PRIMALITY PROOF for Wagstaff numbers! - Page 5

maxal · 2008-10-08, 11:13

Quote:

Originally Posted by AntonVrba

Maxal, Thanks for your input - I realized my errors

I have now corrected them and also demonstrate that the order is indeed 2^p

Version 2.0 is now attached. Please take some time and re-read.

You again make wrong statements about the order. This one, for example:

If $x\in G$ and $x^r = 1$ then the order of $x$ divides $r$ . If $x^s = -1$ then the order of $x$ is $2s$ if and only if $2s = r$ .

As before, let $G$ be the residues modulo 31, $x=6$ , $s=6$ , $r=12$ .
It is true that the order of $x=6$ divides $r=12$ . But while we have the equality $2s = r$ , the order of $x=6$ is not $2s=12$ .
Hence, the second part of your statement above is incorrect.

AntonVrba · 2008-10-08, 11:35

Quote:

Originally Posted by maxal

You again make wrong statements about the order. This one, for example:

If $x\in G$ and $x^r = 1$ then the order of $x$ divides $r$ . If $x^s = -1$ then the order of $x$ is $2s$ if and only if $2s = r$ .

As before, let $G$ be the residues modulo 31, $x=6$ , $s=6$ , $r=12$ .
It is true that the order of $x=6$ divides $r=12$ . But while we have the equality $2s = r$ , the order of $x=6$ is not $2s=12$ .
Hence, the second part of your statement above is incorrect.

Thanks - I realise it now and it is not needed in the Proof, Bruce makes no such claims all he says the order diveds r - so we should not say more than that.

I ammend all statements to the order of x divides r and make no further conclusion the end result stays the same.

More interesting for me is if my argument of Lemma 1 that is if w^(2^p -2n) = 1 then the order of w divides 2^p - this is the one we need to accept if the proof is to be accepted. Possibly we also need to add, similar to complex numbers that that p =2k+1, k odd, which waggstaff indeed are.

regards
Anton

maxal · 2008-10-08, 11:54

Quote:

Originally Posted by AntonVrba

More interesting for me is if my argument of Lemma 1 that is if w^(2^p -2n) = 1 then the order of w divides 2^p - this is the one we need to accept if the proof is to be accepted.

The statement of your lemma sounds weird. What is $b$ there and how it is related to $\alpha$ and $a$ ?
Or do you claim that the order of $a+b\sqrt{c}$ is $2^p$ for all possible $b$ ?!

AntonVrba · 2008-10-08, 12:00

Quote:

Originally Posted by T.Rex

It is possible to use Cycles (instead of a branch of the tree) of the Digraph under $x^2-2$ modulo a Mersenne number Mq for proving that Mq is prime, in q-1 steps (same speed as the original LLT test).

Tony

For Fermets I have not studied.

More intersting is that one can also prove

2^q - 5 , s0 = 6 , s2 = sq
2^q - 3 , s0 = 4 , s2 = sq
2^q - 1 , s0 = (4p-10)/3 , s1=sq ; This is w=-3-1/3
2^q + 3 , s0 = 6 , s2 = sq
2^q + 4 , s0 = 4 , s2 = sq
2^q + 7 , s0 = 5 , s3 = sq

I need to investigate the +- Wq

T.Rex · 2008-10-08, 20:08

OK. I do not understand all, but here are the mistakes I've seen.

a) Use \equiv instead of = anywhere you are using a modulo.
- in the Theorem
- between (1.1) and (1.5)
- and in many other places.

b) Lemma 2
- $h^n(a) = \alpha^{2^n}+\beta^{2^n}$
- In the proof, there is a \ missing before alpha.
- In last line, would be useful to add : $= h^{n+1}(a)$

c) You never show that $h^{n}(6) = S_n$

d) in (1.2) you still have to show that $S_{p-1} \equiv -S_1 \ \pmod{W_p}$ and not $S_{p-1} \equiv S_1 \ \pmod{W_p}$ , I think.

e) a ";" appears at beg of (1.5) and (1.6)

f) Page 3 : should be Lemma 3 instead of Lemma 2.

g) middle of page 3 : the 4 properties of powers of alpha should be with \equiv, not = .

h) end of proof (and not prove) of necessity: the conclusion should be said:
"There is no q <Wp that divides Wp".
Sure that it is the goal, but it helps to remember what was the goal of this part.

i) Proof of sufficiency
$\beta = 3-2 \sqrt{2}$

j) $\alpha^{W_p} = 3^{W_p} +.... + 2^{W_p} {\sqrt{2}}^{W_p} \equiv 3 + 2^{(W_p-1)/2} \times 2 \sqrt{2}$ I think.

