An irreducibility criterion for polynomials over integers

TL;DR

This paper introduces a new criterion for determining the irreducibility of certain integer polynomials, showing they have cyclotomic factors if reducible, and provides practical methods for checking irreducibility and separability.

Contribution

It establishes a novel irreducibility criterion for polynomials with specific coefficient conditions and offers practical procedures for testing irreducibility and separability.

Findings

Polynomials with prime constant term and sum of coefficients have cyclotomic factors if reducible.

A simple irreducibility check for certain trinomials is provided.

Separability criteria for specific quadrinomials are established.

Abstract

In this article, we consider the polynomials of the form $f(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n\in \mathbb{Z}[x],$ where $|a_0|=|a_1|+\dots+|a_n|$ and $|a_0|$ is a prime. We show that these polynomials have a cyclotomic factor whenever reducible. As a consequence, we give a simple procedure for checking the irreducibility of trinomials of this form and separability criterion for certain quadrinomials.

Equations

$$x^4+3x^2+4=(x^2-x+2)(x^2+x+2).$$

$$x^{4n}-(a^2-1)x^{2n}-a^2=(x^n+a)(x^n-a)(x^n+1),$$

$$(x^n-1,x^m-1)=x^{(n,m)}-1.$$

$$(x^{n}+1,x^{m}+1)=\begin{cases}x^{(n,m)}+1&\mbox{ if $e(n)=e(m)$}\\ 1&\mbox{ otherwise.}\end{cases}$$

$$(x^{n}+1,x^{m}-1)=\begin{cases}x^{(n,m/2)}+1&\mbox{ if $e(m)\geq 2e(n)$}\\ 1&\mbox{ otherwise.}\end{cases}$$

$$\prod _{i=1}^s|z_i|=\frac{1}{|d|},$$

$$|a_0|=|-(a_{n_1}{z}_i^{n_1}+a_{n_2}{z}_i^{n_2}+\cdots +a_{n_r}{z}_i^{n_r})|<|a_{n_1}|+|a_{n_2}|+\cdots +|a_{n_r}|$$

$$-a_0=a_{n_1}{\zeta }^{n_1}+a_{n_2}{\zeta }^{n_2}+\cdots +a_{n_r}{\zeta }^{n_r}.$$

$$|a_0|=|a_{n_1}{\zeta }^{n_1}+a_{n_2}{\zeta }^{n_2}+\cdots +a_{n_r}{\zeta }^{n_r}|=\sum _{i=1}^r|a_{n_i}|.$$

$$-a_0=a_{n_i}{\zeta }^{n_i}\left[\left|\frac{a_{n_1}}{a_{n_i}}\right|+\cdots +\left|\frac{a_{n_{i-1}}}{a_{n_i}}\right|+1+\left|\frac{a_{n_{i+1}}}{a_{n_i}}\right|+\cdots +\left|\frac{a_{n_r}}{a_{n_i}}\right|\right],$$

$$n_r{a}_{n_r}{\zeta }^{n_r-1}+\cdots +n_1{a}_{n_1}{\zeta }^{n_1-1}=0.$$

$$x^{n_i}=-\mathrm{sgn}(a_0{a}_{n_i}),1\le i\le r.$$

$$D=(-1{)}^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}{b}^{m-1}{\left[n^{n/d}{b}^{n-m/d}-(-1{)}^{n/d}(n-m{)}^{n-m/d}{m}^{m/d}{a}^{n/d}\right]}^d$$

$$D_f=(-1{)}^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}(p{ϵ}_2{)}^{n-m}{a}^{n-m-1}{\left[n^{n/d}(p{ϵ}_2{)}^{n-m/d}{a}^{m/d}-(-1{)}^{n/d}(n-m{)}^{n-m/d}{m}^{m/d}(b{ϵ}_1{)}^{n/d}\right]}^d,$$

$$n^n(p{ϵ}_2{)}^{n-m}{a}^m=(-1{)}^n(n-m{)}^{n-m}{m}^m(b{ϵ}_1{)}^n.$$

$$f_{c}(x)|(x^{n}+1,x^{m}-1)=\begin{cases}x^{(n,m/2)}+1&\mbox{ if $e(m)\geq 2e(n)$}\\ 1&\mbox{ otherwise}\end{cases}$$

$$f(\zeta )=a{\zeta }^n-b{\zeta }^m+p=-a-b+p=0.$$

$$x^{8k}-x^{7k}-x^k-1=(x^{2k}+1)(x^{3k}-x^k-1)(x^{3k}-x^{2k}-1),\text{ for all }k\in \mathbb{N}.$$

$$n=|(n-m){ϵ}_1{z}^m+(n-r){ϵ}_2{z}^r|\le 2n-m-r$$

Full text

Key Words: Irreducible polynomials, cyclotomic polynomials.

