PRIMALITY PROOF for Wagstaff numbers! - Page 4

T.Rex · 2008-10-07, 22:08

Quote:

Originally Posted by AntonVrba

I would prefer "when the proposed theory is verified"

I just meant to say :
1) I've not read and checked all the proof,
2) I'm only an "amateur" and we need experts to read and check your proof,
3) a tentative proof is said to be a proof once it has been verified by peers and has been published.

I'm sure all these 40 Wagstaff numbres are prime but we cannot say they all are primes before there is a theorem and a verification. And your conjecture becomes a theorem only when there is a proof. And a tentative proof becomes a real proof only once other expert people have read it in depth and have approved it. Otherwise it is not Maths.

Let wait for Bob and others (David, Phil, Richard, ...) to say their word.

But I really hope you've succeeded !

Do you want a brand new 100E banknote with my signature so that you can display it in your office ? But wait for other Math guys to say their word about the proof...

T.

maxal · 2008-10-07, 23:46

Quote:

Originally Posted by AntonVrba

Waggstaff_ver1.1.pdf

Could you please correct all typos, otherwise it is really hard to understand what you meant.
Also, I did not understand what is the statement of Lemma 1 and where its proof starts.

But most important - there is an error in the proof of Theorem 1, where you conclude that the order of $\alpha$ is $2^p - 4$ or $2^p + 4$ . From the equalities right above this conclusion, it follows only that the order of $\alpha$ has the form $4r$ where $r$ is a divisor of $2^{p-2} - 1$ or $2^{p-2} + 1$ .

In general, if $x^{2m} = 1$ and $x^m = -1$ in some group, it only implies that the maximum powers of 2 dividing $2m$ and the order of $x$ must be the same (in other words, $2m$ and the order of $x$ have the same valuation with respect to 2). But this does not imply that the order of $x$ equal $2m$ , unless $m$ is a power of 2.

If you can correct this error without affecting validity of your proof - I would be glad to read the corrected version.

AntonVrba · 2008-10-08, 01:59

Quote:

Originally Posted by maxal

Could you please correct all typos, otherwise it is really hard to understand what you meant.
Also, I did not understand what is the statement of Lemma 1 and where its proof starts.

But most important - there is an error in the proof of Theorem 1, where you conclude that the order of $\alpha$ is $2^p - 4$ or $2^p + 4$ . From the equalities right above this conclusion, it follows only that the order of $\alpha$ has the form $4r$ where $r$ is a divisor of $2^{p-2} - 1$ or $2^{p-2} + 1$ .

In general, if $x^{2m} = 1$ and $x^m = -1$ in some group, it only implies that the maximum powers of 2 dividing $2m$ and the order of $x$ must be the same (in other words, $2m$ and the order of $x$ have the same valuation with respect to 2). But this does not imply that the order of $x$ equal $2m$ , unless $m$ is a power of 2.

If you can correct this error without affecting validity of your proof - I would be glad to read the corrected version.

The statement of Lemma 1 are the equalities (1.5) and (1.6) - I will make that clear in the next revision.

Your objection, yes, that is why I think this proof has opened an unexplored area in number theory - and your objection is the reason why Phil commented "horrific garble" when I tried to explain what I observed, suspected and could not formulate properly. Let me try and explain a bit more and show that it is not an error:

The following statement is commonly used: If $G$ is a finite group then the order of an element is at most the order of the group. If $x\in G$ and $x^r = 1$ then the order of $x$ divides $r$ .

Consider $G(p)$ , $p$ prime it is said that the order is at most $p-1$

Now consider $p=2k+1$ . $k$ odd the above statement is false! as the order of the element $a+Ib$ is $2p-1$ , because $(a+Ib)^p = (a-Ib)$ and $(a+Ib)^{2p} = (a+Ib)$

Thus it would be better to write: If $G$ is a finite group then the order of an element is at most the order of the group. If $x\in G$ and $x^r = x$ then the order of $x$ divides $r-1$

