Does $z ^k+ z^{-k}$ belong to $\Bbb Z[z + z^{-1}]$? Let $z$ be a non-zero element of $\mathbb{C}$.
Does $z^k + z^{-k}$ belong to $\mathbb{Z}[z + z^{-1}]$ for every positive integer $k$?
Motivation:
I came up with this problem from the following question.
Maximal real subfield of $\mathbb{Q}(\zeta )$
 A: Yes. By the fundamental theorem of symmetric polynomials, $x^k+y^k\in\Bbb Z[x,y]^{S_2}$ can be written as a polynomial in $e_1=x+y$ and $e_2=xy$, say $P_k(e_1,e_2)$. Then we have
$$x^k+x^{-k}=P_k(x+x^{-1},1)\in\Bbb Z[x+x^{-1}].$$
We can of course interchange the formal variable $x$ with a specific nonzero complex number as we desire. In fact, this is the power sum $p_k(x,x^{-1})$, and the relationship between the power sums and elementary symmetric polynomials is given recursively by Newton's identities.
For a quick inductive proof of the fundamental theorem,

If $x_n|f$ then $x_1\cdots x_n|f$, and dividing out we are left with a symmetric polynomial of smaller degree than before. Otherwise, write $f(x_1,\cdots,x_{n-1},0)$ as a polynomial $p$ in the elementary symmetric polynomials $\hat{e}_i$ of the first $n-1$ variables, $p(\hat{e}_1,\cdots,\hat{e}_{n-1})$. Now the polynomial $$f(x_1,\cdots,x_n)-p(e_1,\cdots,e_{n-1})$$ is symmetric in all of $x_1,\cdots,x_n$ and evaluates to $0$ at $x_n=0$ ie is divisible by $x_n$. Induct.

which I wrote down here.
A: Yes. It holds for $k=1$; it also holds for $k=2$, since
$$z^2+z^{-2} = (z+z^{-1})^2 - 2\in\mathbb{Z}[z+z^{-1}].$$
Assume that $z^k+z^{-k}$ lie in $\mathbb{Z}[z+z^{-1}]$ for $1\leq k\lt n$. Then, if $n$ is odd, we have:
$$\begin{align*}
z^n+z^{-n} &= (z+z^{-1})^n - \sum_{i=1}^{\lfloor n/2\rfloor}\binom{n}{i}(z^{n-i}z^{-i} + z^{i-n}z^{i})\\
&= (z+z^{-1})^n - \sum_{i=1}^{\lfloor n/2\rfloor}\binom{n}{i}(z^{n-2i}+z^{2i-n}).
\end{align*}$$
and if $n$ is even, we have:
$$\begin{align*}
z^n+z^{-n} &= (z+z^{-1})^n - \binom{n}{n/2} - \sum_{i=1}^{(n/2)-1}\binom{n}{i}(z^{n-i}z^{-i} + z^{i-n}z^{i})\\
&= (z+z^{-1})^n - \binom{n}{n/2} - \sum_{i=1}^{(n/2)-1}\binom{n}{i}(z^{n-2i}+z^{2i-n}).
\end{align*}$$
If $1\leq i\leq \lfloor \frac{n}{2}\rfloor$, then $0\leq n-2i \lt n$, so $z^{n-2i}+z^{2i-n}$ lies in $\mathbb{Z}[z+z^{-1}]$ by the induction hypothesis. Thus, $z^n+z^{-n}$ is a sum of terms in $\mathbb{Z}[z+z^{-1}]$. 
A: Let's go by induction as in @Arturo Magidin's answer.
The result holds for $k=0,1$.
Assume $z^k+z^{-k} \in \mathbb{Z}[z+z^{-1}]$ for $0\le k\le n$.
But
$$z^{n+1}+z^{-(n+1)} = (z^n+z^{-n})(z+z^{-1}) - (z^{n-1} + z^{-(n-1)}),$$
and so $z^{n+1}+z^{-(n+1)} \in \mathbb{Z}[z+z^{-1}]$.