AMS(2010): 11R09, 12D05, 12D10.

1 Introduction

Polynomials of the form $ax^{n}\pm b$ have cyclotomic factors if $a=b.$ W. Ljunggren [4] extended the work of Selmer [9] and showed that a polynomial of form $x^{n}\pm x^{m}\pm 1$ or $x^{n}\pm x^{m}\pm x^{r}\pm 1$ is either irreducible or divisible by a cyclotomic polynomial. The following result provides a criterion for the irreducibility of polynomials with integer coefficients.

Theorem 1** (L. Panitopol, D. Stefänescu [7]).**

Let $f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}\in{\mathbb{Z}}[x]$ be a polynomial such that $|a_{0}|>|a_{1}|+\cdots+|a_{n}|$. If $|a_{0}|$ is a prime or $\sqrt{|a_{0}|}-\sqrt{|a_{n}|}<1$ then $f(x)$ is irreducible in ${\mathbb{Z}}[x]$.

In this paper we show that, if a polynomial of the from $f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}\in{\mathbb{Z}}[x],$ where $|a_{0}|=|a_{1}|+\dots+|a_{n}|$ and $|a_{0}|$ is a prime is reducible, then it is divisible by a cyclotomic polynomial and has exactly one non-reciprocal factor. The condition that $|a_{0}|$ is a prime number is necessary. If $|a_{0}|$ is not a prime number, then $f(x)$ can have any number of non-reciprocal factors. For example,

[TABLE]

More generally, every member of

[TABLE]

where $a\geq 2$ is an integer and $n\geq 1$, has at least two non-reciprocal factors. By use of Capelli’s theorem [2] it can be shown that arbitrary number of non-reciprocal factors can appear for suitable choice of $a$.

But if the absolute value of the product of all the roots of a non-trivial factor of $f(x)$ belongs to $(0,1]$, then one can show that $f(x)$ has a cyclotomic factor even if $|a_{0}|$ is not a prime number.

Theorem 2**.**

Let $f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}\in{\mathbb{Z}}[x],$ where $|a_{0}|=|a_{1}|+\dots+|a_{n}|$. If $g(x)=b_{0}+b_{1}x+\dots+b_{m}x^{m}$ is a factor of $f(x)$ with $0<|b_{0}|\leq|b_{m}|,$ then $g(x)$ is a product of cyclotomic polynomials.

If $|a_{0}|$ is a prime number, then not only such a polynomial $g(x)$ defined in Theorem 2 exists, but also $f(x)$ has exactly one non-reciprocal factor. Further all the roots of $f(x)$ are simple zeros. Before proceeding, by removing zero coefficients if any, we can assume that $f(x)=a_{n_{r}}x^{n_{r}}+\cdots+a_{n_{1}}x^{n_{1}}+a_{0}$ where $a_{0}a_{n_{1}}\cdots a_{n_{r}}\neq 0.$

Theorem 3**.**

Let $f(x)=a_{n_{r}}x^{n_{r}}+\cdots+a_{n_{1}}x^{n_{1}}+a_{0},$ where $|a_{n_{1}}|+\cdots+|a_{n_{r}}|=|a_{0}|$ and $|a_{0}|$ be a prime number. If $f$ is reducible, then $f(x)=f_{c}(x)f_{n}(x)$, where $f_{n}(x)$ is the irreducible non-reciprocal factor of $f(x)$ and $f_{c}(x)$ is the product of all cyclotomic factors of $f(x),$ in particular $f_{c}(x)=\gcd(x^{n_{i}}+\operatorname{sgn}(a_{0}a_{n_{i}})),$ where $1\leq i\leq r$ and $\operatorname{sgn}(x)$ denotes the sign of $x\in{\mathbb{R}}$.

Above theorem generates several other irreducibility criteria for different family of polynomials. Before stating them, we denote the largest even part of $n$ as $e(n)$, i.e., if $n=2^{\alpha}n_{1}$ with $n_{1}$ odd, then $e(n)=2^{\alpha}$.

