Re: [PrimeNumbers] Re: Thanks to all. Proposed binary Prime number test fails.
- --- Mark Rodenkirch <mgrogue@...> wrote:
> This was meant to be sent to the group and I sent it to Jens bySubscirbe!
> 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).
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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 byYGM.
> > 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).
[Non-text portions of this message have been removed]
- --- In firstname.lastname@example.org, "Kevin Acres" <research@...> wrote:
> just wondering ...Yes. Let d be the smallest value so that p divides b^d-1.
> if it holds true that p^2 always divides the
> minimal base^(p-1/x)-1 that p does.
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.
Poohbah of OddPerfect.org
P.S. We still need a few large factors of Vanishing Fermat Quotients