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Re: Prime chains x-->Ax+B [puzzle 11]

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  • djbroadhurst
    ... I tested all with p[3]
    Message 1 of 143 , Dec 4, 2010
      --- In primenumbers@yahoogroups.com,
      "mikeoakes2" <mikeoakes2@...> wrote:

      > Using a non-sieving script, I have gone up to a=1.8*10^10,
      > for all 4 starting p, and have found no 10/12's, just your
      > 9/12 and 2 new ones, viz.
      > [237862725*x - 713588164, [3, 9]]
      > [4939393834*x - 9878787657, [2, 9]]
      > [7484387043*x - 52390709290, [7, 9]]

      I tested all with p[3] < 10^11, finding
      7/12: 2336
      8/12: 60
      9/12: 3 (as above)

      Apologies for setting a puzzle that now seems far too hard :-(

      David
    • 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
        --- 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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