And PLEASE !! use a larger font ! My eyes will have 50 years in 1 week, so it's really difficult to read the paper.

Do you think you'll be finished with the proof so that I'll have this theorem as a nice present for my half-century anniversary, Monday 13th ? after due verifications by peers for sure !

Hey. Where is Bob ?

Tony

axn · 2008-10-08, 20:50

Aren't the "Proof of Necessity" and "Proof of sufficiency" sections reversed?

The part where you show "If the test is true, Wp is prime" should be the proof of sufficiency.

AntonVrba · 2008-10-09, 10:45

Quote:

Originally Posted by T.Rex

And PLEASE !! use a larger font ! My eyes will have 50 years in 1 week, so it's really difficult to read the paper.

Tony

My eyes have 56+ and glasses for last year and I would love a larger typeface and better layout, but as I am a LaTex antitalent - How?

or maybe someone can send me a more suitable template - I use WinEdt

Regards - Anton

maxal · 2008-10-09, 10:55

Anton, put "[12pt]" right after \documentclass in the beginning of your TeX source, so that you will have something like:

Code:

\documentclass[12pt]{article}

T.Rex · 2008-10-09, 12:44

Quote:

Originally Posted by AntonVrba

My eyes have 56+ and glasses for last year and I would love a larger typeface and better layout, but as I am a LaTex antitalent - How?

I too use WinEdt. Last version of Lyx starts to seem to be usable for Math Tex. But I'ven't migrated yet. Still some problems.
I've loocked at a .tex file and I do what Maxal recommends you to do.
56+ ?! Wowww, you're looking younger thanks to your enthousiasm at proving the conjecture !!!
I'm ready to review another version of your proof. But take some time to read it quietly in order to fix all the details that are annoying.
About \equiv and =, I think that it is important to use it in the right place. And it is not always easy...
I would say: $S_n=S_{n-1}^2-2 \ \pmod{W_p}$ , since Sn is computed as a square minus 2 and then modulo Wp. But: $\text{N is prime iff } S_n \equiv S_0 \ \pmod{W_q}$ , saying that the two numbers must be equivalent modulo Wq.
Tony

AntonVrba · 2008-10-09, 16:49

Dear Members

I must appologise (especially to Bob) and withdraw my claim.

My oversight, that Tony Rex has pointed out and needed some time to sink in, is that my claim to the order of $\alpha$ is wrong. The proof of Bruce works because when he states $\omega^{2^p}=1$ and $\omega^{2^{p-1}}=-1$ then the order of $\omega$ is $2^p$ because $2^{p-1}$ has no odd factors and not because it is half of $2^p$ which was my big oversight.

But let us not give up I still have an idea and need to think deeply about it. I thank all for their contribution and kind help.

Tony, have a happy birthday and buy some good wine with the Euro 100.

PS. Is their a way that my withdrawel can be put into the root thread? stops any confusion for a newcomer to the thread.

T.Rex · 2008-10-09, 20:13

Hi Anton,
No problem you failed... this time. It was a very nice battle !
Failures make us stronger, because we learn from our mistakes.
And there is something worst than making mistakes: do nothing.
And thanks to David to have taken time to point where there is a problem.
The goal is to OPEN a track with the cycles of the Digraph. Maybe it is too difficult for now with Wagstaff numbers. Maybe it could be easier with Mersennes or Fermats ? (though there are already primality tests for these numbers)
I hope that other Mathematicians will be interested in working on one of the 3 conjectures, and I will be very happy to pay the 100 € bank-note !
About the wine: Champagne for sure, and a Red Haut-Médoc, and some Sauternes probably.
If you go around Grenoble in the future, just warn me !
And, if you want people to read another paper based on your new idea, don't hesitate !
Regards,
Tony

