# Survey on irreducibility of trinomials

Biswajit Koley, A. Satyanarayana Reddy
Department of Mathematics, Shiv Nadar University, India-201314
(e-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

Let $a,b,c$ be non-zero integers and $f(x)=ax^{n}+bx^{m}+c$ be a trinomial of
degree $n$. We surveyed the irreducibility criteria of $f(x)$ over rational
numbers.

Key Words: Irreducible polynomials, cyclotomic polynomials.
AMS(2010): 11R09, 12D05.

##  1 Introduction

In modern times polynomial factorization has become a fundamental part of
computer algebra systems. The problem of factoring a polynomial into lower
degree polynomials has a long and renowned history. D. Knuth [62] traces back
to Isaac Newton who first attempted to factor a polynomial of arbitrary
degree. In Arithmetica Universalis (1707), Newton gave a method to find linear
and quadratic factors of a polynomial with integer coefficients. Since then
several mathematicians attempted to find irreducibility criteria for different
kinds of polynomials and many other tried to find significant properties of
their factors when reducible. For example, knowing one of its factors is a
cyclotomic polynomial facilitates us to reduce the computational time while
checking the reducibility of a family of polynomials. Before defining the
cyclotomic polynomial, recall that the reciprocal of a polynomial
$f(x)=a_{n}x^{n}+\cdots+a_{1}x+a_{0}$ of degree $n$ is defined as
$\tilde{f}(x)=x^{n}f\left(\frac{1}{x}\right)=a_{0}x^{n}+a_{1}x^{n-1}+\cdots+a_{n}.$
A polynomial $f(x)$ is said to be a reciprocal polynomial if
$f(x)=\pm\tilde{f}(x)$, otherwise it is called non-reciprocal. It follows from
the definition that if $f(x)$ is a reciprocal polynomial, $f(0)\neq 0$, then
$f(x)$ is reducible if and only if $\tilde{f}(x)$ is reducible. If $\alpha$ is
a nonzero root of a reciprocal polynomial $f(x)$, then $\frac{1}{\alpha}$ is
also a root of $f(x)$. One of the widely studied reciprocal families is the
family of cyclotomic polynomials. The $n^{\text{th}}$ cyclotomic polynomial is
defined as

$\Phi_{n}(x)=\underset{(j,n)=1}{\prod\limits_{1\leq j\leq
n}}\left(x-e^{\frac{2\pi ij}{n}}\right),i^{2}=-1.$

The irreducibility problem of binomials $ax^{n}+b,a,b\in{\mathbb{Q}}$ can be
converted to the irreducibility of binomials of the form
$x^{n}-a,a\in{\mathbb{Q}}$. It has been resolved completely. The criterion is
now known as Capelli’s theorem due to Capelli [22].

######  Theorem 1 ([22], [91], [115], [63]).

Let $n\geq 2,a\in{\mathbb{Q}}$. The polynomial $x^{n}-a$ is reducible over
${\mathbb{Q}}$ if and only if either $a=b^{t}$ for some $t|n,t>1$ or
$4|n,a=-4b^{4}$ for some $b\in{\mathbb{Q}}$.

Naturally, the question arises about the irreducibility criterion for
trinomials $f(x)=ax^{n}+bx^{m}+c,$ where $a\in{\mathbb{N}},$
$b,c\in{\mathbb{Z}}\setminus\\{\,0\,\\}$ and $(a,b,c)=1$. Starting from
Eisenstein, many irreducibility criteria appeared for different types of
trinomials. Yet the question is not fully solved. There is no single criterion
which fits to every trinomial. Although several results appeared over time, to
the best of our knowledge there is no survey article on this particular topic.
Therefore, in this article, we tried to collect an exhaustive list of the
irreducibility criteria for trinomials available in the literature.

We compiled the existing irreducibility criteria of trinomials in section 2.
In section 3, we studied the location of zeros of trinomials and their
applications in the factorization of trinomials. The discriminant of
trinomials was discussed in section 4. At the end of this article, we ask a
few questions, answers to those questions will enhance the list of families of
irreducible trinomials. Due to the diversity and intricacies of the theory
applied in the polynomial factorization, we will concentrate on irreducibility
criteria over rational numbers only. Unless specified, throughout the article,
we will assume that all the polynomials have content $1$. By this assumption,
the irreducibility over ${\mathbb{Q}}$ and irreducibility over ${\mathbb{Z}}$
become the same. Further, by ${\epsilon}$ or ${\epsilon}_{i}$, we will always
mean that they belong to the set $\\{\,-1,+1\,\\}$. For in depth study related
to polynomial factorizations and for general references on polynomials, the
reader should refer to [12], [90], and [97].

##  2 Reducibility of trinomials

This section is divided into various subsections depending upon the
irreducibility criteria, which apply to a particular type of trinomials. We
will first discuss those families, whose reducibility nature has been resolved
completely. Later we proceed to compile the results providing a partial answer
to the question of the irreducibility of trinomials.

###  2.1 Reducibility of
$\boldsymbol{x^{n}+b{\epsilon}_{1}x^{m}+{\epsilon}_{2}}$

Let $n,m,b\in{\mathbb{N}}$ and let
$f(x)=x^{n}+b{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ be a trinomial of degree
$n,{\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$. The polynomial $f(x)$ is
reducible if and only if
${\epsilon}_{2}\tilde{f}(x)=x^{n}+b{\epsilon}_{1}{\epsilon}_{2}x^{n-m}+{\epsilon}_{2}$
is reducible. Thus, it is sufficient to check the reducibility of
$x^{n}+b{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ for $1\leq m\leq\frac{n}{2}$.

####  2.1.1 Reducibility of
$\boldsymbol{x^{n}+b{\epsilon}_{1}x+{\epsilon}_{2}}$

Perron [87] is the first mathematician who gave an irreducibility criterion of
a polynomial depending upon the magnitude of the coefficients. He proved that,

Perron’s criterion [87]: Let $f(x)=x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$
be a monic polynomial with integer coefficients and $a_{0}\neq 0$. If

$|a_{n-1}|>1+|a_{n-2}|+\cdots+|a_{1}|+|a_{0}|,$

then $f(x)$ is irreducible over ${\mathbb{Z}}$.

By using Perron’s criterion, the polynomial
${\epsilon}_{2}\tilde{f}(x)=x^{n}+b{\epsilon}_{1}{\epsilon}_{2}x^{n-1}+{\epsilon}_{2}$,
and hence $f(x)=x^{n}+b{\epsilon}_{1}x+{\epsilon}_{2}$ is irreducible for
every $b\geq 3$. If $b$ is either $1$ or $2$, then there are polynomials which
are reducible. For example,

$\displaystyle x^{3}-2x+1=(x-1)(x^{2}+x-1);$ $\displaystyle
x^{8}+{\epsilon}x+1=(x^{2}+{\epsilon}x+1)(x^{6}-{\epsilon}x^{5}+{\epsilon}x^{3}-x^{2}+1).$

Selmer [104] came across the question of reducibility of trinomials of the
form $x^{n}\pm x\pm 1$ while studying continued fractions. With a change of
sign in $x$, it is sufficient to consider $x^{n}+x+1$ and $x^{n}-x-1$ only.
For example, if $n$ is odd and $f(x)=x^{n}-x+1$, then $-f(-x)=x^{n}-x-1.$
Hence, if $n$ is odd, then $f(x)$ is reducible if and only if $-f(-x)$ is
reducible and so on.

If $\alpha_{i}$ is a root of $f(x)=x^{n}+{\epsilon}(x+1)$, then Selmer
considered the following sum

$S_{f}=\sum\limits_{i=1}^{n}\left(\alpha_{i}-\frac{1}{\alpha_{i}}\right).$

Since $S_{f}$ is a symmetric function of roots of $f$, $S_{f}\in{\mathbb{Q}},$
and $S_{f}\in{\mathbb{Z}}$ for $f(x)=x^{n}+{\epsilon}(x+1)$. As any factor of
$f(x)$ over ${\mathbb{Z}}$ has a constant term $\pm 1,$ reducibility of $f(x)$
over ${\mathbb{Z}}$ yields an integer partition of $S_{f}$. Using this fact,
Selmer proved that

######  Theorem 2 (Selmer [104]).

The polynomial $x^{n}+{\epsilon}(x+1)$ is irreducible except for
${\epsilon}=1$ and $n\equiv 2\pmod{3}$. If ${\epsilon}=1$ and $n\equiv
2\pmod{3}$, then $x^{n}+x+1=\Phi_{3}(x)g(x)$, where $g(x)$ is either
identically $1$ or an irreducible polynomial.

Selmer attempted to answer the reducibility problem of trinomials
$x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ for arbitrary values of $m$. By
considering the sum

$S_{f}=\sum\limits_{i=1}^{n}(\alpha_{i}^{m}-\frac{1}{\alpha_{i}^{m}}),$

where $\alpha_{i}$ is a root of $x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}$, he
gave an upper bound for the number of irreducible factors of it.

######  Theorem 3 (Selmer [104]).

Let $n=dn_{1},m=dm_{1},(n_{1},m_{1})=1$ and
$f(x)=x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ be a polynomial of degree
$n\geq 2m$. Then all the roots of $f(x)$ which lie on the unit circle are
roots of unity and $\Phi_{3}(x^{d})$ divides $f(x)$ only when
$n_{1}+m_{1}\equiv 0\pmod{3}.$ Apart from cyclotomic factors, $f(x)$ has at
most $m$ irreducible factors, each is of degree greater than $5$ for $n>7$.

If $b=2$, then a partial answer can be drawn from the work of Schinzel [93] on
the trinomial $x^{n}-2x^{m}+1$. However for complete answer one can refer
either the works of Harrington [48] (see Theorem 14 below) or by that of
Filaseta et al. [40] (see Theorem 9 below). The trinomial
$x^{n}+2{\epsilon}_{1}x+{\epsilon}_{2}$ factors as $(x\pm 1)g(x)$ for some
irreducible polynomial $g(x)\in{\mathbb{Z}}[x]$ and $n\geq 3$. If $n=2,$ then
$x^{2}+2{\epsilon}_{1}x+{\epsilon}_{2}$ is either irreducible or has a
cyclotomic factor. Combiningly we have

######  Theorem 4.

Let $b\in{\mathbb{N}}$ and $f(x)=x^{n}+b{\epsilon}_{1}x+{\epsilon}_{2}$ be a
trinomial of degree $n\geq 2$. Then the followings are true.

  1. (a)

If $b\geq 3$, then $f(x)$ is irreducible.

  2. (b)

If $b=1,2$, then $f(x)$ is reducible if and only if it factor as
$f_{c}(x)f_{n}(x)$, where $f_{c}(x)$ is a product of cyclotomic polynomials,
and $f_{n}(x)\in{\mathbb{Z}}[x]$ is either identically $1$ or an irreducible
polynomial. Further in reducible cases, if $b=1$, $f_{c}(x)$ is either
$\Phi_{3}(x)$ or $\Phi_{6}(x)$, and if $b=2$, then $f_{c}(x)$ is either
$x+1,x-1,(x+1)^{2}$ or $(x-1)^{2}$.

