Irreducibility criterion for certain trinomials

TL;DR

This paper investigates the irreducibility of specific trinomials, providing new criteria and correcting previous results, with a focus on polynomials of the form x^n + ε_1 x^m + ε_2.

Contribution

It introduces an irreducibility criterion for certain trinomials and corrects earlier findings by W. Ljuggren regarding their reducibility and cyclotomic factors.

Findings

Proves irreducibility for x^n + ε_1 x^m + p^k ε_2 when m=1.

Provides cyclotomic factorization and reducibility criteria for x^n + ε_1 x^m + ε_2.

Corrects previous results on the reducibility of these trinomials.

Abstract

In this article we study the irreducibility of polynomials of the form $x^n+\epsilon_1 x^m+p^k\epsilon_2$, $p$ being a prime number. We will show that they are irreducible for $m=1$. We have also provided the cyclotomic factors and reducibility criterion for trinomials of the form $x^n+\epsilon_1x^m+\epsilon_2$, where $\epsilon_i\in \{\, -1,+1\,\}$. This corrects few of the existing results of W. Ljuggren's on $x^n+\epsilon_1x^m+\epsilon_2$.

Equations

$$x^5-x^2+4=(x^2+x+2)(x^3-x^2-x+2);$$

$$x^5-x^4+9=(x^2-3x+3)(x^3+2x^2+3x+3).$$

$$x^{3n}+{ϵ}_1{x}^{2n}+4{ϵ}_1=(x^n+2{ϵ}_1)(x^{2n}-{ϵ}_1{x}^n+2),$$

$$\Phi_{pn}(x)=\begin{cases}\Phi_{n}(x^{p})&\mbox{ if $p|n$}\\ \Phi_{n}(x^{p})/\Phi_{n}(x)&\mbox{ if $p\nmid n$}.\end{cases}$$

$$\left(\prod _{d|n}{\Phi }_d(x),\prod _{d|m}{\Phi }_d(x)\right)=\prod _{d|(n,m)}{\Phi }_d(x).$$

$${\alpha }^n-{\alpha }^m-1=0={\alpha }^n+{\alpha }^{n-m}-1.$$

$${\Phi }_3(x^m)=\prod _{i=0}^a\prod _{d|M}{\Phi }_{2^i{3}^{b+1}d}(x).$$

$$g(x)$$

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

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

$$n^n(p^k{ϵ}_2{)}^{n-m}=(-{ϵ}_1{)}^n(n-m{)}^{n-m}{m}^m$$

$$p^k{ϵ}_2=-(z_1^n+{ϵ}_1{z}_1^m).$$

$$g(x)=x^s{f}_1(x^{-1})f_2(x)=\sum _{i=0}^n{b}_i{x}^i,\text{say}$$

$$\sum _{i=1}^{n-1}{b}_i^2=p^{2k}-p^{2(k-v)}-p^{2v}+2.$$

$$g(x)\tilde{g}(x)=p^k{ϵ}_2{x}^{2n}+b_n{b}_{j_r}{x}^{2n-j_r}+b_0{b}_{j_1}{x}^{n+j_1}+\cdots +p^k{ϵ}_2.$$

$$f(x)x^{n}f(x^{-1})=p^k{ϵ}_2{x}^{2n}+{ϵ}_1{x}^{2n-1}+p^k{ϵ}_1{ϵ}_2{x}^{n+1}+\cdots +p^k{ϵ}_2.$$

Full text

Key Words: Cyclotomic polynomials, irreducible polynomials, reciprocal polynomials

AMS(2010):11R09, 12D05,12E05

Introduction

E.S.Selmer [7] studied the irreducibility of trinomials of the form $x^{n}\pm x^{m}\pm 1$ over ${\mathbb{Q}}$. He provided a complete solution for $m=1.$ Later Ljunggren [4] extended Selmer’s result for all $m>1$ and proved for quadrimonials as well. A version of his result for trinomials is the following.

