## Re: [PrimeNumbers] Re: Thanks to all. Proposed binary Prime number test fails.

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• ... Subscirbe! Phil () ASCII ribbon campaign () Hopeless ribbon campaign / against HTML mail / against gratuitous bloodshed [stolen with
Message 1 of 18 , Jun 3, 2007
--- Mark Rodenkirch <mgrogue@...> wrote:
> This was meant to be sent to the group and I sent it to Jens by
> accident.
>
> If anyone else here has a PowerPC G5, I have a program that could
> search a range of about 1e12 in a day (for a single base).

Subscirbe!

Phil

() ASCII ribbon campaign () Hopeless ribbon campaign
/\ against HTML mail /\ against gratuitous bloodshed

[stolen with permission from Daniel B. Cristofani]

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• ... YGM. --Mark [Non-text portions of this message have been removed]
Message 2 of 18 , Jun 3, 2007
On Jun 3, 2007, at 9:21 AM, Phil Carmody wrote:

> > This was meant to be sent to the group and I sent it to Jens by
> > accident.
> >
> > If anyone else here has a PowerPC G5, I have a program that could
> > search a range of about 1e12 in a day (for a single base).
>
> Subscirbe!

YGM.

--Mark

[Non-text portions of this message have been removed]
• ... Yes. Let d be the smallest value so that p divides b^d-1. First observe that if b^x-1 and b^y-1 and both divisible by p^2 (or any other number), then
Message 3 of 18 , Jun 5, 2007
--- In primenumbers@yahoogroups.com, "Kevin Acres" <research@...> wrote:

> just wondering ...
> if it holds true that p^2 always divides the
> minimal base^(p-1/x)-1 that p does.

Yes. Let d be the smallest value so that p divides b^d-1.

First observe that if b^x-1 and b^y-1 and both divisible by p^2 (or
any other number), then b^(gcd(x,y))-1 is also divisible by p^2 (or
any other number).

Second, observe that b^(pd)-1 is divisible by p^2. To see this,
express b^d as (ps+1), then expand (ps+1)^p with the binomial theorem.

Third, observe that every case of divisibility by p is of the form
b^(kd)-1, and every case of divisibility by p^2 is also a case of
divisiblity by p, so the minimal case of divisibility by p^2 must be
of the form b^(kd)-1.

Finally, if there is any k<p such that b^(kd)-1 is divisible by p^2,
then b^gcd(kd,pd) must also be divisible by p^2, but gcd(kd,pd)=d.

Therefore, in all cases of Vanishing Fermat Quotients, p^2 divides the
same primitive term that p divides.

William
Poohbah of OddPerfect.org

P.S. We still need a few large factors of Vanishing Fermat Quotients
at http://oddperfect.org/FermatQuotients.html
• Earlier, with reference to 29, and 47 the number 61 came up. It is a factor of the 15th Fibonacci number 610 Ring any bells?
Message 4 of 18 , Jun 14, 2007
Earlier, with reference to 29, and 47 the number 61 came up.

It is a factor of the 15th Fibonacci number 610

Ring any bells?
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