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primes of polynomials

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  • gordon_as_number
    Only a comment. If you want to prove a polynomials P(x) is prime, isnt enough to determine the assoiated roots. E.g. P(x)=x^2+2 The number N is P(x) which can
    Message 1 of 14 , Apr 2, 2006
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      Only a comment.

      If you want to prove a polynomials P(x) is prime,
      isnt enough to determine the assoiated roots.

      E.g. P(x)=x^2+2

      The number N is P(x) which can base expanded in
      some base b as

      N=\sum_{i=0}^m c_i b^i .

      Determining the roots and the coefficients I claim
      is order \ln^3(N), which is a rough estimate here
      because I didnt let N vary and I didnt include what
      the value of m is. So you have to polynomials evaluated
      in the different bases b with the same coefficients c_i.
      The number of real c.c. pairs (x-\alpha_i)*(x-\alpha_j)^\star
      = integer or (x-\alpha_i) an integer span the prime factors.
      Arent there theorems that tell you when polynomials have
      non-integer root pairs so that you can make a statement
      towards their construction. I think there are. There
      should be large classes of integers N of the same degree
      and with the same coefficients c_j but with varying base;
      I think these can be constructed in ln^3 N_{max} time with
      N_{max} being the largest integer considered.

      The construction I refer to is given in Fast Factoring
      by Dr. Gordon Chalmers.
    • Phil Carmody
      ... First you need to specify the ring in which you re working. You do not do so. One cannot assume Z[x] given that many of the examples we ve seen have been
      Message 2 of 14 , Apr 2, 2006
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        --- gordon_as_number <gordon_as_number@...> wrote:
        > Only a comment.
        >
        > If you want to prove a polynomials P(x) is prime,

        First you need to specify the ring in which you're working.
        You do not do so.
        One cannot assume Z[x] given that many of the examples we've seen have
        been instead in Q[x].

        > isnt enough to determine the assoiated roots.
        >
        > E.g. P(x)=x^2+2
        >
        > The number N is P(x) which can base expanded in
        > some base b as
        >
        > N=\sum_{i=0}^m c_i b^i .

        Bases ought to be irrelevant. If they're not, you're doing something wrong.

        > Determining the roots and the coefficients I claim
        > is order \ln^3(N), which is a rough estimate here
        > because I didnt let N vary

        Order(funtion(N)) only makes sense if you let N vary.
        You've reached that level of illucidity again...

        > and I didnt include what
        > the value of m is. So you have to polynomials evaluated
        > in the different bases b

        Yup - illucid. Polynomials can be evaluated, but their value is
        independent of any base.

        > with the same coefficients c_i.

        The c_i, as you introduced them, are not coefficients.

        > The number of real c.c. pairs (x-\alpha_i)*(x-\alpha_j)^\star
        > = integer or (x-\alpha_i) an integer span the prime factors.

        With no firm foundation to build on, you've completely lost me now.
        No point in continuing.

        Phil

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