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 $\zeta$). 



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


How to prove this?



Why are most coefficients of these minimal polynomials divisible by $p$?
So $\zeta=-\xi$, where $\xi$ is a primitive $p$'th root of unity. Let $\pi=1-\xi$. The minimal polynomial of $(1+\zeta^n)a$ is a product of terms of the form $X-(1+\zeta^{nj})\alpha\_{ij}$, where $j$ is odd and $i$ runs over some index set (maybe depending on $j$). Note that $j$ is odd because the Galois conjugates of $\zeta$ are the odd powers of $\zeta$. Reducing mod $\pi$, you get a product of terms of the form 
$$
X-(1+(-\xi)^{nj}) \equiv X-(1+(-1)^{nj})\alpha\_{ij} 
\equiv X-(1+(-1)^{n})\alpha\_{ij}\pmod\pi.
$$
So if $n$ is odd, the minimal polynomial of $(1+\zeta^n)a$, reduced mod $\pi$, is a power of $X$, and since the polynomial has $p$-integral rational coefficients, they are all divisible by $p$. (Hmmm... I guess I've assumed that $a$ is $p$-integral. If $a$ has negative $p$-adic valuation, a similar argument should work, where now one may find that the constant term of the minimal poly in $\mathbb Z[X]$ is the one not divisible by $p$.)


