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Re: Quadratic Prime Chains

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  • Paul Underwood
    ... When X is odd the expressions are even. They are not equal to 2, so they are then composite. ... ? factor(x^4-97*x^3+3294*x^2-45458*x+213589) [x^4 -
    Message 1 of 2 , May 25, 2008
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      --- In primenumbers@yahoogroups.com, "aldrich617" <aldrich617@...> wrote:
      >
      > Well, got the official word from D.B.:
      > linear transforms are of no value and
      > egregious factors are annoying. But
      > then again he says that about
      > almost everything. I really thought
      > I was on to something with
      >
      > X^2 - 80X + 1763 --- 79 consecutive primes
      > X^2 - 237X + 14409 --- 76 consecutive primes
      >

      When "X" is odd the expressions are even. They are not equal to 2, so
      they are then composite.

      > but apparently, no.
      >
      > I still want to try to beat the limit though,
      > and use transforms
      > to find a way to make longer prime chains.
      >
      > Can anyone factor
      >
      > X^4 - 97X^3 + 3294X^2 -45458X + 213589
      > (49 consecutive primes)
      >
      > into quadratics?

      ? factor(x^4-97*x^3+3294*x^2-45458*x+213589)
      [x^4 - 97*x^3 + 3294*x^2 - 45458*x + 213589 1]

      Of course this is factored into linear factors over the complex
      numbers ;-)

      >
      > Now that is an equation that really looks like
      > fun to me, and a factorization would be a good
      > place to start.
      >

      Please see:
      http://www.maa.org/editorial/mathgames/mathgames_07_17_06.html

      As far as I can tell, if only monic polynomials[1] with unique
      positive values were permissible, then Euler would have won the
      quadratic race.[2]

      Paul

      [1] http://mathworld.wolfram.com/MonicPolynomial.html
      [2] http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html

      > Aldrich Stevens
      >
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