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Re: [PrimeNumbers] Re: Prime chains x-->Ax+B [puzzle 2]

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  • Phil Carmody
    ... Well, multiplying by 18 and then adding a constant tends to keep your residue modulo 18 pretty constant. Phil
    Message 1 of 143 , Nov 26, 2010
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      --- On Thu, 11/25/10, Kevin Acres <research@...> wrote:
      > [1, 1087, 1, 1, Mat([1087, 1])]
      > [2, 18212081, 5, 1, Mat([18212081, 1])]
      > [3, 346009973, 5, 1, Mat([346009973, 1])]
      > [4, 6246372029, 5, 1, Mat([6246372029, 1])]
      > [5, 112452889037, 5, 1, Mat([112452889037, 1])]
      > [6, 2024170195181, 5, 1, Mat([2024170195181, 1])]
      > [7, 36435081705773, 5, 1, Mat([36435081705773, 1])]
      > [8, 655831488896429, 5, 1, Mat([655831488896429, 1])]
      > [9, 11804966818328237, 5, 1, Mat([11804966818328237, 1])]
      > [10, 212489402748100781, 5, 1, Mat([212489402748100781, 1])]
      > [11, 3824809249484006573, 5, 1, Mat([3824809249484006573, 1])]
      > [12, 68846566490730310829, 5, 1, Mat([68846566490730310829, 1])]
      > [13, 1239238196833163787437, 5, 1, Mat([1239238196833163787437, 1])]
      > [14, 22306287542996966366381, 5, 1, Mat([22306287542996966366381, 1])]
      > [15, 401513175773945412787373, 5, 1, Mat([401513175773945412787373, 1])]
      > [16, 7227237163931017448365229, 5, 0, [43, 1; 4817, 1; 34892107718936409559, 1]]
      > [17, 130090268950758314088766637, 5, 0, [17, 1; 7652368761809312593456861, 1]]
      >
      > BTW both 15/16 chains are all 5 mod 6 expect for the first value
      > which is 1 mod 6.
      >
      > Is there anyone that can tell me if I should read anything
      > in to this or not?

      Well, multiplying by 18 and then adding a constant tends to keep your residue modulo 18 pretty constant.

      Phil
    • djbroadhurst
      ... Suppose that we want to start with a square and get a square. Then we must solve the Diophantine equation y^2 = a*x^2 + b For any pair (a,b), Dario will
      Message 143 of 143 , Jan 7, 2011
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        --- In primenumbers@yahoogroups.com,
        Kevin Acres <research@...> wrote:

        > x=a*x+b either is a square or has a square as a major factor.

        Suppose that we want to start with a square and get a square.
        Then we must solve the Diophantine equation
        y^2 = a*x^2 + b

        For any pair (a,b), Dario will tell us all the solutions:
        http://www.alpertron.com.ar/QUAD.HTM

        David
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