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.
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7$z^k+z^{-k}$ is a polynomial (with integer coeffiecients) of $z+z^{-1}$. In fact it is a Chebyshev polynomial of the second kind. – PAD Jul 25 '12 at 05:43
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What's the reason for the downvotes? – Makoto Kato Jul 27 '12 at 06:05
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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.
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1@MakotoKato yes, as in Arturo's more straightforward answer. I have an affinity for symmetric polynomials though, so I wanted to grab at the chance to mention them. – anon Jul 25 '12 at 04:56
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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 $z^{n+1}+z^{-(n+1)} \in \mathbb{Z}[z+z^{-1}]$.
user26872
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@AlexanderThumm: I use $k=1,2$ as the base case. Arturo showed the result held for $k=2$ and I didn't repeat the argument. – user26872 Jul 25 '12 at 05:33
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1Yes I know. I only wanted to mention, that there would be no argument required if you would start with $k=0,1$. – Alexander Thumm Jul 25 '12 at 05:39
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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 $\mathbb{Z}[z+z^{-1}]$.
Arturo Magidin
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