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Re: Square factors of b^p-1

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  • djbroadhurst
    ... and a sixth, also with p1 = p2: [ 417961, 5699017, 22796069] [1997161, 5699017, 22796069] David
    Message 1 of 81 , Sep 23, 2011
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      --- 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]

      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
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        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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