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• ## Re: Thanks to all. Proposed binary Prime number test fails.

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• ... Keller and Richstein are exhaustive for bases
Message 1 of 18 , May 31, 2007
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--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@...> wrote:

Keller and Richstein are exhaustive for bases < 1000 and exponents <
10^11.

http://www.mscs.dal.ca/~joerg/res/fq.html

OddPerfect.org has collected factors for some of these because Pace
Neilsen needed them in an approach to extending the minimum number of
distinct prime factors for an odd perfect number.

http://oddperfect.org/FermatQuotients.html
• ... 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
Message 1 of 18 , Jun 1, 2007
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Kevin Acres wrote:
> A little bit of pertinent information just arrived in my inbox:
>
> Wilfrid Keller and Jörg Richstein searched the bases
> b = 29, 47, 61
> and found no example of
> b^(p-1) = 1 mod p^2
> for any prime for p < 10^11.

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"

--
Jens Kruse Andersen
• 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
Message 1 of 18 , Jun 1, 2007
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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).

--Mark
• 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 1 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 1 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]

____________________________________________________________________________________
Be a PS3 game guru.
Get your game face on with the latest PS3 news and previews at Yahoo! Games.
http://videogames.yahoo.com/platform?platform=120121
• ... YGM. --Mark [Non-text portions of this message have been removed]
Message 1 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 1 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 1 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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