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Math Help - Problem 46

  1. #1
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    Problem 46

    1)Let f(x) be a monic polynomial* with integer coefficients with \deg f(x) \geq 1. Prove that if the sum of all coefficients and the product of all the complex zeros (counting multiplicity) are both odd then the polynomial has not integer zeros.

    *)Leading term is 1.
    Last edited by ThePerfectHacker; February 28th 2008 at 07:29 PM.
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  2. #2
    Senior Member JaneBennet's Avatar
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    By “complex zeros”, do you mean both complex and real zeros, or just complex non-real zeros?
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    Quote Originally Posted by JaneBennet View Post
    By “complex zeros”, do you mean both complex and real zeros, or just complex non-real zeros?
    I mean all zeros. For example, take x^3+x then the zeros are 0,i,-i. Does saying "zeros in \mathbb{C}" make it better?
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  4. #4
    Member Henderson's Avatar
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    I must not understand the question. My first thought is:
    f(x) = (x-1)(x-3)(x-5) = x^3 - 9x^2 +23x - 15.

    The sum of the zeros is 9, the product of the zeros is 15, and all three zeros are integers.
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  5. #5
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    Thank you. I was inventing this problem, I should have been more careful. I fixed it above. Tell me if it is okay now.
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  6. #6
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    By considering the equation \pmod{2} and the fact that f(1) and f(0) are both odd implies the truth of your statement
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  7. #7
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by SimonM View Post
    By considering the equation \pmod{2} and the fact that f(1) and f(0) are both odd implies the truth of your statement
    how?
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  8. #8
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    a = b \pmod{n} \Rightarrow f(a)=f(b) \pmod{n} (If f(x) is a polynomial)

    If x is a root of a polynomial f(x)=0= 0\pmod{n} which can't happen if both parities are 1 \pmod{n}
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  9. #9
    MHF Contributor chiph588@'s Avatar
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    Proof by contradiction: Assume  f(x) is a polynomial with integer roots.

    First note that  f(0) is the product of all the roots and  f(1) is the sum of coefficients.

    Let  f(x) = (x-r_1) \cdots (x-r_n) , where  r_1, \cdots , r_n are all the roots to  f(x) .

    So  f(0) = \prod_{i = 1}^n (-r_i) and since  f(0) is odd, this implies each  r_i is odd too.

    But  f(1) = \prod_{i = 1}^n (1-r_i) and since  f(1) is odd, this implies that each  1-r_i is odd which implies each  r_i is even.

    Hence a contradiction and therefore there are no integer solutions.
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