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This is equivalent to showing that $\left[ \frac{\zeta -1}{\zeta +1} \right]\left[ \frac{{{\zeta }^{k}}+1}{{{\zeta }^{k}}-1} \right]$ is an integer in the cyclotomic field $\text{Q(}\zeta )$ where $\zeta ={{e}^{2\pi i/n}}$ and gcd(k,n) = 1.

sunshineghh
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1 Answers1

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Note that you can assume without loss of generality that $k$ is odd. Namely, if $k$ is even, then $n$ must be odd and you can replace $k$ by $k + n$.

Now since $gcd(k,n) = 1$, you can find an $r$ such that $kr \equiv 1 \pmod{n}$. Then you expression is the same as \begin{equation} \left[\frac{\zeta - 1}{\zeta + 1}\right]\left[\frac{\zeta^k + 1}{\zeta^k - 1}\right] = \left[\frac{(\zeta^k)^r - 1}{\zeta^k - 1}\right]\left[\frac{\zeta^k + 1}{\zeta + 1}\right] \end{equation} But each of the terms in the above expression factors and you obtain \begin{equation} ((\zeta^k)^{r-1} + (\zeta^k)^{r-2} + ... + 1)(\zeta^{k-1} - \zeta^{k-2} + ... - \zeta + 1) \end{equation} which is an algebraic integer.

jwsiegel
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  • Thank you jwsiegel, your answer is correct and it is possible to go one step further and show that this expression is a (real) unit. To do this just invert the two expressions and expand as shown by jwsiegel. These expressions are scale factors used to describe polygons that result from the outer-billiards map. See dynamicsofpolygons.org. – sunshineghh Feb 09 '18 at 06:57
  • In fact the inverse works fine by just setting $\zeta $ = -$\zeta $ but when n is odd and k is even, keep the replacement k+n instead of k. – sunshineghh Feb 23 '18 at 02:27