####  2.1.2 Reducibility of
$\boldsymbol{x^{n}+b{\epsilon}_{1}x^{m}+{\epsilon}_{2},\,m\geq 2}$

  1. (A)

$\boldsymbol{b\leq 2}$
Let $f(x)=x^{n}+b{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ be a trinomial of degree
$n\geq 2m$ and $b$ is either $1$ or
$2,{\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$. One can get an upper
bound on the number of irreducible factors of
$f(x)=x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ from Theorem 3. In 1960,
Tverberg [112] considered Newton’s formula for power sum of roots and proved a
necessary and sufficient condition for the reducibility of $f$.

######  Theorem 5 (Tverberg [112]).

Suppose $f(x)=x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ is a polynomial of
degree $n\geq 2m$. Then $f(x)$ is irreducible if and only if $f(x)$ has no
root on the unit circle. If $f(x)$ has a root on the unit circle, then it is
the roots of unity and the other factor of $f(x)$ is irreducible.

In the same year, Ljunggren [72] developed an ingenious tool to study the
behavior of the factors of a polynomial $f(x)$ having a small Euclidean norm.
If $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$, then the Euclidean
norm of $f$, denoted by $\|f\|_{2},$ is defined as

$\|f\|_{2}=\left(a_{n}^{2}+a_{n-1}^{2}+\cdots+a_{1}^{2}+a_{0}^{2}\right)^{1/2}.$

Suppose $f(x)=f_{1}(x)f_{2}(x)$ is a non-trivial factorization of $f(x)$.
Ljunggren considered the polynomials $g(x)=\tilde{f}_{1}(x)f_{2}(x)$ and
$\tilde{g}(x)$ so that $g(x)\tilde{g}(x)=f(x)\tilde{f}(x)$. If the value of
$\|f\|_{2}^{2}$ is small, then one can effectively compute the nature of the
factors of $f$. It is interesting to note that $\|f\|^{2}_{2}$ is the
coefficient of $x^{n}$ in $f(x)\tilde{f}(x)$.

Ljunggren applied this method to study trinomials and quadrinomials of the
form $x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ and
$x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}x^{r}+{\epsilon}_{3}$, respectively.
For several other applications of this method, readers can refer [13], [32],
[33], [39], [41], [45]. In the case of trinomials, Ljunggren proved that

######  Theorem 6 (Ljunggren [72]).

If $n=n_{1}d,m=m_{1}d,(n_{1},m_{1})=1,n\geq 2m,$ then the polynomial

$f(x)=x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2},$

is irreducible, apart from the following three cases, where $n_{1}+m_{1}\equiv
0\pmod{3}$:

$n_{1},m_{1}$ both odd, ${\epsilon}_{1}=1$; $n_{1}$ even, ${\epsilon}_{2}=1$;
$m_{1}$ even, ${\epsilon}_{1}={\epsilon}_{2}$.

In these three cases, $f(x)$ is the product of the polynomial
$x^{2d}+{\epsilon}_{1}^{m}{\epsilon}_{2}^{n}x^{d}+1$ and a second irreducible
polynomial.

However, the above factor is not correct in all cases. For example, by Theorem
6, the polynomial $x^{50}-x^{4}-1$ has factor $x^{4}+x^{2}+1$ but they are
divisible by $x^{4}-x^{2}+1$. More generally, if $d$ is even, $d_{1}\equiv
5\pmod{6}$, $d_{2}\equiv 1\pmod{6}$ and $d_{3}$ is odd, then the polynomials
$x^{dd_{1}}+x^{d}-1,x^{dd_{2}}-x^{2d}-1$ and $x^{2^{d_{3}}d}-x^{d}+1$ are
divisible by $x^{2d}+x^{d}+1$ by Theorem 6. But they are divisible by
$x^{2d}-x^{d}+1$. One can check that Ljunggren’s result is true only when
${\epsilon}_{1}={\epsilon}_{2}=1$. We corrected these errors in [65] and
proved that

######  Theorem 7 (Koley & Reddy [65]).

Let $n,m\in{\mathbb{N}},$ $n\geq 2m$ and $m=2^{a}\cdot 3^{b}\cdot m_{1},$
$n-2m=2^{p}\cdot 3^{q}\cdot n_{1},$ where $a,b,p,q\geq 0,(m_{1}n_{1},6)=1.$ If
${\epsilon}_{1}+{\epsilon}_{2}\neq 2$ and
$f(x)=x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ is reducible, then $q>b$ and
one of the following holds:

    1. (a)

${\epsilon}_{1}=-{\epsilon}_{2}=1,a=p$,

    2. (b)

${\epsilon}_{1}=-1,a{\epsilon}_{2}<p{\epsilon}_{2}$.

In each of these cases, the cyclotomic part of $f$ is
$f_{c}(x)=\Phi_{6}(x^{(n,m)})$.

Schinzel [93] studied the polynomials of the form $x^{n}-2x^{m}+1$. If $n=2m,$
then $x^{n}-2x^{m}+1=(x^{m}-1)^{2}$ is a product of cyclotomic polynomials.
For $n\neq 2m$, he proved that

######  Theorem 8 (Schinzel [93]).

Let $n\neq 2m$. The polynomial $x^{n}-2x^{m}+1$ can be factored as
$(x^{(n,m)}-1)g(x)$ where $g(x)\in{\mathbb{Z}}[x]$ is an irreducible non-
reciprocal polynomial except in the following cases:

$\displaystyle x^{7k}-2x^{2k}+1=(x^{k}-1)(x^{3k}+x^{2k}+1)(x^{3k}+x^{k}+1);$
$\displaystyle x^{7k}-2x^{5k}+1=(x^{k}-1)(x^{3k}+x^{2k}+1)(x^{3k}-x^{k}-1),$

for every $k\geq 1$.

If $n=2m,$ then $x^{n}+2x^{m}+1=(x^{m}+1)^{2}$ is a product of cyclotomic
polynomials. On the other hand, if we replace $x$ by $-x$ in Theorem 8, then
one can conclude the following.

If $n$ is even, $m$ is odd, then $x^{n}+2x^{m}+1$ can be factored as
$(x^{(n,m)}+1)g(x)$ where $g(x)$ is an irreducible non-reciprocal polynomial.

Similarly, partial results can be drawn for $x^{n}+2x^{m}-1$ and
$x^{n}-2x^{m}-1$ by replacing $x$ with $-x$ in Theorem 8. Recently Filaseta et
al. [40] gave a complete characterization of the irreducibility of
$x^{n}+2{\epsilon}_{1}x^{m}+{\epsilon}_{2}$.

######  Theorem 9 (Filaseta et al. [40]).

Let $n>m>0$ and $f(x)=x^{n}+2{\epsilon}_{1}x^{m}+{\epsilon}_{2}$. Then the
non-cyclotomic part of $f(x)$ is reducible in the following cases:

$\displaystyle x^{7k}-2x^{2k}+1=(x^{k}-1)(x^{3k}+x^{2k}-1)(x^{3k}+x^{k}+1);$
$\displaystyle x^{7k}-2x^{5k}+1=(x^{k}-1)(x^{3k}+x^{2k}+1)(x^{3k}-x^{k}-1);$
$\displaystyle x^{7k}+2x^{2k}-1=(x^{k}+1)(x^{3k}-x^{2k}+1)(x^{3k}+x^{k}-1);$
$\displaystyle x^{7k}-2x^{5k}-1=(x^{k}+1)(x^{3k}-x^{k}+1)(x^{3k}-x^{k}-1);$
$\displaystyle x^{7k}+2x^{3k}+1=(x^{3k}-x^{2k}+1)(x^{4k}+x^{3k}+x^{2k}+1);$
$\displaystyle x^{7k}+2x^{4k}+1=(x^{3k}-x^{k}+1)(x^{4k}+x^{2k}+x^{k}+1);$
$\displaystyle x^{7k}+2x^{3k}-1=(x^{3k}+x^{2k}-1)(x^{4k}-x^{3k}+x^{2k}+1);$
$\displaystyle x^{7k}-2x^{4k}-1=(x^{3k}-x^{k}-1)(x^{4k}+x^{2k}-x^{k}+1),$

for every $k\geq 1.$

  2. (B)

$\boldsymbol{b\geq 3}$

Let $f(x)=x^{n}+b{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ be a polynomial of degree
$n\geq 2m$ and let $b\geq 3,{\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$.
Recall from the subsection 2.1.1 that these polynomials are irreducible when
$m=1$. Minkusinski and Schinzel [76] considered trinomials of the form
$x^{n}+p{\epsilon}_{1}x^{m}+{\epsilon}_{2},$ where $p$ is an odd prime. They
proved that there are only finitely many such trinomials that are reducible.

######  Theorem 10 (Minkusinski & Schinzel [76]).

If $p$ is an odd prime, then there are only a finite number of ratios
$\frac{n}{m}$ for which $f(x)=x^{n}+p{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ is
reducible.

In 1971, Schinzel [96] asked the following question:
Is it possible to find a reducible trinomial of the form $x^{n}+Ax^{m}+1$ with
$|A|>2,n\neq 2m$ and $n>m>0$?

Coray answered this question in a letter to Schinzel furnishing the following
examples:

$\displaystyle
x^{13}-3x^{4}+1=(x^{3}-x^{2}+1)(x^{10}+x^{9}+x^{8}-x^{6}-\cdots+x^{2}+1);$
$\displaystyle
x^{13}+3x^{7}+1=(x^{4}-x^{3}+1)(x^{9}+x^{8}+x^{7}+x^{6}-x^{4}+x^{3}+1).$

A. Bremner [16] considered the trinomials $x^{n}+Ax^{m}+1$ with $|A|>2$ and
proved more than what Schinzel asked. He showed that there are only finitely
many trinomials with an irreducible cubic factor. More precisely,

######  Theorem 11 (Bremner [16]).

Let $n\geq 2m>0$ and the trinomial $x^{n}+Ax^{m}+1$ has an irreducible factor
of degree three, $0\neq A\in{\mathbb{Z}}$. If $n>3$, then the only
possibilities are the following:

$\displaystyle
x^{4}+2{\epsilon}x+1=(x+{\epsilon})(x^{3}-{\epsilon}x^{2}+x+{\epsilon});$
$\displaystyle x^{5}+x+1=(x^{2}+x+1)(x^{3}-x^{2}+1);$ $\displaystyle
x^{7}-2x^{2}+1=(x-1)(x^{3}+x^{2}-1)(x^{3}+x+1);$ $\displaystyle
x^{7}+2x^{3}+1=(x^{3}-x^{2}+1)(x^{4}+x^{3}+x^{2}+1);$ $\displaystyle
x^{13}-3x^{4}+1=(x^{3}-x^{2}+1)(x^{10}+x^{9}+x^{8}-x^{6}-2x^{5}-2x^{4}-x^{3}+x^{2}+1);$
$\displaystyle
x^{33}+67x^{11}+1=(x^{3}+x+1)(x^{30}-x^{28}-x^{27}+x^{26}+2x^{25}-\cdots-2x^{3}+x^{2}-x+1).$

If $n$ is odd, then by changing the variable $x$ by $-x$, the polynomial
$x^{n}+Ax^{m}-1$ has irreducible cubic factors in the above cases only. If $n$
is even, Lutczyk [96] provided the following example,

$x^{8}+3x^{3}-1=(x^{3}+x-1)(x^{5}-x^{3}+x^{2}+x+1).$

Bremner [16] also gave several examples when $n$ is even. For example,

$x^{6}+(4b^{4}-4b)x^{2}-1=(x^{3}+2bx^{2}+2b^{2}x+1)(x^{3}-2bx^{2}+2b^{2}x-1),$

where $b\neq 0,1,b\in{\mathbb{Z}}$ is an infinite family of polynomials having
a cubic factor. By Perron’s criterion,
$x^{3}+2b^{2}{\epsilon}_{1}x^{2}+2bx+{\epsilon}_{2}$ is irreducible when
$b\neq 0,1$ and hence so is its reciprocal. However, it is not known whether
there are only finitely many such trinomials of the form $x^{n}+Ax^{m}-1$
having a cubic factor when $n$ is even. We, therefore, ask the following
question,

######  Question 12.

Let $n,m,A\in{\mathbb{N}}$, $n\geq 2m,$ $A>2$ and let $n$ be an even integer.
Are there finitely many trinomials of the form $x^{n}\pm Ax^{m}-1$ which have
an irreducible cubic factor?

