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Re: p1, p2 mod q^2

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  • djbroadhurst
    ... 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
    Message 1 of 2 , Oct 2, 2011
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      --- 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
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