# On the Factorization of lacunary polynomials

Michael Filaseta  Mathematics Department
University of South Carolina
Columbia, SC 29208
USA<EMAIL_ADDRESS>
###### Abstract.

This paper addresses the factorization of polynomials of the form
$F(x)=f_{0}(x)+f_{1}(x)x^{n}+\cdots+f_{r-1}(x)x^{(r-1)n}+f_{r}(x)x^{rn}$ where
$r$ is a fixed positive integer and the $f_{j}(x)$ are fixed polynomials in
$\mathbb{Z}[x]$ for $0\leq j\leq r$. We provide an efficient method for
showing that for $n$ sufficiently large and reasonable conditions on the
$f_{j}(x)$, the non-reciprocal part of $F(x)$ is either $1$ or irreducible. We
illustrate the approach including giving two examples that arise from trace
fields of hyperbolic $3$-manifolds.

###### Key words and phrases:

factorization, sparse polynomial, lacunary polynomial, reciprocal polynomial

###### 2020 Mathematics Subject Classification:

Primary 11R09; Secondary 11C08, 12E05, 57M50

Dedicated to the fond and inspirational memory of A. Schinzel.

##  1\. Introduction

The main goal of this paper is to explain how to obtain information about the
factorization of polynomials of the form

(1.1) $F(x)=\sum_{j=0}^{r}f_{j}(x)x^{nj},$

where $r$ is an integer greater than or equal to $1$, the polynomials
$f_{0}(x),\ldots,$ $f_{r}(x)$ are in $\mathbb{Z}[x]$ and $n$ is a sufficiently
large positive integer. In line with the work of A. Schinzel described in the
next paragraph, for a non-zero Laurent polynomial
$F(x)\in\mathbb{Q}[x,x^{-1}]$, we define $\operatorname{J}F$ as $x^{k}F(x)$
where $k\in\mathbb{Z}$ is chosen so that $x^{k}F(x)\in\mathbb{Q}[x]$ and the
constant term of $x^{k}F(x)$ is non-zero. Our particular interest is in
determining general conditions on the polynomials $f_{0}(x),\ldots,f_{r}(x)$
for which the non-reciprocal part of $F(x)$ is either $1$ or irreducible
provided only that $n$ is sufficiently large. The non-reciprocal part of a
polynomial $F(x)\in\mathbb{Z}[x]$ is defined as follows. The reciprocal of a
non-zero polynomial $f(x)\in\mathbb{Q}[x]$ is
$\tilde{f}(x)=\operatorname{J}f(1/x)=x^{\deg f}f(1/x)$. A non-zero polynomial
$f(x)\in\mathbb{Q}[x]$ is _reciprocal_ if $f(x)=\pm\tilde{f}(x)$. The non-
reciprocal part of $f(x)\in\mathbb{Z}[x]$ is $f(x)$ removed of each
irreducible reciprocal factor in $\mathbb{Z}[x]$ with a positive leading
coefficient. The latter is to be interpreted as removing these factors to the
multiplicity to which they appear so that only irreducible non-reciprocal
factors remain. ln Schinzel’s work (cf. [15, 17, 27]), the non-reciprocal part
of $f(x)$ is denoted $\operatorname{L}f(x)$, though we will avoid using this
notation in the remainder of this paper.

This paper is motivated in part by the tremendous amount of work A. Schinzel
produced on the factorization of lacunary (or sparse) polynomials (cf. [13,
14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 32]). In
particular, his early papers on the subject gave a description of how the non-
reciprocal part of a lacunary polynomial $f(x)$ factors, and later he extended
his results to the non-cyclotomic part of a lacunary polynomial $f(x)$ (that
is $f(x)$ removed of all of its cyclotomic polynomial factors) in the case
that $f(x)$ is not reciprocal. The latter was made possible by advancements by
E. Bombieri and U. Zannier described in an appendix by Zannier in [31] (also,
see the work of E. Bombieri , D. W. Masser and U. Zannier [2]).

The paper is also motivated by private communications of the author with S.
Garoufalidis. S. Garoufalidis pointed out the significance of determining the
factorization of polynomials of the form (1.1) to determine the trace field
associated with $\Gamma\subset\text{PSL}_{2}(\mathbb{C})$. In particular, he
pointed out the trace field of the Whitehead link $W_{m/n}$ is
$\mathbb{Q}(x-1/x)$ where $x$ is some root of

$(x(x+1))^{m}x^{4n}-(x-1)^{m}$

(see W. D. Neumann and A. W. Reid [11]); the trace field of the Whitehead link
for the complement of the $(-2,3,3+2n)$ pretzel knot $K_{n}$ is
$\mathbb{Q}(x+1/x)$ where $x$ is some root of

$x^{4n}+\left(-x^{3}+4x^{2}-8+4x^{-2}-x^{-3}\right)x^{2n}+1$

(see M. L. Macasieb and T. W. Mattman [10]); and the trace field under $-a/b$
Dehn fillings of the complement of the $4_{1}$ knot is $\mathbb{Q}(x+1/x)$
where $x$ is some root of

$x^{4b}-x^{2b}-x^{a}-2-x^{-a}-x^{-2b}+x^{-4b}$

(S. Garoufalidis, private communication).

The first of these three Laurent polynomials is easier to approach given
current information in the literature, and S. Garoufalidis and the author [7]
show that if $m$ is a fixed odd positive integer, there is an $N(m)$ such that
if $n\geq N$ is an integer relatively prime to $m$, then the polynomial given
by $(x(x+1))^{m}x^{4n}-(x-1)^{m}$ is $x^{2}+1$ times an irreducible polynomial
(over $\mathbb{Q}$). The other two Laurent polynomials are more difficult to
find literature to exploit. There are two reasons for this. First, if $F(x)$
is one of these two Laurent polynomials, then $\operatorname{J}F(x)$ is
reciprocal. Based on current knowledge, determining the complete factorization
of reciprocal polynomials in a general form is more difficult than the
nonreciprocal case. For example, with regard to Schinzel’s work described
above, he requires a condition that $\operatorname{J}F(x)$ is reciprocal when
discussing the factorization of the non-cyclotomic part of $F(x)$ (see [27,
Theorem 2]). Second, the first of the three Laurent polynomials is the special
case of (1.1) with $r=1$ which has been investigated more fully in the
literature (eg., [6, 8, 12, 29]), whereas the second and third Laurent
polynomials above are cases where $r\geq 2$ in (1.1) for which similar
literature is lacking.

In general, what is reasonable to expect given the current literature on the
subject is that except for exceptional cases, perhaps explicitly laid out, one
can show that $F(x)$ in (1.1) for large $n$ has non-reciprocal part which is
either identically $1$ or irreducible. In the case that $F(x)$ itself is
reciprocal, this would necessarily mean that the non-reciprocal part is $1$
and that each irreducible factor is reciprocal. The latter is of some
significance. Given an arbitrary non-zero polynomial $g(x)\in\mathbb{Q}[x]$,
the polynomial $g(x)\tilde{g}(x)$ is reciprocal. Thus, apriori, given only
that $F(x)$ is reciprocal, we cannot deduce that its irreducible factors are
themselves reciprocal. Knowing the irreducible factors are reciprocal also
provides information about the Galois group of a number field generated by the
roots of an irreducible factor of the polynomial. More precisely, if $f(x)$ is
an irreducible reciprocal polynomial of degree $n>1$, then $n=2m$ for some
integer $m\geq 1$ and the associated Galois group is isomorphic to a subgroup
of all signed permutations on $m$ objects and, hence, has order dividing
$2^{m}m!$ (see [3]).

With the above in mind, our main result is the following.

######  Theorem 1.1.

Let $r\geq 1$ be an integer, and

$F(x,y)=f_{0}(x)+f_{1}(x)y+\cdots+f_{r}(x)y^{r}\in\mathbb{Z}[x,y],$

with

$\gcd{}_{\mathbb{Z}[x]}\big{(}f_{0}(x),f_{1}(x),\ldots,f_{r}(x)\big{)}=1,\quad
f_{0}(0)\neq 0,\quad\text{and}\quad f_{r}(x)\neq 0.$

Let $n$ be sufficiently large. Then the non-reciprocal part of $F(x,x^{n})$ is
not reducible if, whenever we write $n=k\ell+t$, where $k,\ell$ and $t$ are
integers satisfying

(1.2) $k\geq 2\max_{0\leq j\leq r}\\{\deg(f_{j})\\}\quad\text{ and }\quad
0\leq t<\dfrac{k}{r(r+1)}$

the polynomial

$f_{0}(x)+f_{1}(x)x^{t}y^{\ell}+\cdots+f_{r-1}(x)x^{(r-1)t}y^{(r-1)\ell}+f_{r}(x)x^{rt}y^{r\ell}$

is a power of $x$ (possibly $x^{0}$) times an irreducible polynomial in
$\mathbb{Z}[x,y]$, and whenever we write $n=k\ell-t$, where $k,\ell$ and $t$
are integers satisfying (1.2), the polynomial

$f_{0}(x)x^{rt}+f_{1}(x)x^{(r-1)t}y^{\ell}+\cdots+f_{r-1}(x)x^{t}y^{(r-1)\ell}+f_{r}(x)y^{r\ell}$

is a power of $x$ (possibly $x^{0}$) times an irreducible polynomial in
$\mathbb{Z}[x,y]$.

The subscript on the greatest common divisor indicates the it is taken over
the ring $\mathbb{Z}[x]$, so the polynomials
$f_{0}(x),f_{1}(x),\ldots,f_{r}(x)$ are not all divisible by an element of
$\mathbb{Z}[x]$ other than one of the units $\pm 1$. The terminology “not
reducible” instead of “irreducible” is to allow for the possibility that the
non-reciprocal part of $F(x,x^{n})$ is $1$. Note also that the conditions in
the theorem imply $\ell\geq 1$.

For clarification, the work of A. Schinzel beginning with [15] (see also [5])
already gives, in theory, an effective way of determining whether the non-
reciprocal parts of $F(x,x^{n})$ in Theorem 1.1, and even more general
polynomials, are reducible. The goal in this paper is to provide a more
practical approach, a method in particular of efficiently handling polynomials
like the Laurent polynomials described earlier.

In Section 2, we elaborate on a method to show that polynomials in two
variables of the form appearing throughout Theorem 1.1 are irreducible (or a
power of $x$ times an irreducible polynomial). In Section 3, we establish a
preliminary result for Theorem 1.1 along the lines given in [6]. A simpler
variant of Theorem 1.1 and Theorem 1.1 itself will be established in Section
4. In Section 5, we will look at consequences of Theorem 1.1. We provide one
example there involving the non-reciprocal polynomial

$\displaystyle x^{6n}+(x+1)$ $\displaystyle
x^{5n+1}+2x^{4n}+(x^{4}-x^{3}-x^{2}-2x-2)x^{3n-2}$
$\displaystyle\quad+2x^{2n}+(x+1)x^{n-2}+1,$

showing that for $n$ sufficiently large, the polynomial is irreducible. We
then give examples corresponding to the last two Laurent polynomials connected
to trace fields indicated above, and we obtain results showing that, for large
enough exponent variables, the associated polynomials have irreducible factors
that are reciprocal.

Before leaving the introduction, we comment that under the assumption of
Lehmer’s conjecture on the Mahler measure of a non-cyclotomic irreducible
polynomial in $\mathbb{Z}[x]$, each irreducible non-cyclotomic factor of
polynomials of the form (1.1), with
$\gcd(f_{0}(x),f_{1}(x),\ldots,f_{r}(x))=1$, has degree $>cn$, where $c$ is a
positive constant depending only on the polynomials
$f_{0}(x),f_{1}(x),\ldots,f_{r}(x)$ and where $n$ is an arbitrary positive
integer (see [9]). Furthermore, the recent proof of V. Dimitrov [4] on the
Schinzel-Zassenhaus conjecture now implies the same result, but with a
different constant, unconditionally under rather general conditions on the
fixed $f_{j}(x)$. In particular, with $f_{r}(x)$ a power of $x$, as in all the
examples in Section 5, such a result holds unconditionally.

##  2\. Factorization of polynomials in two variables

In this section, we are interested in polynomials of the general form

(2.1) $F(x,y)=\sum_{j=0}^{r}f_{j}(x)y^{j}\in\mathbb{Z}[x,y],\qquad\text{where
}r\geq 1\text{ and }f_{r}(x)\neq 0,$

and determining an irreducibility result for $F(x,y^{n})$ where $n$ is a
positive integer. Suppose $F(x,y)$ is irreducible in $\mathbb{Z}[x,y]$ so
that, in particular, there is no polynomial in $\mathbb{Z}[x]$ other than $\pm
1$ that divides every $f_{j}(x)$. Viewing $F(x,y)$ as a polynomial in $y$, we
let $\alpha$ denote a root of $F(x,y)=0$ (so $F(x,\alpha)=0$) and consider the
field $K=\mathbb{Q}(x)[\alpha]$. An important theorem due to Alfredo Capelli
(see [30]) implies that $F(x,y^{n})$ is reducible in $\mathbb{Z}[x,y]$ if and
only if the polynomial $y^{n}-\alpha$ is reducible in $K[y]$. Equally,
important, Capelli’s work also gives us precise knowledge on when
$y^{n}-\alpha$ factors in $K[y]$. He shows that $y^{n}-\alpha$ factors in
$K[y]$ if and only if at least one of the following holds: (i)  The number $n$
is divisible by a prime $p$ and $\alpha=\beta^{p}$ for some $\beta\in K$. (ii)
The number $n$ is divisible by $4$ and $\alpha=-4\beta^{4}$ for some $\beta\in
K$. Observe that the first result of Capelli’s above implies that in the case
of (i), the polynomial $F(x,y^{p})$ is reducible in $\mathbb{Z}[x,y]$; and in
the case of (ii), the polynomial $F(x,y^{4})$ is reducible in
$\mathbb{Z}[x,y]$. We will determine a small set $\mathcal{P}$ of odd primes
with the property that if $\alpha=\beta^{p}$ for some odd prime $p$ and some
$\beta\in K$, then $p\in\mathcal{P}$. Thus, determining the $n$ for which
$F(x,y^{n})$ is irreducible (or reducible) in $\mathbb{Z}[x,y]$ will be
reduced to checking the factorization of $F(x,y^{p})$ for $p\in\mathcal{P}$
and checking the factorization of $F(x,y^{4})$.

A particularly easy consequence of $\alpha=\beta^{p}$ is that the norm of
$\alpha$ in $K$ over $\mathbb{Q}(x)$ is a $p$th power of an element of
$\mathbb{Q}(x)$. Thus, from (2.1), we see that if $\alpha=\beta^{p}$, then
$(-1)^{r}f_{0}(x)/f_{r}(x)$ is a $p$th power in $\mathbb{Q}(x)$. Hence, an
easy approach to determining a set $\mathcal{P}$ of odd primes $p$ for which
$F(x,y^{p})$ can factor non-trivially is simply to take $\mathcal{P}$ to be
the odd primes $p$ for which the polynomial $(-1)^{r}f_{0}(x)/f_{r}(x)$ is a
$p$th power. However, one case of interest to us is the case where
$(-1)^{r}f_{0}(x)/f_{r}(x)=\pm 1$ which is a $p$th power for every odd prime
$p$. With this in mind, an immediate goal is to determine something more,
perhaps along similar lines, that one can use to narrow the search for primes
$p$ for which $F(x,y^{p})$ factors non-trivially.

