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

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  • mikeoakes2
    I wonder if any mileage can be got from the identity (provided x y):- a*x^n+b*y^n = ((a-b)*lucasV(x+y,x*y,n+1)+(b*x-a*y)*lucasV(x+y,x*y,n))/(x-y) Mike
    Message 1 of 37 , Nov 28, 2009
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      I wonder if any mileage can be got from the identity (provided x<>y):-
      a*x^n+b*y^n =
      ((a-b)*lucasV(x+y,x*y,n+1)+(b*x-a*y)*lucasV(x+y,x*y,n))/(x-y)

      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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