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

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  • j_chrtn
    Hi Mike, What about : a=140 x=2 b=47 y=3 ? jl
    Message 1 of 37 , Nov 27, 2009
      Hi Mike,

      What about :
      a=140
      x=2
      b=47
      y=3

      ?

      jl



      --- In primenumbers@yahoogroups.com, "mikeoakes2" <mikeoakes2@...> wrote:
      >
      > Here is a small puzzle:-
      >
      > Find fixed integers a, b, x, y such that the expression
      > a*x^n+b*y^n
      > is prime for all n in the range 1 <= n <= n_max,
      > where n_max is to be as large as possible.
      >
      > Hint: n_max=9 is certainly achievable.
      >
      > -Mike Oakes
      >
    • 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 7:49 AM
        --- 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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