The following question probably sounds ridiculously trivial but algebra is not my strong point...
Let $z \in \mathbb{C}$, let $p(z)$ be a Laurent polynomial, and let $n$ be some integer strictly greater than $1$. Does the condition that $p(z^n)/p(z)$ be a Laurent polynomial imply that all roots of $p(z)$ must be $1$?
Example: put $p(z) = z-1$, then $$ \frac{p(z^n)}{p(z)} = \frac{z^n-1}{z-1} = \frac{(z-1)(1+z+z^2+\ldots+z^{n-1})}{z-1} = 1+z+z^2+\ldots+z^{n-1} $$ which is Laurent. On the other hand, this is not possible if, say, $p(z) = z - 1/2$ or if $p(z)=(z-1)(z-\alpha)$ for any $\alpha \not= 1$ etc because in those cases $p(z)$ does not divide into $p(z^n)$ without remainder.
If this is correct, does this result have a name? How does one prove this?
Any help warmly received.
The condition that $\frac{p(z^n)}{p(z)}=z^m q(z)$, for some polynomial $q$ and some integer $m$, is really the condition that if $\alpha$ is a nonzero root of $p$ with some multiplicity, then $\alpha^n$ is also a root of $p$ with the same multiplicity.
So if $\alpha$ is a root, so is $\alpha^n, \alpha^{n^2}, \ldots, \alpha^{n^k}, \ldots$, and they all have the same multiplicity. Since $p$ has finite degree, $\alpha^{n^k}=\alpha$ for some $k\geq1$, and $|\alpha|=1$ (but not necessarily $\alpha=1$).
So $p$ must be a product of polynomials of the form $$ p_\alpha(z) = \prod_{j=0}^{k-1}(z-\alpha^{n^j}), \qquad |\alpha|=1, \qquad \alpha^{n^k}=\alpha. $$