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

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  • Phil Carmody
    ... This doesn t fit the tuplet pattern, but would fall to raw gensv . Unfortunately the only wrapper around gensv that works currently is tuplet , not
    Message 1 of 143 , Dec 2, 2010
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      > Puzzle 11: Find a plupluperfect chain of 12, i.e. a
      > triplet
      > (A, B, p[1]) such that A = 1 mod 11, B is coprime to 11,
      > and the iteration p[n+1] = A*p[n] + B yields a chain of
      > 12 increasing primes, p[n], for n = 1 to 12.
      >
      > Comments: Clearly, p[2] = 11 and hence B = 11 - A*p[1], with
      > p[1] = 2, 3, 5 or 7. So this boils down to 4 sub-problems of
      > Carmody-scale difficulty (maybe easy for Phil; harder for mere
      > mortals like Kevin, Mike, or me). I did not yet find a solution.

      This doesn't fit the 'tuplet' pattern, but would fall to raw 'gensv'.
      Unfortunately the only wrapper around 'gensv' that works currently is
      'tuplet', not the raw engine. So I'll have to duck out of this until
      2015, when I finally get gensv working again.

      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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