For an odd prime $p$, let $\zeta:=e^{\frac{\pi i}p}$ and choose $1<n<p$. Further let $q(x)$ and $r(x)$ be integer polynomials such that $r(x)$ has no common factor with $x^n+1$, and $\xi$ any root of unity (not necessarily related to $\zeta$).

I have observed that the minimal polynomial of $$(1+\zeta^n)\frac{q(\xi)}{r(\xi)}$$ always seems to have all its coefficients divisible by $p$, except either the first or the last one.

Likewise for the minimal polynomial of $(1+\zeta^n)a$, where $a$ is any real algebraic number.

How to prove this?

For an odd prime $p$, let $\zeta:=e^{\frac{\pi i}p}$ and choose odd $1<n<p$. Further let $q(x)$ and $r(x)$ be integer polynomials such that $r(x)$ has no common factor with $x^n+1$, and $\xi$ any root of unity (not necessarily related to $\zeta$).

I have observed that the minimal polynomial of $$(1+\zeta^n)\frac{q(\xi)}{r(\xi)}$$ always seems to have all its coefficients divisible by $p$, except either the first or the last one.

Likewise for the minimal polynomial of $(1+\zeta^n)a$, where $a$ is any real algebraic number.

How to prove this?