Does $z ^k+ z^{-k}$ belong to $\Bbb Z[z + z^{-1}]$?

Question

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 $\mathbb{Q}(\zeta )$

Answer 1

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.

Answer 2

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 $z^{n+1}+z^{-(n+1)} \in \mathbb{Z}[z+z^{-1}]$.

Answer 3

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 $\mathbb{Z}[z+z^{-1}]$.

Answer 4

I want to provide an elementary answer to the question that has been closed as a duplicate of this question. I believe it would be helpful for some users of the MSE community. The following solution also answers Prove $z^n + 1/{z^n}$ is rational if $z+1/z$ is rational.

Let $\left(a+\frac1a\right)=\frac pq=2r\in\mathbb Q$, for some $p,q\in\mathbb{Z},\ q\neq0$.

We have $$a^2-2ra+1=0\tag{1}$$ which is quadratic in $a$, and its roots are reciprocal/conjugate of each other. We can write, $$a=r\pm\sqrt{r^2-1}=r(1\pm\sqrt{1-s^2})\tag{2}$$ where $s=1/r\in\mathbb Q$.
Now, $$\begin{aligned}a^n+\frac{1}{a^n}&=r^n\left(\left(1+\sqrt{1-s^2}\right)^n+\left(1-\sqrt{1-s^2}\right)^n\right)\\&=r^n\sum_{k=0}^n\binom{n}{k}\left[1+(-1)^k\right]\left(\sqrt{1-s^2}\right)^k\qquad\text{binomial theorem}\\&=2r^n\displaystyle\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}(1-s^2)^k\in\mathbb{Q}\qquad\text{as }r,s\in\mathbb Q\\&=2\left(\frac{p}{2q}\right)^n\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}\left(\frac{p^2-4q^2}{p^2}\right)^k\in\mathbb Q\qquad\text{as }p,q\in\mathbb Z.\end{aligned}$$

Hence, $\left(a^n+\frac{1}{a^n}\right)$ is rational if $\left(a+\frac1a\right)$ is rational, i.e., $\left(a+\frac1a\right)\in\mathbb Q\implies\left(a^n+\frac{1}{a^n}\right)\in\mathbb Q,\ \forall n\in\mathbb{Z}$.