We suppose as above that $F(x,y)$ is irreducible in $\mathbb{Z}[x,y]$. We also
suppose that at least one of the $f_{j}(x)$ appearing in (2.1) has degree $>0$
and that the coefficient of $y^{r}$ in $F(x,y)$ is a polynomial in $x$
(possibly of degree $0$) having a positive leading coefficient. Suppose
further that $\alpha=\beta^{p}$ as described in (i) above. Recall that
$\beta\in K=\mathbb{Q}(x)[\alpha]$. On the other hand, $\alpha=\beta^{p}$
implies $\alpha\in\mathbb{Q}(x)[\beta]$. Since the degree of the extension
$\mathbb{Q}(x)[\alpha]$ over $\mathbb{Q}(x)$ is $r$, we deduce that the degree
of the extension $\mathbb{Q}(x)[\beta]$ over $\mathbb{Q}(x)$ is $r$. Hence,
there is an irreducible polynomial in $\mathbb{Z}[x,y]$ of degree $r$ in $y$
having $\beta$ as a root, say

(2.2) $G(x,y)=\sum_{j=0}^{r}g_{j}(x)y^{j}\in\mathbb{Z}[x,y],\qquad\text{where
}g_{r}(x)\neq 0.$

As with $F(x,y)$, we note that there is no polynomial in $\mathbb{Z}[x]$ other
than $\pm 1$ that divides every $g_{j}(x)$, and we may suppose that the
coefficient of $y^{r}$ in $G(x,y)$ is a polynomial in $x$ (possibly of degree
$0$) having a positive leading coefficient.

Let $\alpha_{1},\ldots,\alpha_{r}$ be the conjugates of $\alpha$ including
$\alpha$ so that $F(x,\alpha_{j})=0$ for $1\leq j\leq r$. Since $\beta\in
K=\mathbb{Q}(x)[\alpha]$, we can write $\beta$ as a polynomial in $\alpha$
with coefficients in $\mathbb{Q}(x)$, say
$\beta=H(\alpha)\in\mathbb{Q}(x)[\alpha]$. We denote by $\beta_{j}$ the field
conjugate $H(\alpha_{j})$. Observe that $\alpha=\beta^{p}=H(\alpha)^{p}$.
Thus, each $\alpha_{j}$ is a root of $y=H(y)^{p}$; in other words,
$\alpha_{j}=H(\alpha_{j})^{p}$ and, hence, $\alpha_{j}=\beta_{j}^{p}$, for
$1\leq j\leq r$.

We begin here with a description of the basic “idea” behind our approach
connecting $G(x,y)$ to $F(x,y)$ and leading us to a set of odd primes
$\mathcal{P}$ as indicated above. If we view $G(x,y)$ as a polynomial in
$\mathbb{F}_{p}(x)[y]$, and consider the $r$ roots
$\beta^{\prime}_{1},\ldots,\beta^{\prime}_{r}$ of $G(x,y)$ in an extension
$K^{\prime}$ of $\mathbb{F}_{p}(x)$, then

$G(x,y)^{p}=\sum_{j=0}^{r}g_{j}(x)^{p}y^{pj}=g_{r}(x)^{p}\prod_{j=1}^{r}\big{(}y^{p}-(\beta^{\prime}_{j})^{p}\big{)}.$

Consequently, the polynomial $H(x,y)=\sum_{j=0}^{r}g_{j}(x)^{p}y^{j}$ will
have roots $\big{(}\beta^{\prime}_{1}\big{)}^{p}$,
$\ldots,\big{(}\beta^{\prime}_{r}\big{)}^{p}$ in $K^{\prime}$. Thus, in
$\mathbb{F}_{p}[x,y]$, the polynomials $F(x,y)$ and $H(x,y)$ will differ at
most by a factor in $\mathbb{F}_{p}[x]$. With a little more work, one can see
that $F(x,y)=H(x,y)$ in $\mathbb{F}_{p}[x,y]$. We give details for how one can
establish such a result next. Of some importance to us, what follows works not
just for odd primes $p$ but for $p=2$ as well.

Let

$\sigma_{1}=\sum_{j=1}^{r}\alpha_{j},\ \sigma_{2}=\sum_{1\leq i<j\leq
r}\alpha_{i}\alpha_{j},\ \ldots,\ \sigma_{r}=\prod_{j=1}^{r}\alpha_{j},$

denote the elementary symmetric functions of the roots of $F(x,y)$ viewed as a
polynomial in $y$ over $\mathbb{Q}(x)$. Thus,

$\sigma_{1}=-\dfrac{f_{r-1}(x)}{f_{r}(x)},\
\sigma_{2}=\dfrac{f_{r-2}(x)}{f_{r}(x)},\ \ldots,\
\sigma_{r}=(-1)^{r}\,\dfrac{f_{0}(x)}{f_{r}(x)}.$

Let

$\sigma^{\prime}_{1}=\sum_{j=1}^{r}\beta_{j},\ \sigma^{\prime}_{2}=\sum_{1\leq
i<j\leq r}\beta_{i}\beta_{j},\ \ldots,\
\sigma^{\prime}_{r}=\prod_{j=1}^{r}\beta_{j}.$

For each $j\in\\{1,2,\dots,r\\}$, we have
$\sigma^{\prime}_{j}=(-1)^{j}g_{r-j}(x)/g_{r}(x)\in\mathbb{Q}(x)$.

We show next that there exist $w_{i,j}(x)\in\mathbb{Z}[x]$ satisfying

(2.3) $\sigma_{j}=(\sigma^{\prime}_{j})^{p}+\dfrac{p}{g_{r}(x)^{p}}\sum_{0\leq
i<j}g_{r-i}(x)w_{i,j}(x)\quad\text{for }1\leq j\leq r.$

Before turning to the proof of (2.3), we explain our interest in it. Suppose
then for the moment that we know (2.3) holds. We rewrite
$\sigma^{\prime}_{j}=(-1)^{j}g_{r-j}(x)/g_{r}(x)$ to deduce that

$\sigma_{j}=\dfrac{1}{g_{r}(x)^{p}}\bigg{(}(-1)^{pj}g_{r-j}(x)^{p}+p\sum_{0\leq
i<j}g_{r-i}(x)w_{i,j}(x)\bigg{)}\quad\text{for }1\leq j\leq r,$

where we include the $p$ in the exponent $pj$ of $-1$ so that the identity
also holds for $p=2$. Since the $\sigma_{j}$ represent the elementary
symmetric functions in $\alpha_{1},\dots,\alpha_{r}$, the polynomial

$F_{0}(x,y)=\sum_{j=0}^{r}\bigg{(}(-1)^{(p+1)j}g_{r-j}(x)^{p}+(-1)^{j}p\sum_{0\leq
i<j}g_{r-i}(x)w_{i,j}(x)\bigg{)}y^{r-j}$

has roots $\alpha_{1},\dots,\alpha_{r}$. Let $u(x)$ denote an irreducible
polynomial in $\mathbb{Z}[x]$. We claim that $u(x)$ cannot divide every
coefficient of $F_{0}(x,y)$ as a polynomial in $y$. Assume otherwise. Then
looking at the coefficients of $y^{r}$, $y^{r-1}$, $\ldots$, $y^{0}$ for
$F_{0}(x,y)$ in turn, we deduce that $u(x)$ divides each of $g_{r}(x)$,
$g_{r-1}(x)$, $\ldots$, $g_{0}(x)$. This contradicts that no polynomial in
$\mathbb{Z}[x]$ other than $\pm 1$ divides every $g_{j}(x)$. Hence,
$F_{0}(x,y)$ is a polynomial in $y$ with coefficients having no common factors
in $\mathbb{Z}[x]$ other than $\pm 1$. Since $F(x,y)$ also has this property
and the coefficients of $y^{r}$ in both $F_{0}(x,y)$ and $F(x,y)$ are
polynomials in $x$ having a positive leading coefficient, we obtain

$F(x,y)=F_{0}(x,y)\equiv\sum_{j=0}^{r}g_{j}(x)^{p}y^{j}\pmod{p},$

which as noted above, in our discussion of $H(x,y)$, was our immediate goal.

We turn now to the proof of (2.3). For the proof, we recall that any symmetric
polynomial in $\beta_{1},\dots,\beta_{r}$ with coefficients in $\mathbb{Z}$
can be expressed as a polynomial in
$\sigma^{\prime}_{1},\dots,\sigma^{\prime}_{r}$ with coefficients in
$\mathbb{Z}$. In fact, we make use of a particular argument for obtaining the
polynomial in $\sigma^{\prime}_{1},\dots,\sigma^{\prime}_{r}$. For a symmetric
polynomial
$h(\beta_{1},\dots,\beta_{r})\in{\mathbb{Z}}[\beta_{1},\dots,\beta_{r}]$, set
$T=T_{h}$ to be the set of $r$-tuples $(\ell_{1},\dots,\ell_{r})$ with the
coefficient of $\beta_{1}^{\ell_{1}}\cdots\beta_{r}^{\ell_{r}}$ in
$h(\beta_{1},\dots,\beta_{r})$ non-zero. We define the size of $h$ to be
$(k_{1},\dots,k_{r})$ where $(k_{1},\dots,k_{r})$ is the element of $T$ with
$k_{1}$ as large as possible, $k_{2}$ as large as possible given $k_{1}$, etc.
Since $h(\beta_{1},\dots,\beta_{r})$ is symmetric, it follows that
$(\ell_{1},\dots,\ell_{r})\in T$ if and only if each permutation of
$(\ell_{1},\dots,\ell_{r})$ is in $T$. This implies that $k_{1}\geq
k_{2}\geq\cdots\geq k_{r}$. Observe that we can use the notion of size to form
an ordering on the symmetric polynomials in
${\mathbb{Z}}[\beta_{1},\dots,\beta_{r}]$ in the sense that if $h_{1}$ has
size $(k_{1},\dots,k_{r})$ and $h_{2}$ has size
$(k^{\prime}_{1},\dots,k^{\prime}_{r})$, then $h_{1}>h_{2}$ if there is an
$i\in\\{0,1,\dots,r-1\\}$ such that
$k_{1}=k^{\prime}_{1},\dots,k_{i}=k^{\prime}_{i}$, and
$k_{i+1}>k^{\prime}_{i+1}$. Note that the elements of
${\mathbb{Z}}[\beta_{1},\dots,\beta_{r}]$ which have size $(0,0,\dots,0)$ are
precisely the constants (the elements of ${\mathbb{Z}}$).

Since $\alpha_{j}=\beta_{j}^{p}$ for $1\leq j\leq r$, we may view $\sigma_{j}$
as a symmetric polynomial in $\beta_{1},\dots,\beta_{r}$ of size
$(p,\dots,p,0,\dots,0)$, where the number of leading $p$’s is $j$. In general,
we can take a given symmetric polynomial in
${\mathbb{Z}}[\beta_{1},\dots,\beta_{r}]$ of size $(k_{1},\dots,k_{r})$ and
write it in the form

(2.4)
$m\cdot\big{(}\sigma^{\prime}_{1}\big{)}^{k_{1}-k_{2}}\big{(}\sigma^{\prime}_{2}\big{)}^{k_{2}-k_{3}}\cdots\big{(}\sigma^{\prime}_{r-1}\big{)}^{k_{r-1}-k_{r}}\big{(}\sigma^{\prime}_{r}\big{)}^{k_{r}}+s(\beta_{1},\ldots,\beta_{r}),$

where $m\in\mathbb{Z}$ and $s(\beta_{1},\ldots,\beta_{r})$ is a symmetric
polynomial of size less than $(k_{1},\dots,k_{r})$. Here, $m$ is chosen so
that when the expression above is expanded the term corresponding to
$m\beta_{1}^{k_{1}}\beta_{2}^{k_{2}}\cdots\beta_{r}^{k_{r}}$ agrees with the
corresponding term in the given symmetric polynomial. We proceed to do this
for $\sigma_{j}$ which again is of size $(p,\dots,p,0,\dots,0)$. Thus,

$\sigma_{j}=\big{(}\sigma^{\prime}_{j}\big{)}^{p}+s_{1}(\beta_{1},\ldots,\beta_{r}),$

for some symmetric polynomial in ${\mathbb{Z}}[\beta_{1},\dots,\beta_{r}]$. Of
some significance here is the multinomial theorem applied to
$\big{(}\sigma^{\prime}_{j}\big{)}^{p}$ assures us that each coefficient of

$s_{1}(\beta_{1},\ldots,\beta_{r})=\sigma_{j}-\big{(}\sigma^{\prime}_{j}\big{)}^{p}\in{\mathbb{Z}}[\beta_{1},\dots,\beta_{r}]$

is divisible by $p$. Thus, we may write

$s_{1}(\beta_{1},\ldots,\beta_{r})=p\cdot s_{2}(\beta_{1},\ldots,\beta_{r}),$

for some symmetric
$s_{2}=s_{2}(\beta_{1},\ldots,\beta_{r})\in{\mathbb{Z}}[\beta_{1},\dots,\beta_{r}]$.
As the size of $s_{2}$ is necessarily less than the size
$(p,\dots,p,0,\dots,0)$ of $\sigma_{j}$, we deduce that the size of $s_{2}$ is
of the form $(k_{1},\dots,k_{r})$ where $k_{j}<p$. Let $i\in\\{1,2,\dots,j\\}$
be minimal such that $k_{i}<p$.

Suppose first that $i>1$. Then we rewrite $s_{2}(\beta_{1},\ldots,\beta_{r})$
using (2.4) as

$m_{2}\big{(}\sigma^{\prime}_{1}\big{)}^{k_{1}-k_{2}}\big{(}\sigma^{\prime}_{2}\big{)}^{k_{2}-k_{3}}\cdots\big{(}\sigma^{\prime}_{r-1}\big{)}^{k_{r-1}-k_{r}}\big{(}\sigma^{\prime}_{r}\big{)}^{k_{r}}+s_{3}(\beta_{1},\ldots,\beta_{r}),$

where $m_{2}\in\mathbb{Z}$ and where $s_{3}=s_{3}(\beta_{1},\ldots,\beta_{r})$
is a symmetric polynomial in ${\mathbb{Z}}[\beta_{1},\dots,\beta_{r}]$. Since
$k_{1}=p$, we note that the first term above has total degree $k_{1}\leq p$ in
the variables
$\sigma^{\prime}_{1},\sigma^{\prime}_{2},\ldots,\sigma^{\prime}_{r}$. It will
be of some significance that similarly the total degree in each polynomial in
$\sigma^{\prime}_{1},\sigma^{\prime}_{2},\ldots,\sigma^{\prime}_{r}$ below is
$\leq p$ as we proceed. Observe further that $k_{i-1}=p$ and $k_{i}<p$ implies
that $k_{i-1}-k_{i}$ is positive. Thus, in the first of the two terms above,
$\sigma^{\prime}_{i-1}$ appears as a factor. Note also that the size of
$s_{3}$ is less than the size of $s_{2}$ and, in particular, still less than
$(p,\dots,p,0,\dots,0)$. Thus, if the size still has a first component of $p$,
then we can repeat this process, rewriting $s_{3}$ as a multiple of at least
one of $\sigma^{\prime}_{t}$ for $1\leq t\leq i-1$ plus a symmetric polynomial
$s_{4}$ of smaller size, and then rewriting $s_{4}$ as well if the first
component of the size remains $p$, and so on. As the sizes are decreasing, at
some point the summand involving a new symmetric polynomial will have size
$(k^{\prime}_{1},\dots,k^{\prime}_{r})$ with $k^{\prime}_{1}<p$.

