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?

$\begingroup$ Since $P(1)\equiv P(1)\equiv 1\pmod{2}$, $\pm 1$ are not roots of $P.$ $\endgroup$ – SuperMario Nov 2 '18 at 22:28

$\begingroup$ what do you mean by anti palindromic? $a_0+a_1x+\dots+a_nx^n$ where $a_k=a_{nk}$ for all $k=0,1,\dots,n$? $\endgroup$ – Fedor Petrov Nov 2 '18 at 22:53

$\begingroup$ $P(x) = x^nP(1/x)$ $\endgroup$ – SuperMario Nov 2 '18 at 23:18

2$\begingroup$ As @PhilippLampe pointed out in your other post, your definition of "antipalindromic 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. $\endgroup$ – LSpice Nov 2 '18 at 23:31

2$\begingroup$ I'm sure, SuperMario, that you can figure out whether Fedor's polynomial has any roots on the unit circle. $\endgroup$ – Gerry Myerson Nov 3 '18 at 1:41
$P$ 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. 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^{2m2}\Big) \\ & = \mbox{gcd}\Big(1+X+...+X^{2m}, \big(1+X+...+X^{m1}\big)^2\Big)\\ & = \mbox{gcd} \Bigg(\frac{X^{2m+1}1}{X1},\ \Big(\frac{X^m1}{X1}\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.

$\begingroup$ @SuperMario were you taking part in the SMF competition ? $\endgroup$ – charMD Nov 8 '18 at 0:25