Theorem 1**.**

(Ljunggren) Let $f(x)=x^{n}+\epsilon_{1}x^{m}+\epsilon_{2}$ where ${\epsilon}_{j}\in\{-1,+1\}$. Then $f(x)$ has at most one irreducible non-reciprocal factor and a reciprocal factor of $f(x)$ if any, is the product cyclotomic polynomials.

Ljunggren provided the possible cyclotomic factors for trinomials, but they seemed to be incorrect in certain cases. For example, according to Theorem 3 of Ljunggren [4], the polynomials $x^{50}-x^{4}-1$ and $x^{50}+x^{22}-1$ are divisible by $x^{4}+x^{2}+1$ but they are divisible by $x^{4}-x^{2}+1.$ Similarly if $d$ is even, $d_{1}\equiv 5\pmod{6},d_{2}\equiv 1\pmod{6}$ and $d_{3}$ is odd, then $x^{dd_{1}}+x^{d}-1,x^{dd_{2}}-x^{2d}-1$ and $x^{2^{d_{3}}d}-x^{d}+1$ are divisible by $x^{2d}+x^{d}+1$ but they are actually divisible by $x^{2d}-x^{d}+1.$

Based on the above examples, we revisited Ljunggren’s work and corrected those errors. For similar studies and related work, reader can look into  [7], [4], [8], [5], [1].

Immediately the question appears about the reducibility of polynomials of the form $x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}p$, $p$ being a prime. If $p$ is an odd prime, then the polynomials are irreducible directly follows from Proposition $1$ of [6]. Recently, the authors[3] have shown that $x^{n}+{\epsilon}_{1}x^{m}+2{\epsilon}_{2}$ has exactly one irreducible non-reciprocal factor apart from its cyclotomic factors. The method used there doesn’t apply to the polynomials $x^{n}+{\epsilon}_{1}x^{m}+p^{k}{\epsilon}_{2}$ with $k\geq 2$. With a different approach, we will prove that

Theorem 2**.**

Suppose $f(x)=x^{n}+{\epsilon}_{1}x+{\epsilon}_{2}p^{k}$ be a polynomial of degree $n\geq 2$ with $p$ being a prime number and ${\epsilon}_{i}\in\{\,-1,+1\,\},k\geq 2$. Then $f(x)$ is irreducible.

For arbitrary $m$ there are, indeed, polynomials which are reducible. For example,

[TABLE]

More generally,

[TABLE]

for every $n\geq 1$. Although $f(x)$ is reducible for $m>1$, we will show that $f(x)$ can not have more than $k$ factors. More precisely,

Theorem 3**.**

Suppose $p$ is a prime and $f(x)=x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}p^{k}$ with ${\epsilon}_{i}\in\{\,-1,+1\,\}$ be a polynomial of degree $n$ and $k\geq 2$. Then $f(x)$ is a product of atmost $k$ distinct non-reciprocal irreducible polynomials.

The separability of such polynomials has also been considered there. Throughout the paper, we will consider the reducibility over ${\mathbb{Q}}$ (and hence over ${\mathbb{Z}}$) only. If $n$ is a positive integer, we define $e(n)$ as the largest even part of $n$, i.e. $n=2^{a}n_{1}$ with $n_{1}$ odd implies $e(n)=2^{a}$.

Factorization of $x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}$

Let $f(x)=x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ be a polynomial of degree $n$ with ${\epsilon}_{i}\in\{\,-1,+1\,\}$. From Theorem 1 $f(x)$ has a cyclotomic factor whenever it is reducible. To determine the reducibility criterion of $f(x)$, it is, therefore, sufficient to find the cyclotomic factors of $f(x)$. Before we start, we recall a few basic properties of cyclotomic polynomials which will be useful later.

Proposition 4**.**

Suppose $n$ is a positive integer and $\Phi_{n}(x)$ be the $n$th cyclotomic polynomial.