Combining terms arising from above, we see that

$\sigma_{j}=\big{(}\sigma^{\prime}_{j}\big{)}^{p}+p\sum_{t=1}^{i-1}\sigma^{\prime}_{t}s^{\prime}_{t}(\sigma^{\prime}_{1},\ldots,\sigma^{\prime}_{r})+p\,s_{0}(\beta_{1},\ldots,\beta_{r}),$

where each
$s^{\prime}_{t}=s^{\prime}_{t}(\sigma^{\prime}_{1},\ldots,\sigma^{\prime}_{r})$
is a polynomial in
${\mathbb{Z}}[\sigma^{\prime}_{1},\dots,\sigma^{\prime}_{r}]$ and
$s_{0}=s_{0}(\beta_{1},\ldots,\beta_{r})$ is a symmetric polynomial in
${\mathbb{Z}}[\beta_{1},\dots,\beta_{r}]$ of size
$(k^{\prime}_{1},\dots,k^{\prime}_{r})$ with $k^{\prime}_{1}<p$. Note that,
for $1\leq t\leq i-1$, the total degree of $\sigma^{\prime}_{t}s^{\prime}_{t}$
in the variables $\sigma^{\prime}_{1},\dots,\sigma^{\prime}_{r}$ is $\leq p$,
so consequently $s^{\prime}_{t}$ has total degree $<p$. At this point, we
simply repeat the process of using and reusing (2.4) to rewrite $s_{0}$ as a
polynomial in ${\mathbb{Z}}[\sigma^{\prime}_{1},\dots,\sigma^{\prime}_{r}]$.
The significance of having $k^{\prime}_{1}<p$ is that at each stage from this
point on, the size of the symmetric polynomials introduced will have size with
first component $<p$. Observe that if $k_{1}<p$ in the first term

$m\big{(}\sigma^{\prime}_{1}\big{)}^{k_{1}-k_{2}}\big{(}\sigma^{\prime}_{2}\big{)}^{k_{2}-k_{3}}\cdots\big{(}\sigma^{\prime}_{r-1}\big{)}^{k_{r-1}-k_{r}}\big{(}\sigma^{\prime}_{r}\big{)}^{k_{r}}$

of (2.4), then the total degree of this term as a polynomial in
$\sigma^{\prime}_{1},\dots,\sigma^{\prime}_{r}$ is $k_{1}<p$. In other words,
we can repeatedly use (2.4) to rewrite $s_{0}$ as a polynomial
$s^{\prime}_{0}=s^{\prime}_{0}(\sigma^{\prime}_{1},\ldots,\sigma^{\prime}_{r})$
in the variables $\sigma^{\prime}_{1},\dots,\sigma^{\prime}_{r}$ with total
degree $<p$.

Now, we have

(2.5)
$\sigma_{j}=\big{(}\sigma^{\prime}_{j}\big{)}^{p}+p\sum_{t=1}^{i-1}\sigma^{\prime}_{t}s^{\prime}_{t}(\sigma^{\prime}_{1},\ldots,\sigma^{\prime}_{r})+p\,s^{\prime}_{0}(\sigma^{\prime}_{1},\ldots,\sigma^{\prime}_{r}),$

where each $s^{\prime}_{t}$ for $0\leq t\leq i-1$ is a polynomial in
${\mathbb{Z}}[\sigma^{\prime}_{1},\dots,\sigma^{\prime}_{r}]$ with total
degree $<p$ in the variables $\sigma^{\prime}_{1},\dots,\sigma^{\prime}_{r}$.
We make the substitution $\sigma^{\prime}_{t}=(-1)^{t}g_{r-t}(x)/g_{r}(x)$ in
(2.5) to obtain

$\sigma_{j}=\big{(}\sigma^{\prime}_{j}\big{)}^{p}+p\sum_{t=1}^{i-1}\dfrac{g_{r-t}(x)w_{t,j}(x)}{g_{r}(x)^{p}}+\dfrac{p\,w_{0,j}(x)}{g_{r}(x)^{p-1}},$

where each $w_{t,j}(x)$ for $0\leq t\leq i-1$ is in $\mathbb{Z}[x]$. Since
$i\leq j$, we deduce that (2.3) holds. As noted earlier, this leads to

(2.6) $F(x,y)\equiv\sum_{j=0}^{r}g_{j}(x)^{p}y^{j}\pmod{p}.$

The above follows from the condition $\alpha=\beta^{p}$ in (i) and holds for
any prime $p$ satisfying this condition.

Recall that we have required that at least one of the $f_{j}(x)$ appearing in
(2.1) has degree $>0$. We also took the coefficient of $y^{r}$ in $F(x,y)$ to
be a polynomial in $x$ (possibly of degree $0$) having a positive leading
coefficient. This latter condition can be removed at this point since what we
are interested in is the information from (2.6) that each coefficient of
$F(x,y)$ is a $p$th power modulo $p$, and if necessary we can multiply both
sides of (2.6) by $(-1)^{p}$ to obtain a result in the case that the latter
condition does not hold. We define the height of $f_{j}(x)$ to be the maximum
of the absolute values of the coefficients of $f_{j}(x)$. Suppose $p$ is an
odd prime that is greater than the height $H(f_{j})$ for each $f_{j}(x)$. This
is more than we will need and only really require here that $f_{j}(x)$ is not
constant modulo $p$ for some $j$. Thus, for some $j$, the coefficient
$f_{j}(x)$ of $y^{j}$ on the left-hand side of (2.6) viewed modulo $p$ is of
degree $>0$. It follows that the corresponding $g_{j}(x)$ is a non-constant
polynomial modulo $p$ on the right-hand side of (2.6). In this case, the
polynomial $g_{j}(x)^{p}$ modulo $p$ must be of degree at least $p$ which in
turn implies that $\deg f_{j}\geq p$. Thus, we may take $\mathcal{P}$ to be
the set of odd primes $p$ for which

(2.7) $p\leq\max\\{\deg f_{0},\dots,\deg f_{r},H(f_{0}),\dots,H(f_{r})\\}.$

In other words, (2.6) implies (2.7). In particular, with $\mathcal{P}$ so
chosen, if $F(x,y^{p})$ for $p\in\mathcal{P}$ and $F(x,y^{4})$ are all
irreducible, then $F(x,y^{n})$ is irreducible for every positive integer $n$.
Recalling the opening paragraph of this section, we summarize the above as
follows.

######  Theorem 2.1.

Let $F(x,y)$ be given as in (2.1), with $F(x,y)$ irreducible in
$\mathbb{Z}[x,y]$ and with at least one $f_{j}(x)$ non-constant. Let
$\mathcal{P}$ be the set of odd primes $p$ satisfying (2.7). Then $F(x,y^{n})$
is reducible in $\mathbb{Z}[x,y]$ if and only if either $p|n$ and $F(x,y^{p})$
is reducible for some $p\in\mathcal{P}$, or $2|n$ and $F(x,y^{2})$ is
reducible, or $4|n$ and $F(x,y^{4})$ is reducible. In particular, if
$F(x,y^{p})$ for $p\in\mathcal{P}$ and $F(x,y^{4})$ are all irreducible, then
$F(x,y^{n})$ is irreducible in $\mathbb{Z}[x,y]$ for every positive integer
$n$. Furthermore, if $p\in\mathcal{P}\cup\\{2\\}$ and $F(x,y^{p})$ is
reducible, then $F(x,y)$ is necessarily of the form (2.6).

There is some value in looking closer at the case that $\alpha=-4\beta^{4}$ as
in (ii), which we briefly sketch here. We suppose now that
$\alpha=-4\beta^{4}$ and $G(x,y)$ in (2.2) has $\beta$ as a root, where again
$G(x,y)$ is an irreducible polynomial in $\mathbb{Z}[x,y]$ and $g_{r}(x)$ has
a positive leading coefficient. For $1\leq j\leq r$, we define $\sigma_{j}$
and $\sigma^{\prime}_{j}$ as before. Instead of (2.3), we have in this case

(2.8)
$\sigma_{j}=(-4)^{j}(\sigma^{\prime}_{j})^{4}+\dfrac{2^{2j+1}}{g_{r}(x)^{4}}\sum_{0\leq
i<j}g_{r-i}(x)w_{i,j}(x)\quad\text{for }1\leq j\leq r,$

for some polynomials $w_{i,j}(x)\in\mathbb{Z}[x]$. The proof of (2.8) is
analogous to the proof of (2.3). The implications are a bit different, and we
discuss that next.

We will want to take advantage of a little more information on the
$w_{i,j}(x)$. Observe that

$\sigma_{r}=\alpha_{1}\cdots\alpha_{r}=(-4)^{r}\big{(}\beta_{1}^{4}\cdots\beta_{r}^{4}\big{)}=(-4)^{r}(\sigma^{\prime}_{r})^{4}.$

Thus, in regards to (2.8), we can take $w_{i,r}(x)=0$ for $0\leq i<r$.

Recalling $\sigma^{\prime}_{j}=(-1)^{j}g_{r-j}(x)/g_{r}(x)$, we deduce

$F_{1}(x,y)=\sum_{j=0}^{r}\bigg{(}2^{2j}g_{r-j}(x)^{4}+(-1)^{j}2^{2j+1}\sum_{0\leq
i<j}g_{r-i}(x)w_{i,j}(x)\bigg{)}y^{r-j}$

has roots $\alpha_{1},\ldots,\alpha_{r}$. As with $F_{0}(x,y)$, we would like
to determine whether the coefficients of $F_{1}(x,y)$ viewed as a polynomial
in $y$ have any common factors. Analogous to $F_{0}(x,y)$, we see that every
coefficient of $F_{1}(x,y)$ cannot be divisible by an irreducible polynomial
$u(x)$ in $\mathbb{Z}[x]$ if either $u(x)$ is non-constant or $u(x)$ is an odd
prime. In other words, the only irreducible polynomials in $\mathbb{Z}[x]$
that might divide every coefficient of $F_{1}(x,y)$ are $\pm 2$. Suppose
$2^{s}$ divides every coefficient of $F_{1}(x,y)$ for some positive integer
$s$ with $s$ maximal. It follows that $F_{1}(x,y)/2^{s}$ is an irreducible
polynomial in $\mathbb{Z}[x,y]$ having roots $\alpha_{1},\ldots,\alpha_{r}$.
Since $g_{r}(x)$ and $f_{r}(x)$ have a positive leading coefficient, we
necessarily have

$\displaystyle F(x,y)$ $\displaystyle=\dfrac{F_{1}(x,y)}{2^{s}}$
$\displaystyle=\dfrac{1}{2^{s}}\sum_{j=0}^{r}\bigg{(}2^{2j}g_{r-j}(x)^{4}+(-1)^{j}2^{2j+1}\sum_{0\leq
i<j}g_{r-i}(x)w_{i,j}(x)\bigg{)}y^{r-j}.$

Let $e_{j}$ be the non-negative integer for which $2^{e_{j}}$ is the largest
power of $2$ dividing the coefficient of $y^{j}$ in $F_{1}(x,y)$ in
$\mathbb{Z}[x]$. Let $e^{\prime}_{j}$ be the non-negative integer for which
$2^{e^{\prime}_{j}}$ is the largest power of $2$ dividing $g_{j}(x)$ in
$\mathbb{Z}[x]$. In particular, $e_{r}=4e^{\prime}_{r}$. From the formulation
of $F_{1}(x,y)$ above, we see that $e_{j}\geq 2r-2j$ for each
$j\in\\{0,1,\ldots,r\\}$ and if $e^{\prime}_{j}=0$, then $e_{j}=2r-2j$. Also,
the irreducibility of $G(x,y)$ guarantees that some $e^{\prime}_{i}=0$. Given
such an $i$, we deduce $s\leq 2r-2i$, and $2^{2(i-j)}$ divides the coefficient
of $y^{j}$ in $F(x,y)$ for $0\leq j<i$. In particular, if the maximal such $i$
is $\geq 1$, then $f_{0}(x)$ is divisible by $4$. Suppose the only such $i$ is
$i=0$. Then either $s<2r$ and $f_{0}(x)$ is divisible by $2$ or $s=2r$. We
consider the case that $s=2r$. From $s=2r$, $e^{\prime}_{0}=0$ and
$w_{i,r}(x)=0$ for $0\leq i<r$, we have $f_{0}(x)=g_{0}(x)^{4}$. If $r$ is
odd, then $s\equiv 2\pmod{4}$. As $e_{r}=4e^{\prime}_{r}\equiv 0\pmod{4}$, we
see that $f_{r}(x)$ is divisible by $4$ in this case.

Thus far we have seen that either $f_{0}(x)$ is divisible by $2$ or $f_{r}(x)$
is divisible by $4$ unless $r$ is even, $s=2r$, $e^{\prime}_{0}=0$, and
$e^{\prime}_{j}\geq 1$ for $1\leq j\leq r$. Furthermore, in this case, $s=2r$
implies that $2^{2r}$ divides each coefficient of $y^{j}$ in $F_{1}(x,y)$ and,
in particular, $2^{2r}$ divides $g_{r}(x)^{4}$. Thus, $e^{\prime}_{r}\geq
r/2$. We will be interested in the particular case that $r=4$, so suppose
$f_{0}(x)$ is not divisible by $2$ and $f_{r}(x)$ is not divisible by $4$,
giving us the above conclusions with $r=4$. We make use of the explicit
formula in this case that

$\displaystyle\sigma_{1}$ $\displaystyle=\sum_{j=1}^{4}(-4\beta_{j}^{4})$
$\displaystyle=-4\big{(}(\sigma^{\prime}_{1})^{4}-4(\sigma^{\prime}_{1})^{2}(\sigma^{\prime}_{2})+4(\sigma^{\prime}_{1})(\sigma^{\prime}_{3})+2(\sigma^{\prime}_{2})^{2}-4(\sigma^{\prime}_{4})\big{)},$

which leads to the more precise formulation of the polynomial

$\displaystyle
4g_{3}(x)^{4}-16g_{3}(x)^{2}g_{2}(x)g_{4}(x)+16g_{3}(x)g_{1}(x)g_{4}(x)^{2}$
$\displaystyle+8g_{2}(x)^{2}g_{4}(x)^{2}-16g_{0}(x)g_{4}(x)^{3}$

for the coefficient of $y^{r-1}=y^{3}$ in $F_{1}(x,y)$. This polynomial is
required to be divisible by $2^{2r}=2^{8}$ in $\mathbb{Z}[x]$. As
$e^{\prime}_{4}\geq 4/2=2$ and $e^{\prime}_{j}\geq 1$ for $1\leq j\leq 3$, we
see that each term beyond the first term is divisible by $2^{9}$ in
$\mathbb{Z}[x]$, so the first term $4g_{3}(x)^{4}$ must be divisible by
$2^{8}$. It follows that $e^{\prime}_{3}\geq 2$. We deduce that the
coefficient of $y^{3}$ in $F_{1}(x,y)$ is divisible by $2^{9}$ so that the
coefficient of $y^{3}$ in $F(x,y)$ is divisible by $2$. Hence, we have that in
the case that $\alpha=-4\beta^{4}$ as in (ii) and $r=4$, then either
$f_{0}(x)$ or $f_{3}(x)$ is divisible by $2$ or $f_{4}(x)$ is divisible by
$4$.

To summarize, in the case $r=4$, we have shown that if $F(x,y^{4})$ is
reducible, then either $F(x,y^{2})$ is reducible and (2.6) holds or one of
$f_{0}(x)$ or $f_{3}(x)$ is divisible by $2$ or $f_{4}(x)$ is divisible by
$4$. For general odd $r$, we have also seen that if $F(x,y^{4})$ is reducible,
then either $F(x,y^{2})$ is reducible and (2.6) holds or $f_{0}(x)$ is
divisible by $2$ or $f_{4}(x)$ is divisible by $4$.

We end this section with an example that will play a role later in our paper.
Example. Let

$F(x,y)=1-x(x+1)y-2x^{2}y^{2}-x^{2}(x+1)y^{3}+x^{4}y^{4}$

so that $F(x,y)$ is of the form (2.1) with $f_{0}(x)=1$, $f_{1}(x)=-x(x+1)$,
$f_{2}(x)=-2x^{2}$, $f_{3}(x)=-x^{2}(x+1)$ and $f_{4}(x)=x^{4}$. One checks
directly that $F(x,y)$ and $F(x,y^{4})$ are irreducible. Since
$f_{1}(x)=-x(x+1)$, we see that for every odd prime $p$, the congruence in
(2.6) does not hold. Theorem 2.1 now implies that $F(x,y^{n})$ is irreducible
for every positive integer $n$.