Corollary 4**.**

Suppose $f(x)=a_{n_{r}}x^{n_{r}}+\cdots+a_{n_{1}}x^{n_{1}}+a_{0}\in{\mathbb{Z}}_{+}[x]$ be a polynomial with $a_{n_{1}}+\dots+a_{n_{r}}=a_{0}$ and $a_{0}$ is a prime number. Then $f(x)$ is irreducible if and only if there exist distinct $i,j$ such that $e(n_{i})\neq e(n_{j}).$

Corollary 4 provides an extensive class of irreducible polynomials. The simplest among them is: If $p$ is a number and $n_{1},n_{2},\ldots,n_{p}$ are positive integers with $(n_{1},n_{2},\ldots,n_{p})=1$, then $x^{n_{1}}+x^{n_{2}}+\cdots+x^{n_{p}}+p$ is irreducible.

On the other hand, if we have two consecutive exponents then the irreducibility criterion for such polynomials become easier.

Corollary 5**.**

Suppose $f(x)=a_{n_{r}}x^{n_{r}}+a_{n_{r-1}}x^{n_{r-1}}+\cdots+a_{0}$ be a polynomial with $n_{r}=n_{r-1}+1$. Let $|a_{n_{1}}|+\cdots+|a_{n_{r}}|=|a_{0}|$ and $|a_{0}|$ is a prime number. Then $f(x)$ is reducible if and only if $f(1)=0$ or $f(-1)=0$.

The proof of this Corollary directly follows from Theorem 3 along with the fact that $(x^{n}\pm 1,x^{n-1}\pm 1)$ is either $1$ or $x\pm 1$. The condition $n_{r}=n_{r-1}+1$ can be replaced by $n_{j}=n_{j-1}+1$ for $2\leq j\leq r$.

In Section 3 we present an explicit irreducibility criterion for trinomials of the form $ax^{n}+b\epsilon_{1}x^{m}+p\epsilon_{2}$ with $a+b=p,$ where $a,b,p\in{\mathbb{Z}}_{+},$ $p$ is a prime number and $\epsilon_{i}\in\{-1,1\}$. In the end, we determine the separability of quadrinomials of form $x^{n}+\epsilon_{1}x^{m}+\epsilon_{2}x^{r}+\epsilon_{3},\epsilon_{i}\in\{-1,1\}$ over ${\mathbb{Q}}$, the set of rational numbers.

2 Proofs

In this section, we give proof for Theorem 3 and Corollary 4. With suitable modifications in the proof of Theorem 3, one can prove Theorem 2. Suppose $n,m$ are positive integers. Then it is known that

[TABLE]

The greatest common divisor for one or two positive signs in the above is as follows:

Lemma 6**.**

Suppose $n,m$ are positive integers. Then

1. (a)

[TABLE] 2. (b)

[TABLE]

Proof.

Let $n=2^{\alpha}n_{1}$ and $m=2^{\beta}m_{1}$ be prime factorizations of $n$ and $m$ respectively. Then the proof can be carried out from the identities: $x^{n}+1=\prod\limits_{d|n_{1}}\Phi_{2^{\alpha+1}d}(x)$ and $x^{m}-1=\prod\limits_{i=0}^{\beta}\prod\limits_{d|m_{1}}\Phi_{2^{i}d}(x).$ We omit the details. ∎

.

Proof of Theorem 3: Suppose $f(x)=g(x)h(x)$ be a non-trivial factorization of $f(x).$ Since $|a_{0}|=|g(0)||h(0)|$ is a prime number, without loss of generality we can assume $|h(0)|=|a_{0}|.$ Suppose $g(x)=f_{1}(x)f_{2}(x)\cdots f_{l}(x),$ where $f_{i}(x)$ is an irreducible factor of $g(x).$ Now we claim that for every $i,$ $f_{i}(x)$ is a cyclotomic polynomial. It is sufficient to prove that $f_{1}(x)$ is a cyclotomic polynomial.

It is clear that $f_{1}(0)=\pm 1.$ Let $\deg(f_{1}(x))=s$ and $z_{1},z_{2},\ldots,z_{s}$ be roots of $f_{1}(x)$. Then

[TABLE]

where $d$ is the leading coefficient of $f_{1}(x)$. If $|z_{i}|<1$ for some $i$, then

[TABLE]

contradicts the hypothesis. So all of the roots of $f_{1}(x)$ lies in the region $|z|\geq 1$. But from Equation (1) we have $\prod_{i}|z_{i}|\leq 1.$ Consequently $d=\pm 1.$ From Kronecker’s theorem $f_{1}(x)$ is a cyclotomic polynomial. Next we will prove $g(x)=\gcd(x^{n_{i}}+\operatorname{sgn}(a_{0}a_{n_{i}})),$ where $1\leq i\leq r.$

