--- 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