##  3\. A theorem connecting single and multivariate polynomials

To help with the statements of the results below and the proofs to follow, we
discuss notation here. The expression $a\operatorname{\>mod}k$ will denote the
unique integer $b$ in $[0,k)$ for which $a\equiv b\pmod{k}$. If $u$ is a real
number, $\lfloor u\rfloor$ will denote the greatest integer $\leq u$, $\lceil
u\rceil$ the least integer $\geq u$, and $\|u\|$ the minimal distance from $u$
to an integer. We will use $\\{u\\}$ to denote $u-\lfloor u\rfloor$ unless it
is clear from the context that $\\{u\\}$ refers to a set consisting of the
single element $u$. In the case that $f(x)$ is a polynomial, $\|f\|$
represents the $2$-norm defined as the squareroot of the sum of the squares of
the coefficients of $f(x)$. As before, the reciprocal of the polynomial $f(x)$
is $\tilde{f}(x)=x^{\deg f}f(1/x)$, and the non-reciprocal part of
$f(x)\in\mathbb{Z}[x]$ is $f(x)$ removed of each irreducible reciprocal factor
in $\mathbb{Z}[x]$ with a positive leading coefficient.

######  Theorem 3.1.

Let $F(x)=\sum_{j=0}^{r}a_{j}x^{d_{j}}\in\mathbb{Z}[x]$, where
$0=d_{0}<d_{1}<\cdots<d_{r}$ and $a_{r}a_{0}\neq 0$. Let
$\varepsilon\in(0,1/4]$, and let $k_{0}$ be a real number $\geq 2$. Set
$\kappa=\lceil 1/\varepsilon\rceil$. Suppose that

$\deg
F\geq\max\bigg{\\{}\dfrac{\kappa^{2N-2}}{2}\bigg{(}\varepsilon-\dfrac{1}{\kappa}\bigg{)}^{-1},k_{0}\big{(}\kappa^{N-1}+\varepsilon\big{)}\bigg{\\}},\
\text{ where }N=2\,\|F\|^{2}+2r-5.$

If the non-reciprocal part of $F(x)$ is reducible in $\mathbb{Z}[x]$, then
there exists an integer $k$ in the interval $[k_{0},(\deg F)/(1-\varepsilon))$
satisfying:

  * (i)

For each $j\in\\{0,1,\dots,r\\}$, the number $d_{j}\operatorname{\>mod}k$ is
in

$[0,\varepsilon k)\cup((1-\varepsilon)k,k).$

  * (ii)

If $\overline{d}_{j}$ and $\ell_{j}$ are defined by

$\overline{d}_{j}=(d_{j}+\lfloor\varepsilon
k\rfloor)\operatorname{\>mod}k\quad\text{and}\quad d_{j}+\lfloor\varepsilon
k\rfloor=k\ell_{j}+\overline{d}_{j}$

and $G(x,y)=\sum_{j=0}^{r}a_{j}x^{\overline{d}_{j}}y^{\ell_{j}}$, then
$x^{-M}G(x,y)$ is reducible in $\mathbb{Z}[x,y]$, where $M$ is a non-negative
integer chosen as large as possible with the constraint that
$x^{-M}G(x,y)\in\mathbb{Z}[x,y]$.

The above result is very similar to Theorem 2 in [6] and follows along very
similar lines. As the result nevertheless is different, we present the details
below.

Before beginning the proof, we observe that $k<d_{r}/(1-\varepsilon)$ implies

$k-1-\lfloor\varepsilon k\rfloor<k(1-\varepsilon)<d_{r}.$

Hence, $d_{r}+\lfloor\varepsilon k\rfloor\geq k$ so that $\ell_{r}\geq 1$.
Thus, $G(x,y)$ depends on $y$.

To establish Theorem 3.1, we make use of a seemingly unconnected combinatorial
result of interest in itself. Fix $\varepsilon>0$. Suppose that
$v_{1},v_{2},\dots,v_{\rho}$ are distinct non-negative integers written in
increasing order and that we wish to determine an integer $k\geq 2$ satisfying
the property that for each $j\in\\{1,2,\dots,\rho\\}$, we have $v_{j}\equiv
v^{\prime}_{j}\pmod{k}$ for some $v^{\prime}_{j}\in(-\varepsilon k,\varepsilon
k)$. In [6], the case $\varepsilon=1/4$ was of particular importance and
focused on there. Observe that the value $k=\lfloor
v_{\rho}/\varepsilon\rfloor+1$ satisfies the property stated above for $k$,
but we will want a smaller value of $k$. We show that for each $\rho$, there
exists a $V(\rho)=V(\rho,\varepsilon)$ such that if $v_{\rho}\geq V(\rho)$,
then there is a $k\in[2,v_{\rho}/(1-\varepsilon))$ such that for each
$j\in\\{1,2,\dots,\rho\\}$, we have $v_{j}\equiv v^{\prime}_{j}\pmod{k}$ for
some $v^{\prime}_{j}\in(-\varepsilon k,\varepsilon k)$.

######  Lemma 3.2.

Let $\varepsilon\in(0,1)$, let $\rho$ be a positive integer, and let $k_{0}$
be a real number $\geq 2$. Set $\kappa=\lfloor 1/\varepsilon\rfloor+1$ and

$V(\rho)=\max\bigg{\\{}\dfrac{\kappa^{2\rho-2}}{2}\bigg{(}\varepsilon-\dfrac{1}{\kappa}\bigg{)}^{-1},k_{0}\big{(}\kappa^{\rho-1}+\varepsilon\big{)}\bigg{\\}}.$

Let $v_{1},v_{2},\dots,v_{\rho}$ be non-negative integers satisfying
$v_{1}<v_{2}<\cdots<v_{\rho}$ and $v_{\rho}\geq V(\rho)$. Then there exists an
integer $k\in[k_{0},v_{\rho}/(1-\varepsilon))$ such that
$v_{j}\operatorname{\>mod}k$ is in $[0,\varepsilon k)\cup((1-\varepsilon)k,k)$
for each $j\in\\{1,2,\dots,\rho\\}$.

###### Proof.

For $\rho=1$, the conditions imply $v_{\rho}\geq k_{0}$ and we can take
$k=v_{\rho}$. We consider now the case that $\rho>1$. Let $D$ be an integer
with

$1\leq
D\leq\dfrac{\sqrt{2v_{\rho}}}{\kappa^{\rho-1}}\,\bigg{(}\varepsilon-\dfrac{1}{\kappa}\bigg{)}^{1/2}.$

The condition $v_{\rho}\geq V(\rho)$ implies that such an integer $D$ exists.
We will show that one may take

(3.1)
$k\in\bigg{(}\dfrac{v_{\rho}}{\kappa^{\rho-1}D+\varepsilon},\dfrac{v_{\rho}}{D-\varepsilon}\bigg{)}.$

In particular, the lemma will follow by choosing $D=1$. Note that the
conditions in the lemma, $D=1$ and $\rho>1$ imply that the interval in (3.1)
has length $>1$, so such a $k$ exists.

Let $x_{j}=v_{j}/v_{\rho}$ for $j\in\\{1,2,\dots,\rho\\}$. We want to show
that there is an integer $k$ satisfying (3.1) and integers
$w_{1},w_{2},\dots,w_{\rho}$ such that

(3.2) $|v_{j}-w_{j}k|<\varepsilon k\qquad\text{ for }1\leq j\leq\rho.$

We explain next how the Dirichlet box principle implies that there is an
integer $d$ satisfying $D\leq d\leq\kappa^{\rho-1}D$, $D|d$ and

(3.3) $\|dx_{j}\|\leq\dfrac{1}{\kappa}\qquad\text{ for }1\leq j\leq\rho-1.$

For each $d^{\prime}\in\\{D,2D,3D,\dots,(\kappa^{\rho-1}+1)D\\}$, we define
the point

$P(d^{\prime})=\big{(}\\{d^{\prime}x_{1}\\},\\{d^{\prime}x_{2}\\},\dots,\\{d^{\prime}x_{\rho-1}\\}\big{)}.$

Each such point has its coordinates in the interval $[0,1)$. For each of the
$\kappa^{\rho-1}$ choices of $u_{j}\in\\{0,1,\ldots,\kappa-1\\}$, where $1\leq
j\leq\rho-1$, we consider the cube

$\displaystyle\mathcal{C}(u_{1},\dots,u_{\rho-1})=\\{(t_{1},t_{2},\dots,t_{\rho-1}):u_{j}/\kappa\leq
t_{j}<$ $\displaystyle(u_{j}+1)/\kappa$ $\displaystyle\text{ for }1\leq
j\leq\rho-1\\}.$

The Dirichlet box principle implies that there are $d_{1}^{\prime}$ and
$d_{2}^{\prime}$ in

$\\{D,2D,\ldots,(\kappa^{\rho-1}+1)D\\}$

with $d_{1}^{\prime}>d_{2}^{\prime}$ such that the points $P(d_{1}^{\prime})$
and $P(d_{2}^{\prime})$ are both in the same cube $C(u_{1},\dots,u_{\rho-1})$.
Set

$d=d_{1}^{\prime}-d_{2}^{\prime}\in\\{D,2D,\ldots,\kappa^{\rho-1}D\\}.$

For every $j\in\\{1,2,\dots,\rho-1\\}$, we obtain

$dx_{j}=d_{1}^{\prime}x_{j}-d_{2}^{\prime}x_{j}=\lfloor
d_{1}^{\prime}x_{j}\rfloor-\lfloor
d_{2}^{\prime}x_{j}\rfloor+\\{d_{1}^{\prime}x_{j}\\}-\\{d_{2}^{\prime}x_{j}\\}.$

It follows that each such $dx_{j}$ is within

$|\\{d_{1}^{\prime}x_{j}\\}-\\{d_{2}^{\prime}x_{j}\\}|\leq\dfrac{1}{\kappa}$

of the integer $\lfloor d_{1}^{\prime}x_{j}\rfloor-\lfloor
d_{2}^{\prime}x_{j}\rfloor$. This establishes (3.3). Clearly, (3.3) holds with
$j=\rho$ as well. Observe that the upper bound on $D$ above implies that

$d\leq\sqrt{2v_{\rho}}\,\bigg{(}\varepsilon-\dfrac{1}{\kappa}\bigg{)}^{1/2}.$

For $1\leq j\leq\rho$, take $w_{j}$ to be the nearest integer to $dx_{j}$.

For the moment, suppose $w_{j}\neq 0$ (so that $w_{j}\geq 1$) for each
$j\in\\{1,2,\dots,\rho\\}$. Then (3.2) follows provided

$\dfrac{k}{v_{\rho}}\in\bigcap_{1\leq
j\leq\rho}\bigg{(}\dfrac{x_{j}}{w_{j}+\varepsilon},\dfrac{x_{j}}{w_{j}-\varepsilon}\bigg{)}.$

For each $j\in\\{1,2,\dots,\rho\\}$, since $w_{j}\leq d$, we deduce from (3.3)
that

$\dfrac{d+(1/\kappa)}{d+\varepsilon}\geq\dfrac{w_{j}+(1/\kappa)}{w_{j}+\varepsilon}\geq\dfrac{dx_{j}}{w_{j}+\varepsilon}$

and

$\dfrac{d-(1/\kappa)}{d-\varepsilon}\leq\dfrac{w_{j}-(1/\kappa)}{w_{j}-\varepsilon}\leq\dfrac{dx_{j}}{w_{j}-\varepsilon}.$

Hence, (3.2) holds provided

(3.4)
$\dfrac{k}{v_{\rho}}\in\bigg{(}\dfrac{d+(1/\kappa)}{d(d+\varepsilon)},\dfrac{d-(1/\kappa)}{d(d-\varepsilon)}\bigg{)}.$

The length of the interval on the right is

$\dfrac{2}{(d^{2}-\varepsilon^{2})}\bigg{(}\varepsilon-\dfrac{1}{\kappa}\bigg{)}>\dfrac{2}{d^{2}}\bigg{(}\varepsilon-\dfrac{1}{\kappa}\bigg{)}\geq\dfrac{1}{v_{\rho}}$

so that $k$ exists satisfying (3.4). Observe that (3.4) and the definition of
$d$ imply (3.1) holds.

Now, suppose some $w_{j}=0$ with $j\in\\{1,2,\dots,\rho\\}$. We again choose
$k$ so that (3.4) holds. For each $w_{j}\neq 0$, the above argument gives
$|v_{j}-w_{j}k|<\varepsilon k$ as in (3.2). On the other hand, if $w_{j}=0$,
then (3.3) and the definitions of $w_{j}$, $x_{j}$, and $k$ imply

$\kappa dv_{j}\leq
v_{\rho}<k\dfrac{d\big{(}d+\varepsilon\big{)}}{d+(1/\kappa)}\leq\kappa\varepsilon
dk.$

Hence, (3.2) holds for such $w_{j}$ as well, completing the proof. ∎

###### Proof of Theorem 3.1..

Suppose that the non-reciprocal part of $F(x)$ is reducible. As described in
detail in the proof of Theorem 2 of [6], there exist non-reciprocal
polynomials $u(x)$ and $v(x)$ in $\mathbb{Z}[x]$ such that $F(x)=u(x)v(x)$.
Let $W(x)=u(x)\tilde{v}(x)$. The polynomials $F(x)$, $\widetilde{F}(x)$,
$W(x)$, and $\widetilde{W}(x)$ are distinct polynomials of degree $d_{r}$ with
any two having greatest common divisor of degree $<d_{r}$. Note that

(3.5)
$F(x)\widetilde{F}(x)=u(x)v(x)\tilde{u}(x)\tilde{v}(x)=W(x)\widetilde{W}(x).$

By comparing the coefficient of $x^{d_{r}}$ on the left and right sides of
(3.5), we deduce that $\|F\|^{2}=\|W\|^{2}$.

We write $W(x)$ in the form $W(x)=\sum_{j=0}^{s}b_{j}x^{e_{j}}$ where the
$b_{j}$ are non-zero and $0=e_{0}<e_{1}<\cdots<e_{s}=d_{r}$. Then
$\|W\|^{2}=\|F\|^{2}$ implies $s\leq\|F\|^{2}-1$. Consider the set

$\displaystyle T=\\{d_{1},d_{2},$
$\displaystyle\dots,d_{r}\\}\cup\\{d_{r}-d_{1},d_{r}-d_{2},\dots,d_{r}-d_{r-1}\\}$
$\displaystyle\cup\\{e_{1},e_{2},\dots,e_{s-1}\\}\cup\\{e_{s}-e_{1},e_{s}-e_{2},\dots,e_{s}-e_{s-1}\\}.$

Observe that $|T|\leq 2\,\|F\|^{2}+2r-5$. We use the lower bound on
$d_{r}=\deg F$ in the statement of the theorem together with Lemma 3.2 to
deduce that there exists an integer $k\in[k_{0},d_{r}/(1-\varepsilon))$ such
that $t\operatorname{\>mod}k$ is in $[0,\varepsilon
k)\cup((1-\varepsilon)k,k)$ for every $t\in T$. Fix such an integer $k$.

