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Math Help - Polynomial question.

  1. #1
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    Polynomial question.

    I have f(x)=x^n+a_1x^{n-1}+...+a_n. With integer coefficients.

    I know that f(0) and f(1) are odd numbers.

    I have to prove that f(x) has no roots over Q.


    ///
    From the fact that f(0) is odd I know that a_n is odd...

    Which approach should I use to finish the proof?
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  2. #2
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    Hello andreas

    Just a question : is it f(x)=x^n+a_1x^{n-1}+...+a_n
    or f(x)=a_0x^n+a_1x^{n-1}+...+a_n
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hello andreas

    Just a question : is it f(x)=x^n+a_1x^{n-1}+...+a_n
    or f(x)=a_0x^n+a_1x^{n-1}+...+a_n

    It is f(x)=x^n+a_1x^{n-1}+...+a_n
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  4. #4
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    Quote Originally Posted by andreas View Post
    It is f(x)=x^n+a_1x^{n-1}+...+a_n
    OK
    Suppose \frac{p}{q} is a root of f (with p and q integers and fraction in its lowest terms)
    Then
    f(\frac{p}{q})\;=\;0

    (\frac{p}{q})^n+a_1(\frac{p}{q})^{n-1}+...+a_n\;=\;0

    Multiplying by q^n
    p^n+a_1p^{n-1}q+...+a_nq^n\;=\;0
    a_1p^{n-1}q+...+a_nq^n\;=\;-p^n
    q dividing the first member divides the second one (p^n)
    q and p having no common factor then q=1
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  5. #5
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    See this and look at last post.
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  6. #6
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    could you fix the link?
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  7. #7
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    Quote Originally Posted by andreas View Post
    I have f(x)=x^n+a_1x^{n-1}+...+a_n. With integer coefficients.

    I know that f(0) and f(1) are odd numbers.

    I have to prove that f(x) has no roots over Q.
    if f has a rational root, it has to be an integer because f \in \mathbb{Z}[x] is monic. suppose f(k)=0 for some k \in \mathbb{Z}. now k has to be odd since k(k^{n-1} + a_1k^{n-2} + \cdots +a_{n-1})=-a_n=-f(0).

    on the other hand f(x)=(x-1)g(x)+f(1), for some g \in \mathbb{Z}[x]. thus (k-1)g(k)=-f(1), which is impossible because f(1) is odd and k-1 is even. \Box
    Last edited by NonCommAlg; November 29th 2008 at 11:30 PM.
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