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

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• ... 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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