11footnotetext: Corresponding Author email(s<EMAIL_ADDRESS><EMAIL_ADDRESS>
2010MSC: Primary 12E05; 30C15; 11C08; 26C10; 26D05; 30A10
_Keywords_ : Polynomials; Integer coefficients; Location of zeros; Bounds;
Inequalities, Irreducibility

# Irreducibility via location of zeros

Jitender Singh1,∗ 0000-0003-3706-8239 Department of Mathematics, Guru Nanak
Dev University, Amritsar-143005<EMAIL_ADDRESS>and  Sanjeev Kumar2
0000-0001-6882-4733 Department of Mathematics, SGGS College, Sector-26,
Chandigarh-160019<EMAIL_ADDRESS>
###### Abstract.

In this paper, we obtain several new classes of irreducible polynomials having
integer coefficients whose zeros lie inside an open disk around the origin or
outside a closed annular region in the complex plane. Such irreducible
polynomials are devised by imposing Perron–type sufficiency conditions on
their coefficients.

##  1\. Introduction.

The classical irreducibility criteria due to Schönemann (1846), Eisenstein
(1850), Dumas (1906), and Perron (1907) have become paradigm for testing
irreducibility of polynomials having rational coefficients. In fact
irreducible polynomials having integer coefficients can be used to construct
finite-fields, a fundamental tool in cryptography and computer science. The
realm revealing riveting facts about the irreducibility of polynomials over
prescribed domains has always been the cradle of such crucial classical
results which for decades have witnessed cogent extensions and
generalizations. Such irreducibility criteria have exhibited a close affinity
to prime numbers and primality as is evident from the illustrious
Buniakowski’s conjecture of 1854 which asserts that if $f$ is an irreducible
polynomial having integer coefficients such that the elements in the set
$f(\mathbb{N})$ have no common factors other than $\pm 1$, then the set
$f(\mathbb{N})$ contains infinitely many prime numbers. The converse of
Buniakowski’s conjecture holds affirmatively via primality.

Another classical irreducibility result due to A. Cohn [1, p. 133] states that
if a prime number can be expressed in base 10 as $\sum_{i=0}^{m}a_{i}10^{i}$
for some positive integer $m$, then the polynomial $\sum_{i=0}^{m}a_{i}z^{i}$
is irreducible in $\mathbb{Z}[z]$. Cohn’s result was then generalized to
arbitrary base by Brillhart et al. [2] and further in Bonciocat et al. [3]. In
[4], Murty utilized the information about location of zeros of the given
polynomial in the complex plane as a tool to provide an elementary proof of
Cohn’s irreducibility criterion. Interestingly, one of the main results of
Murty [4], generalized by Girstmair [5] apprised of a strong converse of
Buniakowski’s conjecture. Thus, the information about location of zeros of a
given polynomial having integer coefficients can play a significant role in
deducing irreducibility. In view of this, Perron (1907) proved that if
$f=a_{0}+a_{1}z+a_{2}z^{2}+\cdots+a_{m-2}z^{m-2}+a_{m-1}z^{m-1}+z^{m}\in\mathbb{Z}[z]$,
$m\geq 2$ is such that

$\displaystyle|a_{m-1}|$ $\displaystyle>$ $\displaystyle
1+|a_{m-2}|+|a_{m-3}|+\cdots+|a_{1}|+|a_{0}|,$

then $f$ is irreducible in $\mathbb{Z}[z]$. As an application of Rouché’s
Theorem (see[6, p. 153]), the aforementioned inequality reveals that exactly
$m-1$ zeros of $f$ lie in the interior of the unit disk $|z|\leq 1$ in the
complex plane and the remaining one zero of $f$ lies outside the unit disk.
The irreducibility of $f$ can be deduced as follows. On the contrary, assume
that $f(z)=f_{1}(z)f_{2}(z)$ for nonconstant polynomials $f_{1}$ and $f_{2}$
in $\mathbb{Z}[z]$. Then all zeros of one of $f_{1}$ or $f_{2}$ lie in the
interior of the unit circle in the complex plane. Assume that each zero
$\theta$ of $f_{1}$ satisfies $|\theta|<1$. Then expressing $f_{1}$ as
$f_{1}(z)=\prod_{\theta}(z-\theta)$, where the product is over all zeros of
$f_{1}$, we have

$\displaystyle 1\leq|f_{1}(0)|=\prod_{\theta}|\theta|<1,$

which is a contradiction, and so, $f$ must be irreducible. On these lines, one
has the following class of irreducible polynomials.

