below q=5000 against 33 s^q+7 which also can be proven prime with the theory presented, primes in close proximity of Mersennes can now also be investigated

2008-10-07, 20:28	#29
T.Rex Feb 2004 France 2²·229 Posts	Clarifications... Hi Anton, I'm a little lost with the notation... (I'm used with the notation of Ribenboim or Williams). However, in (1.1), you say : $h^p(6) -h^n(2) = 0$ And then later you use $h^n(a)=\alpha^n+\beta^n \text{ where } \alpha+\beta=a$ and where $\alpha$ and $\beta$ are the roots of $X^2-aX+1=0$ . Am I too tired, or should you not use the same names $\alpha$ and $\beta$ to the roots of $X^2-6X+1=0$ and $X^2-2X+1=0$ like you do on top of page 2 ? Or maybe you wanted to write: (1.1) $h^p(6) - h^2(6) = 0$ ? $a$ should'nt be the same (6) in all this part ? (or maybe I too need a sleep...) T.

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2008-10-07, 18:28	#27
uigrad Aug 2008 126₈ Posts	Wow, I'm pulling for you too Anton. It looks like a really big story, if everything checks out! Once the conjecture is proved (if indeed it is correct), I expect we'll start a Wagstaff equivalent to GIMPS. I'd like to nominate a name for it: Wagstaff Internet Massive Prime Search Ok, so maybe that wouldn't be best. Well, I'm jumping the gun anyway.