On the Middle Coefficient of a Cyclotomic Polynomial

TL;DR

This paper presents an elementary proof that the middle coefficient of cyclotomic polynomials is either zero or an odd integer, depending on whether n is a power of two, clarifying a key property of these polynomials.

Contribution

It offers a simple, elementary proof of a specific property of cyclotomic polynomial coefficients, improving understanding of their structure.

Findings

Middle coefficient is zero if n is a power of 2.

Middle coefficient is an odd integer for other n ≥ 3.

Provides a clearer proof of a known property.

Abstract

In this article, we provide a short and elementary proof of the following result: For $n \geq 3$ the middle coefficient of $\Phi_n(x)$ is either zero (when $n$ is a power of $2$) or an odd integer.

Equations

$${\Phi }_n(x)=\prod _{(k,n)=1}\left(x-e^{2\pi ik/n}\right).$$

$$\prod _{i|n,i>1}{\Phi }_i(x)=\frac{x^n-1}{x-1},$$

$$\prod _{i|n,i>1}{\Phi }_i(-1)=\frac{(-1{)}^n-1}{(-1)-1}=1.$$

$$\qquad\qquad\Phi_{n}(x)=f_{n}(x+x^{-1})\cdot x^{\phi(n)/2}\qquad\mbox{($n\geq 3$),}$$

$$f_3(x)=x+1,$$

$$f_4(x)=x,$$

$$f_5(x)=x^2+x-1,$$

$$f_6(x)=x-1,$$

$$f_7(x)=x^3+x^2-2x-1,$$

$$f_8(x)=x^2-2.$$

$${\Phi }_n(x)=\left[(x+x^{-1}{)}^k+a_{k-1}(x+x^{-1}{)}^{k-1}+\cdots \pm 1\right]\cdot x^k.$$

$$\sum _{i=2j}{a}_i\left(\genfrac{}{}{0ex}{}{i}{i/2}\right)\pm 1=\sum _j{a}_{2j}\left(\genfrac{}{}{0ex}{}{2j}{j}\right)\pm 1.$$

$$\left(\genfrac{}{}{0ex}{}{2j}{j}\right)=\left(\genfrac{}{}{0ex}{}{2j-1}{j-1}\right)+\left(\genfrac{}{}{0ex}{}{2j-1}{j}\right)=2\left(\genfrac{}{}{0ex}{}{2j-1}{j}\right).$$

Full text

On the Middle Coefficient of a Cyclotomic Polynomial

Gregory P. Dresden

The cyclotomic polynomials $\Phi_{n}$ for $n=1,2,3,\dots$ (familiar to every student of algebra) are the minimal polynomials for the primitive $n$th roots of unity:

[TABLE]

Clearly $\Phi_{n}$ has degree $\phi(n)$, where $\phi$ signifies Euler’s totient function. These monic polynomials can be defined recursively as $\Phi_{1}(x)=x-1$ and $\displaystyle\prod_{i|n}\Phi_{i}(x)=x^{n}-1$ for $n>1$. The first few are easily calculated to be $x-1,\ x+1,\ x^{2}+x+1,\ x^{2}+1,\dots$ . For these and other basic facts, see an algebra text such as [5].

While it might appear that the coefficients of the cyclotomic polynomials are always $\pm 1$, the presence of $2x^{7}$ in $\Phi_{105}(x)$ shows that this is not invariably the case (and indeed is a good counterexample for those students who insist that the “law of small numbers”  is universally valid; see [4] for further discussion). Naturally, much work has been done on the values of the coefficients of $\Phi_{n}(x)$. One amazing fact worthy of mention is that every integer appears as a coefficient in some cyclotomic polynomial (see [1], [8]).

In this article, we provide a short and elementary proof of the following result:

Theorem 1. For $n\geq 3$ the middle coefficient of $\Phi_{n}(x)$ is either zero (when $n$ is a power of $2$) or an odd integer.

A similar result can be found in [6], where Lam and Leung directly calculate the middle coefficient of $\Phi_{pq}(x)$ for distinct primes $p$ and $q$ and show it to be $\pm 1$. This had been done earlier by Beiter [2] for the case of distinct odd primes. Both papers rely on the partition of $\phi(pq)/2$ into $rp+sq$. In contrast, our proof uses only some very basic facts about minimal polynomials. We also point out that for $n\not=pq$ the polynomial $\Phi_{n}(x)$ could indeed have a middle coefficient different from $1$ or $-1$. The first such occurence is at $n=385$ (giving a middle coefficient of $-3$), after which one sees $5$ at $n=4785$, followed by $-7$ at $n=7735$, and $19$ at $n=11305$. All these values of $n$ are square-free products of small odd primes, which is alluded to in [8].

Before proceeding with the proof of Theorem 1, we do a bit of preliminary work. The first lemma establishes a useful fact about $\Phi_{n}(x)$.

