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$$b=\left \lfloor (\sqrt[3]{28}-3)^{-n} \right \rfloor$$

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

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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.

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Note that $3=\sqrt[3]{27}$, so $\sqrt[3]{28}-3=\frac{28-27}{\sqrt[3]{28^2}+\sqrt[3]{28\cdot 27}+\sqrt[3]{27^2}}$, so we are looking at $\left\lfloor \left( \sqrt[3]{28^2}+\sqrt[3]{28\cdot 27}+\sqrt[3]{27^2}\right)^n\right \rfloor$ Now experimentation says the number inside the floor is just a tiny bit less than a multiple of $3$ for $n \gt 1$. Usually this would be proved by finding a $c$ such that $\left( \sqrt[3]{28^2}+\sqrt[3]{28\cdot 27}+\sqrt[3]{27^2}\right)^n + c^n$ is an integer and a multiple of $3$ and $|c| \lt 1$

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    $\begingroup$ Shouldn't it be ${\color{red}+}\sqrt[3]{28\cdot 27}$? $\endgroup$ Dec 2 '13 at 21:58
  • $\begingroup$ @PabloRotondo: yes, it should. Thanks $\endgroup$ Dec 2 '13 at 21:59

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