######  Theorem A.

For each natural number $m\geq 2$, the polynomial

$\displaystyle f(z)=z^{m}-z^{m-1}-\cdots-z-1$

is irreducible in $\mathbb{Z}[z]$.

###### Proof.

Observe that $f(1)=-(m-1)<0$ and $f(2)=1>0$ so that a real zero $\zeta$ of $f$
lies between $1$ and $2$. If $n$ is odd, then $\zeta$ is the only real zero of
$f$. If $m$ is even, then $f$ has another real zero which lies between $-1$
and $0$ since then $f(-1)=1$ and $f(0)=-1$. In either case, exactly one real
zero $\zeta$ of $f$ satisfies $1<\zeta<2$. Consequently, one has
$f(z)=(z-\zeta)g(z)$, where the degree $m-1$ polynomial $g\in\mathbb{R}[z]$ is
given by

(1)
$g=\frac{1}{\zeta}+\Bigl{(}\frac{1}{\zeta}+\frac{1}{\zeta^{2}}\Bigr{)}z+\cdots+\Bigl{(}\frac{1}{\zeta}+\frac{1}{\zeta^{2}}+\cdots+\frac{1}{\zeta^{n-1}}\Bigr{)}z^{n-2}+z^{n-1}.$

By an easily derivable result of Aziz and Mohammad [10, Theorem B], we observe
that each zero $\theta$ of $g$ satisfies

$\displaystyle|\theta|\leq\max_{1\leq k\leq
m-1}\Bigl{\\{}\frac{\sum_{j=1}^{k}\zeta^{j}}{1+\sum_{j=1}^{k}\zeta^{j}}\Bigr{\\}}=1-\frac{1}{\zeta^{m}}<1.$

Thus, except $\zeta$, all other zeros of $f$ lie in the interior of the unit
disk $|z|<1$ in the complex plane. Now the irreducibility of $f$ is immediate.
∎

Perron [7] provided multivariate version of his irreducibility criterion for
polynomials over arbitrary fields. An elegant proof of the multivariate
version of Perron’s irreducibility criterion has been provided by Bonciocat
[8]. Recently, in [9], Singh and Garg proved an interesting factorization
result on a class of polynomials having integer coefficients, which also
generalize the irreducibility criterion of Perron. In this paper, we shall
approach irreducibility by devising some Perron–type sufficiency conditions on
the coefficients of the given polynomial having integer coefficients. Recall
that a polynomial having integer coefficients is called primitive if the
greatest common divisor of its coefficients is 1. Our main results are the
following:

######  Theorem 1.

Let $f=a_{0}+a_{1}z+\cdots+a_{m}z^{m}\in\mathbb{Z}[z]$ be primitive. Suppose
there exists a positive real number $\alpha$ such that

$\displaystyle|a_{m}|\alpha^{m}>|a_{0}|+|a_{1}|\alpha+\cdots+|a_{m-1}|\alpha^{m-1}.$

Further, if there exist natural numbers $n$ and $d$ satisfying $n\geq\alpha+d$
for which $|f(n)|/d$ is prime, or $|f(n)|/d$ is a prime-power coprime to
$|f^{\prime}(n)|$, then $f$ is irreducible in $\mathbb{Z}[z]$.

######  Theorem 2.

Let $f=a_{0}+a_{1}z+\cdots+a_{m}z^{m}\in\mathbb{Z}[z]$ be primitive. Suppose
there exist positive real numbers $\alpha$ and $\beta$ with $\alpha<\beta$ and
an index $j\in\\{0,1,\ldots,m\\}$ such that

$\displaystyle|a_{j}|\alpha^{j}>\sum_{i\neq j,~{}i=0}^{m}|a_{i}|\beta^{i}.$

Further, if there exist natural numbers $n$ and $d$ satisfying $\beta-d\geq
n\geq\alpha+d$ for which $|f(n)|/d$ is prime, or $|f(n)|/d$ is a prime-power
coprime to $|f^{\prime}(n)|$, then $f$ is irreducible in $\mathbb{Z}[z]$.

