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Math Help - prove f is a rational polynomial

  1. #1
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    prove f is a rational polynomial

    f(x) is polynomial with complex coefficients. \forall n\in Z, f(n)is integer, prove: coefficients of f(x) are rational numbers, and give some examples about rational case.

    ### Prove:

    ---
    * consider coefficients are integers, of course f(n) are integers.

    * consider coefficients are rationals, we have f(x)=\frac{1}{2}x(x+1), two consecutive integer can be divided by 2, there must be one even number.

    How about the cases?:


    * real coeffs

    * complex coeffs

    ---
    And can you give me some more examples?
    Last edited by integer; October 7th 2013 at 02:20 AM.
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  2. #2
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    Re: prove f is a rational polynomial

    The following ideas are based on this article (PDF). It turns out that every integer-valued polynomial p(x) has the form

    p(x) = a_0+a_1\binom{x}{1}+ \dots+ a_n\binom{x}{n} (*)

    Indeed, the coefficients (omitting the constant term, which is zero) of polynomials \binom{x}{1}, \binom{x}{2}, ..., \binom{x}{n} form a triangular matrix, so these polynomials form a basis, i.e., every polynomial of degree ≤ n with no constant term can be expressed as a linear combination of these polynomials, perhaps with complex coefficients. So, every polynomials from ℂ[x] of degree ≤ n can be written in the form (*) with complex a_k. Next prove by induction on k that a_k\in\mathbb{Z}.
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