Define $\overline{d}_{j}$ and $\ell_{j}$ as in the theorem, and define
$\overline{e}_{j}$ and $m_{j}$, for $0\leq j\leq s$, similarly by

$\overline{e}_{j}=(e_{j}+\lfloor\varepsilon
k\rfloor)\operatorname{\>mod}k\qquad\text{and}\qquad e_{j}+\lfloor\varepsilon
k\rfloor=km_{j}+\overline{e}_{j}.$

For $0\leq j\leq r$, we also define $\overline{d}^{\prime}_{j}$ and
$\ell^{\prime}_{j}$ by

$\overline{d}^{\prime}_{j}=(d_{r}-d_{j}+\lfloor\varepsilon
k\rfloor)\operatorname{\>mod}k,\qquad\text{and}\qquad
d_{r}-d_{j}+\lfloor\varepsilon
k\rfloor=k\ell^{\prime}_{j}+\overline{d}^{\prime}_{j};$

and, for $0\leq j\leq s$, we also define $\overline{e}^{\prime}_{j}$ and
$m^{\prime}_{j}$ by

$\overline{e}^{\prime}_{j}=(e_{s}-e_{j}+\lfloor\varepsilon
k\rfloor)\operatorname{\>mod}k\qquad\text{and}\qquad
e_{s}-e_{j}+\lfloor\varepsilon
k\rfloor=km^{\prime}_{j}+\overline{e}^{\prime}_{j}.$

Since $t\operatorname{\>mod}k$ is in $[0,\varepsilon
k)\cup((1-\varepsilon)k,k)$ for every $t\in T$, we see that the numbers
$\overline{d}_{j}$ and $\overline{d}^{\prime}_{j}$, for $0\leq j\leq r$, and
the numbers $\overline{e}_{j}$ and $\overline{e}^{\prime}_{j}$, for $0\leq
j\leq s$, all lie in the interval $[0,2\varepsilon k)$. Define
$G_{1}(x,y)=G(x,y)$ (as in the statement of the theorem),

$G_{2}(x,y)=\sum_{j=0}^{r}a_{j}x^{\overline{d}^{\prime}_{j}}y^{\ell^{\prime}_{j}},\qquad
H_{1}(x,y)=\sum_{j=0}^{s}b_{j}x^{\overline{e}_{j}}y^{m_{j}},$

and

$H_{2}(x,y)=\sum_{j=0}^{s}b_{j}x^{\overline{e}^{\prime}_{j}}y^{m^{\prime}_{j}}.$

Hence,

(3.6) $\displaystyle\begin{split}G_{1}(x,x^{k})=x^{\lfloor\varepsilon
k\rfloor}F(x),\qquad G_{2}(x,x^{k})=x^{\lfloor\varepsilon
k\rfloor}\widetilde{F}(x),\phantom{ssld}\\\\[5.0pt]
H_{1}(x,x^{k})=x^{\lfloor\varepsilon k\rfloor}W(x)\qquad\text{and}\qquad
H_{2}(x,x^{k})=x^{\lfloor\varepsilon k\rfloor}\widetilde{W}(x).\end{split}$

Corresponding to (3.5), we establish next that

(3.7) $G_{1}(x,y)G_{2}(x,y)=H_{1}(x,y)H_{2}(x,y).$

Recall that we are working with the condition that $\varepsilon\leq 1/4$ so
that $2\varepsilon k\leq k/2$. Expanding the left-hand side of (3.7), we
obtain an expression of the form $\sum_{j=0}^{J}g_{j}(x)y^{j}$ where possibly
some $g_{j}(x)$ are $0$ but otherwise $\deg g_{j}<k$ for each $j$. Since

$x^{2\lfloor\varepsilon
k\rfloor}F(x)\widetilde{F}(x)=G_{1}(x,x^{k})G_{2}(x,x^{k})=\sum_{j=0}^{J}g_{j}(x)x^{kj},$

we see that the terms in each $g_{j}(x)x^{kj}$ correspond precisely to the
terms in the expansion of $x^{2\lfloor\varepsilon
k\rfloor}F(x)\widetilde{F}(x)$ having degrees in the interval $[kj,k(j+1))$.
Similarly, writing the right-hand side of (3.7) in the form
$\sum_{j=0}^{J^{\prime}}h_{j}(x)y^{j}$, we get each $h_{j}(x)$ is either $0$
or has degree $<k$ and the terms in $h_{j}(x)x^{kj}$ correspond to the terms
in the expansion of $x^{2\lfloor\varepsilon k\rfloor}W(x)\widetilde{W}(x)$
having degrees in the interval $[kj,k(j+1))$. We see now that (3.7) is a
consequence of (3.5).

From (3.5) and (3.6), given any two of $G_{1}(x,y)$, $G_{2}(x,y)$,
$H_{1}(x,y)$ and $H_{2}(x,y)$, each one will have a factor different from $x$
that does not divide the other. The theorem now follows from (3.7). ∎

##  4\. Reduction to polynomials in one variable

In this section, we will be interested in obtaining information about the
factorization of $F(x,x^{n})$ in $\mathbb{Z}[x]$ given $F(x,y)$ as in (2.1).
As suggested by the previous section, the methods we employ here give us
results about the factorization of the non-reciprocal part of $F(x,x^{n})$.

We restrict ourselves to the case that

$\gcd{}_{\mathbb{Z}[x]}\big{(}f_{0}(x),f_{1}(x),\ldots,f_{r}(x)\big{)}=1,\quad
f_{0}(0)\neq 0,\quad\text{and}\quad f_{r}(x)\neq 0.$

The above restrictions are minor; in particular, we can simply factor out an
element of $\mathbb{Z}[x]$ from $F(x,y)$ if either of the first two conditions
do not hold and proceed from there. To clarify, we allow for the possibility
that $f_{j}(0)=0$ for some or all $j\in\\{1,2,\ldots,r\\}$ and $f_{j}(x)=0$
for some or all $j\in\\{1,2,\ldots,r-1\\}$.

Let

(4.1) $F_{0}(x)=F(x,x^{n})=\sum_{j=0}^{r}f_{j}(x)x^{nj},$

where we consider here

(4.2) $n>\max_{0\leq j\leq r-1}\\{\deg f_{j}\\}.$

For the purposes of using Theorem 3.1, we write

$F_{0}(x)=\displaystyle\sum_{j=0}^{m}a_{j}x^{d_{j}},$

where $0=d_{0}<d_{1}<\cdots<d_{m}$, the numbers $n,2n,\ldots,rn$ are included
in the set $\\{d_{0},d_{1},\dots,d_{m}\\}$, and possibly the coefficients
$a_{j}$ corresponding to $x^{n},x^{2n},\ldots,x^{rn}$ are $0$ but otherwise
$a_{j}\neq 0$. With this understanding, there exist non-negative integers
$\rho_{0},\ldots,\rho_{r}$ such that

$\displaystyle f_{0}(x)=\sum_{j=0}^{\rho_{0}}a_{j}x^{d_{j}},\quad
f_{1}(x)=\sum_{j=\rho_{0}+1}^{\rho_{1}}a_{j}x^{d_{j}-n},$
$\displaystyle\quad\ldots,\quad
f_{r}(x)=\sum_{j=\rho_{r-1}+1}^{\rho_{r}}a_{j}x^{d_{j}-rn},$

where

$d_{0}=0,\ d_{\rho_{0}+1}=n,\ d_{\rho_{1}+1}=2n,\ \ldots,\
d_{\rho_{r-1}+1}=rn.$

As noted, some or all of the coefficients
$a_{\rho_{0}+1},a_{\rho_{1}+1},\ldots,a_{\rho_{r-1}+1}$ may be $0$.

Set

$k_{0}=2\max_{0\leq j\leq
r}\\{\deg(f_{j})\\}\quad\text{and}\quad\varepsilon=\dfrac{1}{2r+2}\leq\dfrac{1}{4}.$

Choose $n$ sufficiently large so that the inequality on $\deg F_{0}$ holds in
Theorem 3.1. Suppose further that the non-reciprocal part of $F_{0}(x)$ is
reducible. Fix $k$ as in Theorem 3.1. For $j\in\\{0,\ldots,r\\}$, define
$\overline{d_{j}}$ and $\ell_{j}$ as in Theorem 3.1. Since $k\geq
2\deg(f_{0})$, we have that $d_{j}+\lfloor\varepsilon k\rfloor<k$ for
$j\in\\{0,1,\ldots,\rho_{0}\\}$. Hence, each of $\ell_{0},\ell_{1},$
$\ldots,\ell_{\rho_{0}}$ is 0. We show next that, for each
$i\in\\{1,2,\ldots,r\\}$, the numbers
$\ell_{\rho_{i-1}+1},\ell_{\rho_{i-1}+2},\ldots,\ell_{\rho_{i}}$ are all
equal. Assume that $\ell_{s}\neq\ell_{\rho_{i}}$ for some
$s\in\\{\rho_{i-1}+1,\rho_{i-1}+2,\ldots,\rho_{i}-1\\}$. The ordering on the
$d_{j}$ and the definition of the $\ell_{j}$ imply that
$\ell_{s}<\ell_{\rho_{i}}$. By the way $\overline{d_{j}}$ is defined, we know
that each $\overline{d_{j}}$ is in $[0,2\varepsilon k)\subseteq[0,k/2)$.
Hence, by the definition of $\ell_{j}$, we obtain

$d_{\rho_{i}}-d_{s}=k(\ell_{\rho_{i}}-\ell_{s})+\big{(}\overline{d_{\rho_{i}}}-\overline{d_{s}}\big{)}>k-\frac{k}{2}=\frac{k}{2}\geq\deg(f_{i}).$

On the other hand, from the definition of the $\rho_{j}$, we see that

$d_{\rho_{i}}-d_{s}\leq d_{\rho_{i}}-d_{\rho_{i-1}+1}\leq\deg(f_{i}).$

This apparent contradiction implies then that
$\ell_{\rho_{i-1}+1},\ldots,\ell_{\rho_{i}}$ are all equal.

Since the $d_{j}$ are increasing, the equality of the numbers
$\ell_{\rho_{i-1}+1},\ldots,\ell_{\rho_{i}}$ implies that

$\overline{d_{\rho_{i-1}+1}}=\min\big{\\{}\,\overline{d_{\rho_{i-1}+1}},\overline{d_{\rho_{i-1}+2}},\ldots,\overline{d_{\rho_{i}}}\,\big{\\}}\quad\text{for
}1\leq i\leq r.$

Similarly,

$\overline{d_{0}}=\min\big{\\{}\,\overline{d_{0}},\overline{d_{1}},\ldots,\overline{d_{\rho_{0}}}\,\big{\\}}=\lfloor\varepsilon
k\rfloor.$

Set

$M_{0}=\min_{0\leq j\leq
m}\big{\\{}\overline{d_{j}}\big{\\}}=\min\\{\,\overline{d_{0}},\overline{d_{\rho_{0}+1}},\ldots,\overline{d_{\rho_{r-1}+1}}\,\\}.$

In the notation of Theorem 3.1, $M\geq M_{0}$. Define

$t_{i}=\overline{d_{\rho_{i-1}+1}}-M_{0}\ \ \ \text{(for $1\leq i\leq
r$),}\quad\text{and}\quad t_{0}=\overline{d_{0}}-M_{0}=\lfloor\varepsilon
k\rfloor-M_{0}.$

Thus, $t_{0},t_{1},\ldots,t_{r}$ are non-negative integers $<2\varepsilon k$
with at least one equal to $0$.

Let $G(x,y)$ be defined as in Theorem 3.1. Thus,

(4.3) $\displaystyle G(x,y)$
$\displaystyle=\sum_{j=0}^{m}a_{j}x^{\overline{d_{j}}}y^{\ell_{j}}$
$\displaystyle=\sum_{j=0}^{\rho_{0}}a_{j}x^{\overline{d_{j}}}+\sum_{j=\rho_{0}+1}^{\rho_{1}}a_{j}x^{\overline{d_{j}}}y^{\ell_{\rho_{1}}}+\cdots+\sum_{j=\rho_{r-1}+1}^{\rho_{r}}a_{j}x^{\overline{d_{j}}}y^{\ell_{\rho_{r}}}.$

For $0\leq j\leq\rho_{0}$, we have

$\overline{d_{j}}=d_{j}+\lfloor\varepsilon k\rfloor=d_{j}+t_{0}+M_{0}.$

For $1\leq i\leq r$ and $\rho_{i-1}+1\leq j\leq\rho_{i}$, we have

(4.4) $\displaystyle\overline{d_{j}}$ $\displaystyle=d_{j}+\lfloor\varepsilon
k\rfloor-k\ell_{\rho_{i}}=d_{j}+\overline{d_{\rho_{i-1}+1}}-d_{\rho_{i-1}+1}$
$\displaystyle=d_{j}-d_{\rho_{i-1}+1}+t_{i}+M_{0}=d_{j}-in+t_{i}+M_{0}.$

Observe that, for $1\leq i\leq r$, we obtain from (4.4) and the definition of
$t_{i}$ that

(4.5) $in+\lfloor\varepsilon
k\rfloor=k\ell_{\rho_{i}}+t_{i}+M_{0}\qquad\text{with
}t_{i}+M_{0}\in[0,2\varepsilon k).$

As $\ell_{\rho_{0}}=0$ and $t_{0}+M_{0}=\lfloor\varepsilon k\rfloor$, we see
that (4.5) also holds for $i=0$. Substituting (4.4) and then (4.5) into (4.3),
we deduce

$\displaystyle G(x,x^{k})$
$\displaystyle=f_{0}(x)x^{t_{0}+M_{0}}+f_{1}(x)x^{t_{1}+M_{0}}x^{k\ell_{\rho_{1}}}+\cdots+f_{r}(x)x^{t_{r}+M_{0}}x^{k\ell_{\rho_{r}}}$
$\displaystyle=x^{\lfloor\varepsilon k\rfloor}F_{0}(x).$

Fix $i\in\\{1,\ldots,r\\}$. From (4.5), we obtain and

(4.6) $(i-1)n+\lfloor\varepsilon
k\rfloor=k\ell_{\rho_{i-1}}+t_{i-1}+M_{0}\qquad\text{with}\qquad
t_{i-1}+M_{0}\in[0,2\varepsilon k).$

By multiplying equation (4.6) by $i$ and (4.5) by $i-1$ and taking a
difference, we deduce

(4.7) $\lfloor\varepsilon
k\rfloor=k(i\ell_{\rho_{i-1}}-(i-1)\ell_{\rho_{i}})+d^{\prime}_{i}$

where

$d^{\prime}_{i}=i(t_{i-1}+M_{0})-(i-1)(t_{i}+M_{0})\in\big{(}{-2}(r-1)\varepsilon
k,2r\varepsilon k\big{)}.$

As $\varepsilon=1/(2r+2)$, we deduce that
$k(i\ell_{\rho_{i-1}}-(i-1)\ell_{\rho_{i}})$ is a multiple of $k$ in the
interval $(-k,k)$, and therefore $0$. Thus,
$i\ell_{\rho_{i-1}}=(i-1)\ell_{\rho_{i}}$ for each $i\in\\{1,\ldots,r\\}$.
Considering $i=1,2,\ldots,r$ in turn, we deduce

$\ell_{\rho_{0}}=0\quad\text{and}\quad\ell_{\rho_{i}}=i\ell_{\rho_{1}}\quad\text{for
}1\leq i\leq r.$

Define $\ell=\ell_{\rho_{1}}$. Recall that after the statement of Theorem 3.1,
we showed that some $\ell_{j}\geq 1$. Hence, $\ell\geq 1$.

From $i\ell_{\rho_{i-1}}=(i-1)\ell_{\rho_{i}}$, (4.7) and the definition of
$d^{\prime}_{i}$, we deduce

(4.8) $it_{i-1}-(i-1)t_{i}+M_{0}=\lfloor\varepsilon k\rfloor\quad\text{for
}1\leq i\leq r.$

Taking $i=j+1$ and $i=j$ in this equation, subtracting the first from the
second, and dividing by $j$, we obtain

$2t_{j}=t_{j+1}+t_{j-1}\quad\text{ for }1\leq j\leq r-1.$

In other words, $t_{j}$ is the average of $t_{j-1}$ and $t_{j+1}$ for $1\leq
j\leq r-1$. Hence, the minimum value of $t_{i}$ occurs for $i=0$ or $i=r$.
Recall that the $t_{i}$ are non-negative with at least one equal to $0$.
Therefore, either $t_{0}=0$ or $t_{r}=0$. Define

$t=\begin{cases}t_{1}&\text{if }t_{0}=0\\\ t_{r-1}&\text{if
}t_{r}=0.\end{cases}$

Since $t_{j}$ is the average of $t_{j-1}$ and $t_{j+1}$ for $1\leq j\leq r-1$,
we deduce that either $t_{i}=it$ for $0\leq i\leq r$ or $t_{i}=(r-i)t$ for
$0\leq i\leq r$.

Combining the above, we obtain that one of

(4.9)
$x^{M_{0}}\big{(}f_{0}(x)+f_{1}(x)x^{t}x^{k\ell}+\cdots+f_{r-1}(x)x^{(r-1)t}x^{(r-1)k\ell}+f_{r}(x)x^{rt}x^{rk\ell}\big{)}$

and

(4.10)
$x^{M_{0}}\big{(}f_{0}(x)x^{rt}+f_{1}(x)x^{(r-1)t}x^{k\ell}+\cdots+f_{r-1}(x)x^{t}x^{(r-1)k\ell}+f_{r}(x)x^{rk\ell}\big{)}$

is equal to $x^{\lfloor\varepsilon k\rfloor}F_{0}(x)$. In the first case
above, (4.8) with $i=1$ implies $M_{0}=\lfloor\varepsilon k\rfloor$. In the
second case, (4.8) with $i=r$ implies $M_{0}=\lfloor\varepsilon k\rfloor-rt$.

