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I want to show,

if i know that $\cos(\vartheta)=\frac{1}{3}$ than $\cos(n\vartheta)=\frac{a_n}{3^n}$ for $n\in \mathbb{N}$, where $a_n \in \mathbb{Z}$,$3 \nmid a_n $

My approach was to do it by induction. For n=1 it is clear. Than i used $\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$ with $\alpha=n\vartheta$ and $\beta=\pm \vartheta$. I added the results and got $\cos((n+1)\vartheta)=2\cos(\vartheta)\cos(n\vartheta)-\cos((n-1)\vartheta)$.

I get then

$\cos((n+1)\vartheta)=2\frac{1}{3}\frac{a_n}{3^n}-\cos((n-1)\vartheta) = \frac{2a_n-\cos((n-1)\vartheta)3^{n+1}}{3^{n+1}}$

I don't know how to argue that $2a_n-\cos((n-1)\vartheta)3^{n+1} \in \mathbb{Z}$ and that $3 \nmid 2a_n-\cos((n-1)\vartheta)3^{n+1}$.

Thanks in advance

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If $a_n = 3^n \cos(n \vartheta)$, your equation $\cos((n+1) \vartheta) = 2 \cos(\vartheta) \cos(n \vartheta) - \cos((n-1) \vartheta)$ can be written as $$ a_{n+1} = 2 a_n - 9 a_{n-1} $$ and then it's obvious that if $a_n$ and $a_{n-1}$ are integers, so is $a_{n+1}$, and if $a_{n} \not\equiv 0 \bmod 3$ then $a_{n+1} \not\equiv 0 \bmod 3$.

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