This is in continuation of the question asked in this earlier post [here](https://mathoverflow.net/questions/314083/reference-request-for-anti-palindromic-polynomials). Given an anti palindromic polynomial of degree $n$ with odd coefficients, does it have roots on the unit circle? 



Roots of anti-palindromic polynomial if coefficients are odd.
$P$ 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 $P(\lambda) = 0$.


$ $


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](https://math.stackexchange.com/questions/2985667/even-polynomial-having-multiple-with-odd-coefficients). 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.


