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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 ???

edited Dec 30 2010 at 22:11

Mark Schwarzmann
2,182●1●13●27

asked Dec 30 2010 at 21:23

Luis H Gallardo
1,253●3●7

Lavender Honey · Accepted Answer · 2010-12-30 22:16:11Z UTC

5

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

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.

edited Dec 30 2010 at 22:16

answered Dec 30 2010 at 21:51

Lavender Honey
6,024●30●69

	thanks for example – Luis H Gallardo Dec 30 2010 at 22:10
	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 2011 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 2011 at 15:38

sign of values of an integer polynomial on roots of unity

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