######  Theorem 3.

Let $f=a_{0}+a_{1}z+\cdots+a_{m}z^{m}\in\mathbb{Z}[z]$ be primitive. Suppose
there exists positive real numbers $\alpha$ and $\beta$ with $\alpha<\beta$
and an index $j\in\\{0,1,\ldots,m\\}$ such that

$\displaystyle|a_{j}|\alpha^{j}>\bigl{(}{\beta}/{\alpha}\bigr{)}^{m-j}\sum_{i\neq
j,~{}i=0}^{m}|a_{i}|\alpha^{i}.$

Further, if there exist natural numbers $n$ and $d$ satisfying $\beta-d\geq
n\geq\alpha+d$ for which $|f(n)|/d$ is prime, or $|f(n)|/d$ is a prime-power
coprime to $|f^{\prime}(n)|$, then $f$ is irreducible in $\mathbb{Z}[z]$.

######  Remark.

If we take $j=m$ in Theorem 3, then the condition on the coefficients of $f$
becomes $|a_{m}|\alpha^{m}>\sum_{i\neq j,~{}i=0}^{m}|a_{i}|\alpha^{i}$, which
holds for all $\beta>\alpha$. In view of this, once an $n$ with
$n\geq\alpha+d$ is found for which $|f(n)|/d$ is prime or a prime-power, we
can take $\beta=n+d$ so that $\beta-d\geq n$ can always be satisfied. Thus,
for $j=m$, Theorem 3 reduces to Theorem 1.

######  Example 1.

For $k\geq m+2\geq 4$ and prime $p$ with $n=p^{2}>p+d$, $1\leq d\leq p(p-1)$;
the polynomial

$\displaystyle V_{d}=-p^{2}+z\pm p^{k-m}dz^{m}$

satisfies the hypothesis of Theorem 1 with

$\displaystyle\alpha$ $\displaystyle=$ $\displaystyle
p,~{}a_{0}=-p^{2},~{}a_{1}=1,a_{i}=0~{}(i=2,3,\ldots,m-1),~{}a_{m}=\pm
p^{k-m}d,$ $\displaystyle n$ $\displaystyle=$ $\displaystyle p^{2}\geq
p+d=\alpha+d,~{}V_{d}(p^{2})=\pm p^{k+m}d,~{}V_{d}^{\prime}(p^{2})\equiv 1\mod
p,$

so that $\gcd(|V_{d}(p^{2})|/d,|V_{d}^{\prime}(p^{2})|)=1$, and

$\displaystyle|a_{m}|\alpha^{m}=p^{k}d\geq
p^{3}>p^{2}+p=\sum_{i=0}^{m-1}|a_{i}|\alpha^{i}.$

So, the polynomial $V_{d}$ is irreducible in $\mathbb{Z}[z]$.

######  Example 2.

Consider the polynomial

$\displaystyle W_{d}=(z-p)+(z-p)^{2}+\cdots+(z-p)^{m-1}\pm(p^{2k-2}d)z^{m},$

for $k\geq m\geq 2$ and $p\geq 1+d$. Here, $a_{i}=\sum_{j=i}^{m-1}{j\choose
i}(-p)^{j-i}$ for $i=0,1,\ldots,m-1$. Further, $a_{m}=\pm p^{2k-2}d$,
$\alpha=1$, and $n=p\geq 1+d$. We find that $W_{d}(p)=\pm p^{2k+m-2}d$,
$W_{d}^{\prime}(p)\equiv 1\mod p$. These along with the fact that $p^{2}>1+p$
yield the following:

$\displaystyle|a_{m}|\alpha^{m}=\frac{p^{2k}d}{p^{2}}\geq\frac{(p^{2})^{m}d}{p^{2}}>\frac{(1+p^{2})^{m}}{p^{2}}>(1+p^{2})\frac{(1+p^{2})^{m-1}-1}{1+p^{2}-1}>\sum_{i=0}^{m-1}|a_{i}|\alpha^{i}.$

Since $a_{m-1}=1$, it follows that $W_{d}$ is a primitive polynomial. By
Theorem 1, the polynomial $W_{d}$ is irreducible in $\mathbb{Z}[z]$.