Let $\zeta$ be a primitive $t^{\text{th}}$ root of unity with $f(\zeta)=0$. Then

[TABLE]

Taking modulus on both sides

[TABLE]

From triangle inequality, the last two equations hold if and only if the ratio of any two part is a positive real number. Therefore, $a_{n_{r}}\zeta^{n_{r}-n_{i}}/a_{n_{i}}=|a_{n_{r}}\zeta^{n_{r}-n_{i}}/a_{n_{i}}|$ gives $\zeta^{n_{r}-n_{i}}=\operatorname{sgn}(a_{n_{r}}a_{n_{i}})$ for $1\leq i\leq r-1$. From (2), we have

[TABLE]

so that $\zeta^{n_{i}}=-\operatorname{sgn}(a_{0}a_{n_{i}})$. From $\zeta^{n_{r}-n_{i}}\zeta^{n_{i}}$, one gets the last equation. Remaining all the equations satisfied by $\zeta$ can be drawn from these $r$ equations. Conversely, if $\zeta$ satisfy each of the $r$ equations $x^{n_{i}}+\operatorname{sgn}(a_{0}a_{n_{i}})=0$ then $f(\zeta)=0$. It remains to show that each cyclotomic factor has multiplicity one. If not, suppose $\zeta$ satisfies the $r$ equations $x^{n_{i}}+\operatorname{sgn}(a_{0}a_{n_{i}})=0,1\leq i\leq r$ and $f(\zeta)=0,f^{\prime}(\zeta)=0$. Then the last relation gives

[TABLE]

Using those $r$ equations satisfied by $\zeta$, we will end up with the relation $n_{r}|a_{n_{r}}|+\cdots+|a_{n_{1}}|n_{1}=0$, which is not possible. ∎

Remark 7**.**

Let $g(x)$ be the product of all cyclotomic factors of $f(x),$ where $f(x)$ is as stated in Theorem 2. Then similar to Theorem 3, it can be seen that $g(x)$ is the greatest common divisor of the equations

[TABLE]

.

Proof of Corollary 4: If $g(x)=h(x^{d})\in{\mathbb{Z}}[x]$, $d\geq 1$, then $g(x)$ has a cyclotomic factor if and only if $h(x)$ has a cyclotomic factor. Thus it is sufficient to prove the result for polynomials whose exponents are relatively prime. The result follows from Theorem 3 and Lemma 6. ∎

3 Applications

We apply the above theory to trinomials of the form $f(x)=ax^{n}+b\epsilon_{1}x^{m}+p\epsilon_{2},$ where $a,b,p$ are positive integers, $a+b=p$, $p$ is a prime and $\epsilon_{i}\in\{-1,1\}$.

A. Schinzel[8] had found out the cyclotomic factors of trinomials. For the present family of polynomials, we will give the irreducibility criterion in a more simpler form. Similar studies related to trinomials can be found on [9],[10],[1],[5].

From Theorem 3, $f(x)$ is separable over ${\mathbb{Q}}$. However, by using the discriminant formula for trinomials, one can prove the separability of $f(x)$ over ${\mathbb{Q}}$, where $|f(0)|$ is prime, in a wider range. For example, it is known that

Theorem 8** (C.R. Greenfield, D. Drucker [3]).**

The discriminant of the trinomial $x^{n}+ax^{m}+b$ is

[TABLE]

where $d=(n,m)$.

Theorem 9**.**

Let $a,b,p\in{\mathbb{N}}$, $b\leq p$ and $p$ be a prime number. Then $f(x)=ax^{n}+b\epsilon_{1}x^{m}+p\epsilon_{2},$ where $\epsilon_{i}\in\{-1,1\}$ is separable over ${\mathbb{Q}}$.

Proof.

From Theorem 8, the discriminant of $f$ is

[TABLE]

where $d=(n,m)$. Since the polynomial $x^{v}-\alpha,0\neq\alpha\in{\mathbb{C}},$ is separable over ${\mathbb{Q}}$, it is sufficient to consider $d=(n,m)=1$. $f(x)$ has multiple root if and only if $D_{f}=0$, i.e.,

[TABLE]

Since $d=1$ and $n\geq 3$, it is not possible that $b=1$. If $b>1$, $n|b$ and $p|m$ or $p|n-m$ give $m,n-m<n\leq b\leq p$, a contradiction. If $n=2$, then $f(x)=ax^{2}+b\epsilon_{1}x+p\epsilon_{2}$ has multiple roots if and only if $p|b$ and $p=4\epsilon_{2}a$, contradicting the fact that $p$ is a prime number. ∎

To determine the cyclotomic factors of trinomials, it can readily be observed that if $\epsilon_{1}=1,\epsilon_{2}=-1$ then $f(1)=0$. In fact $f(x)/x^{(n,m)}-1$ is an irreducible non-reciprocal polynomial by Theorem 3 and 9. We will consider the remaining cases below.

