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Re: Complex a*x^n+b*y^n puzzle

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  • mikeoakes2
    ... Here is the output from complexification of yesterday s simple-minded pari-gp script, for n_max=9:- a=3 + 3*I, b=2 + 3*I, x=2, y=1 - 2*I n=1 u=14 + 5*I
    Message 1 of 37 , Nov 28, 2009
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      --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
      >
      > Complex a*x^n+b*y^n puzzle: Find 4 Gaussian integers a,b,x,y
      > such that x and y are not units and a*x^n+b*y^n is a Gaussian
      > prime, for 1 <= n <= n_max, with n_max as large as possible.

      Here is the output from complexification of yesterday's simple-minded pari-gp script, for n_max=9:-

      a=3 + 3*I, b=2 + 3*I, x=2, y=1 - 2*I
      n=1 u=14 + 5*I uu=221
      n=2 u=18 - 5*I uu=349
      n=3 u=-4 - 5*I uu=41
      n=4 u=-38 + 75*I uu=7069
      n=5 u=64 + 295*I uu=91121
      n=6 u=558 + 455*I uu=518389
      n=7 u=1276 - 85*I uu=1635401
      n=8 u=722 - 1485*I uu=2726509
      n=9 u=-3016 - 625*I uu=9486881

      (u is your expression a*x^n+b*y^n, uu is its norm.)

      Mike
    • djbroadhurst
      ... There is only rational prime of the form a^2 + b^2 that yields precisely 4 distinct Gaussian primes, namely 2 = 1^2 + 1^2. If a^2 + b^2 is an odd rational
      Message 37 of 37 , Dec 5, 2009
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        --- In primenumbers@yahoogroups.com,
        "Robdine" <robdine@...> wrote:

        > any rational prime that can be represented by the sum
        > of 2 squares (a^2+b^2) will define 4 gaussian primes

        There is only rational prime of the form a^2 + b^2 that yields
        precisely 4 distinct Gaussian primes, namely 2 = 1^2 + 1^2.

        If a^2 + b^2 is an odd rational prime, we have
        8 [sic] asociates of the Gaussian prime z = a + I*b,
        since we may mulitply it and its conjugate z = a - I*b
        by the 4 units I, -I, -1, 1, obtaining 8 distinct
        Gaussian integers that are prime.

        David
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