We focus now on $G(x,y)$. We have that $G(x,y)$ is either

$x^{M_{0}}\big{(}f_{0}(x)+f_{1}(x)x^{t}y^{\ell}+\cdots+f_{r-1}(x)x^{(r-1)t}y^{(r-1)\ell}+f_{r}(x)x^{rt}y^{r\ell}\big{)}$

or

$x^{M_{0}}\big{(}f_{0}(x)x^{rt}+f_{1}(x)x^{(r-1)t}y^{\ell}+\cdots+f_{r-1}(x)x^{t}y^{(r-1)\ell}+f_{r}(x)y^{r\ell}\big{)}.$

We deduce that $M$ in Theorem 3.1 is $M_{0}$ in the first case, but the value
of $M$ depends on whether $f_{r}(0)=0$ in the second case. If $f_{r}(0)\neq
0$, then $M=M_{0}$ in this case as well.

In any case, to deduce the irreducibility of the the non-reciprocal part of
$F(x,x^{n})$ for $n$ sufficiently large, Theorem 3.1 implies that it suffices
to show that both

(4.11)
$f_{0}(x)+f_{1}(x)x^{t}y^{\ell}+\cdots+f_{r-1}(x)x^{(r-1)t}y^{(r-1)\ell}+f_{r}(x)x^{rt}y^{r\ell}$

and

(4.12)
$f_{0}(x)x^{rt}+f_{1}(x)x^{(r-1)t}y^{\ell}+\cdots+f_{r-1}(x)x^{t}y^{(r-1)\ell}+f_{r}(x)y^{r\ell},$

other than possibly a factor of a power of $x$, are irreducible as a
polynomial in two variables for every non-negative integer $t$.

We summarize this as follows.

######  Theorem 4.1.

Let $F(x,y)$ be as in (2.1) with

$\gcd{}_{\mathbb{Z}[x]}\big{(}f_{0}(x),f_{1}(x),\ldots,f_{r}(x)\big{)}=1\quad
f_{0}(0)\neq 0,\quad\text{and}\quad f_{r}(x)\neq 0.$

Let $n$ be sufficiently large. Then the non-reciprocal part of $F(x,x^{n})$ is
not reducible if, for every non-negative integer $t$ and every positive
integer $\ell$, the polynomials (4.11) and (4.12) are a power of $x$ (possibly
$x^{0}$) times an irreducible polynomial in $\mathbb{Z}[x,y]$.

Theorem 4.1 is close to what we stated for Theorem 1.1 in the introduction.
However, if we are not interested in establishing that $F(x,x^{n})$ is
irreducible for all $n$ sufficiently large but rather for $n$ sufficiently
large of a certain form, then Theorem 1.1 is valuable and needs a little more
justification. We begin by showing $n=k\ell\pm t$.

First, consider $\eqref{polyoneeq}$. We recall the discussion after (4.9) and
(4.10). In the case of $\eqref{polyoneeq}$, we have $M_{0}=\lfloor\varepsilon
k\rfloor$ in (4.9). The values of $t$ and $\ell$ were determined by

$t=t_{1}=\overline{d_{\rho_{0}+1}}-M_{0}=\overline{d_{\rho_{0}+1}}-\lfloor\varepsilon
k\rfloor\quad\text{and}\quad\ell=\ell_{\rho_{1}}=\ell_{\rho_{0}+1}.$

Further, $d_{\rho_{0}+1}=n$. From Theorem 3.1, we deduce

$n+\lfloor\varepsilon k\rfloor=d_{\rho_{0}+1}+\lfloor\varepsilon
k\rfloor=k\ell_{\rho_{0}+1}+\overline{d_{\rho_{0}+1}}=k\ell+t+\lfloor\varepsilon
k\rfloor.$

Thus, $n=k\ell+t$. In the notation of Theorem 3.1 and (4.1),
$F_{0}(x)=F(x,x^{n})$ and the value of $n$ comes from the expression
$f_{1}(x)x^{n}$ that one obtains by replacing $y$ with $x^{k}$ in
$\eqref{polyoneeq}$. In the case of $\eqref{polytwoeq}$, we have

$0=t_{r}=\overline{d_{\rho_{r-1}+1}}-M_{0}=\overline{d_{\rho_{r-1}+1}}-\lfloor\varepsilon
k\rfloor+rt\quad\text{and}\quad\ell_{\rho_{r}}=r\ell.$

Thus,

$\displaystyle rn+\lfloor\varepsilon k\rfloor$
$\displaystyle=d_{\rho_{r-1}+1}+\lfloor\varepsilon
k\rfloor=k\ell_{\rho_{r-1}+1}+\overline{d_{\rho_{r-1}+1}}$
$\displaystyle=k\ell_{\rho_{r}}+\lfloor\varepsilon k\rfloor-
rt=r(k\ell-t)+\lfloor\varepsilon k\rfloor.$

Hence, in this case, $n=k\ell-t$, and the value of $n$ can be viewed as coming
from the expression $f_{1}(x)x^{n}$ that one obtains by dividing the
polynomial in $\eqref{polytwoeq}$ by $x^{rt}$ and replacing $y$ with $x^{k}$.

The condition on $k$ in Theorem 1.1 follows from our choice of $k\geq k_{0}$.
Recall that the $t_{j}$ are non-negative integers $<2\varepsilon k$ and that
$\varepsilon=1/(2r+2)$. As $t_{0},t_{1},\ldots,t_{r}$ are the same as
$0,t,2t,\ldots,rt$ in some order, we deduce $t<k/(r(r+1))$.

This completes the justification of Theorem 1.1.

##  5\. Applications

In this section, we give three examples using a couple different approaches to
show how one can apply Theorem 1.1. The first example is meant just to
demonstrate the power of the approach for non-reciprocal polynomials. The last
two examples come from the Laurent polynomials discussed in the introduction.
Recalling the first one listed there was handled in [7], the last two examples
here address the two other Laurent polynomials beginning with the last one.

Example 1. Let $n\geq 2$ be an integer, and set

$\displaystyle G(x)=x^{6n}+(x+1)$ $\displaystyle
x^{5n+1}+2x^{4n}+(x^{4}-x^{3}-x^{2}-2x-2)x^{3n-2}$
$\displaystyle\quad+2x^{2n}+(x+1)x^{n-2}+1.$

One can check that if $n=4$, the polynomial factors as a product of two
irreducible non-reciprocal polynomials. For $n$ sufficiently large, we show
that $G(x)$ has no reciprocal irreducible factors and the non-reciprocal part
of $G(x)$ is irreducible. Then it will follow that $G(x)$ is irreducible for
all $n$ sufficiently large.

Assume $u(x)$ is an irreducible reciprocal factor of $G(x)$, and let $\alpha$
be a root of $u(x)$. Since $G(0)\neq 0$, we see that $\alpha\neq 0$, and the
definition of $u(x)$ being reciprocal implies that $1/\alpha$ is a root of
$u(x)$. We deduce that $\alpha$ is a root of $G(x)$ and of
$\widetilde{G}(x)=x^{\deg G}G(1/x)$. Since

$\displaystyle\widetilde{G}(x)=x^{6n}+(x+1)$ $\displaystyle
x^{5n+1}+2x^{4n}+(-2x^{4}-2x^{3}-x^{2}-x+1)x^{3n-2}$
$\displaystyle\quad+2x^{2n}+(x+1)x^{n-2}+1,$

we deduce that $\alpha$ is a root of

$G(x)-\widetilde{G}(x)=(3x^{4}+x^{3}-x-3)x^{3n-2}.$

Therefore, $u(x)$ is an irreducible factor of

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

One checks that $\pm 1$ are not roots of $G(x)$. Furthermore, since
$G(x)\in\mathbb{Z}[x]$ is monic, the irreducible polynomial $3x^{2}+x+3$
cannot be a factor of $G(x)$. We obtain a contradiction. Thus, $G(x)$ has no
reciprocal irreducible factors.

Let $n$ be sufficiently large. Let

$F(x,y)=\sum_{j=0}^{6}f_{j}(x)y^{j},$

where

$\displaystyle f_{0}(x)=1,\quad f_{1}(x)=x+1,\quad f_{2}(x)=2x^{4},$
$\displaystyle f_{3}(x)=x^{4}(x^{4}-x^{3}-x^{2}-2x-2),$ $\displaystyle
f_{4}(x)=2x^{8},\quad f_{5}(x)=x^{11}(x+1)\quad\text{and}\quad
f_{6}(x)=x^{12}.$

Then $F(x,x^{n-2})=G(x)$.

Our goal now is to apply Theorem 1.1 or, for this example, Theorem 4.1. Let

$F_{1,t}(x,y)=\sum_{j=0}^{6}f_{j}(x)x^{jt}y^{j}\quad\text{ and }\quad
F_{2,t}(x,y)=\sum_{j=0}^{6}f_{j}(x)x^{(6-j)t}y^{j}.$

To show $F_{1,t}(x,y^{\ell})$ and $F_{2,t}(x,y^{\ell})$ are a power of $x$
times an irreducible polynomial for all non-negative integers $t$ and positive
integers $\ell$, we apply Theorem 2.1. Given the condition in Theorem 2.1 that
$F(x,y)$ is irreducible in $\mathbb{Z}[x,y]$, we want to first justify that
$F_{1,t}(x,y)$ and $F_{2,t}(x,y)$ are a power of $x$ times an irreducible
polynomial in $\mathbb{Z}[x,y]$ for all non-negative integers $t$ (so that we
are actually applying Theorem 2.1 to $F_{1,t}(x,y)$ and $F_{2,t}(x,y)$ divided
by an appropriate power of $x$). Observe that

$F_{1,t}\big{(}x,y/x^{t}\big{)}=F_{1,0}(x,y)\quad\text{ and }\quad
F_{2,t}\big{(}x,x^{t}y\big{)}=x^{6t}F_{2,0}(x,y).$

As $f_{0}(x)=1$, we see that there are no irreducible polynomials in
$\mathbb{Z}[x]$ dividing all the coefficients of $F_{1,t}(x,y)$ viewed as a
polynomial in $y$. Also, the polynomial $x$ is the only possible irreducible
polynomial in $\mathbb{Z}[x]$ dividing all the coefficients of $F_{2,t}(x,y)$
viewed as a polynomial in $y$. Thus, for $j\in\\{1,2\\}$, each irreducible
factor of $F_{j,t}(x,y)$ in $\mathbb{Z}[x,y]$ other than $x$ has positive
degree in $y$. Therefore, $F_{j,t}(x,y)$ is a power of $x$ times an
irreducible polynomial if and only if $F_{j,0}(x,y)$ is, which follows from
the above in $\mathbb{Q}(x)[y]$ and then by Gauss’s Lemma for
$\mathbb{Z}[x,y]$. One checks directly that the two polynomials $F_{1,0}(x,y)$
and $F_{2,0}(x,y)$ are irreducible. Therefore, $F_{1,t}(x,y)$ and
$F_{2,t}(x,y)$ are a power of $x$ times an irreducible polynomial in
$\mathbb{Z}[x,y]$ for all non-negative integers $t$

Since $f_{1}(x)=x+1$, we see that (2.6) with $F$ replaced by $F_{1,t}$ and
with $F$ replaced by $F_{2,t}$ does not hold for each prime $p$. To finish
applying Theorem 2.1, it remains to show $F_{1,t}(x,y^{4})$ and
$F_{2,t}(x,y^{4})$ are a power of $x$ times an irreducible polynomial for all
non-negative integers $t$.

Write $t=4q+r$ where $q\in\mathbb{Z}$ and $r\in\\{0,1,2,3\\}$. Observe that

$F_{1,t}\big{(}x,(y/x^{q})^{4}\big{)}=F_{1,r}(x,y^{4})\quad\text{ and }\quad
F_{2,t}\big{(}x,(x^{q}y)^{4}\big{)}=x^{24q}F_{2,r}(x,y^{4}).$

The only possible common irreducible factor in $\mathbb{Z}[x]$ among the
coefficients of $F_{1,t}(x,y)$ and $F_{2,t}(x,y)$ as polynomials in $y$ is $x$
and only in the case of $F_{2,t}(x,y)$. For $j\in\\{1,2\\}$, a factorization
of $F_{j,t}(x,y^{4})$ in $\mathbb{Q}(x)[y]$ into polynomials of degree $>0$ in
$y$ will correspond to a factorization of $F_{j,r}(x,y^{4})$ in
$\mathbb{Q}(x)[y]$ into polynomials of degree $>0$ in $y$, and vice versa. We
deduce that $F_{j,t}(x,y^{4})$ is a power of $x$ times an irreducible
polynomial in $\mathbb{Z}[x,y]$ if and only if $F_{j,r}(x,y^{4})$ is a power
of $x$ times an irreducible polynomial in $\mathbb{Z}[x,y]$. As
$r\in\\{0,1,2,3\\}$ and $j\in\\{1,2\\}$, one can check directly that the eight
polynomials $F_{j,r}(x,y^{4})$ are a power of $x$ times an irreducible
polynomial in $\mathbb{Z}[x,y]$.

Given $G(x)$ has no irreducible reciprocal factors, Theorem 1.1 now implies
that $G(x)$ is irreducible for all $n$ sufficiently large.

Example 2. In the introduction, we mentioned the Laurent polynomial

$x^{4b}-x^{2b}-x^{a}-2-x^{-a}-x^{-2b}+x^{-4b}$

associated with the trace field under $-a/b$ Dehn fillings of the complement
of the $4_{1}$ knot. Here, $a$ and $b$ are coprime positive integers. For
fixed $a$ and sufficiently large $b$ or for fixed $b$ and sufficiently large
$a$, one can apply the prior material to obtain information on the
factorization of the non-reciprocal part of the polynomial
$\operatorname{J}(x^{4b}-x^{2b}-x^{a}-2-x^{-a}-x^{-2b}+x^{-4b})$.

We illustrate this here by considering the case that $a$ is fixed and $b$ is
sufficiently large. Then we are interested in the factorization of

$\displaystyle F_{0}(x)$
$\displaystyle=x^{8b}-x^{6b}-x^{4b+a}-2x^{4b}-x^{4b-a}-x^{2b}+1$
$\displaystyle=x^{4a}x^{4(2b-a)}-x^{3a}x^{3(2b-a)}-(x^{a}+1)^{2}x^{a}x^{2(2b-a)}-x^{a}x^{2b-a}+1.$

We define

$F(x,y)=f_{0}(x)+f_{1}(x)y+f_{2}(x)y^{2}+f_{3}(x)y^{3}+f_{4}(x)y^{4},$

where

$\displaystyle f_{0}(x)=1,\ f_{1}(x)=-x^{a},\ f_{2}(x)=-(x^{a}+1)^{2}x^{a},$
$\displaystyle f_{3}(x)=-x^{3a},\ \text{ and }\ f_{4}(x)=x^{4a}.$

Thus, with $n=2b-a$, we have $F(x,x^{n})=F_{0}(x)$. Further, define

$F_{1}(x,y)=f_{0}(x)+f_{1}(x)x^{t}y+f_{2}(x)x^{2t}y^{2}+f_{3}(x)x^{3t}y^{3}+f_{4}(x)x^{4t}y^{4}$

and

$F_{2}(x,y)=f_{0}(x)x^{4t}+f_{1}(x)x^{3t}y+f_{2}(x)x^{2t}y^{2}+f_{3}(x)x^{t}y^{3}+f_{4}(x)y^{4}.$

For Theorem 1.1, we want to determine whether the polynomials
$F_{1}(x,y^{\ell})$ and $F_{2}(x,y^{\ell})$ are a power of $x$ times an
irreducible polynomial in $\mathbb{Z}[x,y]$.