###  2.2 Reducibility of
$\boldsymbol{x^{n}+b{\epsilon}_{1}x^{m}+c{\epsilon}_{2},\,c>1}$

Suppose $f(x)=x^{n}+b{\epsilon}_{1}x^{m}+c{\epsilon}_{2}$ is a trinomial of
degree $n$, where
$b,c\in{\mathbb{N}},c>1,{\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$.
Unlike in the case of $x^{n}+b{\epsilon}_{1}x^{m}+{\epsilon}_{2}$, here
${\epsilon}_{2}\tilde{f}(x)$ is not of the form of $f(x)$. So, we have to
consider the reducibility of $f(x)$ for $1\leq m\leq n-1$. Nagell [79] gave
the following irreducibility criterion for trinomials with a large middle
coefficient.

######  Theorem 13 (Nagell [79]).

Let $b,c\in{\mathbb{N}}$ and let
$f(x)=x^{n}+b{\epsilon}_{1}x^{m}+c{\epsilon}_{2}$ be a polynomial of degree
$n$. Then $f(x)$ is irreducible if

  1. (a)

$b>1+c^{n-1}$ and

  2. (b)

if $d|n,d>1,$ then $c$ is not a $d^{\text{th}}$ power. In particular, $c>1$.

Nagell’s criterion is weaker than that of Perron’s criterion in the case of
$m=n-1$. If $m=n-1,$ then the second condition of Nagell’s theorem becomes
redundant.

####  2.2.1 Reducibility of
$\boldsymbol{x^{n}+b{\epsilon}_{1}x^{n-1}+c{\epsilon}_{2},\,c>1}$

J. Harrington [48] asked the values of $n$ and $b$ for which

$f(x)=x^{n}+b(x^{n-1}+x^{n-2}+\cdots+x+1)$

is irreducible. If $b=1,$ then $f(x)$ is a product of cyclotomic polynomials.
For $b=-1$, one can see from Theorem 8 and

$x^{n}-2x^{n-1}+1=(x-1)(x^{n-1}-x^{n-2}-\cdots-x-1),$

that $f(x)$ is irreducible. However, the answer is not known for $|b|\geq 2$.
Harrington studied the trinomials of the form
$x^{n}+b{\epsilon}_{1}x^{n-1}+c{\epsilon}_{2}$ to answer these questions and
proved that

######  Theorem 14 (Harrington [48]).

Let $n\geq 3,b,c\in{\mathbb{N}},nb\neq 9,b\neq c,b\geq 2$ and $c\leq 2(b-1)$.
If the trinomial $f(x)=x^{n}+b{\epsilon}_{1}x^{n-1}+c{\epsilon}_{2}$ is
reducible, then $f(x)=(x\pm 1)f_{n}(x),$ where $f_{n}(x)\in{\mathbb{Z}}[x]$ is
a non-reciprocal irreducible polynomial.

Theorem 14 is no longer true if $c>2(b-1)$. For example,

$\displaystyle
x^{6}+2{\epsilon}x^{5}+4=(x^{3}-2x+2{\epsilon})(x^{3}+2{\epsilon}x^{2}+2x+2{\epsilon});$
$\displaystyle x^{7}+2x^{6}+9=(x^{3}-3x+3)(x^{4}+2x^{3}+3x^{2}+3x+3).$

By Perron’s criterion, if $c<b-1,$ then
$x^{n}+b{\epsilon}_{1}x^{n-1}+c{\epsilon}_{2}$ is irreducible. Hence,

######  Theorem 15.

Let $n\geq 4,b\geq 2,c\in{\mathbb{N}},$ and let
$f(x)=x^{n}+b{\epsilon}_{1}x^{n-1}+c{\epsilon}_{2}$ be a polynomial of degree
$n$.

  1. (a)

If $c<b-1,$ then $f(x)$ is irreducible.

  2. (b)

If $b-1\leq c\leq 2(b-1),b\neq c$ and $f(x)$ is reducible, then $f(x)=(x\pm
1)f_{n}(x),$ where $f_{n}(x)\in{\mathbb{Z}}[x]$ is a non-reciprocal
irreducible polynomial.

If $n$ is even, then $x^{n}+bx^{n-1}+b-1=(x+1)f_{n}(x)$ shows that the lower
bound of $c$ can not be improved further. For $b=c,$ Harrington conjectured
that

######  Conjecture 16 (Harrington [48]).

Let $n,b\in{\mathbb{N}}$, $b\geq 2$ and
$f(x)=x^{n}+b{\epsilon}_{1}x^{n-1}+b{\epsilon}_{2}$. If $f(x)\neq x^{2}+4x+4$
and $f(x)\neq x^{2}-4x+4,$ then $f(x)$ is irreducible.

####  2.2.2 Reducibility of
$\boldsymbol{x^{2m}+b{\epsilon}_{1}x^{m}+c{\epsilon}_{2},\,c>1}$

Let $b,c\in{\mathbb{N}},c>1$ and
$f(x)=x^{2m}+b{\epsilon}_{1}x^{m}+c{\epsilon}_{2}$ be a polynomial of degree
$2m$. From Nagell’s criterion, $f(x)$ is irreducible whenever $b>1+c^{2m-1}$
and $c\neq u^{d},$ where $d|2m,d>1,u\in{\mathbb{N}}$. Also note that if
$b^{2}-4c{\epsilon}_{2}$ is a square, then
$x^{2}+b{\epsilon}_{1}x+c{\epsilon}_{2}$ is reducible and hence
$x^{2m}+b{\epsilon}_{1}x^{m}+c{\epsilon}_{2}$ is a product of two $m$ degree
polynomials.

A. Schinzel studied the problem of reducibility of trinomials extensively in
[98], [99], [100]. He even considered the factorization problem of trinomials
over an arbitrary field and provided many necessary and sufficient conditions
for several families of trinomials. One such criterion is

######  Theorem 17 (Schinzel [98]).

Let $K$ be any field of characteristic different from $2$ and
$b,c\in{\mathbb{N}}$. Then the polynomial
$x^{2m}+b{\epsilon}_{1}x^{m}+c{\epsilon}_{2}$ is reducible over $K$ if and
only if either $b^{2}-4c{\epsilon}_{2}$ is a square in $K$ or there is a prime
$p$ dividing $m$ such that $x^{2p}+b{\epsilon}_{1}x^{p}+c{\epsilon}_{2}$ is
reducible over $K$ or $4|m$ and $x^{8}+b{\epsilon}_{1}x^{4}+c{\epsilon}_{2}$
is reducible over $K$.

It is, therefore, sufficient to find the reducibility condition of
$x^{2p}+b{\epsilon}_{1}x^{p}+c{\epsilon}_{2}$ to completely determine the
reducibility of $x^{2m}+b{\epsilon}_{1}x^{m}+c{\epsilon}_{2},$ where $p$ is a
prime dividing $m$. In the course of determining a Galois group of a family of
polynomials, Filaseta et al. [38] proved that

######  Theorem 18 (Filaseta et al. [38]).

Let $p$ be an odd prime and $u\geq 2$. Then $x^{2p}+{\epsilon}_{1}x^{p}+u^{p}$
is irreducible over ${\mathbb{Q}}$.

Later they separately considered the reducibility of trinomials of the form
$x^{2p}+b{\epsilon}_{1}x^{p}+c{\epsilon}_{2},$ where $b,c\in{\mathbb{N}}$ and
$p$ is a prime number in [37]. They noticed that if
$b^{2}-4c{\epsilon}_{2}=d^{2}$ or $c=b^{2}=d^{2p},p\geq 5$ or
$c=\frac{b^{2}}{2}=2^{p}d^{2p},p\geq 3$ or
$c=\frac{b^{2}}{3}=3^{p}d^{2p},p\geq 5,$ for some integer $d\in{\mathbb{Z}}$,
then $x^{2p}+b{\epsilon}_{1}x^{p}+c{\epsilon}_{2}$ is reducible over
${\mathbb{Z}}$. It has been proved in [37] that if $b,c$ do not satisfy any of
these conditions, then $x^{2p}+b{\epsilon}_{1}x^{p}+c{\epsilon}_{2}$ is
irreducible over ${\mathbb{Z}}$ provided

$p\nmid(181^{2}-1)\underset{q|b}{\prod\limits_{q\text{ prime }}}(q^{2}-1).$

However, there are irreducible polynomials even when $p$ divides the above
product. For example, $x^{4}+2x^{2}+3$ is irreducible. Here, $p=b=2,c=3;$
$b,c$ do not satisfy any of the above conditions and the product is
$(181^{2}-1)(2^{2}-1)$ divisible by $2.$

####  2.2.3 Reducibility of
$\boldsymbol{x^{n}+b{\epsilon}_{1}x+c{\epsilon}_{2},\,c>1}$

Suppose $f(x)=x^{n}+b{\epsilon}_{1}x+c{\epsilon}_{2}$ is a polynomial of
degree $n\geq 2,$ where $b\in{\mathbb{N}}$ and $c\geq
2,{\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$. Serret [105] and Ore [81]
studied the irreducibility of trinomials of the form $x^{n}-x+c{\epsilon}_{2}$
independently. By the use of reducibility of polynomials modulo a prime, they
gave an irreducibility criterion for them.

######  Theorem 19 (Serret [105], Ore [81]).

Let $c\in{\mathbb{N}}$ and $f(x)=x^{p}-x+c{\epsilon}_{2}$ be a polynomial of
prime degree $p$. Then $f(x)$ is irreducible if $p\nmid c$.

There are irreducible polynomials with $p|c$. For example, $x^{5}-x+10$ and
$x^{2}-x+2$ both are irreducible. Irreducible polynomials can be found for
composite values of $n$ as well. For example,

$\displaystyle x^{4}-x+2;\,\,$ $\displaystyle x^{15}-x+3;$ $\displaystyle
x^{10}-x+6;\,\,$ $\displaystyle x^{6}-x+18,$

are all irreducible. A.I. Bonciocat and N.C. Bonciocat [11] gave an
irreducibility criterion for arbitrary polynomials with constant term
divisible by large prime power. A version of their result in case of
trinomials is

######  Theorem 20 (Bonciocat & Bonciocat [11]).

Let $a,b,c\in{\mathbb{N}},$ $p$ be a prime number, and $p\nmid bc$, $r\geq
0,k\geq 1$. The polynomial $ax^{n}+bp^{r}{\epsilon}_{1}x+p^{k}c{\epsilon}_{2}$
is irreducible if

$p^{k}>bp^{2r}+c^{n-1}ap^{nr}.$

It is not possible to improve Theorem 20 for arbitrary values of $a,b,c$. For
example, let $f(x)=x^{2}+2x-8$. Then
$a=b=c=1,p=n=2,r=1,k=3,2^{k}=2^{2r}+2^{nr}$ and $f(x)=(x-2)(x+4)$.

