For the following, let $m=2^{r_{0}}p_{1}^{r_{1}}\cdots p_{n}^{r_{n}}\geq 10$ be an even integer (so that $r_{0}\geq 1$), and let $\omega=e^{2\pi i/m}$ be a primitive $m$-th root of unity. Furthermore, let $c_{0},\dots,c_{\lfloor{\frac{m-1}{4}}\rfloor}\in\mathbb{Z}$ be such that
$$2^{c_{0}}(\omega+\omega^{-1})^{c_{1}}(\omega^{2}+\omega^{-2})^{c_{2}}\cdots(\omega^{\lfloor{\frac{m-1}{4}}\rfloor}+\omega^{-\lfloor{\frac{m-1}{4}}\rfloor})^{c_{\lfloor{\frac{m-1}{4}}\rfloor}}=1.\;\;\;(**)$$
Can one find some set of relations (say $R$) among the $\{c_{i}\}$ such that $(**)$ holds if and only if the relations $R$ hold?
Here is what I can show so far: Note that for each divisor $d$ of $\frac{m}{2}$ with $d<\frac{m-1}{4}$, we have that $$\prod_{\substack{1\le k\le \lfloor{\frac{m-1}{4}}\rfloor \\ (k,\frac{m}{2})=d}}(\omega^{k}+\omega^{-k})^{2}=\prod_{\substack{1\le k\le \lfloor{\frac{m-1}{4}}\rfloor \\ (k,\frac{m}{2})=d}}(\omega^{2k}+\omega^{-2k}+2)=\prod_{\substack{1\le k\le \lfloor{\frac{m-1}{4}}\rfloor \\ (k,\frac{m}{2})=d}}(1+\omega^{2k})(1+\omega^{-2k})=\prod_{(k,\frac{m}{2})=d}(1+\omega^{2k}).$$ But note that $$\prod_{(k,\frac{m}{2})=d}(1+\omega^{2k})=N_{\mathbb{Q}(\omega^{2d})/\mathbb{Q}}(1+\omega^{2d}).$$ Since $$1+\omega^{2d}=\frac{1-\omega^{4d}}{1-\omega^{2d}},$$ we see that $$N_{\mathbb{Q}(\omega^{2d})/\mathbb{Q}}(1+\omega^{2d})=\frac{N_{\mathbb{Q}(\omega^{2d})/\mathbb{Q}}(1-\omega^{4d})}{N_{\mathbb{Q}(\omega^{2d})/\mathbb{Q}}(1-\omega^{2d})}=\frac{N_{\mathbb{Q}(\omega^{2d})/\mathbb{Q}(\omega^{4d})}(N_{\mathbb{Q}(\omega^{4d})/\mathbb{Q}}(1-\omega^{4d}))}{N_{\mathbb{Q}(\omega^{2d})/\mathbb{Q}}(1-\omega^{2d})}=\frac{(N_{\mathbb{Q}(\omega^{4d})/\mathbb{Q}}(1-\omega^{4d}))^{[\mathbb{Q}(\omega^{2d}):\mathbb{Q}(\omega^{4d})]}}{N_{\mathbb{Q}(\omega^{2d})/\mathbb{Q}}(1-\omega^{2d})}.$$ From this we can deduce that $$N_{\mathbb{Q}(\omega^{2d})/\mathbb{Q}}(1+\omega^{2d})=\left\{ \begin{array}{ll} 1 & \mbox{if } \frac{m}{4d} \text{ is divisible by at least two primes}, \\ 2 & \mbox{if } \frac{m}{2d}=2^{t},\;t\geq 2, \\ p & \mbox{if } \frac{m}{2d}=2p^{t},\;t\geq 1. \\ \end{array} \right.$$ From our calculation, we see that $$\prod_{\substack{1\le k\le \lfloor{\frac{m-1}{4}}\rfloor \\ (k,\frac{m}{2})=d}}(\omega^{k}+\omega^{-k})=\pm\sqrt{N_{\mathbb{Q}(\omega^{2d})/\mathbb{Q}}(1+\omega^{2d})}.$$ In fact, we must have that $$\prod_{\substack{1\le k\le \lfloor{\frac{m-1}{4}}\rfloor \\ (k,\frac{m}{2})=d}}(\omega^{k}+\omega^{-k})=\sqrt{N_{\mathbb{Q}(\omega^{2d})/\mathbb{Q}}(1+\omega^{2d})},$$ since $$\prod_{\substack{1\le k\le \lfloor{\frac{m-1}{4}}\rfloor \\ (k,\frac{m}{2})=d}}(\omega^{k}+\omega^{-k})=\prod_{\substack{1\le k\le \lfloor{\frac{m-1}{4}}\rfloor \\ (k,\frac{m}{2})=d}}(2\cos(\frac{2k\pi}{m})$$ and $0<k<\frac{m}{4}$ implies that $0<\cos(\frac{2k\pi}{m})<1$ for each $k$. Therefore $$\prod_{\substack{1\le k\le \lfloor{\frac{m-1}{4}}\rfloor \\ (k,\frac{m}{2})=d}}(\omega^{k}+\omega^{-k})=\left\{ \begin{array}{ll} 1 & \mbox{if } \frac{m}{4d} \text{ is divisible by at least two primes}, \\ \sqrt{2} & \mbox{if } \frac{m}{2d}=2^{t},\;t\geq 2, \\ \sqrt{p} & \mbox{if } \frac{m}{2d}=2p^{t},\;t\geq 1. \\ \end{array} \right.$$ What this calculation implies is that if the $\{c_{i}\}$ satisfy the property that $c_{k}=c_{k'}$ for all $1\le k,k'\le\lfloor\frac{m-1}{4}\rfloor$ with $(k,m/2)=(k',m/2)$ (let's call this Property S), then equation $(**)$ holds if and only if:
- $c_{0}+2\sum_{t=3}^{r_{0}}c_{\frac{m}{2^{t}}}=0$, and
- $\sum_{t=1}^{r_{i}}c_{\frac{m}{4p^{t}}}=0$ for all $i=1,\dots,n$.
My gut feeling is that Property S always holds if $(**)$ holds, since in particular $(**)$ would imply that the expression $$2^{c_{0}}(\omega+\omega^{-1})^{c_{1}}(\omega^{2}+\omega^{-2})^{c_{2}}\cdots(\omega^{\lfloor{\frac{m-1}{4}}\rfloor}+\omega^{-\lfloor{\frac{m-1}{4}}\rfloor})^{c_{\lfloor{\frac{m-1}{4}}\rfloor}}$$ is invariant under the action of $\text{Gal}(\mathbb{Q}(\omega+\omega^{-1})/\mathbb{Q})$, which shuffles around the multiplicands within each divisibility class, possibly adding a $\pm$. However I'm unable to show that this indeed implies Property S.
Is there way to show that $(**)$ implies Property S, possibly using extra symmetries in the expressions involved? Or perhaps another approach to this problem altogether to obtain the desired set of relations R? Thanks!