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Let $n>0$ be an even integer divisible by $4$

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

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

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

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

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$(1+x+x^2+x^3+x^4+x^5+ \ldots + x^{n-1})-2x^{n/2}$.

EDIT: It is customary after someone answers your question not to change the question. As it happens, the slightest possible change to my answer still provides an answer to the latest version of your question (at the time of this edit). The little icon produced for my username (produced completely randomly as it happens) reflects my opinion of the practice of changing the question after it has been answered.

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  • $\begingroup$ Let $C=circ(r_0,r_1, \ldots ,r_{n-1})$ be the circulant matrix with first row $r_0, \ldots ,r_{n-1}$. The $R(\omega)$ are the eigenvalues of $C$. We may consider the special case when $C$ is orthogonal in order to see different signs taken by the eigenvalues. $\endgroup$
    – Luis H Gallardo
    Commented Jan 6, 2011 at 10:13
  • $\begingroup$ @unknown mannekin pisse: Assume that $C$ is also a Hadamard matrix. I am afraid not to yet understand why when the circulant matrix $C$ is symmetric or close to be symmetric, the associated quadratic form $Q$ transforming $x \to x^T Cx$ (or something alike when $C$ is not symmetric) must have a very special signature, of the type $(1,*)$ or $(-1,*)$, (that forces $n=4$) while normally (for more non-circulant Hadamard's) the signature should be of the type $ ((n+\sqrt{n})/2,(n-\sqrt{n})/2)$. $\endgroup$
    – Luis H Gallardo
    Commented Jan 21, 2011 at 15:38

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