For $j\in\\{1,2\\}$, we write $F_{j}(x,y)=x^{k_{j}}G_{j}(x,y)$ where
$k_{j}\geq 0$ is an integer and $G_{j}(x,y)\in\mathbb{Z}[x,y]$ is not
divisible by $x$. Since $f_{0}(x)=1$, we see that $k_{1}=0$. We want to apply
Theorem 2.1 to $G_{j}(x,y)$, so we show first that $G_{j}(x,y)$ is
irreducible. In the example at the end of Section 2, we saw that the
polynomial

$1-x(x+1)y^{n}-2x^{2}y^{2n}-x^{2}(x+1)y^{3n}+x^{4}y^{4n}$

is irreducible for every positive integer $n$. By a change of variables, we
see then that

$\displaystyle 1-y(y+1)$ $\displaystyle
x^{a}-2y^{2}x^{2a}-y^{2}(y+1)x^{3a}+y^{4}x^{4a}$
$\displaystyle=1-x^{a}y-(x^{a}+1)^{2}x^{a}y^{2}-x^{3a}y^{3}+x^{4a}y^{4}$
$\displaystyle=f_{0}(x)+f_{1}(x)y+f_{2}(x)y^{2}+f_{3}(x)y^{3}+f_{4}(x)y^{4}$

is irreducible for every positive integer $a$. Observe that this is the same
as $F_{1}(x,y)$ and $F_{2}(x,y)$ with $t=0$. We take advantage of this
momentarily.

Assume $G_{1}(x,y)=F_{1}(x,y)=U(x,y)V(x,y)$ where $U(x,y)$ and $V(x,y)$ are
nonunits in $\mathbb{Z}[x,y]$. Since $f_{0}(x)=1$, the polynomials $f_{0}(x)$,
$f_{2}(x)x^{2t}$ and $f_{4}(x)x^{4t}$ have no common factor in
$\mathbb{Z}[x]$, so the degrees of $U(x,y)$ and $V(x,y)$ in the variable $y$
are each at least $1$. Since

$\displaystyle f_{0}(x)+f_{1}(x)y+f_{2}(x)y^{2}+f_{3}(x)y^{3}+f_{4}(x)y^{4}$
$\displaystyle=G_{1}(x,y/x^{t})$ $\displaystyle=U(x,y/x^{t})\,V(x,y/x^{t}),$

we see that $f_{0}(x)+f_{1}(x)y+f_{2}(x)y^{2}+f_{3}(x)y^{3}+f_{4}(x)y^{4}$ is
reducible in $\mathbb{Q}(x)[y]$ and, hence, in $\mathbb{Z}[x,y]$. However, as
noted above, $f_{0}(x)+f_{1}(x)y+f_{2}(x)y^{2}+f_{3}(x)y^{3}+f_{4}(x)y^{4}$ is
irreducible independent of the choice of $a\geq 1$. Thus, we obtain a
contradiction, and $G_{1}(x,y)$ is irreducible.

Assume now $x^{k_{2}}G_{2}(x,y)=F_{2}(x,y)=x^{k_{2}}U(x,y)V(x,y)$ where
$U(x,y)$ and $V(x,y)$ are nonunits in $\mathbb{Z}[x,y]$. As $x$ is not a
factor of $G_{2}(x,y)=U(x,y)V(x,y)$ and $f_{0}(x)=1$, again we have the
degrees of $U(x,y)$ and $V(x,y)$ in the variable $y$ are each at least $1$. In
this case,

$\displaystyle x^{4t}\big{(}f_{0}(x)+f_{1}(x)y+f_{2}(x)y^{2}+f_{3}(x)y^{3}$
$\displaystyle+f_{4}(x)y^{4}\big{)}=F_{2}(x,x^{t}y)$
$\displaystyle\quad=x^{k_{2}}U(x,x^{t}y)\,V(x,x^{t}y),$

so that again we deduce
$f_{0}(x)+f_{1}(x)y+f_{2}(x)y^{2}+f_{3}(x)y^{3}+f_{4}(x)y^{4}$ is reducible in
$\mathbb{Q}(x)[y]$, leading to a contradiction. Thus, $G_{2}(x,y)$ is
irreducible.

Now, consider $G_{1}(x,y)=F_{1}(x,y)$. Recalling Theorem 1.1, in the case of
$F_{1}(x,y^{\ell})$, we have $n=k\ell+t$. Also,
$F_{0}(x)=F(x,x^{n})=F_{1}(x,x^{k\ell})$ so that $n=2b-a$.

Assume that $F_{1}(x,y^{4})$ is irreducible but $F_{1}(x,y^{\ell})$ is
reducible for some positive integer $\ell$. By Theorem 2.1, there must then be
an odd prime $p$ dividing $\ell$ for which $F_{1}(x,y^{p})$ is reducible.
Further, as stated in Theorem 2.1, we obtain (2.6) with $F$ there replaced by
$F_{1}$. Observe that if $p|a$ and $p|t$, then in fact (2.6) holds with $F$
replaced by $F_{1}$. However, then we have $p|a$, $p|t$ and $p|\ell$ so that
$p|(k\ell+t)$ which is equivalent to $p|n$. As $n=2b-a$ and $p$ is an odd
prime dividing $a$, we would have $p|b$, contrary to the condition given that
$a$ and $b$ are coprime positive integers. Thus, either $p\nmid a$ or $p\nmid
t$. Using (2.6) with $F$ replaced by $F_{1}$, we see that $f_{1}(x)x^{t}$ is a
$p$th power modulo $p$ and, consequently, that $p|(a+t)$. Also, we have that
$f_{2}(x)x^{2t}$ is a $p$th power modulo $p$, so its degree as a polynomial
modulo $p$, which is $3a+2t$, must be divisible by $p$. On the other hand, $p$
dividing both $a+t$ and $3a+2t$ implies $p|a$ and $p|t$, giving a
contradiction.

For the case of $F_{1}(x,y)$, we are left with determining whether
$F_{1}(x,y^{4})$ is reducible. Observe that if $F_{1}(x,y^{4})$ is reducible,
the remarks after Theorem 2.1 immediately give us that $F_{1}(x,y^{2})$ is
reducible and (2.6) holds with $F$ there replaced by $F_{1}$ and $p=2$. From
(2.6), we deduce that $f_{1}(x)x^{t}$ and $f_{2}(x)x^{2t}$ are squares modulo
$2$ so that $a+t$ and $a+2t$ are even. Hence, we have that both $a$ and $t$
are even. We make particular use of the latter.

We write $t$ in the form $2q$, where $q$ is a non-negative integer. Since
$f_{0}(x)=1$, we see that if $F_{1}(x,y^{2})$ is reducible, then so is
$F_{1}(x,(y/x^{q})^{2})$. As $F_{1}(x,(y/x^{q})^{2})$ takes the form of
$F_{1}(x,y^{2})$ with $t=0$, we can restrict our attention to determining
whether $F_{1}(x,y^{2})$ is reducible with $t=0$.

Set $t=0$. Let $W(x,y)=F_{1}(x,y^{2})$. We provide here an alternative
approach from prior material in this paper for establishing that $W(x,y)$ is
irreducible. The degree of $F_{1}(x,y^{2})$ in $y$ is $8$. We consider
explicit values of $y\in\mathbb{Z}$ for which $W(x,y)$ can be checked to be
irreducible in $\mathbb{Z}[x]$. For example, consider

$W(x,2)=2^{8}x^{4a}-2^{6}x^{3a}-2^{4}(x^{a}+1)^{2}x^{a}-2^{2}x^{a}+1.$

We write $W(x,2)=w(x^{a})$ where in this case

$w(x)=256x^{4}-80x^{3}-32x^{2}-20x+1.$

One checks that the quartic $w(x)$ is irreducible in $\mathbb{Z}[x]$ (using
Maple, for example). By Capelli’s Theorem, $w(x^{a})$ can be reducible only if
$p|a$ for some prime $p$ and $w(x^{p})$ is reducible or $4|a$ and $w(x^{4})$
is reducible. A computation shows that $w(x^{4})$ is irreducible, so this
second case does not happen. In the first case, Capelli’s theorem implies we
furthermore can assume that if $\gamma$ is a root of $w(x)$, then $\gamma$ is
a $p$th power in $\mathbb{Q}(\gamma)$ so that the norm of $\gamma$ is a $p$th
power in $\mathbb{Q}$. The norm of a root of $w(x)$ is $1/256=(1/2)^{8}$.
Hence, in this case $p=2$. As we have already verified that $w(x^{4})$ is
irreducible, we know $w(x^{2})$ is irreducible. Thus, $W(x,2)$ is irreducible
for all integers $a\geq 1$.

The idea now is to repeat the above process to obtain the irreducibility of
$W(x,k)$ for $17$ different $k\in\mathbb{Z}$ and for all positive integers
$a$. We used $k\in\\{2,3,\ldots,18\\}$. Let $w_{k}(x)$ be the analog to $w(x)$
above so that $w_{k}(x^{a})=W(x,k)$. For each $k$, we verified computationally
the irreducibility of $w_{k}(x^{a})$ for $a\in\\{1,3,4\\}$. As the norm of a
root of $w_{k}(x)$ is $1/k^{8}$ and this can be a $p$th power for
$k\in\\{2,3,\ldots,18\\}$ only for $p\in\\{2,3\\}$, we deduce as above that
$W(x,k)$ is irreducible for each $k\in\\{2,3,\ldots,18\\}$ and for all
positive integers $a$.

Assume that there is a positive integer $a$ for which $W(x,y)$ is reducible in
$\mathbb{Z}[x,y]$. As the constant term of $W(x,y)$ is $1$, we deduce that
$W(x,y)=F_{1}(x,y^{2})=U(x,y)V(x,y)$ where $U(x,y)$ and $V(x,y)$ are in
$\mathbb{Z}[x,y]$ and each of degree $\geq 1$ in $y$. Denote these degrees by
$d_{U}$ and $d_{V}$, respectively. For each $k\in\\{2,3,\ldots,18\\}$, the
irreducibility of $W(x,k)=U(x,k)V(x,k)$ in $\mathbb{Z}[x]$ implies that one of
$U(x,k)$ or $V(x,k)$ is $\pm 1$. Since $W(x,y)=U(x,y)V(x,y)$, we have
$d_{U}+d_{V}=8$. The pigeon-hole principle gives us that there is a subset $S$
of $\\{2,3,\ldots,18\\}$ such that one of the following holds:

  * •

The size of $S$ is $d_{U}+1$ and $U(x,k)=1$ for all $k\in S$.

  * •

The size of $S$ is $d_{U}+1$ and $U(x,k)=-1$ for all $k\in S$.

  * •

The size of $S$ is $d_{V}+1$ and $V(x,k)=1$ for all $k\in S$.

  * •

The size of $S$ is $d_{V}+1$ and $V(x,k)=-1$ for all $k\in S$.

We consider the first possibility above, noting the argument in each case is
similar. Writing $S=\\{k_{0},k_{1}\ldots,k_{s}\\}$ and
$U(x,y)=u_{0}(x)+u_{1}(x)y+\cdots+u_{s}(x)y^{s}$ where $s=d_{U}$, we obtain

$u_{0}(x)+u_{1}(x)k_{j}+\cdots+u_{s}(x)k_{j}^{s}=1\quad\text{for }0\leq j\leq
s.$

Thus,

$\begin{pmatrix}1&k_{0}&\cdots&k_{0}^{s}\\\ 1&k_{1}&\cdots&k_{1}^{s}\\\
\vdots&\vdots&\ddots&\vdots\\\
1&k_{s}&\cdots&k_{s}^{s}\end{pmatrix}\begin{pmatrix}u_{0}(x)\\\ u_{1}(x)\\\
\vdots\\\ u_{s}(x)\end{pmatrix}=\begin{pmatrix}1\\\ 1\\\ \vdots\\\
1\end{pmatrix}.$

The matrix above is a Vandermonde matrix, and its determinant is non-zero.
Hence, there are unique $u_{0}(x),u_{1}(x),\ldots,u_{s}(x)$ satisfying the
matrix equation above. Since the values $u_{0}(x)=1$ and $u_{j}(x)=0$ for
$1\leq j\leq s$ satisfy the equation, we deduce that $U(x,y)=1$, contradicting
that $d_{U}\geq 1$. Therefore, $W(x,y)$ is irreducible in $\mathbb{Z}[x,y]$
for all positive integers $a$.

So far we have shown that $F_{1}(x,y^{\ell})$ is irreducible in
$\mathbb{Z}[x,y]$ for all integers $t\geq 0$ and $\ell\geq 0$ for which
$n=2b-a=k\ell+t$. We now turn to $F_{2}(x,y^{\ell})$ where the integers $t\geq
0$ and $\ell\geq 0$ that we are interested in satisfy $n=2b-a=k\ell-t$. We
give a similar argument to that just given for $F_{1}(x,y)$. In the case of
$F_{2}(x,y)$, we have $F_{2}(x,y)=x^{k_{2}}G_{2}(x,y)$ and $k_{2}$ may be
positive (and will be unless $t=0$). We are interested in establishing that
$F_{2}(x,y^{\ell})$ is a power of $x$ times an irreducible polynomial. In
other words, we want to show $G_{2}(x,y^{\ell})$ is irreducible. We note that
the explicit values of $f_{j}(x)$ give that

$k_{2}=\begin{cases}4t&\text{if }0\leq t\leq a/2\\\ a+2t&\text{if }a/2<t\leq
3a/2\\\ 4a&\text{if }t>3a/2.\end{cases}$

Fix an odd prime $p$ satisfying (2.7). To simplify notation, we set $u=k_{2}$.
Observe that $u$ is independent of $p$ and $u\leq 4t$. Assume that
$F_{2}(x,y^{p})/x^{u}=G_{2}(x,y^{p})$ is a reducible polynomial. By Theorem
2.1, we obtain that (2.6) holds. Looking at the constant term and the
coefficients of $y$ and $y^{2}$ in $F_{2}(x,y^{p})/x^{u}$, we see that
$p|(4t-u)$, $p|(a+3t-u)$ and $p|(3a+2t-u)$. These imply $p|a$, $p|t$ and
$p|u$. As in the case of $F_{1}(x,y^{p})$, this leads to $p$ dividing both $a$
and $b$, contradicting that $a$ and $b$ are relatively prime.

Defining $u$ as above, we assume now that
$F_{2}(x,y^{4})/x^{u}=G_{2}(x,y^{4})$ is reducible. As with $F_{1}(x,y)$, we
can deduce from the comments after Theorem 2.1 that $F_{2}(x,y^{2})/x^{u}$ is
reducible. Taking $p=2$ in our discussion above about $F_{2}(x,y^{p})/x^{u}$,
we deduce $2|t$, $2|u$ and $2|a$ (but not necessarily that $2|b$). Writing
$t=2q$, we see that $F_{2}(x,(x^{q}y)^{2})/x^{8q}$ is of the form
$F_{2}(x,y^{2})$ with $t=0$. Furthermore, as the coefficients of
$F_{2}(x,y^{2})/x^{u}$ as a polynomial in $y$ have no common irreducible
factor in $\mathbb{Z}[x]$, we can conclude that if $F_{2}(x,y^{2})/x^{u}$ is
reducible, then $F_{2}(x,y^{2})$ with $t=0$ is reducible.

With $t=0$, we have

$F_{2}(x,y)=F_{1}(x,y)=F(x,y).$

As we have already established in this case that $F_{1}(x,y^{2})$ is
irreducible for all positive integers $a$, we deduce that $F_{2}(x,y^{2})$ is
irreducible for all positive integers $a$.

