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 ???

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.

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 ???

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 ???