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

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  • djbroadhurst
    ... Congrats to Phil for scoring 11/12, twice: {A=[519815699116440,1188368874866394];q=2; for(k=1,#A,a=A[k];b=11-q*a;p=q;if(a%11==1,
    Message 1 of 143 , Dec 4, 2010
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      --- In primenumbers@yahoogroups.com,
      Phil Carmody <thefatphil@...> wrote:

      > > but then it did indeed start to get tough.
      > Corroborated: (and here, as 2 are for free, 7=9)
      ...
      > 9 519815699116440
      ...
      > 9 1188368874866394

      Congrats to Phil for scoring 11/12, twice:

      {A=[519815699116440,1188368874866394];q=2;
      for(k=1,#A,a=A[k];b=11-q*a;p=q;if(a%11==1,
      for(n=1,12,print1(isprime(p)" ");p=a*p+b));print());}

      1 1 1 1 1 1 1 1 1 1 1 0
      1 1 1 1 1 1 1 1 1 1 1 0

      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
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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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