not divide

**maria18** · April 24th 2009, 12:59 PM

Hello, need your help thanks

Prove that if $n\in\mathbb{N}^{*}$ then $6\nmid\lfloor\big(\root3\of{28}-3\big)^{-n}\rfloor$

**Opalg** · April 25th 2009, 11:57 PM

Originally Posted by maria18

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.

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