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Re: Square factors of b^p-1

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  • djbroadhurst
    ... These have larger q: 1783447^48649379 = 1 mod 97298759^2 25659449^48649379 = 1 mod 97298759^2 2076259^48687659 = 1 mod 97375319^2 34543973^48687659 = 1 mod
    Message 1 of 81 , Oct 3, 2011
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      --- In primenumbers@yahoogroups.com,
      "djbroadhurst" <d.broadhurst@...> wrote:

      > ... the most lucrative case, p1=p2=(q-1)/2 ...
      >
      > 25638913^35364419 = 1 mod 70728839^2
      > 26450899^35364419 = 1 mod 70728839^2

      These have larger q:

      1783447^48649379 = 1 mod 97298759^2
      25659449^48649379 = 1 mod 97298759^2

      2076259^48687659 = 1 mod 97375319^2
      34543973^48687659 = 1 mod 97375319^2

      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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