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Re: Square factors of b^p-1

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  • djbroadhurst
    ... In the course of this Mike Oakes found 8 pairs of Wieferich pairs not recorded Michael Mosinghoff, who condidered only q = 1 mod 4 when q 10^7 and hence
    Message 1 of 81 , Oct 3, 2011
      --- In primenumbers@yahoogroups.com,
      "mikeoakes2" <mikeoakes2@...> wrote:

      > > Puzzle 3a: Find primes b1 < p1 < q, b2 < p2 < q, b1 < b2, s.t.
      > > b1^p1 = b2^p2 = 1 mod q^2, (with no limits on q^2).
      >
      > For b1, b2, q < 10^7.5, the complete list of such q is:-
      > 555383
      > 1767407
      > 2103107
      > 2452757
      > 7400567
      >
      > 12836987
      > 14668163
      > 15404867
      > 16238303
      > 19572647
      > 22796069 [found by DB]
      > 25003799
      > 26978663
      > 27370727
      >
      > Each (q-1) is of the form 2^m*p, and p1 = p2 in each case.

      In the course of this Mike Oakes found 8 pairs of
      Wieferich pairs not recorded Michael Mosinghoff,
      who condidered only q = 1 mod 4 when q > 10^7 and
      hence missed SG pairs (p,q=2*p+1).

      In the format [q,m,p,[b1,b2]] these new
      double Wieferich results are:

      [12836987, 1, 6418493, [2061197, 4631743]]
      [14668163, 1, 7334081, [3692407, 7145591]]
      [15404867, 1, 7702433, [2824163, 3123217]]
      [16238303, 1, 8119151, [639167, 1005581]]
      [19572647, 1, 9786323, [3204463, 7873399]]
      [25003799, 1, 12501899, [59273, 4701583]]
      [26978663, 1, 13489331, [2589553, 8582417]]
      [27370727, 1, 13685363, [5002507, 5356201]]

      Moreover M.O must have found many more
      isolated Wieferich pairs with q > 10^7.
      Perhaps these might be communicated to M.M ?

      David
    • Walter Nissen
      Hi , all , Not even draft yet , let alone published : http://upforthecount.com/math/nnp.html still needs a lot of work . But , I have updated the top of the
      Message 81 of 81 , Nov 29, 2011
        Hi , all ,

        Not even draft yet , let alone published :
        http://upforthecount.com/math/nnp.html
        still needs a lot of work .
        But , I have updated the top of the main table :
        http://upforthecount.com/math/nnp2.txt
        with complete factorizations now extending through
        np ( 90 ) .

        When is np ( n ) prime ?
        So far , only for 1 , 2 and 3 .
        When n mod 6 = 4 , np ( n ) has at least 3 algebraic
        factors , 3 and n^2 + n + 1 ( twice ) .
        But what about the residue ( primitive part ? ) ?
        When is this prime ?
        n t n^n + (n+1)^(n+1)
        4 4 3 * 7 * 7 * 23
        16 6 3 * 7 * 7 * 13 * 13 * 34041259347101651
        Any others ?

        Cheers ,

        Walter
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