Theorem 10**.**

Let $f(x)=ax^{n}+b\epsilon_{1}x^{m}+p\epsilon_{2}$, where $a,b,p\in{\mathbb{N}}$, $p$ be a prime number, $\epsilon_{i}\in\{-1,1\}$ and $a+b=p$. Then $f(x)$ is irreducible except in the following cases:

1. (a)

$\epsilon_{1}=-1,\epsilon_{2}=1;e(n)<e(m)$, 2. (b)

$\epsilon_{1}=\epsilon_{2}=-1;e(n)>e(m)$, 3. (c)

$\epsilon_{1}=\epsilon_{2}=1;e(n)=e(m)$,

and in each of the above cases $f_{c}(x)=x^{(n,m)}+1$.

Proof.

Suppose $f(x)=ax^{n}-bx^{m}+p$ be reducible. From Theorem 3 the cyclotomic factors of $f(x)$ divide the equations $x^{n}+1=0$ and $x^{m}-1=0$. From Lemma 6, we have,

[TABLE]

Let $n=2^{\alpha}n_{1},m=2^{\beta}m_{1}$ be the prime factorization of $n$ and $m$ respectively, and $d=(n,m/2)=2^{\min(\alpha,\beta)}(n_{1},m_{1})$. If $\alpha\geq\beta,$ then $f(x)=f_{1}(x^{2^{\beta}})$ with $f_{1}(x)$ irreducible by Corollary 4. Therefore, $f(x)$ is reducible if and only if $\alpha<\beta$, that is, $e(n)<e(m)$. Also, if $e(n)<e(m)$ then $(n,m/2)=(n,m)$. Conversely, let $\zeta$ be a root of $x^{d}+1$ where $d=(n,m)$. Then

[TABLE]

From Theorem 3 or Theorem 9, we get $f_{c}(x)=x^{(n,m)}+1$. In a similar fashion, part (b), (c) can be proved. We omit the details. ∎

W. Ljunggren[4] has proved that the quadrinomials of the form $g(x)=x^{n}+\epsilon_{1}x^{m}+\epsilon_{2}x^{r}+\epsilon_{3}$ can have at most one irreducible non-reciprocal factor apart from its cyclotomic factors. However, Mills [6] proved that Ljunggren’s result is true except for four families in which corresponding quadrinomials factor as a product of two non-reciprocal irreducible polynomials apart from its cyclotomic factors. For example,

[TABLE]

Therefore, in either case, determination of the separability of quadrinomials is same as to check separability of its cyclotomic parts. There are quadrinomials which are not separable. For example, $x^{4}+x^{3}+x+1=(x+1)^{2}(x^{2}-x+1).$

Theorem 11**.**

Suppose $f(x)=x^{n}+\epsilon_{1}x^{m}+\epsilon_{2}x^{r}+\epsilon_{3}$ be a polynomial of degree $n$, $\epsilon_{i}\in\{\,1,-1\,\}$. Then $f(x)$ is separable if $f(\pm 1)\neq 0$.

Proof.

Without loss of generality, we can take $n\leq m+r$. Otherwise we replace $f(x)$ by $\epsilon_{3}x^{n}f(x^{-1})$. Let $h(x)=(f(x),f^{\prime}(x))$. Then $h(x)|nf(x)-xf^{\prime}(x)=(n-m)\epsilon_{1}x^{m}+(n-r)\epsilon_{2}x^{r}+n\epsilon_{3}$. If $n<m+r$, then all the roots of $nf(x)-xf^{\prime}(x)$ lies outside the unit circle. For if $|z|\leq 1$ be a root of $nf(x)-xf^{\prime}(x)$, then

[TABLE]

contradicts the fact that $n<m+r$. Since all the roots of $nf(x)-xf^{\prime}(x)$ lies in the region $|z|>1$, we have $|h(0)|>1$. This contradicts the fact that $h(0)|f(0)$.

Let $n=m+r$. From [4] and [6], $h(x)$ is a cyclotomic polynomial. Therefore, it is sufficient to consider $(m,r)=1$. Then $nf(x)-xf^{\prime}(x)$ satisfies Theorem 2 and by Remark 7, if $h(x)\neq 1$ then $h(x)$ divides either $x+1$ or $x-1$. ∎