1. (a)

Let $p$ be a prime. Then

[TABLE] 2. (b)

Define $D_{n}^{m}=\{\,d\in{\mathbb{N}}\mid\operatorname{lcm}(m,d)=mn\,\}$. Then $\Phi_{n}(x^{m})=\prod\limits_{d\in D_{n}^{m}}\Phi_{d}(x)$. 3. (c)

If $p$ is a prime and $(p,n)=1$ then $\prod\limits_{d|p^{\gamma}n}\Phi_{d}(x)=\prod\limits_{i=0}^{\gamma}\prod\limits_{d|n}\Phi_{p^{i}d}(x).$ In particular, $x^{n}+1=x^{2n}-1/x^{n}-1=\prod\limits_{d|2n,d\nmid n}\Phi_{d}(x)=\prod\limits_{d|n}\Phi_{2d}(x)$. 4. (d)

If $n,m$ are positive integers, then

[TABLE]

Considering the elementary nature we omit the detailed proof. One can look into Thangadurai [9] for the same.

The polynomial $f(x)=x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ is reducible if and only if ${\epsilon}_{2}x^{n}f(x^{-1})=x^{n}+{\epsilon}_{1}{\epsilon}_{2}x^{n-m}+{\epsilon}_{2}$ is reducible. Therefore, for a given $n$ it is sufficient to consider the reducibility of polynomials $f(x)=x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ with $n\geq 2m$. Throughout the section, we will consider $m=2^{a}\cdot 3^{b}\cdot M,n-2m=2^{p}\cdot 3^{q}\cdot N$ as the prime factorization of $m$ and $n-2m$ respectively.

Theorem 5**.**

Let $f(x)=x^{n}-x^{m}-1$ be reducible. Then $q>b,e(m)>e(n-2m)$ and $f(x)$ is divisible by $\Phi_{6}(x^{(n,m)})$.

Proof.

Since $f(x)$ is reducible, from Theorem 1 $f(x)$ has a reciprocal factor. Consequently, there exits an ${\alpha}\in{\mathbb{C}},$ where ${\alpha}\neq\pm 1,0$ such that both ${\alpha}$ and $\frac{1}{{\alpha}}$ are roots of $f(x).$ That is

[TABLE]

In other words, ${\alpha}$ is a root of the polynomial $x^{n-2m}+1.$ This evenually implies $f(x)$ is irreducible for $n=2m$. So we need to consider $n>2m$ for the remaining part. Since $\alpha$ satisfies $x^{n-2m}+1=0$ and $x^{n}-x^{m}-1=0$, it would satisfy $x^{2m}+x^{m}+1=0$. In particular, $\alpha$ is a root of $g(x)=\gcd(x^{n-2m}+1,x^{2m}+x^{m}+1).$

From Proposition 4, it can be seen that $x^{2m}+x^{m}+1=\prod\limits_{d\in D_{3}^{m}}\Phi_{d}(x),$ where $D_{3}^{m}=\{d\in{\mathbb{N}}|\operatorname{lcm}(m,d)=3m\}.$ If we consider the prime factorizations of $m$ and $n-2m$, then we have $x^{n-2m}+1=\prod\limits_{i=0}^{q}\prod\limits_{d|N}\Phi_{2^{p+1}3^{i}d}(x)$ and

[TABLE]

Let $n$ be odd so that $n-2m$ odd or equivalently $p=0.$ Hence

[TABLE]

If $q\geq b+1$ then $n=2^{a+1}3^{b}M+3^{q}N=3^{b}u_{3}$ where $u_{3}$ odd and $3\nmid u_{3}$.

Also, $(n-2m,m)=(n,m)=3^{b}d_{1}$ gives $\prod\limits_{d|d_{1}}\phi_{2.3^{b+1}d}(x)=\prod\limits_{d|d_{1}}\phi_{6.3^{b}d}(x)=\Phi_{6}(x^{(n,m)}).$

On the other hand, if $n$ is even then $n-2m$ is even and $p\geq 1$. Then

[TABLE]

If $a\geq p+1,q\geq b+1$ then $n=2^{a+1}3^{b}M+2^{p}3^{q}N=2^{p}3^{b}u_{4}$ where $(u_{4},6)=1$. ∎

Corollary 6**.**

If $n=2^{a}3^{b}$ with $a+b>0$ then $x^{n}-x^{m}-1$ is irreducible for every $m<n$.