####  Reducibility of
$\boldsymbol{x^{n}+p^{r}{\epsilon}_{1}x+p^{k}{\epsilon}_{2}}$

Let $p$ be a prime number and
$f(x)=x^{n}+p^{r}{\epsilon}_{1}x+p^{k}{\epsilon}_{2}$ be a trinomial of degree
$n\geq 2,$ $r,k\geq 0,{\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$. If
$a=b=c=1$ in Theorem 20, then

The polynomial $x^{n}+p^{r}{\epsilon}_{1}x+p^{k}{\epsilon}_{2}$ is irreducible
if

$p^{k}>p^{2r}+p^{nr},\quad\text{ where }k\geq 1,r\geq 0.$

There are irreducible polynomials that do not satisfy the condition of Theorem
20. For example, $x^{4}+2{\epsilon}_{1}x+2^{4}{\epsilon}_{2}$ is irreducible
while $2^{4}<2^{2}+2^{4}$.

It is possible to find more irreducible families of polynomials of the form
$x^{n}+p^{r}{\epsilon}_{1}x+p^{k}{\epsilon}_{2}$, where $p$ is a prime number
and $r,k\geq 0$. The case $k=0$ has already been solved in
$x^{n}+b{\epsilon}_{1}x+{\epsilon}_{2}$ and $k=1,r\geq 1$ case follows from
Eisenstein’s criterion.

Eisenstein’s criterion [34]: Let $p$ be a prime number and let
$f(x)=a_{n}x^{n}+\cdots+a_{1}x+a_{0}\in{\mathbb{Z}}[x]$ be a polynomial of
degree $n$. If $p|a_{i}$ for $0\leq i\leq n-1$, and $p\nmid a_{n},p^{2}\nmid
a_{0}$, then $f(x)$ is irreducible over ${\mathbb{Z}}$.

The case $k=1,r=0$ and $p$ is an odd prime follows from Theorem 20. In [66]
(see Theorem 25 below), we have shown that if
$x^{n}+{\epsilon}_{1}x+2{\epsilon}_{2}$ is reducible, then it is a product of
cyclotomic factors and an irreducible non-reciprocal polynomial. If $k\geq 2$
and $r=0$, then it is irreducible by Theorem 20. An alternate proof of this
particular case can also be found in [65]. For $k\geq 2,r=1$, we will show
that

######  Theorem 21.

Let $p$ be a prime number and $k\geq 1,n\geq 2$. Then
$x^{n}+p{\epsilon}_{1}x+p^{k}{\epsilon}_{2}$ is irreducible.

###### Proof.

Suppose $f(x)$ is reducible. Let $f(x)=f_{1}(x)f_{2}(x)$ be a proper
factorization of $f(x)$, where

$\displaystyle f_{1}(x)=x^{s}+c_{1}x^{s-1}+\cdots+c_{s-1}x+c_{s};$
$\displaystyle f_{2}(x)=x^{t}+d_{1}x^{t-1}+\cdots+d_{t-1}x+d_{t},\quad t+s=n.$

Comparing the coefficient of $x$ in $f(x)=f_{1}(x)f_{2}(x)$, we have

$\displaystyle c_{s-1}d_{t}+c_{s}d_{t-1}=p{\epsilon}_{1},$ (1)

Reducing the coefficient modulo $p$, we get
$f_{p}(x)=x^{n}=(f_{1})_{p}(x)(f_{2})_{p}(x)$. This is possible only when
$p|c_{i},d_{j}$ for every $i$ and $j$. In other words, $p^{2}$ divides the
left hand side of (1), which is not the case for the right hand side of (1).
Hence, $x^{n}+p{\epsilon}_{1}x+p^{k}{\epsilon}_{2}$ is irreducible for every
$k\geq 1,n\geq 2$. ∎

If $k\geq 2$ and $r\geq 2$, then there are polynomials which are reducible.
For example,

$\displaystyle x^{5}-9x-27=(x^{2}+3x+3)(x^{3}-3x^{2}+6x-9);$ $\displaystyle
x^{5}+81x+243=(x^{2}+3x+9)(x^{3}-3x^{2}+27);$ $\displaystyle
x^{8}+81x-81=(x^{2}-3x+3)(x^{6}+3x^{5}+6x^{4}+9x^{3}+9x^{2}-27).$

The complete characterization of the case $k\geq 2$ and $r\geq 2$ is not
known. The condition of Theorem 20, $p^{k}>p^{2r}+p^{nr}$ is equivalent to
$k>nr$ except for $k=nr+1,p=2,n=2$. If $p=2,n=2$ and $k=nr+1$, then
$f(x)=x^{2}+2^{r}{\epsilon}_{1}x+2^{2r+1}{\epsilon}_{2}$ is reducible only
when ${\epsilon}_{2}=-1,$ and in that case
$f(x)=(x+2^{r-1}(3+{\epsilon}_{1}))(x-2^{r-1}(3-{\epsilon}_{1}))$.

On the other hand, the polynomial
$f(x)=x^{n}+p^{r}{\epsilon}_{1}x+p^{k}{\epsilon}_{2}$ is irreducible over
${\mathbb{Z}}$ if and only if
$f(p^{k}x)=p^{k}{\epsilon}_{2}(p^{k(n-1)}{\epsilon}_{2}x^{n}+p^{r}{\epsilon}_{1}{\epsilon}_{2}x+1)$
is irreducible over ${\mathbb{Q}}$, that is, if and only if
$p^{k(n-1)}{\epsilon}_{2}x^{n}+p^{r}{\epsilon}_{1}{\epsilon}_{2}x+1$ is
irreducible over ${\mathbb{Z}}$. The polynomial
$\tilde{g}(x)=p^{k(n-1)}{\epsilon}_{2}x^{n}+p^{r}{\epsilon}_{1}{\epsilon}_{2}x+1$
is irreducible over ${\mathbb{Z}}$ if and only if
$g(x)=x^{n}+p^{r}{\epsilon}_{1}{\epsilon}_{2}x^{n-1}+p^{k(n-1)}{\epsilon}_{2}$
is irreducible over ${\mathbb{Z}}$. From Perron’s criterion, the polynomial
$g(x)$ is irreducible if $p^{r}>p^{k(n-1)}+1$. Since $k\geq 2$, this is same
as $r>k(n-1)$.

Therefore, if we summarize all these results for
$x^{n}+p^{r}{\epsilon}_{1}x+p^{k}{\epsilon}_{2}$, then it can be put in the
following theorem and table below.

######  Theorem 22.

Let $p$ be a prime number, $k,r\geq 0$ and let
$f(x)=x^{n}+p^{r}{\epsilon}_{1}x+p^{k}{\epsilon}_{2}$ be a polynomial of
degree $n\geq 2$.

  1. (a)

If $f(x)$ is reducible and either $r\in\\{\,0,1\,\\}$ or $k\in\\{\,0,1\,\\}$,
then $f(x)=f_{c}(x)f_{n}(x)$, where $f_{c}(x)$ is a product of cyclotomic
polynomials and $f_{n}(x)$ is either $1$ or a non-reciprocal irreducible
polynomial.

  2. (b)

If $k\geq 2,r\geq 2$ and $f(x)\neq x^{2}+2^{r}{\epsilon}_{1}x-2^{2r+1},$ then
$f(x)$ is irreducible except possibly for $\frac{k}{n}\leq r\leq k(n-1)$.

$k$ | $r$ | $f(x)$ | Irreducible | Reducible
---|---|---|---|---
$0$ | $0$ | $x^{n}-x-1$ | $n\geq 2$ |
|  | $x^{n}+x-1$ | $n\not\equiv 5\pmod{6}$ | $n\equiv 5\pmod{6},$
|  |  |  | $f(x)=\Phi_{6}(x)f_{n}(x)$
|  | $x^{n}+{\epsilon}x+1$ | $n\not\equiv 2\pmod{6}$ | $n\equiv 2\pmod{6},$
|  |  |  | $f(x)=\Phi_{3}({\epsilon}x)f_{n}(x)$
$0$ | $\geq 1$ | $x^{n}+{\epsilon}_{1}p^{r}x+{\epsilon}_{2}$ | $p^{r}>2$ |
|  | $x^{n}+2x-1$ | $n\geq 2$ |
|  | $x^{n}-2x-1$ | $n\equiv 0\pmod{2}$ | $n\equiv 1\pmod{2},$
|  |  |  | $f(x)=(x+1)f_{n}(x)$
|  | $x^{n}-2x+1$ |  | $n\geq 3,f(x)=(x-1)f_{n}(x)$
|  |  |  | $n=2,f(x)=(x-1)^{2}$
|  | $x^{n}+2x+1$ | $n\equiv 1\pmod{2}$ | $n\equiv 0\pmod{2},$
|  |  |  | $n\geq 3,f(x)=(x+1)f_{n}(x)$
|  |  |  | $n=2,f(x)=(x+1)^{2}$
$1$ | $0$ | $x^{n}+{\epsilon}_{1}x+p{\epsilon}_{2}$ | $p\geq 3$ |
|  | $x^{n}+x-2$ |  | $f(x)=(x-1)f_{n}(x)$
|  | $x^{n}-x+2$ | $n\geq 2$ |
|  | $x^{n}+x+2$ | $n\equiv 0\pmod{2}$ | $n\equiv 1\pmod{2},$
|  |  |  | $f(x)=(x+1)f_{n}(x)$
|  | $x^{n}-x-2$ | $n\equiv 1\pmod{2}$ | $n\equiv 0\pmod{2},$
|  |  |  | $f(x)=(x+1)f_{n}(x)$
$1$ | $\geq 1$ | $x^{n}+p^{r}{\epsilon}_{1}x+p^{k}{\epsilon}_{2}$ | $n\geq 2$ |
$\geq 2$ | $\leq 1$ | $x^{n}+p^{r}{\epsilon}_{1}x+p^{k}{\epsilon}_{2}$ | $n\geq 2$ |
| $\geq 2$ | $x^{n}+p^{r}{\epsilon}_{1}x+p^{k}{\epsilon}_{2}$ | According to Theorem 22

######  Question 23.

Let $p$ be a prime number, $r,k,n\geq 2.$ For what values of $r$, where
$\frac{k}{n}\leq r\leq k(n-1),$ the polynomial
$x^{n}+p^{r}{\epsilon}_{1}x+p^{k}{\epsilon}_{2},{\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$
is irreducible over ${\mathbb{Z}}$?

