p1, p2 mod q^2
- If 0 < b1 < p1 < q
and 0 < b2 < p2 < q
b1^p1 = 1 mod q^2
b2^p2 = 1 mod q^2
What can we conclude about relation between p1 and p2?
Consider the case where q is prime.
It must also be true that
b1^p1 = 1 mod q
b2^p2 = 1 mod q.
and since q is prime, p1 divides (q-1) and
p2 divides (q-1).
If p1 is not equal to p2, then it must be that
p1*p2 divides (q-1).
which means that q would have to be much larger, perhaps out of bounds
- --- In firstname.lastname@example.org,
Kermit Rose <kermit@...> wrote:
> b1^p1 = 1 mod q^2Indeed. Moreover, (b1,q) and (b2,q) are Wieferich pairs
> b2^p2 = 1 mod q^2
> If p1 is not equal to p2, then it must be that
> p1*p2 divides (q-1).
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.