Since the proof for the remaining three families are almost same, instead of duplicating we state them without proof. The detailed proof can be carried out by using Proposition 4 and Theorem 5.

Theorem 7**.**

Let $f(x)=x^{n}+{\epsilon}_{1}x^{m}-{\epsilon}_{1}$ be reducible with ${\epsilon}_{1}\in\{\,-1,+1\,\}$. Then $f(x)$ is divisible by $\Phi_{6}(x^{(n,m)})$ and the following holds:

1. (a)

${\epsilon}_{1}=1,e(m)=e(n-2m),q>b$; 2. (b)

${\epsilon}_{1}=-1,e(m)<e(n-2m),q>b$.

Theorem 8**.**

Let $f(x)=x^{n}+x^{m}+1$ be reducible. Then $f(x)$ is divisible by $\Phi_{3}(x^{(n,m)})$ and either of the following holds necessarily

1. (a)

$n=2m,M>1$; 2. (b)

$n\neq 2m,q>b$.

If we summarize all the results of this section, it fits perfectly within the below table.

If $n=2m$

then $x^{n}\pm x^{m}-1$ are irreducible. And $x^{n}-x^{m}+1=\Phi_{6}(x^{m})$, $x^{n}+x^{m}+1=\Phi_{3}(x^{m})$ are reducible or irreducible according to Proposition 4(b).

If $n\neq 2m$

then

1. 1.

If $m$ is odd then $x^{n}-x^{m}-1$ is irreducible. 2. 2.

If $m+n$ is odd then $x^{n}+x^{m}-1$ is irreducible. 3. 3.

If $n$ is odd then $x^{n}-x^{m}+1$ is irreducible.

The following table summarizes the irreducibility of all polynomials for $n\neq 2m.$ Suppose $m=2^{a}\cdot 3^{b}\cdot M$, $n-2m=2^{p}\cdot 3^{q}\cdot N.$ And $F$ is one of the nontrivial reciprocal factor of $x^{n}\pm x^{m}\pm 1.$

$m$ $n$ $x^{n}-x^{m}-1$ $x^{n}+x^{m}-1$ $x^{n}-x^{m}+1$ $x^{n}+x^{m}+1$

even odd reducible if $q>b$ irreducible irreducible reducible if $q>b$

$F=\Phi_{6}(x^{(n,m)})$

$F=\Phi_{3}(x^{(n,m)})$

even even reducible reducible reducible same as above

if $q>b,a>p$ if $a=p$, $q>b$ if $p>a,q>b$

$F=\Phi_{6}(x^{(n,m)})$ $F=\Phi_{6}(x^{(n,m)})$ $F=\Phi_{6}(x^{(n,m)})$

odd even irreducible irreducible same as above same as above

odd odd irreducible same as even-even irreducible same as above

Factorization of $x^{n}+{\epsilon}_{1}x^{m}+p^{k}{\epsilon}_{2}$

Suppose $f(x)=x^{n}+{\epsilon}_{1}x^{m}+p^{k}{\epsilon}_{2}$ be a polynomial of degree $n$ with ${\epsilon}_{i}\in\{\,-1,+1\,\},k\geq 2$. We will first prove the separability of such polynomials using discriminant. It is known that

Theorem 9**.**

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

[TABLE]

where $d=(n,m)$.

Theorem 10**.**

Let $p$ be a prime. The polynomial $f(x)=x^{n}+{\epsilon}_{1}x^{m}+p^{k}{\epsilon}_{2}$ is separable over ${\mathbb{Q}}$, ${\epsilon}_{i}\in\{\,-1,+1\,\}$.

Proof.

By Theorem 9, the discriminant of $f(x)$ is

[TABLE]

with $d=(n,m)$. Since $f(x)$ is separable over ${\mathbb{Q}}$ if and only if $f(x^{d})$ is separable, it is sufficient to consider $d=1$. $f(x)$ has multiple root if and only if $D_{f}=0$. Then from (1), we have

[TABLE]

which is not possible as $d=1$ and $p$ being a prime. ∎

Theorem 11**.**

Let $p$ be a prime and $f(x)=x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}p^{k}$ be a polynomial of degree $n$ with $k\geq 2$. Then $f(x)$ has all its root on the region $|z|>1$.