####  2.2.4 Reducibility of
$\boldsymbol{x^{n}+b{\epsilon}_{1}x^{m}+p^{k}{\epsilon}_{2},\,k\geq 1}$

Instead of restricting the value of $m$ and the middle coefficient, let
$f(x)=x^{n}+b{\epsilon}_{1}x^{m}+p^{k}{\epsilon}_{2}$ be an $n$ degree
trinomial, where $p$ is a prime number,
$b,k\in{\mathbb{N}},{\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$. If $k=1$
and $p|b$, then $f(x)$ is irreducible due to Eisenstein’s criterion. If $k=1$
and $p\nmid b$, there are only finitely many $b$ values for which $f(x)$ is
reducible. On the other hand, for $k\geq 2$, the polynomial is irreducible
provided $p^{k}$ is sufficiently large.

  1. (A)

$\boldsymbol{k=1}$

By Nagell’s criterion, $x^{n}+b{\epsilon}_{1}x^{m}+p{\epsilon}_{2}$ is
irreducible whenever $b>p^{n-1}+1$. If the constant term is a large prime
number, then Panitopol and Stefänescu [86] gave an effective criterion for the
irreducibility of polynomials.

######  Theorem 24 (Panitopal & Stefänescu [86]).

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

Thus, $x^{n}+b{\epsilon}_{1}x^{m}+p{\epsilon}_{2}$ is irreducible for every
prime $p>b+1$. Theorem 24 fails to answer the irreducibility of
$x^{n}+{\epsilon}_{1}x^{m}+2{\epsilon}_{2}$. We have considered the equality
condition in Theorem 24 and proved that

######  Theorem 25 (Koley & Reddy [66]).

Let
$f(x)=a_{n_{r}}x^{n_{r}}+a_{n_{r-1}}x^{n_{r-1}}+\cdots+a_{n_{1}}x^{n_{1}}+a_{0}\in{\mathbb{Z}}[x],r\geq
2$, where $a_{i}\neq 0$ for every $i$ and let
$|a_{n_{1}}|+\cdots+|a_{n_{r-1}}|+|a_{n_{r}}|=|a_{0}|$ be a prime number. If
$f(x)$ is reducible, then it is the product of an irreducible non-reciprocal
polynomial and distinct cyclotomic polynomials. Furthermore, the product of
all of its cyclotomic factors is given by
$f_{c}(x)=(x^{n_{r}}+\operatorname{sgn}(a_{0}a_{n_{r}}),x^{n_{r-1}}+\operatorname{sgn}(a_{0}a_{n_{r-1}}),\ldots,x^{n_{1}}+\operatorname{sgn}(a_{0}a_{n_{1}})),$
where $\operatorname{sgn}(x)$ denotes the sign of $x\in{\mathbb{R}}$.

Theorem 25 is not true if the constant term is not a prime number. For
example,

$\displaystyle x^{4}+3x^{2}+4=(x^{2}-x+2)(x^{2}+x+2);$ (2) $\displaystyle
2x^{3}+11x^{2}+8x-21=(x-1)(x+3)(2x+7).$

From Theorem 25, the polynomial
$x^{n}+(p-1){\epsilon}_{1}x^{m}+p{\epsilon}_{2}$ has only one non-reciprocal
irreducible factor apart from its cyclotomic factors. By Eisenstein’s
criterion, if $b\equiv 0\pmod{p}$, then
$x^{n}+b{\epsilon}_{1}x^{m}+p{\epsilon}_{2}$ is irreducible. Therefore,
combining all of them, one can see that

######  Theorem 26.

Let $p$ be a prime, $b\in{\mathbb{N}}$ and let
$f(x)=x^{n}+b{\epsilon}_{1}x^{m}+p{\epsilon}_{2}$ be a polynomial of degree
$n$. Then $f(x)$ is irreducible except possibly for

$p-1\leq b\leq p^{n-1}+1,\quad b\not\equiv 0\pmod{p}.$

If $b=p-1$ and $f(x)$ is reducible, then $f(x)=f_{c}(x)f_{n}(x)$, where
$f_{c}(x)=(x^{n}+{\epsilon}_{2},x^{m}+{\epsilon}_{1}{\epsilon}_{2})$ is the
cyclotomic part of $f(x)$ and $f_{n}(x)\in{\mathbb{Z}}[x]$ is a non-reciprocal
irreducible polynomial.

  2. (B)

$\boldsymbol{k\geq 2}$

For $m=1$, we have the irreducibility criterion given in Theorem 20. If $m=2$,
then A.I. Bonciocat and N.C. Bonciocat provided a criterion similar to Theorem
20.

######  Theorem 27 (Bonciocat & Bonciocat [11]).

Let $a,b,c\in{\mathbb{N}}$ and $p$ be a prime number, $p\nmid bc,r\geq 0,k\geq
1$. The polynomial
$f(x)=ax^{n}+bp^{r}{\epsilon}_{1}x^{2}+p^{k}c{\epsilon}_{2}$ is irreducible if
$k$ is odd and

$p^{k}>bcp^{3r}+c^{n-1}ap^{nr}.$

The theorem is not true if $k$ is an even integer. For example, let
$f(x)=x^{4}+7x^{2}+2^{4}$. Then $a=1=c,b=7,n=k=4,p=2,r=0$ and $2^{4}>7+1$.
However, $f(x)=(x^{2}-x+4)(x^{2}+x+4)$.

When $k$ is an even integer, Jonassen [56] studied trinomials of the form
$x^{n}+{\epsilon}_{1}x^{m}+4{\epsilon}_{2}$. By applying a refinement of
Ljunggren’s method, he proved that

######  Theorem 28 (Jonassen [56]).

Suppose $n>m$ are positive integers. Then the polynomial
$x^{n}+{\epsilon}_{1}x^{m}+4{\epsilon}_{2}$ is irreducible except in the
following cases:

$\displaystyle x^{3k}+{\epsilon}_{1}x^{2k}+4{\epsilon}_{1}$
$\displaystyle=(x^{k}+2{\epsilon}_{1})(x^{2k}-{\epsilon}_{1}x^{k}+2);$
$\displaystyle x^{5k}+{\epsilon}_{1}x^{2k}-4{\epsilon}_{1}$
$\displaystyle=(x^{3k}+{\epsilon}_{1}x^{2k}-x^{k}-2{\epsilon}_{1})(x^{2k}-{\epsilon}_{1}x^{k}+2);$
$\displaystyle x^{11k}+{\epsilon}_{1}x^{4k}+4{\epsilon}_{1}$
$\displaystyle=(x^{5k}-x^{3k}-{\epsilon}_{1}x^{2k}+2{\epsilon}_{1})(x^{6k}+x^{4k}+{\epsilon}_{1}x^{3k}+x^{2k}+2),$

for any $k\geq 1$ and the factors on the right hand side are irreducible in
each cases.

If $c=1,r=0$ in both Theorem 20 and Theorem 27, then the conditions read
$p^{k}>a+b$, which is similar to that of Theorem 24. Because of this reason,
we considered polynomials of the form
$f(x)=a_{n}x^{n}+\cdots+a_{3}x^{3}+p^{k}{\epsilon},$ where $p$ is a prime
number, $p\nmid a_{3}a_{n}$ and $k\geq 2$ in [64]. To frame the results of
[64] conveniently, we introduce the following sets: Let
$n>m>0,a,b\in{\mathbb{N}}$ and $p$ be a prime number. We define

${\mathcal{S}}_{m}=\\{\,ax^{n}+b{\epsilon}_{1}x^{m}+p^{k}{\epsilon}_{2}\mid
p\nmid ab,k\geq 2,{\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}\,\\}$

and for $m\geq 2,\,{\mathcal{S}}_{m}^{\prime}=\\{\,f\in{\mathcal{S}}_{m}\mid
k\not\equiv 0\pmod{m}\,\\}$. Then it has been shown that

######  Theorem 29 (Koley & Reddy [64]).

Let
$f(x)=ax^{n}+b{\epsilon}_{1}x^{3}+p^{k}{\epsilon}_{2}\in{\mathcal{S}}_{3}^{\prime}$
be a trinomial of degree $n\geq 4$. If $p^{k}>a+b$, then $f(x)$ is
irreducible.

The condition $p\nmid ab$ in the definition of ${\mathcal{S}}_{m}$ cannot be
dropped. For example,

$2x^{4}-3x^{3}-81=(x-3)(2x^{3}+3x^{2}+9x+27).$

Further, the necessity of the condition $3\nmid k$ can be seen from the
following examples:

$\displaystyle
x^{4}+4{\epsilon}x^{3}+27=(x+3{\epsilon})^{2}(x^{2}-2{\epsilon}x+3);$ (3)
$\displaystyle 2x^{5}-3x^{3}+27=(x^{2}-3x+3)(2x^{3}+6x^{2}+9x+6).$

Later we consider the case $p^{k}=a+b$ for
$f(x)=ax^{n}+b{\epsilon}_{1}x^{m}+p^{k}{\epsilon}_{2},m\leq 3$ and proved that

######  Theorem 30 (Koley & Reddy [64]).

Let
$f(x)=ax^{n}+b{\epsilon}_{1}x^{m}+p^{k}{\epsilon}_{2}\in{\mathcal{S}}_{1}\cup{\mathcal{S}}_{2}^{\prime}\cup{\mathcal{S}}_{3}^{\prime}$
be a polynomial of degree $n$. If $p^{k}=a+b$ and $f(x)$ is reducible, then
$f(x)=f_{c}(x)f_{n}(x),$ where
$f_{c}(x)=(x^{n}+{\epsilon}_{2},x^{m}+{\epsilon}_{1}{\epsilon}_{2})$ is the
product of all cyclotomic factors of $f$ and $f_{n}(x)$ is an irreducible non-
reciprocal polynomial. In particular, $f(x)$ is irreducible except for the
following families of polynomials:

$f(x)$ | $f_{c}(x)$
---|---
$ax^{2u+1}+bx+p^{k}{\epsilon}_{2}$ | $x+{\epsilon}_{2}$
$ax^{2u}-bx-p^{k}$ | $x+1$
$ax^{2(2u+1)}+bx^{2}+p^{k}{\epsilon}_{2}$ | $x^{2}+{\epsilon}_{2}$
$ax^{2u+1}-bx^{2}+p^{k}$ | $x+1$
$f(x)$ | $f_{c}(x)$
---|---
$ax^{2^{2t}(2u+1)}-bx^{2}-p^{k}$ | $x^{2}+1$
$ax^{2(2u+1)}+bx^{2}+p^{k}{\epsilon}_{2}$ | $x^{2}+{\epsilon}_{2}$
$ax^{2u+1}+bx^{3}+p^{k}{\epsilon}_{2}$ | $x^{(2u+1,3)}+{\epsilon}_{2}$
$ax^{2^{t}(2u+1)}-bx^{3}-p^{k}$ | $x^{(2u+1,3)}+1$

for every $u,t\geq 1$.

Example (2) shows that Theorem 30 is not true for
$f(x)\in{\mathcal{S}}_{2}\setminus{\mathcal{S}}_{2}^{\prime}$. The following
example illustrate that it is neither true for
$f(x)\in{\mathcal{S}}_{3}\setminus{\mathcal{S}}_{3}^{\prime}$ as well

$3x^{4}-5x^{3}-8=(x+1)(x-2)(3x^{2}-2x+4).$

###  2.3 Reducibility of
$\boldsymbol{ax^{n}+b{\epsilon}_{1}x^{m}+c{\epsilon}_{2}}$

Suppose $f(x)=ax^{n}+b{\epsilon}_{1}x^{m}+c{\epsilon}_{2}$ is a trinomial of
degree $n$, where $a,b,c\in{\mathbb{N}}$. If $a,c>1$, then there are very few
results available related to the reducibility criterion. Theorem 20 and
Theorem 27 gives an irreducibility criterion when the constant term is
divisible by a large prime power. A. Schinzel [94] found the conditions on
coefficients so that an arbitrary trinomial is divisible by a cyclotomic
polynomial.

