# Roots of anti-palindromic polynomial if coefficients are odd.

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?

• Since $P(1)\equiv P(-1)\equiv 1\pmod{2}$, $\pm 1$ are not roots of $P.$ Nov 2, 2018 at 22:28
• what do you mean by anti palindromic? $a_0+a_1x+\dots+a_nx^n$ where $a_k=-a_{n-k}$ for all $k=0,1,\dots,n$? Nov 2, 2018 at 22:53
• $P(-x) = x^nP(1/x)$ Nov 2, 2018 at 23:18
• 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. Nov 2, 2018 at 23:31
• I'm sure, SuperMario, that you can figure out whether Fedor's polynomial has any roots on the unit circle. Nov 3, 2018 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^{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.