# sign of values of an integer polynomial on roots of unity

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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$(1+x+x^2+x^3+x^4+x^5+ \ldots + x^{n-1})-2x^{n/2}$.
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. –  Luis H Gallardo Jan 6 '11 at 10:13
@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 $((n+\sqrt{n})/2,(n-\sqrt{n})/2)$. –  Luis H Gallardo Jan 21 '11 at 15:38