######  Theorem 31 (Schinzel [94]).

Let $a,b,c$ are non-zero integers, $0<m<n,d=(m,n),n=dn_{1},m=dm_{1}$. If
$f_{c}(x)$ is the cyclotomic part of $f(x)=ax^{n}+bx^{m}+c$, then

$f_{c}(x)=\begin{cases}x^{2d}+{\epsilon}_{1}^{m_{1}+n_{1}}{\epsilon}_{2}^{m_{1}}x^{d}+1,&\mbox{
if $c={\epsilon}_{1}a={\epsilon}_{2}b,n_{1}+m_{1}\equiv
0\pmod{3},{\epsilon}_{1}^{m_{1}}={\epsilon}_{2}^{n_{1}}$;}\\\
(x^{d}-(-{\epsilon}_{1})^{m_{1}}{\epsilon}_{1}{\epsilon}_{2})^{2},&\mbox{ if
$c={\epsilon}_{1}a+{\epsilon}_{2}b,(-{\epsilon}_{1})^{m_{1}}=(-{\epsilon}_{2})^{n_{1}},an{\epsilon}_{1}+bm{\epsilon}_{2}=0$;}\\\
x^{d}-(-{\epsilon}_{1})^{m_{1}}{\epsilon}_{1}{\epsilon}_{2},&\mbox{ if
$c={\epsilon}_{1}a+{\epsilon}_{2}b,(-{\epsilon}_{1})^{m_{1}}=(-{\epsilon}_{2})^{n_{1}},an{\epsilon}_{1}+bm{\epsilon}_{2}\neq
0$;}\\\ 1&\mbox{ otherwise.}\end{cases}$

Schinzel also proved that there are only finitely many trinomials
$ax^{n}+bx^{m}+c$ whose non-cyclotomic part is reducible. More precisely,

######  Theorem 32 (Schinzel [94]).

For any non-zero integers $a,b,c$ there exist two effectively computable
constants $A(a,b,c)$ and $B(a,b,c)$ such that if $n>m>0$ and the non-
cyclotomic part of $ax^{n}+b^{m}+c$ is reducible, then

  1. (a)

$\frac{n}{(n,m)}\leq A(a,b,c)$,

  2. (b)

there exists integers $\alpha$ and $\beta$ such that
$\frac{m}{\beta}=\frac{n}{\alpha}$ is integral, $0<\beta<\alpha\leq B(a,b,c)$
and if $g(x)$ is an irreducible non-reciprocal factor of
$ax^{\alpha}+bx^{\beta}+c$ then $g(x^{n/\alpha})$ is an irreducible non-
reciprocal factor of $ax^{n}+bx^{m}+c$.

All the trinomials that we came across so far have at most two non-reciprocal
irreducible factors. We recall the two sets ${\mathcal{S}}_{m}$ and
${\mathcal{S}}_{m}^{\prime}$ introduced earlier. From Theorem 27, if
$f(x)\in{\mathcal{S}}_{2}^{\prime}$ and $p^{k}>a+b$, then $f(x)$ is
irreducible. We have shown in [64] that if
$f(x)\in{\mathcal{S}}_{2}\setminus{\mathcal{S}}_{2}^{\prime}$ and $p^{k}>a+b$,
then $f(x)$ can have at most two non-reciprocal irreducible factors. Apart
from the polynomials given in Theorem 28, many polynomials having two
irreducible non-reciprocal factors and $k$ even. For example, if $p$ is a
prime and $k\geq 1,a\geq 1$, then

$\displaystyle x^{4}+(2\cdot
p^{k}-a^{2})x^{2}+p^{2k}=(x^{2}+ax+p^{k})(x^{2}-ax+p^{k})\in{\mathcal{S}}_{2}\setminus{\mathcal{S}}_{2}^{\prime}$

is a product of two irreducible non-reciprocal factors. Later we extend it and
proved that if $f(x)\in{\mathcal{S}}_{3}\setminus{\mathcal{S}}_{3}^{\prime}$
and $p^{k}>a+b,$ then $f(x)$ can have at most three non-reciprocal irreducible
factor. The sharpness of this result follows from example (3). However, the
number of non-reciprocal irreducible factors of $ax^{n}+bx^{m}+c$ is
unbounded. For example,

$x^{2k}-2^{k+1}x^{k}+2^{2k}=(x^{k}-2^{k})^{2}=\left(\prod\limits_{d|k}2^{\varphi(d)}\Phi_{d}\left(\frac{x}{2}\right)\right)^{2},$
(4)

where $2^{\varphi(d)}\Phi_{d}\left(\frac{x}{2}\right)\in{\mathbb{Z}}[x]$ is an
irreducible non-reciprocal polynomial for every $d\in{\mathbb{N}}$. Over time,
several other questions were asked related to the irreducibility of
trinomials. For example, Schinzel [103] asked:

######  Question 33.

Does there exist an integer $K,$ independent of the polynomial, such that
every trinomial in ${\mathbb{Q}}[x]$ has an irreducible factor in
${\mathbb{Q}}[x]$ with at most $K$ terms?

Mrs. H. Smyczek [103] showed that if $K$ exists, then $K\geq 6$. Bremner [15]
improved to $K\geq 8$ by furnishing an exact trinomial of the form
$x^{14}+bx^{2}+c,$ having precisely two factors each of length $8$. However,
the question remains open as of today.

Inspired by the work of Bremner [15], several authors tried to characterize
trinomials as having a factor of fixed length. For further studies in this
direction, one can look into [24], [49], [68], [70],[71], [89], [92],
[95],[98], [99], [125].

Let $0<n_{1}\leq n_{2}\leq\cdots\leq n_{k}$. A trinomial
$x^{n}+bx^{m}+c\in{\mathbb{Q}}[x]$ is said to have reducibility type
$(n_{1},n_{2},\ldots,n_{k}),$ if it is a product of irreducible polynomials in
${\mathbb{Q}}[x]$ of degree $n_{1},n_{2},\ldots,n_{k}$. For example, if we
take $k=15$ in (4), then

$x^{30}-2^{16}x^{15}+2^{30}=(x-2)^{2}(x^{2}+2x+4)^{2}(x^{4}+2x^{3}+4x^{2}+8x+16)^{2}(x^{8}-2x^{7}+\cdots+32x^{3}-128x+256)^{2}$

has a reducibility type $(1,1,2,2,4,4,8,8)$. A. Bremner and M. Ulas [17]
studied possible reducibility type for $n\leq 10$ and $m\leq 3$. They showed
that if $m=1,$ then the reducibility type $(1,1,1,1)$ is impossible. It has
been conjectured that

######  Conjecture 34 (Bremner & Ulas [17]).

Let $b,c\in{\mathbb{N}}$ and $f(x)=x^{n}+b{\epsilon}_{1}x+c{\epsilon}_{2}$ be
a trinomial of degree $n\geq 4$. Then the reducibility type $(1,1,1,n-3)$
doesn’t exist.

Suppose $f(x)\in{\mathbb{Q}}[x]$ is a polynomial which divides infinitely many
trinomials. Can $f(x)$ divide a linear or quadratic polynomial in $x^{r}$ for
some $r\geq 1$? For example, $\Phi_{3}(x)$ divides $x^{3k+2}+x+1$ for every
$k\geq 1$, by Theorem 2, and $\Phi_{3}(x)|\Phi_{3}(x^{r})=x^{2r}+x^{r}+1$ for
every $r\geq 1,(r,3)=1$. Is this true for arbitrary $f(x)\in{\mathbb{Q}}[x]?$
Posner and Rumsey [88] investigated this question and subsequent development
can be found in [46], [102].

Till now we were confined to mainly questions regarding trinomials over field
of rational numbers. Irreducibility of trinomials over arbitrary field is not
that much common in the literature. Let ${\mathbb{K}}$ be an algebraic number
field and $f(x)=x^{n}+bx^{m}+c\in{\mathbb{K}}[x]$ has a linear or quadratic
factor $g(x)$. Schinzel [101] studied reducibility of the part
$h(x)=\frac{f(x)}{g(x^{(n,m)})}$ and gave several necessary and sufficient
condition for reducibility of $h(x)$ over ${\mathbb{K}}$. For irreducibility
of trinomials over the function field, we refer to [98], [99], [100].

The reducibility of trinomials over finite field has also been studied
extensively. There is a classical result, known as modulo $p$-test ($p$ being
a prime number), which relates the irreducibility of polynomials over a finite
field and the irreducibility of polynomials over ${\mathbb{Q}}$.
Modulo $p$-test([43], Theorem 17.3):  Let $p$ be a prime and suppose that
$f(x)\in{\mathbb{Z}}[x]$ with $\deg(f(x))\geq 1$. Let $\tilde{f}(x)$ be the
polynomial in ${\mathbb{Z}}_{p}[x]$ obtained from $f(x)$ by reducing all the
coefficients of $f(x)$ modulo $p$. If $\tilde{f}(x)$ is irreducible over
${\mathbb{Z}}_{p}$ and $\deg(\tilde{f}(x))=\deg(f(x))$, then $f(x)$ is
irreducible over ${\mathbb{Q}}$.

For a general overview of polynomial factorization over finite field, the
reader can consider [82], [118]. Swan [110] proved that if ${\mathbb{F}}$ is a
field of odd characteristic and $f(x)\in{\mathbb{F}}[x]$ is a polynomial of
degree $n$, having no repeated roots, then $r\equiv n\pmod{2}$ if and only if
discriminant (see section 4 for definition) of $f(x)$ is a square in
${\mathbb{F}}$, where $r$ is the number of irreducible factors of $f(x)$.
Dickson [28] also proved this result independently. L. Carlitz [23] considered
the trinomial $g(x)=x^{2q^{n}+1}+x^{q^{n}-1}+1$ and proved that the degree of
every irreducible factor of $g(x)$ over ${\mathbb{F}}_{q}$ divides either $2n$
or $3n$, where $q$ is a prime power.

The distribution of irreducible trinomials over ${\mathbb{F}}_{3}$ has been
studied by O. Ahmadi [2]. It has been proved that if $n$ is even, $q$ is an
odd prime power, then $x^{n}+bx^{m}+c$ is irreducible over ${\mathbb{F}}_{q}$
if and only if $x^{n}-bx^{m}+c$ is irreducible over ${\mathbb{F}}_{q}$. Later
Ahmadi proved that if $4|n,$ then $x^{n}-x^{m}+1$ is irreducible over
${\mathbb{F}}_{3}$. A necessary condition for the irreducibility of trinomials
over a finite field is given by J. Gathen [121]:

######  Theorem 35 (Gathen [121]).

Let $p$ be an odd prime,
$f(x)=x^{n}+bx^{m}+c,b,c\in{\mathbb{F}}_{p^{r}}\backslash\\{\,0\,\\},r\geq
1,n>m\geq
1,d=(n,m),n_{1}=n/d,m_{1}=m/d,k_{1}=p(p^{r}-1),k_{2}=\operatorname{lcm}(4,k_{1}).$
Then the discriminant of $f$ and the property that $f$ is squarefree with odd
number of irreducible factors depends only on the following residues:

$n\pmod{k_{2}};\quad m\pmod{k_{1}};\quad n_{1}\pmod{p^{r}-1}\quad\text{ and
}\quad m_{1}\pmod{p^{r}-1}.$

For further study on the factorization of trinomials over finite field, reader
should refer to [1], [6], [10], [35], [42], [120], [122].

