sign of special integral polynomials over roots of unity

Question

Let $h>0$ be a positive odd integer. Let $n=4h^2.$

Let $R(t)=r_0+r_1t+ \cdots + r_{n-1}t^{n-1}$

be a polynomial with integer coefficients in $\{-1,1\}$ such that

$R(\omega)$ is a nonzero integer for all complex $\omega \notin ${-1,1} such that

(a)

$$ \omega^n=1 $$

and

(b)

$$ R(1) = 2h $$

Can we deduce that all these integers $R(\omega)$ have the same sign ???

reason: R(t) is the ``representer" polynomial of a circulant $-1,1$ matrix $C$ of order $n$ with first row $(r_0, \ldots,r_{n-1})$. I am trying to understand what happens when all the eigenvalues of $C$ (i.e., the $R(\omega)$) are real.

Answer 1

Following the advice of the unknown: $$ R(t)=\frac{1-x^{16}}{1-x}-12x^8 $$ is a counter-example to your sign expectation. You can easily generalize this to any degree $n=4h^2$.

Edit. Igor brings to my attention the fact that the coefficients of $R(t)$ has to be all $\pm1$. For this situation one (including the author) could simply use brute force: if $n=16$, the polynomial of the desired form is a linear combination of the polynomials $(1-x^{16})/(1-x^j)$, $(1-x^8)/(1-x^k)$ and $x^8$. Just pick all possible integer combinations with the $\pm1$ restriction on the coefficients and check their values at the 16th roots of unity. This would also suggest how to proceed for the general $n$.

I turn my "answer" to the community wiki mode, so please feel free (if necessary) to add further details.