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Re: F = 20z^2 - n and (5r^2 -F)/n

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  • djbroadhurst
    ... Explanation: I had solved Aldrich s issquare problem (5*r^2 - F)/11 = a^2, with F = 20*z^2 - 11, by setting r = 3*p, a = 2*p, where p =
    Message 1 of 7 , Feb 28, 2010
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      --- In primenumbers@yahoogroups.com,
      Kevin Acres <research@...> wrote:

      > Was the hint that generous or did I miss the point entirely?

      The hint was indeed very generous and my point was this:

      > I conjecture that this generous hint gives the only answer.

      Explanation: I had solved Aldrich's "issquare problem"
      (5*r^2 - F)/11 = a^2, with F = 20*z^2 - 11,
      by setting r = 3*p, a = 2*p, where
      p = 63304192701203884454712276208334615126937940298786013924413
      is, most crucially, a prime.

      Then F = p^2 is the square of a prime. Thus I conjecture that
      (5*r^2 - F)/11 is never a square for 2*z < r < 3*p,
      though it is certainly a square for r = 2*z and for r = 3*p.
      As Kevin noticed, 3*p is very close to 6*sqrt(5)*z, since
      z^2 = (p^2 + 11)/20.

      Puzzle: (p^2 + 11)/20 is a square for the primes
      p = 3, 13, 67, 4253, 21587, 6950947 ...
      Find the 17th prime in this sequence.

      Comment: Extra credit will be gained for proving
      that your answer is correct.

      David
    • djbroadhurst
      ... Kevin sent me a correct answer, in decimal format. But for that he needed 454 characters. The solution can be written in less than 20 characters, using a
      Message 2 of 7 , Mar 2, 2010
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        --- In primenumbers@yahoogroups.com,
        "djbroadhurst" <d.broadhurst@...> wrote:

        > Puzzle: (p^2 + 11)/20 is a square for the primes
        > p = 3, 13, 67, 4253, 21587, 6950947 ...
        > Find the 17th prime in this sequence.

        Kevin sent me a correct answer, in decimal format. But for
        that he needed 454 characters. The solution can be written
        in less than 20 characters, using a form that OpenPFGW
        readily understands. So the puzzle is still open, for a
        compact answer.

        Meanwhile, congratulations to Kevin, for a valid result

        David
      • djbroadhurst
        ... Kevin has now found a reasonably compact answer as a linear combination of 3 Fibonacci numbers. In fact the answer can be written in terms of only 2:
        Message 3 of 7 , Mar 3, 2010
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          --- In primenumbers@yahoogroups.com,
          "djbroadhurst" <d.broadhurst@...> wrote:

          > Puzzle: (p^2 + 11)/20 is a square for the primes
          > p = 3, 13, 67, 4253, 21587, 6950947 ...
          > Find the 17th prime in this sequence.

          Kevin has now found a reasonably compact answer
          as a linear combination of 3 Fibonacci numbers.
          In fact the answer can be written in terms of only 2:

          Solution [in 19 characters]:

          F(2166)/2+3*F(2168)

          Proof [by exhaustion, with primality proving]:

          P(n,s) = fibonacci(6*n)/2 + 3*fibonacci(6*n+2*s);
          {c=0;for(n=0,361,forstep(s=-1,1,2,p=P(n,s);
          if(issquare((p^2+11)/20)&&isprime(p),c++;
          print1(s*n" "))));print([c]);}

          0 -1 1 -3 3 5 -11 15 -45 -47 -65 75 105 253 -303 -359 361 [17]

          Comments: Dario Alpern's http://www.alpertron.com.ar/QUAD.HTM
          may be used to prove that my Fibonacci method is exhaustive.
          A probable continuation of the sequence of primes is as follows:

          F(3966)/2+3*F(3964), 829
          F(5334)/2+3*F(5336), 1116
          F(8682)/2+3*F(8684), 1816
          F(15066)/2+3*F(15068), 3150
          F(15438)/2+3*F(15436), 3227
          F(41166)/2+3*F(41168), 8604
          F(50874)/2+3*F(50876), 10633
          F(114702)/2+3*F(114704), 23972
          F(117978)/2+3*F(117980), 24657

          with the number of digits of each PRP given, after the comma.
          The last 3 have been consigned to Henri's repository of PRPs.

          David
        • Kevin Acres
          ... My solution, in 29 characters, is: 3/2*F(2169)+5*F(2164)+F(2161) David s sequence, including the non-prime values, starting with the 13,67 pair can be
          Message 4 of 7 , Mar 4, 2010
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            At 03:15 PM 4/03/2010, djbroadhurst wrote:

            > > Puzzle: (p^2 + 11)/20 is a square for the primes
            > > p = 3, 13, 67, 4253, 21587, 6950947 ...
            > > Find the 17th prime in this sequence.
            >
            >Kevin has now found a reasonably compact answer
            >as a linear combination of 3 Fibonacci numbers.
            >In fact the answer can be written in terms of only 2:
            >
            >Solution [in 19 characters]:
            >
            >F(2166)/2+3*F(2168)

            My solution, in 29 characters, is:

            3/2*F(2169)+5*F(2164)+F(2161)

            David's sequence, including the non-prime values, starting with the
            13,67 pair can be reproduced by the following:

            Let f1=F(12*n+9), f2=F(12*n+4), f3=F(12*n+1), a=f1+2*f2 and o=1/2*f1+3*f2+f3

            Then the lower of each pair is a-o and the higher is a+o or:

            1/2*f1-f2-f3 and 3/2*f1+5*f2+f3

            In the case of the p454 just substitute n for 180 in the above assignments.

            The pari/gp script to generate the sequence follows:

            for(i=0,10,
            f1=fibonacci(12*i+9);
            f2=fibonacci(12*i+4);
            f3=fibonacci(12*i+1);
            o=1/2*f1+3*f2+f3;
            a=f1+2*f2;
            print([a-o,a+o]);
            );

            I strayed a little from the puzzle by solving for 180z^2-p^2-11
            before trying to re-create the sequence from 13/67 onwards.

            Once again, Dario's page at: http://www.alpertron.com.ar/QUAD.HTM was
            of great help.


            Kevin.
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