Proof.

Let $z_{1}$ be a root of $f(x)$ with $|z_{1}|\leq 1$. Then $f(z_{1})=0$ gives

[TABLE]

Taking modulus on both side gives $p^{k}=|z_{1}^{n}+{\epsilon}_{1}z_{1}^{m}|\leq|z_{1}|^{n}+|z_{1}|^{m}\leq 2$, which contradicts the fact that $p$ is a prime number and $k\geq 2$. Hence all the roots of $f(x)$ lies in the region $|z|>1$. ∎

By using this theorem, we will prove Theorem 3.

.

Proof of Theorem 3: Suppose $f(x)=\prod\limits_{i=1}^{t}f_{i}(x)$ be non-trivial factorization of $f(x)$, where each $f_{i}(x)$ is irreducible. Since $f(x)$ is a monic polynomial, we assume that each $f_{i}(x)$ is monic. From Theorem 10 $f(x)$ being separable, $f_{i}(x)\neq f_{j}(x)$ for $i\neq j$. By using Theorem 11, from $|f(0)|=p^{k}=\prod\limits_{i=1}^{t}|f_{i}(0)|$, we have $|f_{i}(0)|\geq p$. In other words, $t\leq k$ and consequently they are non-reciprocal. ∎

Now we will prove the irreducibility of $x^{n}+{\epsilon}_{1}x+{\epsilon}_{2}p^{k}$ for every $k\geq 2$.

.

Proof of Theorem 2: If $f(x)=x^{2}+{\epsilon}_{1}x+p^{k}{\epsilon}_{2}$ is reducible then all of its roots are integers only, by Rational Root theorem. But $u(u+{\epsilon}_{1})=-{\epsilon}_{2}p^{k}$ is not possible for any integer $u$.

Suppose $n\geq 3$ and $f(x)$ is reducible. Let $f(x)=f_{1}(x)f_{2}(x)$ be a non-trivial factorization of $f(x)$ with $\deg(f_{1})=s$. Without loss of generality, we assume that both $f_{1}(x),f_{2}(x)$ are monic polynomials. From Theorem 11 we have $|f_{i}(0)|\geq p$. Since $f_{1}(0)f_{2}(0)=p^{k}{\epsilon}_{2}$, let $|f_{1}(0)|=p^{v}$ and $|f_{2}(0)|=p^{k-v}$ for some $v\geq 1$. We consider the following two polynomials

[TABLE]

and $\tilde{g}(x)=\sum\limits_{i=0}^{n}b_{n-i}x^{i}$. The way the polynomials has been defined, we have $f_{1}(0)=b_{n}$ and $f_{2}(0)=b_{0}$. Let $b_{0}=p^{v}{\epsilon}_{2}^{\prime}$ with $|{\epsilon}_{2}^{\prime}|=1$. Since $g(x)\tilde{g}(x)=x^{n}f(x)f(x^{-1})$, comparing the coefficients of $x^{n}$, we get

[TABLE]

Suppose there are $r$ number of non-zero $b_{i}$’s, say $0<j_{r}<j_{r-1}<\cdots<j_{1}<n$ such that $b_{j_{l}}\neq 0$. Then $g(x)=b_{n}x^{n}+b_{j_{1}}x^{j_{1}}+\text{middle terms}+b_{j_{r}}x^{j_{r}}+b_{0}$ and

[TABLE]

Whereas

[TABLE]

Since $n\geq 3$, the second largest term in (3) is $x^{2n-1}$ and has coefficient ${\epsilon}_{1}$. The second largest term in (2) is either $x^{2n-j_{r}}$ or $x^{n+j_{1}}$ or both. That is either $j_{r}=1$ or $j_{1}=n-1$ or $n=j_{r}+j_{1}=1+(n-1)$ respectively. In all these cases, the corresponding coefficient is divisible by $p$ which is impossible. Therefore $f(x)$ has to be irreducible. ∎