##  3 Location of zeros of trinomials

The location of zeros of a polynomial sometimes helps us in determining the
irreducibility of the polynomial. For example, if a monic polynomial
$f(x)\in{\mathbb{Z}}[x],f(0)\neq 0$ has exactly one root in $|z|>1$ and all
the remaining roots are in $|z|<1$, then $f(x)$ is irreducible. For if
$f(x)=f_{1}(x)f_{2}(x)$ in ${\mathbb{Z}}[x]$, then either $f_{1}(x)$ or
$f_{2}(x)$ has all its root in $|z|<1$, contradicting the fact
$f_{1}(x),f_{2}(x)\in{\mathbb{Z}}[x]$. Similarly, if a monic polynomial
$f(x)\in{\mathbb{Z}}[x],f(0)\neq 0$ has $k$ roots in $|z|>1$ and remaining
roots are in $|z|<1$, then $f(x)$ can have at most $k$ irreducible factors. In
this section, we will discuss about the location of zeros of trinomials.
Unless specified otherwise, throughout the section. it is assumed that all the
coefficients are non-zero real numbers. Kennedy [61] specified the region in
which all the roots of the trinomial $x^{n}+bx^{m}+c$ lie.

######  Theorem 36 (Kennedy [61]).

If $\alpha$ is the positive real root of the equation $x^{n}+|b|x^{m}-|c|=0$
and $\beta$ is the positive real root of $x^{n}-|b|x^{m}-|c|=0$, then the
roots of the trinomials $x^{n}+bx^{m}+c$ lies in the annulus
$\alpha\leq|z|\leq\beta$.

However, finding a positive real root of $x^{n}+bx^{m}-|c|$ is not an easy
task to accomplish. Because of this reason, Kennedy gave a weaker but easily
computable bound.

######  Theorem 37 (Kennedy [61]).

If $\alpha$ and $\beta$ are the positive real roots of the equation
$x^{n-m}=|c|^{n-m/n}-|b|$ and $x^{n-m}=|c|^{n-m/n}+|b|$ respectively, then the
roots of $x^{n}+bx^{m}+c$ lie in the annulus $\alpha<|z|<\beta$.

Suppose $a,b,c$ are non-zero real numbers, $|a|+|c|<|b|$, and
$f(x)=ax^{n}+bx^{m}+c$. For $|z|=1$,

$|az^{n}+bz^{m}+c-bz^{m}|=|az^{n}+c|\leq|a|+|c|<|b|=|bz^{m}|.$

From Rouché’s theorem, $f(x)=ax^{n}+bx^{m}+c,$ $|a|+|c|<|b|$ has $m$ number
zeros in $|z|<1$. If $|a|+|c|=|b|$, then the location of zeros has been
calculated by Dilcher et al. [29]. They considered trinomials of the form
$bx^{n}-ax^{m}+a-b$ with $a>b>0$ and proved that

######  Theorem 38 (Dilcher et al. [29]).

Suppose $a>b>0$ are real numbers. The number of zeros of
$f(x)=bx^{n}-ax^{m}+a-b$ strictly inside the unit circle is $m-\gcd(n,m)$ if
$\frac{a}{b}\geq\frac{n}{m}$, and $m$ if $\frac{a}{b}<\frac{n}{m}$.

M. A. Brilleslyper and L.E. Schaubroeck [18] counted the number of roots in
the interior of the unit circle for the trinomial $x^{n}+x^{m}-1$. They proved
that if $m=1$, then there are exactly $2\lfloor{\frac{n}{6}}\rfloor+1$ roots
in the interior of the unit circle, where $\lfloor{x}\rfloor$ denotes the
greatest integer less than or equal to $x$. They further conjectured that

######  Conjecture 39 (Brilleslyper & Schaubroeck [18]).

Let $n,m\in{\mathbb{N}}$ and $(n,m)=1$. Then the number of roots in the
interior of the unit circle for the polynomial $x^{n}+x^{m}-1$ is

$2\Bigl{\lfloor}{\frac{n+m-1}{6}}\Bigr{\rfloor}+1.$

The remaining results available in the literature focus on the zeros of
trinomials of the form $ax^{n}+bx+c$. For example, Nicolas and Schinzel [80]
studied the location of zeros of $x^{n+1}-ax+(a-1),$ and later the results
have been improved further by Hernane and Nicolas [51]. The zeros of
$(a-2)x^{n}+(a-1)x-b$, $a>2$ has been studied in [3]. Fejér [36] and
Szeg$\hat{o}$ [111] independently showed that the trinomial $cx^{n}-x+1$ has
root both in the regions $|z-1|\geq 1$ and $|z-1|\leq 1$, for any non-zero
real number $c$. Joyal et.al. [58] proved that if $n\geq 3,$ then the
polynomial $cx^{n}-x+1$ has a root outside every circle that passes through
the origin.

P. Vassilev [119] considered trinomials with complex coefficients. He proved
that the trinomial $z^{n}-pz-1$ has only one root in the interval $(0,1)$ when
$p<0$. Recently A. Melman [75] studied the problem of the location of zeros of
trinomials extensively and improved many existing known results.

The number of roots of a trinomial over a finite field has been estimated in
[60]. For similar studies over finite fields, one may look into [27], [55],
[116], [117].

##  4 Discriminant of trinomials

Suppose $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}\in{\mathbb{Z}}[x]$
is a polynomial of degree $n$. Let $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$
be the roots of $f(x)$. The discriminant of $f(x)$ is defined as

$D_{f}=a_{n}^{2n-2}\prod\limits_{i<j}(\alpha_{i}-\alpha_{j})^{2}=(-1)^{\binom{n}{2}}a_{n}^{2n-2}\prod\limits_{i\neq
j}(\alpha_{i}-\alpha_{j}).$

The quantity $(-1)^{\binom{n}{2}}a_{n}D_{f}$ is called the resultant of
$f(x)$. More precisely, the resultant of $f(x)$, sometimes denoted by
$\operatorname{Res}(f,f^{\prime})$, is defined as

$\operatorname{Res}(f,f^{\prime})=(-1)^{\binom{n}{2}}a_{n}^{2n-1}\prod\limits_{i<j}(\alpha_{i}-\alpha_{j})^{2}=a_{n}^{2n-1}\prod\limits_{i\neq
j}(\alpha_{i}-\alpha_{j}),$

where $f^{\prime}(x)=na_{n}x^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+a_{1}$ is the
formal derivative of $f(x)$.

From the definition, it follows that $f(x)\in{\mathbb{Z}}[x]$ has multiple
roots if and only if $D_{f}=0$. If $f(x)\in{\mathbb{Z}}[x]$ has multiple
roots, then $f(x)$ is said to be a non-separable polynomial over
${\mathbb{Q}}$. Otherwise, it is said to be a separable polynomial over
${\mathbb{Q}}$.

There is another concept, called the discriminant of an algebraic number
field. Suppose ${\mathbb{K}}$ is an algebraic number field and
${\mathcal{O}}_{{\mathbb{K}}}$ is the ring of integers of ${\mathbb{K}}$. Let
$b_{1},b_{2},\ldots,b_{n}$ be an integral basis of
${\mathcal{O}}_{{\mathbb{K}}}$ and let
$\\{\,\sigma_{1},\ldots,\sigma_{n}\,\\}$ be the set of embeddings of
${\mathbb{K}}$ into ${\mathbb{C}}$. Then the discriminant  of ${\mathbb{K}}$,
denoted by $\Delta_{{\mathbb{K}}}$, is defined as

$\Delta_{{\mathbb{K}}}=\det(B)^{2},\,\,\qquad\mbox{ where
$B=(\sigma_{i}(b_{j}))_{i,j}$.}$

######  Example.

Let $d$ be a square-free integer and let ${\mathbb{K}}={\mathbb{Q}}(\sqrt{d})$
be a quadratic field. Then the discriminant of ${\mathbb{K}}$ is (see [74] for
details)

$\Delta_{{\mathbb{K}}}=\begin{cases}d,&\mbox{ if $d\equiv 1\pmod{4};$}\\\
4d,&\mbox{ if $d\equiv 2,3\pmod{4}.$}\end{cases}$

The discriminant of a polynomial and discriminant of an algebraic number field
are closely related. Let $f(x)$ be a monic irreducible polynomial with a root
$\alpha$ and ${\mathbb{K}}={\mathbb{Q}}(\alpha)$. Then the discriminant of
$f(x)$ is

$D_{f}=i(\alpha)^{2}\Delta_{{\mathbb{K}}},\,\,\qquad\mbox{ where
$i(\alpha)=\\#[{\mathcal{O}}_{{\mathbb{K}}}:{\mathbb{Z}}[\alpha]].$}$

If $D_{f}$ is square-free, then $i(\alpha)^{2}=1$ would imply that
${\mathcal{O}}_{{\mathbb{K}}}={\mathbb{Z}}[\alpha],$ and hence
${\mathcal{O}}_{{\mathbb{K}}}$ has a basis in powers of $\alpha$. In this
case, ${\mathbb{K}}$ is said to be a monogenic field. There are several
advantages of knowing a field to be monogenic. For example, let ${\mathbb{K}}$
be a monogenic field. Then the Galois closure ${\mathbb{L}}$ of ${\mathbb{K}}$
has Galois group $S_{n}$ and ${\mathbb{L}}$ is everywhere unramified over its
subfield ${\mathbb{Q}}(\sqrt{D_{f}})$. Therefore, the knowledge of the square-
freeness of the discriminant of a polynomial is useful. However, it is very
difficult to provide a formula for the discriminant of a polynomial with more
than three non-zero terms. To the best of our knowledge, only quadrinomials
for which the discriminant is known as of today are of the form
$x^{n}+t(x^{2}+bx+c)$ (see [85] for details).

The reader can refer to Chapter 10 of [7] for a detailed literature review on
the discriminants of trinomials up to 1980. P. Lefton [69] proved the
discriminant formula of an arbitrary trinomial by using a polynomial and its
derivatives along with the properties of roots of unity. He showed that the
discriminant of $ax^{n}+bx^{m}+c$ is

$D=(-1)^{\binom{n}{2}}a^{n-m-1}c^{m-1}(n^{n/d}a^{m/d}c^{(n-m)/d}-(-1)^{n/d}(n-m)^{(n-m)/d}m^{m/d}b^{n/d})^{d},$

where $d=(n,m)$. D. Drucker and D. Goldschmidt [31] developed an ingenious
method to compute the determinant of a matrix. By using this method,
G.R.Greenfield and D. Drucker [44] proved that

######  Theorem 40 (Greenfield & Drucker [44]).

The discriminant of the trinomial $x^{n}+bx^{m}+c$ is

$D=(-1)^{\binom{n}{2}}c^{m-1}\left[n^{n/d}c^{n-m/d}-(-1)^{n/d}(n-m)^{n-m/d}m^{m/d}b^{n/d}\right]^{d},$

where $d=(n,m)$.