2008-10-08, 20:08	#49
T.Rex Feb 2004 France 2²×229 Posts	My comments about v2.0 OK. I do not understand all, but here are the mistakes I've seen. a) Use \equiv instead of = anywhere you are using a modulo. - in the Theorem - between (1.1) and (1.5) - and in many other places. b) Lemma 2 - $h^n(a) = \alpha^{2^n}+\beta^{2^n}$ - In the proof, there is a \ missing before alpha. - In last line, would be useful to add : $= h^{n+1}(a)$ c) You never show that $h^{n}(6) = S_n$ d) in (1.2) you still have to show that $S_{p-1} \equiv -S_1 \ \pmod{W_p}$ and not $S_{p-1} \equiv S_1 \ \pmod{W_p}$ , I think. e) a ";" appears at beg of (1.5) and (1.6) f) Page 3 : should be Lemma 3 instead of Lemma 2. g) middle of page 3 : the 4 properties of powers of alpha should be with \equiv, not = . h) end of proof (and not prove) of necessity: the conclusion should be said: "There is no q <Wp that divides Wp". Sure that it is the goal, but it helps to remember what was the goal of this part. i) Proof of sufficiency $\beta = 3-2 \sqrt{2}$ j) $\alpha^{W_p} = 3^{W_p} +.... + 2^{W_p} {\sqrt{2}}^{W_p} \equiv 3 + 2^{(W_p-1)/2} \times 2 \sqrt{2}$ I think. And PLEASE !! use a larger font ! My eyes will have 50 years in 1 week, so it's really difficult to read the paper. Do you think you'll be finished with the proof so that I'll have this theorem as a nice present for my half-century anniversary, Monday 13th ? after due verifications by peers for sure ! Hey. Where is Bob ? Tony

2008-10-09, 10:55	#52
maxal Feb 2005 2²×3²×7 Posts	Anton, put "[12pt]" right after \documentclass in the beginning of your TeX source, so that you will have something like: Code: \documentclass[12pt]{article}

2008-10-09, 16:49	#54
AntonVrba Jun 2005 2·7² Posts	Withdraw My Claim Dear Members I must appologise (especially to Bob) and withdraw my claim. My oversight, that Tony Rex has pointed out and needed some time to sink in, is that my claim to the order of $\alpha$ is wrong. The proof of Bruce works because when he states $\omega^{2^p}=1$ and $\omega^{2^{p-1}}=-1$ then the order of $\omega$ is $2^p$ because $2^{p-1}$ has no odd factors and not because it is half of $2^p$ which was my big oversight. But let us not give up I still have an idea and need to think deeply about it. I thank all for their contribution and kind help. Tony, have a happy birthday and buy some good wine with the Euro 100. PS. Is their a way that my withdrawel can be put into the root thread? stops any confusion for a newcomer to the thread. Last fiddled with by AntonVrba on 2008-10-09 at 17:15

2008-10-09, 20:13	#55
T.Rex Feb 2004 France 1110010100₂ Posts	Many battles, one victory. Hi Anton, No problem you failed... this time. It was a very nice battle ! Failures make us stronger, because we learn from our mistakes. And there is something worst than making mistakes: do nothing. And thanks to David to have taken time to point where there is a problem. The goal is to OPEN a track with the cycles of the Digraph. Maybe it is too difficult for now with Wagstaff numbers. Maybe it could be easier with Mersennes or Fermats ? (though there are already primality tests for these numbers) I hope that other Mathematicians will be interested in working on one of the 3 conjectures, and I will be very happy to pay the 100 € bank-note ! About the wine: Champagne for sure, and a Red Haut-Médoc, and some Sauternes probably. If you go around Grenoble in the future, just warn me ! And, if you want people to read another paper based on your new idea, don't hesitate ! Regards, Tony

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2008-10-08, 20:50	#50
axn Jun 2003 2²·3·421 Posts	Aren't the "Proof of Necessity" and "Proof of sufficiency" sections reversed? The part where you show "If the test is true, Wp is prime" should be the proof of sufficiency.