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Prove that: $$6 \not\left|\ \left\lfloor\frac 1 {(\sqrt[3]{28} - 3)^{n}}\right\rfloor \ (n \in Z^+)\right.$$ ($\lfloor x\rfloor$ = largest integer not exceeding $x$)

I am very bad as English and number theory, please help me

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What is the meaning of the symbol of three dots vertical with a line through it? Does it mean to show left side is not equal to 6, or not a multiple of 6, or something? –  coffeemath Jan 22 '13 at 9:10
It mean 'not divisible' –  TNA277 Jan 22 '13 at 9:11
TNA227: Thanks. Now it's an interesting problem. –  coffeemath Jan 22 '13 at 9:25
You may use $\nmid$(\nmid) to replace what you are using($\not{\vdots}$)... –  Shane Chern Jan 22 '13 at 9:35
@TNA277 - where is this question from? –  nbubis Jan 22 '13 at 15:57

2 Answers 2

Let $x=1/(\root3\of{28}-3)$. Then $\root3\of{28}=3+x^{-1}$. Cubing, $28=27+27x^{-1}+9x^{-2}+x^{-3}$, which says $x^3-27x^2-9x-1=0$. If we let $y$ and $z$ be the conjugates of $x$, and let $a_n=x^n+y^n+z^n$, then $a_n$ is an integer for all $n$, $a_n$ is the integer closest to $x^n$ (since $y^n$ and $z^n$ go to zero, quickly), and $a_n$ satisfies the recurrence $a_n=27a_{n-1}+9a_{n-2}+a_{n-3}$. Now you can figure out the initial conditions (that is, the values of $a_0,a_1,a_2$) and then you'll be in a position to use the recurrence to work on the residue of $a_n$ modulo $6$. If you look a little more closely at $y^n$ and $z^n$, you may find that $a_n=[x^n]$, I'm not sure. Anyway, there's some work to be done, but this looks like a promising approach.

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I like your method of getting the polynomial for $x$; mine is more brute force. –  robjohn Jan 22 '13 at 15:32

If we set $\eta=\sqrt[3]{28}$ and $\omega=\dfrac1{\eta-3}=\dfrac{\eta^3-27}{\eta-3}=\eta^2+3\eta+9$, then, working $\bmod\ \eta^3-28$: $$ \begin{align} \omega^0&=1\\ \omega^1&=9+3\eta+\eta^2\\ \omega^2&=249+82\eta+27\eta^2\\ \omega^3&=6805+2241\eta+738\eta^2 \end{align}\tag{1} $$ Solving the linear equations involved yields $$ \omega^3-27\omega^2-9\omega-1=0\tag{2} $$ Looking at the critical points of $x^3-27x^2-9x-1$, we see that it has one real root and two complex conjugate roots. The real root is $\omega\stackrel.=27.3306395$, and since the product of all the roots is $1$, the absolute value of the two conjugate roots is less than $\frac15$.

Let $\omega_0=\omega$ and $\omega_1$ and $\omega_2=\overline{\omega}_1$ be the roots of $x^3-27x^2-9x-1=0$. Symmetric functions and the coefficients of $(2)$ yield $$ \begin{align} a_0=\omega_0^0+\omega_1^0+\omega_2^0&=3\\ a_1=\omega_0^1+\omega_1^1+\omega_2^1&=27\\ a_2=\omega_0^2+\omega_1^2+\omega_2^2&=747\quad=27^2-2(-9) \end{align}\tag{3} $$ and, because each $\omega_k$ satisfies $(2)$, $$ a_n=27a_{n-1}+9a_{n-2}+a_{n-3}\tag{4} $$ Because $|\omega_1|=|\omega_2|<\frac15$, $|\,a_n-\omega^n\,|\le\frac2{5^n}$. Also, $(3)$ and $(4)$ show that $a_n\equiv3\pmod{6}$.

Therefore, $\omega^0=1$ and for $n\ge1$, $$ \lfloor\omega^n\rfloor\in\{2,3\}\pmod{6}\tag{5} $$

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