## Re: [PrimeNumbers] Re: Largest PRP Found!?

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• William, Now this makes more sense to me. Thanks 1M for the lesson. I will see what full sense I can make of this email. At the end I should be able to
Message 1 of 14 , Dec 22, 2003
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William,
Now this makes more sense to me. Thanks 1M for the lesson. I will see what full sense I can make of this email. At the end I should be able to understand Aurifeuillian factors fully.
Thanks again...

elevensmooth <elevensmooth@...> wrote:
Some details were still wrong. I got the middle
exponent of one Aurifeuillian factor wrong, and
I selected the wrong small Aurifeuillian factor
as a divisor of the large one. Here's yet
another try to get this right:

C=148330
X=2^C+1
N=X^4-2
N is PRP for 2, 29, 43, 101
(N+1) = (X^2+1)(X+1)(X-1)
(X^2+1)/2 = (2^296659 + 2^148330 + 1)
This is an Aurifeuillian factor of (2^593318 + 1)
593318=11*53938
(2^593318 + 1) has algebraic factors, including (2^53938 + 1).
(2^53938 + 1) has Aurifeuillian factors
(2^26969 - 2^13485 + 1) and (2^26969 + 2^13485 + 1)
(2^26969 - 2^13485 + 1) is a factor of (X^2+1)/2

To verify the small Aurifeuillian factor divides the
large one without using a big number package, let
z=2^13484. The large factor is
(2048z^22 + 64z^11 + 1). The small factor is
(2z^2 � 2z + 1). The polynomial factorization can be
confirmed, for example, at this web site:

http://icm.mcs.kent.edu/research/facdemo.html

Sorry for the continuing string of errors.

William
--
ElevenSmooth: Distributed Factoring of 2^3326400-1
http://www.ElevenSmooth.com

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