# sign of special integral polynomials over roots of unity

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.

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Why are you interested in these polynomials? This question is extremely similar to your previous one, and this question is probably too localized for MO. – J.C. Ottem Dec 30 '10 at 22:48
thanks again, have you un example with n >4 ? – Luis H Gallardo Dec 30 '10 at 23:00
I have not looked at the question properly, but on first glance agree with J.C. Ottem. Without better motivation, these questions just seem somewhat random, and generated on a whim rather than for a coherent purpose. – Yemon Choi Dec 30 '10 at 23:11
reason added to question – Luis H Gallardo Dec 31 '10 at 0:03
I agree with you wadim... – Luis H Gallardo Dec 31 '10 at 10:21

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 believe the question asks for $\pm 1$ coefficients, so this is not a counterexample. – Igor Rivin Dec 31 '10 at 2:38
$16 = 4 \cdot 2^2$ i.e., $h=2$ that is not odd – Luis H Gallardo Dec 31 '10 at 7:54
Luis, you are kidding! There was already a post which has recognized 2 to be the oddest integer. :-) Too many constraints. Please try yourself to modify everything for $4\cdot3^2$. – Wadim Zudilin Dec 31 '10 at 9:56