--- In

primenumbers@yahoogroups.com,

Kermit Rose <kermit@...> wrote:

> b1^p1 = 1 mod q^2

> b2^p2 = 1 mod q^2

> If p1 is not equal to p2, then it must be that

> p1*p2 divides (q-1).

Indeed. Moreover, (b1,q) and (b2,q) are Wieferich pairs

with the same q. Such conjunctions are rare.

Even when we find one, the probability that it solves

our problem, with p2 > p1 > 2, is low.

If q = 2*k*p1*p2 + 1, we might guess the probability as

(p1/(q-1))*(p2/(q-1)) = 1/(2*k*(q-1))

which is why this feat has not yet been accomplished.

But that does not mean that it cannot be done.

David