# Thread: not divide

1. ## not divide

Hello, need your help thanks

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

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

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

$\displaystyle (X-3)^{-n} + (\omega X-3)^{-n} + (\omega^2X-3)^{-n}$
. . . . .\displaystyle \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 $\displaystyle 3*9^n + 84k$ for some integer k, and is therefore an odd multiple of 3.

Next, notice that $\displaystyle \omega X-3 = -(3+\tfrac12X) + \tfrac{\sqrt3}2Xi$, and $\displaystyle X > 3$. Hence $\displaystyle |\omega X-3|>3\sqrt3$ and so $\displaystyle |(\omega X-3)^{-n}|<3^{-3n/2}$, and similarly $\displaystyle |(\omega^2 X-3)^{-n}|<3^{-3n/2}$. So both these quantities are very small (certainly less than 1/2). It follows from the previous paragraph that $\displaystyle (X-3)^{-n}$ is within distance less than 1 from an odd multiple of 3. So its integer part cannot be a multiple of 6.