Vanishing of permanent of a Vandermonde matrix [Edited]

Question

Does there exist an explicit criterion (or a good sufficient condition) for proving that a Vandemonde matrix: $$A(x_1,\dots,x_n):=\left[ \begin{array}{llll}1 & x_1 &\dots& x_1^{n-1}\\ 1 & x_2 &\dots& x_2^{n-1}\\&&\vdots&\\ 1& x_n &\dots& x_n^{n-1}\end{array}\right]$$ has nonzero permanent? ($x_1,\dots,x_n$ are complex numbers) I'm especially interested in the case where $x_i$'s are roots of unity.

Added. Here is an explicit conjecture I want to prove:

If $x_1,\dots,x_n$ are in $\mu_N$ (not necessarily different) and $n$ is coprime to $N$ then $A(x_1,\dots,x_n)\neq 0$.

The conjecture is trivially true when $N$ is a prime and $n<N$, since $A(x_1,\dots,x_n)$ is a sum of $n!$, $N$th roots of unity and $N$ dose not divide $n!$.

Edit. I found counterexamples for the above conjecture for a prime $N$ and every $n>N$ s.t. $N\nmid (n+1)$. Explicitly $A(x_1,\dots,x_n)=0$ when all of $x_i$'s are 1 except (1 + $n$ mod $N$) of them which are equal to a primitive $N$th root of unity.

Answer 1

If the $x_i$ are pairwise different roots of unity of odd order (does this case suffice for your application?) then you can show that the permanent is non-zero by arguing that its norm to $\mathbb{Z}$ has to be an odd rational integer.

More precisely, suppose all $x_i \in \mu_N$ different, where the level $N$ is odd, and fix one of the prime ideal divisors $\mathfrak{p}$ of $2$ for $\mathbb{Z}(\mu_N)$. (This choice may be avoided by applying the norm to $\mathbb{Z}$.) Since $\prod_{\zeta \in \mu_N \setminus \{1\}} (1-\zeta) = N \equiv 1 \mod{2}$, no non-zero difference of $\mu_N$ elements is in $\mathfrak{p}$, and hence the Vandermonde determinant $V(x_1,\ldots,x_n) = \prod_{i < j}(x_i-x_j) \notin \mathfrak{p}$. As the permanent $A(x_1,\ldots,x_n) \equiv V(x_1,\ldots,x_n) \mod{\mathfrak{p}}$, it must be non-zero.