Let $(n,m)=1$ and $f(x)=x^{n}\pm x^{m}\pm 1$. From Theorem 40, the
discriminant of $f$ is

$D_{f}=\pm n^{n}\pm(n-m)^{n-m}m^{m}.$

Recently D. Boyd et al. [14] attempted to find the square-free values of this
discriminant. They found several families of polynomials whose discriminant
has a square factor. For example, if $n\equiv 2\pmod{6},$ then
$\frac{(n^{2}-n+1)^{2}}{9}$ divides $n^{n}-(n-1)^{n-1}$. That is, the
discriminant of $x^{n}-x-1$ has a square-factor when $n\equiv 2\pmod{6}$.
However, such a discriminant is sporadic. If $n\leq 1000$, then there are only
$6$ such values of $n$, $n\in\\{\,130,257,487,528,815,897\,\\}$. It can be
seen that $83^{2}|130^{130}+129^{129}$ and $59^{2}$ divide
$n^{n}+(-1)^{n}(n-1)^{n-1}$ for the remaining values of $n$. They conjectured
that

######  Conjecture 41 (Boyd et al. [14]).

The set of positive integers $n$ such that $n^{n}+(-1)^{n}(n-1)^{n-1}$ is
square-free has density $0.9934466\ldots$, correct to that many decimal
places.

They obtain $0.99344674$ as an upper bound for this density. I.E. Shparlinski
[107] gave a lower bound of the above density.

If the $abc$ conjecture is true, then Anirban Mukhopadhyay et al. [77] showed
that for every odd $n\geq 5$, there is a positive proportion of pairs $(b,c)$
for which the trinomial $x^{n}+bx+c$ is irreducible and has square-free
discriminant. By using square-sieve and bounds of character sums, I.E.
Shparlinski [106] proved a weaker but unconditional version of this result. K.
Kedlaya[59] gave a method to construct a monic irreducible polynomial
$f(x)\in{\mathbb{Z}}[x]$ of degree $n\geq 2$ having exactly $r$ real roots,
$0\leq r\leq n,$ with a square-free discriminant.

On the other hand, discriminants of algebraic number fields have also been
explored by several mathematicians. Suppose $\alpha$ is a root of the
irreducible polynomial $x^{n}+bx+c$ and ${\mathbb{K}}={\mathbb{Q}}(\alpha)$.
K. Komatsu [67] found the discriminant of ${\mathbb{K}}$ as well as an
integral basis for ${\mathbb{K}}$. Later, P. Llorente et al. [73] improved the
results of Komatsu.

B. K. Spearman and K.S. Williams [109] restricted to trinomials
$f(x)=x^{5}+ax+b,$ whose Galois group is a dihedral group of order $10$. If
$\alpha$ is a root of $f$ and ${\mathbb{K}}={\mathbb{Q}}(\alpha)$, then they
give an explicit formula of $\Delta_{{\mathbb{K}}}$ in terms of $a$ and $b$.
A. Alaca and Ş. Alaca [4] utilized a method described by Ş. Alaca [5] and
found a $p$-integral base of ${\mathbb{K}}$ for every prime $p$, which in turn
helps them to provide a formula for $\Delta_{{\mathbb{K}}}$, where
${\mathbb{K}}={\mathbb{Q}}(\alpha),\alpha$ being a root of the irreducible
polynomial $x^{5}+bx+c$.

We have seen that the polynomial $x^{n}-x-1$ is irreducible with discriminant
$n^{n}+(-1)^{n}(n-1)^{n-1}$ and has Galois group $S_{n}$. Boyd et al. [14]
showed that there are certain values of $n$ (see above) for which the
discriminant of $x^{n}-x-1$ has a square factor. J. Lagarias(see problem
99:10, [78]) asked

Is there a monogenic field whose Galois group is $S_{n}$ for every $n\geq 5$?

K. Kedlaya [59] answered this question affirmatively through a different
approach. Boyd et al. [14] answered this by using square freeness of
discriminant. For similar studies in this direction, we refer to [113], [114],
[123], [124].

Galois group of trinomials of different forms has been considered by several
authors. One may consult [9], [19], [20], [21], [25], [26], [50], [57], [83],
[84], [108] for further studies. For example, H. Osada [83] proved that the
Galois group of $x^{n}-x-1$ over ${\mathbb{Q}}$ is $S_{n}$ for all $n\geq 2$.
Later Osada [84] gave some number-theoretic conditions on the coefficients
$b,c$ so that $x^{n}+bx^{m}+c$ has Galois group either $S_{n}$ or $A_{n}$ over
${\mathbb{Q}}$.

In contrast, a class of trinomials has been determined by A. Bishnoi and S.A.
Khanduja [9] having Galois group $S_{n}$. In particular, they showed that if
$n\geq 8,\frac{n}{2}<p<n-2$ and
$|c^{n-p}+(-1)^{n+1}(\frac{n-p}{p})^{n-p}(n!)^{p}b^{n}|$ is not a perfect
square, then $x^{n}+n!p^{n}bx^{p}+n!c$ has Galois group $S_{n},$ $p$ being a
prime number. One such polynomial is $x^{6}+864x^{5}+720$.

The discriminant of trinomials has been explored over finite fields as well.
The discriminant of a trinomial divisible by a square over a finite field has
been entreated in [47]. Suppose ${\mathbb{F}}$ is a finite field of order
$q=p^{n}$ for some prime $p$. Then a polynomial $f(x)\in{\mathbb{F}}[x]$ is
said to be a permutation polynomial if $c\mapsto f(c)$ is a permutation from
${\mathbb{F}}$ to itself. Permutation trinomials has been studied in [8],
[30], [52], [53], [54].

##  5 Open Problems

We have already mentioned several conjectures, open problems, and asked a few
questions in respective places. In this section, we would like to ask few more
questions related to the irreducibility of trinomials and locations of zeros
of trinomials which may give impetus to the reader to work further in this
direction.

Recall that the reducibility nature of
$x^{n}+2{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ has been resolved by Theorem 9. If
$p$ is an odd prime, from Theorem 10, there are only finitely many values of
$\frac{n}{m}$ for which $x^{n}+p{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ is
reducible.

######  Question 42.

Let $p$ be an odd prime, ${\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$ and
$n>m\geq 1$. What are the values of $n$ and $m$ for which
$x^{n}+p{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ is reducible?

From Theorem 32, there are only finitely many trinomials
$x^{n}+b{\epsilon}_{1}x^{m}+c{\epsilon}_{2}$ whose non-cyclotomic part is
reducible. However, it is not known whether there are only finitely many
reducible trinomials of the form $x^{n}+b{\epsilon}_{1}x^{m}+c{\epsilon}_{2}$.
In particular,

######  Question 43.

Let $n>m\geq 1,b\in{\mathbb{N}},$
${\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$. Are there finitely many
reducible trinomials of the form $x^{n}+b{\epsilon}_{1}x^{m}+{\epsilon}_{2}$?

From Theorem 6 and Theorem 7, the trinomial
$x^{n}+{\epsilon}_{1}x^{n-1}+{\epsilon}_{2}$ is reducible only when
$n=6t+2,{\epsilon}_{1}=-{\epsilon}_{2}=-1$ or
$n=6t+5,{\epsilon}_{1}={\epsilon}_{2}=-1,$ $t\in{\mathbb{N}}$. If $b\geq 2$,
then the reducibility of $x^{n}+b{\epsilon}_{1}x^{n-1}+c{\epsilon}_{2},b\neq
c$ has been determined under the certain condition on $c$ in Theorem 15. For
$b=c\geq 2$, Harrington conjectured (recall Conjecture 16) that apart from two
polynomials, $x^{n}+b{\epsilon}_{1}x^{n-1}+b{\epsilon}_{2}$ is irreducible.

######  Question 44.

Let $c\in{\mathbb{N}}$ and $c>1$,
${\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$. What are the values of $c$
for which $x^{n}+{\epsilon}_{1}x^{n-1}+c{\epsilon}_{2}$is irreducible?

More generally, we ask the following question

######  Question 45.

Let $b,c\in{\mathbb{N}},b,c\geq 2,$ and $c>2(b-1),$
${\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$. For what values of $b$ and
$c$, the polynomial $x^{n}+b{\epsilon}_{1}x^{n-1}+c{\epsilon}_{2}$ is
irreducible?

Filaseta et al. [38] gave a sufficient condition for the irreducibility of
$x^{2p}+b{\epsilon}_{1}x^{p}+c{\epsilon}_{2}$. There are only finitely many
values of $p$ for which the irreducibility question remains open.

######  Question 46.

Let $p$ be a prime number, $b,c\in{\mathbb{N}}$,
${\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$ and

$p|(181^{2}-1)\underset{q|b}{\prod\limits_{q\text{prime}}}(q^{2}-1).$

If $b,c$ do not satisfy any of the following conditions

$b^{2}-4c{\epsilon}_{2}=d^{2}\text{ or }c=b^{2}=d^{2p},p\geq 5\text{ or
}c=\frac{b^{2}}{2}=2^{p}d^{2p},p\geq 3\text{ or
}c=\frac{b^{2}}{3}=3^{p}d^{2p},p\geq 5$

for some $d\in{\mathbb{Z}},$ then which are the families among
$x^{2p}+b{\epsilon}_{1}x^{p}+c{\epsilon}_{2}$ irreducible?

In Theorem 26, the irreducibility of
$x^{n}+b{\epsilon}_{1}x^{m}+p{\epsilon}_{2}$ has been discussed except for
$p+1\leq b\leq p^{n-1}+1,p\nmid b$.

######  Question 47.

Let $p$ be a prime number, $b\in{\mathbb{N}}$, and let
$f(x)=x^{n}+b{\epsilon}_{1}x^{m}+p{\epsilon}_{2}.$ For what values of $b$,
$p+1\leq b\leq p^{n-1}+1,p\nmid b$, the polynomial $f(x)$ is irreducible?

The number of roots of $x^{n}+x-1$ which lies inside the unit circle has been
counted in [18] and they conjectured (see Conjecture 39) the exact number of
roots within the unit circle for arbitrary values of $m$.

######  Question 48.

Let $n>m\geq 1$ and ${\epsilon}_{1},{\epsilon}_{2}\in\\{\,-1,+1\,\\}$. How
many roots of $x^{n}+{\epsilon}_{1}x^{m}+{\epsilon}_{2}$ lies in the interior
of the unit circle?

More generally, one can look for polynomials of the form
$f(x)=ax^{n}+bx^{m}+c,$ where $a,b,c$ are non-zero real numbers. From Theorem
38 and related discussion, one can find the number of roots of $f(x)$ that are
inside the unit circle provided $|b|\geq|a|+|c|$.

######  Question 49.

Let $a,b,c\in{\mathbb{R}}\backslash\\{\,0\,\\}$ and $n>m\geq 1.$ How many
roots of $ax^{n}+bx^{m}+c$ lies in the interior of the unit circle?

Acknowledgement: We thank the reviewer for providing valuable suggestions and
comments on an earlier version of the manuscript.

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