Lemma 1. If $n\geq 3$ and is odd, then $\Phi_{n}(-1)=1$.

Proof. For $n\geq 3$,

[TABLE]

so (since $n$ odd)

[TABLE]

Also, $\Phi_{3}(-1)=1$. By a simple induction argument we conclude that $\Phi_{n}(-1)=1$ whenever $n\geq 3$ and is odd.

Next we review some basic information. We use $\zeta_{n}$ to signify a primitive $n$th root of unity (that is, $\zeta_{n}=e^{2\pi ik/n}$ for some $k$ relatively prime to $n$), and $f_{n}(x)$ to denote the minimal polynomial of $\zeta_{n}+\zeta_{n}^{-1}$ (recall that the minimal polynomial of an algebraic complex number $\alpha$ is the monic polynomial $p(x)$ in ${{Q}}[x]$ of smallest degree such that $p(\alpha)=0$). It is not hard to show using elementary methods (see [7]) that $f_{n}$ has integer coefficients and that when $n\geq 3$ the degree of $f_{n}$ is half that of $\Phi_{n}(x)$. In fact,

[TABLE]

because (after simplifying the right-hand side) the polynomials on both sides of (1) are monic, are of degree $\phi(n)$, and have $\zeta_{n}$ as a root. The first few such polynomials $f_{n}$ (for $n\geq 3$) are easy to derive from (1) and read as follows:

[TABLE]

[TABLE]

[TABLE]

From this, we see that the constant term in $f_{n}$ is not always $\pm 1$ (equivalently, $\zeta_{n}+\zeta_{n}^{-1}$ is not necessarily an algebraic unit, meaning a unit in the ring of algebraic integers). However, by doing a careful comparison of the $f_{n}$ with the Chebyshev polynomials, Carlitz and Thomas [3] showed that when $n\geq 3$ and $n$ is not divisible by $4$, the constant term in $f_{n}(x)$ is either $1$ or $-1$. For the sake of completeness, we provide a nonelementary, but much shorter, proof of this fact.

Lemma 2. If $n\geq 3$ and $n\not\equiv 0\ (\mbox{\rm mod }4)$, then $\zeta_{n}+\zeta_{n}^{-1}$ is an algebraic unit.

Proof. Let $m=n$ for $n$ odd and $m=n/2$ for $n$ even. Note that $m$ is itself odd and $m\geq 3$. Note as well that ${\zeta_{n}}^{2}$ is a primitive $m$th root of unity (and thus a root of $\Phi_{m}(x)$). Then ${\zeta_{n}}^{2}+1$ is a root of $\Phi_{m}(x-1)$, which is a monic polynomial with constant term $\Phi_{m}(-1)=1$ (by Lemma 1). It follows that ${\zeta_{n}}^{2}+1$ is an algebraic unit, as is $\zeta_{n}$. Thus, $\zeta_{n}+\zeta_{n}^{-1}=({\zeta_{n}}^{2}+1)/\zeta_{n}$ is likewise an algebraic unit.

We are now ready to bring everything together.

Proof of Theorem 1. If $n=2^{k}$, then $\Phi_{n}(x)=x^{2^{k-1}}+1$, a polynomial with zero as its middle coefficient. We proceed assuming that $n$ is not a power of $2$.

Note that if $\zeta$ is a primitive $4k$th root of unity, then $\zeta^{2}$ is a primitive $2k$th root of unity. Since $\phi(4k)=2\phi(2k)$, we know that $\Phi_{4k}(x)=\Phi_{2k}(x^{2})$. Since the middle coefficient of $\Phi_{2k}(x^{2})$ is the same as that of $\Phi_{2k}(x)$, we can further assume without loss of generality that $4$ does not divide $n$.

Now letting $f_{n}(x)$ be the minimal polynomial of $\zeta_{n}+\zeta_{n}^{-1}$, we know from Lemma 2 that $f_{n}$ has constant coefficient $\pm 1$. Thus, we can write $f_{n}(x)=x^{k}+a_{k-1}x^{k-1}+\cdots+a_{1}x\pm 1$ (for $k=\phi(n)/2$), and so from equation (1) we obtain

[TABLE]

The middle coefficient of $\Phi_{n}(x)$ is the coefficient of the $x^{k}$ term in (2) (recall, $k=\phi(n)/2$). This number is simply the sum of the constant terms appearing in each expression $a_{i}(x+x^{-1})^{i}$ in (2), plus the final $\pm 1$. The constant term in $a_{i}(x+x^{-1})^{i}$ is either zero (for $i$ odd) or $a_{i}{i\choose{i/2}}$ (for $i$ even). As a result, the middle coefficient of $\Phi_{n}(x)$ is

[TABLE]

By a familiar identity,

[TABLE]

Thus the middle coefficient of $\Phi_{n}(x)$ is odd when $n$ is not a power of $2$.