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Math Help - Putnam Problem

  1. #1
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    Putnam Problem

    Hi guys,

    I am currently stuck on this Putnam problem and was wondering if anyone had some tips/strategies to attack it?

    Determine all polynomials that have only real roots and all coefficients are +/-1?
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  2. #2
    Behold, the power of SARDINES!
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    Since \pm 1 is rational the rational roots theorem should be of some help

    Rational root theorem - Wikipedia, the free encyclopedia
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  3. #3
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    Hello, arsenicbear!

    Determine all polynomials that have only real roots and all coefficients are \pm1.

    I started scribbling and came up a surprising answer.
    . . (Or am I totally wrong?)


    There are two linear polynomials: . \begin{array}{ccc}x + 1 \\ x - 1 \end{array}

    There are two cubic polynomials: . \begin{array}{ccc}x^3-x^2 - x + 1 \\ x^3 + x^2 - x - 1  \end{array}


    I did not include x^2-1

    It has a zero-coefficient: . x^2 + {\bf 0}x - 1


    Did I miss any?

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Did I miss any?
    There are two quadratic polynomials x^2+x-1 and x^2-x-1. I can't find any of degree 4 or higher, but then again I have no idea how to prove that there aren't any.
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  5. #5
    MHF Contributor ebaines's Avatar
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    Here are two more quadratics:  -x^2 \pm x +1
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  6. #6
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    This problem has been bugging me. Since I could make no progress on it, I eventually resorted to a Google search for a solution, which I found here.

    We may assume that the leading coefficient is +1. The sum of the squares of the roots of x^n + a_1 x^{n-1} + \ldots + a_n is a_1^2 - 2a_2 . The product of the squares of the roots is a_n^2 . Using the AGM inequality we have \frac{a_1^2 - 2a_2}n \geqslant \sqrt[n]{a_n^2}.

    Since the coefficients are 1 that inequality is (1 2)/n ≥ 1, hence n ≤ 3.

    Clever, huh? No wonder I couldn't do it.

    So there are two of these polynomials in each of degree 1, 2, 3, together with their negatives, and no others, giving a total of 12 such polynomials.
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