This is in continuation of the question asked in this earlier post here. Given an anti palindromic polynomial of degree $n$ with odd coefficients, does it have roots on the unit circle?
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$\begingroup$ Since $P(1)\equiv P(-1)\equiv 1\pmod{2}$, $\pm 1$ are not roots of $P.$ $\endgroup$– StudentCommented Nov 2, 2018 at 22:28
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$\begingroup$ what do you mean by anti palindromic? $a_0+a_1x+\dots+a_nx^n$ where $a_k=-a_{n-k}$ for all $k=0,1,\dots,n$? $\endgroup$– Fedor PetrovCommented Nov 2, 2018 at 22:53
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$\begingroup$ $P(-x) = x^nP(1/x)$ $\endgroup$– StudentCommented Nov 2, 2018 at 23:18
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2$\begingroup$ As @PhilippLampe pointed out in your other post, your definition of "anti-palindromic polynomial" is not the usual one. Given this, if your definition is really the one you mean to use, it would be a good idea to include it in your posts so people don't keep having to ask. $\endgroup$– LSpiceCommented Nov 2, 2018 at 23:31
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2$\begingroup$ I'm sure, SuperMario, that you can figure out whether Fedor's polynomial has any roots on the unit circle. $\endgroup$– Gerry MyersonCommented Nov 3, 2018 at 1:41
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1 Answer
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$P$ has no roots in $\mathbb{U}$. Ad absurdum, assume that there exists $P \in \mathbb{Z}[X]$ with degree $n \ge 1$, such that :
$P(-X) = X^nP\Big(\frac{1}{X}\Big)$,
all the coefficients of $P$ from degree $0$ to $n$ are odd
there exists $\lambda \in \mathbb{U}$ such that $P(\lambda) = 0$.
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Note that $P(-\lambda) = \lambda^n P(1/\lambda) = \lambda^n P(\bar{\lambda}) = 0$. There exists $Q \in \mathbb{Z}[X]$ an irreducible divisor of $P$ such that $Q(\lambda) = 0$. Distinguish two cases :
$Q(-\lambda) = 0$. Then, as $Q$ is irreducible, $Q$ divides $Q(-X)$, and conversely, $Q(-X)$ divides $Q$. Thus $Q(-X) = \pm Q(X)$. Hence, there exists $R \in \mathbb{Z}[X]$ such that $Q(X) = R(X^2)$, and $R(X^2) \mid P$.
$Q(-\lambda) \neq 0$. Then there exists $Q_2$ another irreducible divisor of $P$, such that $Q_2(-\lambda) = 0$. Like before, $Q$ divides $Q_2(-X)$, and $Q_2(-X)$ divides $Q$. As $QQ_2$ divides $P$, we get that $Q(X)Q(-X)$ divides $P$.
The idea for the rest of the proof comes from Rafay Ashary. We will regroup these two cases into one. Let us consider $P^*$, $Q^*$ and $R^*$ representatives of $P,Q,R$ in $\mathbb{Z}/2\mathbb{Z}[X]$. The leading coefficient of $P$ is odd, thus the same goes for $Q$ and $R$, so $P^*$, $Q^*$, $R^*$ are non constant. Moreover it is easy to see that $Q^*(X)Q^*(-X) = Q^*(X)^2$, and $R^*(X^2) = R^*(X)^2$.
In both case we have a non constant polynomial $T \in \mathbb{Z}/2\mathbb{Z}[X]$ such that $T^2$ divides $P^* = \sum \limits_{k=0}^n X^k.$. Note that by plugging $-X$ in $P(-X)=X^nP(1/X)$, it is obvious that $n = 2m$ is even. Hence we found that $\sum \limits_{k=0}^{2m} X^k$ is not squarefree. However, \begin{align*}\mbox{gcd}\Big(\sum \limits_{k=0}^{2m} X^k, \big(\sum \limits_{k=0}^{2m} X^k\big)'\Big) & = \mbox{gcd}\Big(1+X+...+X^{2m}, 1+X^2+...+X^{2m-2}\Big) \\ & = \mbox{gcd}\Big(1+X+...+X^{2m}, \big(1+X+...+X^{m-1}\big)^2\Big)\\ & = \mbox{gcd} \Bigg(\frac{X^{2m+1}-1}{X-1},\ \Big(\frac{X^m-1}{X-1}\Big)^2\Bigg) = 1 \end{align*}
This is absurd, and hence, if $P$ is antisymmetric as defined above, and has all its coefficients odd, then $P$ has no roots on the unit circle.
