# Prove $6 \nmid {28} – 3 \right)^{-n}]$

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 $x$)

#### Solutions Collecting From Web of "Prove $6 \nmid {28} – 3 \right)^{-n}]$"

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.

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 $n\ge1$,
$$\lfloor\omega^n\rfloor\in\{2,3\}\pmod{6}\tag{5}$$