We therefore have that any polynomial $F_{1}(x,y)$ or $F_{2}(x,y)$ arising
from $F_{0}(x)$ is a power of $x$ times an irreducible polynomial in
$\mathbb{Z}[x,y]$. Theorem 1.1 now implies that the non-reciprocal part of
$F_{0}(x)$ is not reducible for each fixed positive integer $a$ and every
sufficiently large positive integer $b$ relatively prime to $a$. The phrase
“not reducible” is deliberate here, as $F_{0}(x)$ is a reciprocal polynomial
and cannot have exactly one irreducible non-reciprocal factor. The fact that
the non-reciprocal part of $F_{0}(x)$ is not reducible implies then that the
non-reciprocal part of $F_{0}(x)$ is identically $1$. In other words, we
deduce that $F_{0}(x)$ is a product of irreducible reciprocal polynomials (for
fixed $a$ and sufficiently large $b$ relatively prime to $a$).

Example 3. We consider

$F(x)=x^{4n}+\left(-x^{3}+4x^{2}-8+4x^{-2}-x^{-3}\right)x^{2n}+1,$

associated with the trace field of the Whitehead link for the complement of
the $(-2,3,3+2n)$ pretzel knot $K_{n}$. This polynomial $F(x)$ is in
$\mathbb{Z}[x]$ for all $n\geq 2$, and one easily checks that $(x-1)^{2}$ is a
factor. A computation suggests that the other factor is always irreducible.
Observe that the polynomial
$x^{2n}+\left(-x^{3}+4x^{2}-8+4x^{-2}-x^{-3}\right)x^{n}+1$ with more general
exponents does not have this property. For example, for $n=7$, this polynomial
factors as

$(x^{4}+x^{3}+3x^{2}+x+1)(x^{6}-x^{5}+2x^{3}-x+1)(x-1)^{2}(x+1)^{2}.$

It is reasonable to obtain a similar result below for this more general
polynomial; however, some adjustments would be needed as the analog to (5.1)
below is reducible for $r=0$ so that one should look at the factors that arise
instead and apply Theorem 2.1 to them.

We show here that, for $n$ sufficiently large, $F(x)$ is a product of
reciprocal irreducible polynomials. We rewrite

$F(x)=f_{0}(x)+f_{1}(x)x^{n-2}+f_{2}(x)x^{2(n-2)}+f_{3}(x)x^{3(n-2)}+f_{4}(x)x^{4(n-2)},$

where

$\displaystyle f_{0}(x)=1,\ f_{1}(x)=0,\
f_{2}(x)=-x^{7}+4x^{6}-8x^{4}+4x^{2}-x,$ $\displaystyle f_{3}(x)=0,\ \text{
and }\ f_{4}(x)=x^{8}.$

To apply Theorem 1.1, we want to determine whether the polynomials
$F_{1}(x,y^{\ell})$ and $F_{2}(x,y^{\ell})$ are a power of $x$ times an
irreducible polynomial in $\mathbb{Z}[x,y]$, where

$F_{1}(x,y)=f_{0}(x)+f_{2}(x)x^{2t}y^{2}+f_{4}(x)x^{4t}y^{4}$

and

$F_{2}(x,y)=f_{0}(x)x^{4t}+f_{2}(x)x^{2t}y^{2}+f_{4}(x)y^{4}.$

We make use of Theorem 2.1.

First, we justify that $F_{1}(x,y)$ and $F_{2}(x,y)$ are a power of $x$ times
an irreducible polynomial in $\mathbb{Z}[x,y]$ for all non-negative integers
$t$. In other words, we justify that

$F_{j}(x,y)=x^{k_{j}}G_{j}(x,y),$

where $k_{j}\geq 0$ is an integer and $G_{j}(x,y)\in\mathbb{Z}[x,y]$ is
irreducible for $j\in\\{1,2\\}$. Assume that $F_{1}(x,y)=U(x,y)V(x,y)$, where
$U(x,y)$ and $V(x,y)$ are nonunits in $\mathbb{Z}[x,y]$. Since $f_{0}(x)=1$,
we see that each of $U(x,y)$ and $V(x,y)$ has degree at least $1$ in $y$.
Thus, $F_{1}(x,y/x^{t})=U(x,y/x^{t})V(x,y/x^{t})$ provides a non-trivial
factorization of $f_{0}(x)+f_{2}(x)y^{2}+f_{4}(x)y^{4}$ in $Q(x)[y]$. However,
one can check directly that

$f_{0}(x)+f_{2}(x)y^{2}+f_{4}(x)y^{4}=1+(-x^{7}+4x^{6}-8x^{4}+4x^{2}-x)y^{2}+x^{8}y^{4}$

is irreducible in $Q(x)[y]$, giving a contradiction. So $F_{1}(x,y)$ is
irreducible in $\mathbb{Z}[x,y]$; in other words, one can take $k_{1}=0$ and
$G_{1}(x,y)=F_{1}(x,y)$. Observe that $F_{2}(x,y)$ is divisible by $x^{8}$ for
$t\geq 4$, so $k_{2}\neq 0$ in general. Since $k_{2}$ is in fact the exponent
on the largest power of $x$ dividing every coefficient of $F_{2}(x,y)$ viewed
as a polynomial in $y$, we see that $k_{2}=2t$ if $t<4$ and $k_{2}=8$ if
$t\geq 4$. With this value of $k_{2}$, assume now that
$F_{2}(x,y)=x^{k_{2}}U(x,y)V(x,y)$, where $U(x,y)$ and $V(x,y)$ are nonunits
in $\mathbb{Z}[x,y]$. Given the definition of $k_{2}$ and that $f_{0}(x)=1$,
we see that each of $U(x,y)$ and $V(x,y)$ has degree at least $1$ in $y$. In
this case, we see that

$x^{4t}\big{(}f_{0}(x)+f_{2}(x)y^{2}+f_{4}(x)y^{4}\big{)}=F_{2}(x,x^{t}y)=x^{k_{2}}U(x,x^{t}y)V(x,x^{t}y)$

leads to a non-trivial factorization of $f_{0}(x)+f_{2}(x)y^{2}+f_{4}(x)y^{4}$
in $Q(x)[y]$, which again is a contradiction. Thus,
$F_{2}(x,y)=x^{k_{2}}G_{2}(x,y)$, where $G_{2}(x,y)$ is irreducible in
$\mathbb{Z}[x,y]$. We are now ready to apply Theorem 2.1 to $G_{j}(x,y)$.

Let $g(x)=-x^{6}+4x^{5}-8x^{3}+4x-1$, so $f_{2}(x)=xg(x)$. For each odd prime
$p$, observe that $g(0)\equiv-1\pmod{p}$ and the degree of $g(x)$ is $6$
modulo $p$. Hence, $f_{2}(x)$ times a power of $x$ cannot be a $p$th power
modulo $p$ if $p>7$. A direct check shows that $g(x)$ is furthermore not a
$p$th power modulo $p$ for $p\in\\{3,5\\}$, and hence the same holds for
$f_{2}(x)$ times a power of $x$. It follows that $G_{1}(x,y)$ and $G_{2}(x,y)$
do not satisfy (2.6) for any odd prime $p$.

To finish applying Theorem 2.1, we are left with considering the factorization
of $G_{1}(x,y^{4})$ and $G_{2}(x,y^{4})$. We write $t=4q+r$, where
$r\in\\{0,1,2,3\\}$. Then

$G_{1}(x,y^{4})=F_{1}(x,y^{4})=f_{0}(x)+f_{2}(x)x^{8q+2r}y^{8}+f_{4}(x)x^{16q+4r}y^{16}.$

Assume $G_{1}(x,y^{4})=U(x,y)V(x,y)$ with $U(x,y)$ and $V(x,y)$ in
$\mathbb{Z}[x,y]$ with neither $\pm 1$. As
$\gcd{}_{\mathbb{Z}[x]}\big{(}f_{0}(x),f_{2}(x),f_{4}(x)\big{)}=1$, each of
$U(x,y)$ and $V(x,y)$ has degree at least $1$ in $y$. Observe that

(5.1)
$U(x,y/x^{q})V(x,y/x^{q})=f_{0}(x)+f_{2}(x)x^{2r}y^{8}+f_{4}(x)x^{4r}y^{16}.$

We deduce that $f_{0}(x)+f_{2}(x)x^{2r}y^{8}+f_{4}(x)x^{4r}y^{16}$ is
reducible in $\mathbb{Q}(x)[y]$ and, hence, in $\mathbb{Z}[x,y]$. As
$r\in\\{0,1,2,3\\}$, this is easily checked by a computation. As the $4$
polynomials $f_{0}(x)+f_{2}(x)x^{2r}y^{8}+f_{4}(x)x^{4r}y^{16}$ for
$r\in\\{0,1,2,3\\}$ are irreducible in $\mathbb{Z}[x,y]$, we conclude that
$F_{1}(x,y^{4})$ is irreducible.

The substitution $t=4q+r$, where $r\in\\{0,1,2,3\\}$ in $F_{2}(x,y^{4})$ gives

$F_{2}(x,y^{4})=f_{0}(x)x^{16q+4r}+f_{2}(x)x^{8q+2r}y^{8}+f_{4}(x)y^{16}.$

Note that $F_{2}(x,y^{4})$ is divisible by $f_{4}(x)=x^{8}$ if $q\geq 1$. We
are wanting to know if $G_{2}(x,y^{4})$ is an irreducible polynomial in
$\mathbb{Z}[x,y]$. Assume that
$F_{2}(x,y^{4})=x^{k_{2}}G_{2}(x,y^{4})=x^{k_{2}}U(x,y)V(x,y)$ with $U(x,y)$
and $V(x,y)$ in $\mathbb{Z}[x,y]$ with neither $\pm 1$ and neither divisible
by $x$. As before, each of $U(x,y)$ and $V(x,y)$ has degree at least $1$ in
$y$. Also,

$x^{k_{2}}U(x,x^{q}y)V(x,x^{q}y)=x^{16q}\big{(}f_{0}(x)x^{4r}+f_{2}(x)x^{2r}y^{8}+f_{4}(x)y^{16}\big{)},$

and we deduce that $f_{0}(x)x^{4r}+f_{2}(x)x^{2r}y^{8}+f_{4}(x)y^{16}$ factors
as a power of $x$ times a product of two polynomials in $\mathbb{Z}[x,y]$,
each with degree $\geq 1$ in $y$. We check that this is not the case so that
$G_{2}(x,y^{4})$ is irreducible.

Hence, Theorem 2.1 implies $F_{1}(x,y^{\ell})$ and $F_{2}(x,y^{\ell})$ are a
power of $x$ times an irreducible polynomial in $\mathbb{Z}[x,y]$ for every
positive integer $\ell$. Then Theorem 1.1 implies that the non-reciprocal part
of $F(x)$ is not reducible for $n$ sufficiently large. Since $F(x)$ is a
reciprocal polynomial, we conclude that $F(x)$ is a product of irreducible
reciprocal polynomials.

### Acknowledgements

The author thanks Stavros Garoufalidis for discussions which helped motivate
this study. He is also grateful for the encouragement and interest Andrzej
Schinzel expressed to him on this research during a conversation in Poznań,
Poland, in 2017.

## References

  * [1]
  * [2] E. Bombieri, D. Masser and U. Zannier, _Anomalous subvarieties—structure theorems and applications_ , Int. Math. Res. Not. IMRN 2007, no. 19, Art. ID rnm057, 33 pp.
  * [3] S. Davis, W. Duke and X. Sun, _Probabilistic Galois theory of reciprocal polynomials_ , Exposition. Math. 16 (1998), 263–270.
  * [4] V. Dimitrov, _A proof of the Schinzel-Zassenhaus conjecture on polynomials_ , arXiv:1912.12545.
  * [5] Michael Filaseta, _On the factorization of polynomials with small Euclidean norm_ , In Number theory in progress, Vol. 1 (Zakopane-Kościelisko, 1997), pages 143–163, de Gruyter, Berlin, 1999.
  * [6] M. Filaseta, K. Ford and S. Konyagin, _On an irreducibility theorem of A. Schinzel associated with coverings of the integers_ , Illinois J. Math. 44 (2000), 633–643.
  * [7] M. Filaseta and S. Garoufalidis, _Factorization of polynomials in hyperbolic geometry and dynamics_ , in preparation.
  * [8] M. Filaseta, H. Li, F. Patane and D. Skabelund, _On the irreducibility of the non-reciprocal part of polynomials of the form $f(x)x^{n}+g(x)$_, Acta Arith. 196 (2020), 187–201.
  * [9] S. Garoufalidis and B. Jeon, _On the trace fields of hyperbolic Dehn fillings_ , arXiv:2103.00767.
  * [10] M. L. Macasieb and T. W. Mattman, _Commensurability classes of $(-2,3,n)$ pretzel knot complements_, Algebr. Geom. Topol. 8 (2008), 1833–1853.
  * [11] W. D. Neumann and A. W. Reid, _Arithmetic of hyperbolic manifolds_ , Topology ’90 Conference, Columbus, OH, 1990, _Ohio State Univ. Math. Res. Inst. Publ._ , Vol. 1, de Gruyter, Berlin, 1992, 273–310.
  * [12] William Sawin, Mark Shusterman and Michael Stoll, _Irreducibility of polynomials with a large gap_ , Acta Arith. 192 (2020), 111-139
  * [13] A. Schinzel, _Reducible lacunary polynomials_ , _Seminar on number theory, 1983–1984_ , Talence, (1983/1984), Univ. Bordeaux I, Talence, 1984, Exp. No. 29, 5.
  * [14] A. Schinzel, _Reducibility of lacunary polynomials_ , _1969 Number Theory Institute_ , Proc. Sympos. Pure Math., Vol. XX, State Univ. New York, Stony Brook, N.Y. 1969, Amer. Math. Soc., Providence, R.I. (1971), 135–149.
  * [15] A. Schinzel, _Reducibility of lacunary polynomials I_ , Acta Arith., 16 (1969/70), 123–159.
  * [16] A. Schinzel, _Reducibility of lacunary polynomials II_ , Acta Arith., 16 (1969/70), 371–392.
  * [17] A. Schinzel, _Reducibility of lacunary polynomials III_ , Acta Arith., 34 (1977/78), 227–266.
  * [18] A. Schinzel, _Reducibility of lacunary polynomials IV_ , Acta Arith., 43 (1984), 313–315.
  * [19] A. Schinzel, _Reducibility of lacunary polynomials V_ , Acta Arith., 43 (1984), 425–440.
  * [20] A. Schinzel, _Reducibility of lacunary polynomials VI_ , Acta Arith., 47 (1986), 277–293.
  * [21] A. Schinzel, _Reducibility of lacunary polynomials VII_ , Monatsh. Math., 102 (1986) 309–337.
  * [22] A. Schinzel, _Errata: “Reducibility of lacunary polynomials VII” [Monatsh. Math. 102 (1986), 309–337]_ , Acta Arith., 53 (1989), 95
  * [23] A. Schinzel, _Reducibility of lacunary polynomials VIII_ , Acta Arith., 50 (1988), 91–106.
  * [24] A. Schinzel, _Reducibility of lacunary polynomials IX_ , _New advances in transcendence theory_ , Durham, 1986, Cambridge Univ. Press, Cambridge, 1988, 313–336.
  * [25] A. Schinzel, _Reducibility of lacunary polynomials X_ , Acta Arith., 53 (1989), 47–97.
  * [26] A. Schinzel, _Reducibility of lacunary polynomials XI_ , Acta Arith., 57 (1991), 165–175.
  * [27] A. Schinzel, _Reducibility of lacunary polynomials XII_ , Acta Arith., 90 (1999), 273–289.
  * [28] A. Schinzel, _Corrigendum to the paper “Reducibility of lacunary polynomials XII”_ , Acta Arith., 188 (2019), 99\.
  * [29] A. Schinzel, _Reducibility of polynomials and covering systems of congruences_ , Acta Arith. 13 (1967), 91–101.
  * [30] A. Schinzel, _Selected topics on polynomials_ , University of Michigan Press, Ann Arbor, Mich., 1982.
  * [31] A. Schinzel, _Polynomials with special regard to reducibility_ , Encyclopedia of Mathematics and its Applications 77, Cambridge University Press, 2000.
  * [32] A. Schinzel and J. Wójcik, _A note on the paper “Reducibility of lacunary polynomials I”_ , Acta Arith., 19 (1971), 195–201.