Now we can continue: consider the the element $a+b\sqrt c$ , and and $(a+b\sqrt c)^r = (a-b\sqrt c)^n$ and $(a+b\sqrt c)^{2r} = (a+b\sqrt c)^{2n}$ and $(a+b\sqrt c)^{q} \neq (a+b\sqrt c)$ for [tex]q<2r[t/ex]

I ask what is the order of the $a+b\sqrt c$ ? - my answer it divides $2r-2n-1$

Now consider the the element $a+b\sqrt c$ , and and $(a+b\sqrt c)^r = (a-b\sqrt c)^{-n}$ and $(a+b\sqrt c)^{2r} = (a+b\sqrt c)^{-2n}$ and $(a+b\sqrt c)^{q} \neq (a+b\sqrt c)$ for $q<2r$

I ask what is the order of the $a+b\sqrt c$ ? - my answer it divides $2r+2n-1$

Above explanation is exactly the result obtained and thus will appear in ver1.1

This, I hope, answers your objection and you will now agree that it is not an error.

maxal - thanks for raising this.

Did you find any further typos not documented here?

I trust above is not "horrific garble" and is food for thought for the experts

regards

ixfd64 · 2008-10-08, 02:22

You should consider getting your paper peer reviewed by a professional organization, such as the AMS. You might also want to submit your paper to arXiv.

Anyways, good luck! I have a strong feeling that you're about to make mathematical history!

maxal · 2008-10-08, 02:38

Quote:

Originally Posted by AntonVrba

Now consider $p=2k+1$ . $k$ odd the above statement is false! as the order of the element $a+Ib$ is $2p-1$ , because $(a+Ib)^p = (a-Ib)$ and $(a+Ib)^{2p} = (a+Ib)$

First off, I think you meant $(a+Ib)^p = -(a+Ib)$ and $(a+Ib)^{2p} = (a+Ib)^2$ . At least in the paper you have equalities of this type.

Second, from $(a+Ib)^p = -(a+Ib)$ and $(a+Ib)^{2p} = (a+Ib)^2$ (actually, the later equality is a consequence of the first; similarly to (1.3) in the paper that follows from (1.4)), it follows that the order of $(a+Ib)$ divides $2(p-1)$ but not $p-1$ . But it does not imply that the order equal $2(p-1)$ .

Let me point you the error more precisely:

You wrote "...the order of $\alpha$ ... cannot be less than $(2^{p-1}-2)^2$ or $(2^{p-1}+2)^2$ by the second."
Here is another typo, you supposedly mean $2\cdot (2^{p-1}-2)$ or $2\cdot (2^{p-1}+2)$ .

The "second" here refers to " $\alpha^{2^{p-1}-2}=-1$ or $\alpha^{2^{p-1}+2}=-1$ ." But it does not imply that the order of $\alpha$ cannot be smaller than $2\cdot (2^{p-1}-2)$ or $2\cdot (2^{p-1}+2)$ .

It seems that you appeal to the statement like:
If $x^m=-1$ then the order of $x$ cannot be smaller than $2m$ .
This statement is wrong, unless $m$ is a power of 2 (which is not your case - you have $m=2^{p-1}\pm 2$ ).

For example, $6^{15}\equiv -1\pmod{31}$ but the order of 6 modulo 31 equals 6 which is smaller than 2*15=30.

AntonVrba · 2008-10-08, 05:44

Quote:

Originally Posted by AntonVrba

I ask what is the order of the $a+b\sqrt c$ ? - my answer it divides $2r-2n-1$

I ask what is the order of the $a+b\sqrt c$ ? - my answer it divides $2r+2n-1$

regards

My answers where false the answer is 2r

I will explain later

AntonVrba · 2008-10-08, 05:53

Quote:

Originally Posted by ixfd64

You should consider getting your paper peer reviewed by a professional organization, such as the AMS. You might also want to submit your paper to arXiv.

Anyways, good luck! I have a strong feeling that you're about to make mathematical history!

That is my ultimate aim but lets cross that bridge when the paper is ready - but I have no access to professional advice here in Dubai, so University Internet is my only resource - even if one gets a public hammering.

English is my second language, so if anybody spots linquistics or presentational mistakes I would gladly accept contribution to correct these.

