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Re: Square factors of b^p-1

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  • mikeoakes2
    ... Using 4 days pari-GP at 3.6Ghz I have solidified these results. For b1, b2, q
    Message 1 of 81 , Oct 3, 2011
      --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
      >
      > --- In primenumbers@yahoogroups.com,
      > "djbroadhurst" <d.broadhurst@> 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).
      > >
      > > Here are 5 such pairs of triples:
      > >
      > > [132529, 277691, 555383]
      > > [211811, 277691, 555383]
      > >
      > > [134507, 883703, 1767407]
      > > [720901, 883703, 1767407]
      > >
      > > [292933, 1051553, 2103107]
      > > [385079, 1051553, 2103107]
      > >
      > > [434243, 613189, 2452757]
      > > [470711, 613189, 2452757]
      > >
      > > [1161143, 3700283, 7400567]
      > > [2768789, 3700283, 7400567]
      >
      > and a sixth, also with p1 = p2:
      >
      > [ 417961, 5699017, 22796069]
      > [1997161, 5699017, 22796069]

      Using 4 days pari-GP at 3.6Ghz I have solidified these results.

      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.

      Mike
    • 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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