Search the web
Sign In
New User? Sign Up
openpfgw · Co-ordination of the OpenPFGW project
? Already a member? Sign in to Yahoo!

Yahoo! Groups Tips

Did you know...
Want to share photos of your group with the world? Add a group photo to Flickr.

Best of Y! Groups

   Check them out and nominate your group.
Having problems with message search? Fill out this form to ensure your group is one of the first to be migrated to the new message search system.

Messages

  Messages Help
Advanced
Wagstaff Numbers - Proof for a Primality Test   Message List  
Reply | Forward Message #1893 of 2071 |
Hello fellow group members,

After a long time I rear my head by posting a proof for a Wagstaff
number primality test.

This test is similar to the well known Lucas Lehmer test and aslo
uses the iteration s -> S^2-2. Let p>3 and prime, W_p=(1/3)(2^p+1)
then primality is given if and only if S_p = S_2 for S_0 = 6.

The complete proof in a pdf file is in the file section of this forum
(last entry in the list) or you could pick it up on below URL.
http://www.mersenneforum.org/showpost.php?p=144516&postcount=1

I would appreciate feedback especially from David as I really dipped
deep into theory.

I really hope we can move some entries from Henri Lifchitz's list.

best regards
Anton







Sun Oct 5, 2008 7:23 am

antonvrba
Offline Offline
Send Email Send Email

Forward
Message #1893 of 2071 |
Expand Messages Author Sort by Date

Hello fellow group members, After a long time I rear my head by posting a proof for a Wagstaff number primality test. This test is similar to the well known...
Anton Vrba
antonvrba
Offline Send Email
Oct 5, 2008
9:02 am

... This would be better on the primenumbers list. ... "Consider an element w ... that has no solution w^k == 1" is meaningless. Skipping everything before it,...
Phil Carmody
thefatphil
Offline Send Email
Oct 5, 2008
9:25 am

... 1) As Phil noted, there is horrific garbling of the meaning of "order". 2) The use of "i" for sqrt(-1) is meaningless, since W_p = 3 mod 4. 3) Anton has...
David Broadhurst
djbroadhurst
Offline Send Email
Oct 5, 2008
5:07 pm

... I meant that there is no such "i" in Z_{W_p}. In Z_{W_p}[sqrt(2)], we make take i = 2^((p-1)/2)*sqrt(2), so that i^2+1 = 3*W_p = 0 mod W_p. David...
David Broadhurst
djbroadhurst
Offline Send Email
Oct 5, 2008
6:28 pm

Thank you Phil and thank you David for pointing out my mistakes I cannot transfer to the other forum as my privallages are not enough to upload files. Instead...
Anton Vrba
antonvrba
Offline Send Email
Oct 7, 2008
7:54 pm

the correct file version is 2.0 ... enough ... of ... equality ... the ... a ... and ... by...
Anton Vrba
antonvrba
Offline Send Email
Oct 8, 2008
12:39 pm

... I append the message that I sent to Anton, showing why Bruce's argument works for Mersenne primes, but not for Wagstaff primes. It seems from ...
David Broadhurst
djbroadhurst
Offline Send Email
Oct 9, 2008
6:13 pm

... Seems is the incorrect word - Has withdrawn is correct I did type an apology to be posted on this forum - It seems I did not hit the send button! best...
Anton Vrba
antonvrba
Offline Send Email
Oct 10, 2008
7:49 am

My oversight, 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 $\omega^{2^{p-1}}=-1$...
Anton Vrba
antonvrba
Offline Send Email
Oct 10, 2008
7:57 am
Advanced

Copyright © 2009 Yahoo! Inc. All rights reserved.
Privacy Policy - Terms of Service - Guidelines - Help