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    not divide

    Hello, need your help thanks

    Prove that if n\in\mathbb{N}^{*} then 6\nmid\lfloor\big(\root3\of{28}-3\big)^{-n}\rfloor
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  2. #2
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    Quote Originally Posted by maria18 View Post
    Prove that if n\in\mathbb{N}^{*} then 6\nmid\lfloor\big(\root3\of{28}-3\big)^{-n}\rfloor
    If x^3 = 28 then (x-3)(x^2+3x+9) = x^3-27 = 28-27=1. Therefore (x-3)^{-1} = 9+3x+x^2, and
    (x-3)^{-n} = (9+3x+x^2)^n = 9^n + a_1x + a_2x^2 + \ldots + a_{2n}x^{2n}, a polynomial in x of degree 2n with integer coefficients.

    Let \omega = e^{2\pi i/3} = -\tfrac12 + \tfrac{\sqrt3}2i, a complex cube root of unity. Note that 1+\omega+\omega^2 = 0. Let X = \sqrt[3]{28}. Then the complex cube roots of 28 are X,\ \omega X and \omega^2X. Therefore

    (X-3)^{-n} + (\omega X-3)^{-n} + (\omega^2X-3)^{-n}
    . . . . . \begin{aligned}= 3*9^n + {}& (1+\omega+\omega^2)a_1X + (1+\omega^2+\omega)a_2X^2 + 3a_3X^3 \\\quad&+ (1+\omega+\omega^2)a_4X^4 + (1+\omega^2+\omega)a_5X^5 + 3a_6X^6 + \ldots\\ = 3*9^n + {}&3*28(a_3 + a_6X^3 + \ldots).\end{aligned}

    This is a (real) integer of the form 3*9^n + 84k for some integer k, and is therefore an odd multiple of 3.

    Next, notice that \omega X-3 = -(3+\tfrac12X) + \tfrac{\sqrt3}2Xi, and X > 3. Hence |\omega X-3|>3\sqrt3 and so |(\omega X-3)^{-n}|<3^{-3n/2}, and similarly |(\omega^2 X-3)^{-n}|<3^{-3n/2}. So both these quantities are very small (certainly less than 1/2). It follows from the previous paragraph that (X-3)^{-n} is within distance less than 1 from an odd multiple of 3. So its integer part cannot be a multiple of 6.
    Last edited by Opalg; April 26th 2009 at 02:48 AM. Reason: corrected typo
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