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Thread: Polynomial question.

  1. #1
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    Polynomial question.

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

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

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


    ///
    From the fact that $\displaystyle f(0)$ is odd I know that $\displaystyle a_n$ 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 $\displaystyle f(x)=x^n+a_1x^{n-1}+...+a_n$
    or $\displaystyle f(x)=a_0x^n+a_1x^{n-1}+...+a_n$
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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 $\displaystyle f(x)=x^n+a_1x^{n-1}+...+a_n$
    or $\displaystyle f(x)=a_0x^n+a_1x^{n-1}+...+a_n$

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

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

    Multiplying by q^n
    $\displaystyle p^n+a_1p^{n-1}q+...+a_nq^n\;=\;0$
    $\displaystyle a_1p^{n-1}q+...+a_nq^n\;=\;-p^n$
    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 $\displaystyle f(x)=x^n+a_1x^{n-1}+...+a_n$. With integer coefficients.

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

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

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