I truely wish we are witnessing some history here

regards
Anton

henryzz · 2008-10-08, 06:30

Quote:

Originally Posted by T.Rex

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.

i fogot about that
i have written some code using gmp and have proved the next one prime as long as this conjecture is true

AntonVrba · 2008-10-08, 07:56

Quote:

Originally Posted by maxal

Let me point you the error more precisely:

.

Maxal you are correct, I have seen my error and it is being corrected.

Thanks for all the input

T.Rex · 2008-10-08, 07:58

For those who are interesting to know from where this story comes, here are the 2 conjectures I made for Mersenne numbers in this thread of the forum:

1) 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).

2) It is possible to use Cycles of the Digraph under $x^2-2$ modulo a Mersenne number Mq for proving that Mq is not prime, in (q-1)/n steps (n>1) (and thus faster than the LLT test).

Anton generalized first conjecture for Wagstaff numbers, providing at least a new PRP test for Wagstaff numbers, and he is on the way to proof it (it's my best wish for him) and to provide a (brand new) primality test !

But look at the second conjecture and think about how much it could (even with simply n=2) speed up the GIMPS (or WIMPS) project (if composite Mersenne numbers no often verify this property, for sure) !
It would be enough to prove something like this PRP test: $M_q \text{ is prime } \Rightarrow \ S_{(q-1)/n} \equiv S_0 \text{ , for some } n>1 \text{ dividing } q-1 \text{ and for some seed } S_0, \text{ with: } S_n \equiv S_{n-1}^2-2 \ \pmod{M_q}$ .
Proving this "necessity" part should not be so difficult, if/when Anton succeeds in showing that his technic succeeds in proving his conjecture about Wagstaff numbers, and once we know how to find the appropriate seed...

Think also that, with such kind of PRP test, it could drastically speed up the study of $F_{33}$ , the 33th Fermat number, which requires today about 5000 years of computation with a 16 cores machines to prove it is prime or composite (and about only 6-12 months only in 15 years from now).

Tony

AntonVrba · 2008-10-08, 09:22

Quote:

Originally Posted by maxal

If you can correct this error without affecting validity of your proof - I would be glad to read the corrected version.

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.

2008-10-08, 07:58	#43
T.Rex Feb 2004 France 2²·229 Posts	Use cycles instead of the tree of a LLT Digraph For those who are interesting to know from where this story comes, here are the 2 conjectures I made for Mersenne numbers in this thread of the forum: 1) 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). 2) It is possible to use Cycles of the Digraph under $x^2-2$ modulo a Mersenne number Mq for proving that Mq is not prime, in (q-1)/n steps (n>1) (and thus faster than the LLT test). Anton generalized first conjecture for Wagstaff numbers, providing at least a new PRP test for Wagstaff numbers, and he is on the way to proof it (it's my best wish for him) and to provide a (brand new) primality test ! But look at the second conjecture and think about how much it could (even with simply n=2) speed up the GIMPS (or WIMPS) project (if composite Mersenne numbers no often verify this property, for sure) ! It would be enough to prove something like this PRP test: $M_q \text{ is prime } \Rightarrow \ S_{(q-1)/n} \equiv S_0 \text{ , for some } n>1 \text{ dividing } q-1 \text{ and for some seed } S_0, \text{ with: } S_n \equiv S_{n-1}^2-2 \ \pmod{M_q}$ . Proving this "necessity" part should not be so difficult, if/when Anton succeeds in showing that his technic succeeds in proving his conjecture about Wagstaff numbers, and once we know how to find the appropriate seed... Think also that, with such kind of PRP test, it could drastically speed up the study of $F_{33}$ , the 33th Fermat number, which requires today about 5000 years of computation with a 16 cores machines to prove it is prime or composite (and about only 6-12 months only in 15 years from now). Tony

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2008-10-08, 02:22	#37
ixfd64 Bemusing Prompter "Danny" Dec 2002 California 2390₁₀ Posts	You should consider getting your paper peer reviewed by a professional organization, such as the AMS. You might also want to submit your paper to arXiv. Anyways, good luck! I have a strong feeling that you're about to make mathematical history!