TITLE: Proof that b is not divisible by 6
QUESTION [5 upvotes]: $$b=\left \lfloor (\sqrt[3]{28}-3)^{-n} \right \rfloor$$
The brackets mean that the number is the largest integer smaller than $(\sqrt[3]{28}-3)^{-n} $
Proof that b is never divisible by 6.
I have no idea how to even start, I thought to do it by induction, but that didn't even remotely work :(
Anybody keen to help out

REPLY [3 votes]: Hint: I claim
$$ (\sqrt[3]{28} - 3)^{-n} + (\omega \sqrt[3]{28} - 3)^{-n} + (\omega^2 \sqrt[3]{28} - 3)^{-n} $$
is an integer, where $\omega$ is a primitive cube root of unity.