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

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• I did indeed look through all of the interesting links that you sent. I m just wondering if I can get the time to work through the information and see if it
Message 1 of 18 , Jun 1, 2007
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I did indeed look through all of the interesting links that you sent. I'm
just wondering if I can get the time to work through the information and see
if it holds true that p^2 always divides the minimal base^(p-1/x)-1 that p
does.

Kevin.

-----Original Message-----
From: Jens Kruse Andersen [mailto:jens.k.a@...]
Sent: 02 June 2007 01:12

[...]

Did you look at the first link I gave you:
http://www.fermatquotient.com/FermatQuotienten/FermQ_Sort
It includes those bases searched to 5.074*10^12 (with no odd solutions),
and much more.
It was updated 3 days ago and appears to be the current records.
http://www.fermatquotient.com/ says:
"http://www.fermatquotient.com/FermatQuotienten/News
Neues und Rekorde / News and records / Update 29.05.2007"
• ... I suspect it s provable, but I can t locate my copy of Hardy and Wright tonight to skim for inspiration. The minimal value is the column called order at
Message 2 of 18 , Jun 2, 2007
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--- In primenumbers@yahoogroups.com, "Kevin Acres" <research@...> wrote:

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

I suspect it's provable, but I can't locate my copy of Hardy and
Wright tonight to skim for inspiration.

The minimal value is the column called "order" at
http://oddperfect.org/FermatQuotients.html , so it was easy to verify
that p^2 divided the minimal value for all of those.

William
Poohbah of oddperfect.org
• ... Subscirbe! Phil () ASCII ribbon campaign () Hopeless ribbon campaign / against HTML mail / against gratuitous bloodshed [stolen with
Message 3 of 18 , Jun 3, 2007
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--- 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 4 of 18 , Jun 3, 2007
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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 5 of 18 , Jun 5, 2007
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--- 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 6 of 18 , Jun 14, 2007
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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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