######  Example 3.

Consider the polynomial

$X_{d}=(p^{m}-p)+z\pm(p^{1+2m+j}d)z^{j}-z^{m},~{}1\leq d\leq p-1,~{}2\leq
j\leq m-1,$

where $j$, $m\geq 3$ and $d$ are positive integers and $p$ is a prime number.
Here, $a_{0}=p^{m}-p$, $a_{1}=1$, $a_{i}=0$ for $i\not\in\\{0,1,j,m\\}$,
$a_{j}=\pm p^{1+2m+j}d$, and $a_{m}=-1$. Taking $\alpha=1/p$, $\beta=p^{2}$,
and $n=p$, we have

$\displaystyle|a_{j}|\alpha^{j}$ $\displaystyle=$ $\displaystyle
p^{1+2m}d>p^{m}-p+p^{2}+p^{2m}=\sum_{i\neq j,~{}i=0}^{m}|a_{i}|\beta^{i},$
$\displaystyle\alpha+d$ $\displaystyle<$ $\displaystyle 1+d\leq
n=p<p^{2}-d=\beta-d.$

Further, the polynomial $X_{d}$ is primitive, and $|X_{d}(p)|/d=p^{1+2(m+j)}$
and ${X_{d}}^{\prime}(p)\equiv 1\mod p$. By Theorem 2, the polynomial $X_{d}$
is irreducible in $\mathbb{Z}[z]$.

######  Example 4.

Consider the polynomial

$Y_{d}=(p^{m}-p)+z\pm(p^{k}d)z^{j}-z^{m};~{}1\leq d\leq p-1,~{}2\leq j\leq
m-1,$

where $k\geq 3m-2j$, $m\geq 3$, $j$, and $d$ are positive integers and $p$ is
a prime number. Here, $a_{0}=p^{m}-p$, $a_{1}=1$, $a_{i}=0$ for
$i\not\in\\{0,1,j,m\\}$, $a_{j}=\pm p^{k}d$, and $a_{m}=-1$. Taking
$\alpha=1$, $\beta=p+d$, and $n=p$, we have

$\displaystyle|a_{j}|\alpha^{j}$ $\displaystyle=$ $\displaystyle p^{k}d\geq
p^{3m-2j}>(p+d)^{m-j}(p^{m}-p+2)=\bigl{(}{\beta}/{\alpha}\bigr{)}^{m-j}\sum_{i\neq
j,~{}i=0}^{m}|a_{i}|\alpha^{i},$ $\displaystyle\alpha+d$ $\displaystyle=$
$\displaystyle 1+d\leq n=p=\beta-d.$

Further, the polynomial $Y_{d}$ is primitive, and $|Y_{d}(p)|/d=p^{k+j}$ and
${Y_{d}}^{\prime}(p)\equiv 1\mod p$. By Theorem 3, the polynomial $Y_{d}$ is
irreducible in $\mathbb{Z}[z]$.

##  2\. Proof of Theorems 1–3.

###### Proof of Theorem 1.

Let $|f(n)|/d=p^{k}$ for some prime $p$ and positive integer $k$. If
$|z|\geq\alpha$, then in view of the hypothesis, we have
$|a_{m}|\alpha^{m}>\sum_{j=0}^{m-1}|a_{j}|\alpha^{j}$. Consequently, we have

$\displaystyle|f(z)|\geq|z|^{m}\Bigl{(}|a_{m}|-\sum_{i=0}^{m-1}|a_{i}||z|^{-(m-i)}\Bigr{)}\geq\alpha^{m}\Bigl{(}|a_{m}|-\sum_{i=0}^{m-1}|a_{i}|\alpha^{-(m-i)}\Bigr{)}>0,$

which shows that each zero $\theta$ of $f$ satisfies $|\theta|<\alpha$. Now
assume on the contrary that $f(z)=f_{1}(z)f_{2}(z)$ for nonconstant
polynomials $f_{1}$ and $f_{2}\in\mathbb{Z}[z]$. Since we have

$\displaystyle\pm p^{k}d=f(n)=f_{1}(n)f_{2}(n),$

at least one of $|f_{1}(n)|$ or $|f_{2}(n)|$ is divisible by $p$. Assume that
$p$ divides $|f_{2}(n)|$. Firstly, let us suppose that $p$ does not divide
$|f_{1}(n)|$. Then $p^{k}$ divides $|f_{2}(n)|$ so that $p^{k}\leq|f_{2}(n)|$.
Then we have $|f_{1}(n)|\leq d$. If $c~{}(\neq 0)$ is the leading coefficient
of $f_{1}$, then

$\displaystyle f_{1}(n)=c\prod_{\theta}(n-\theta),$

where the product runs over all zeros $\theta$ of $f_{1}$. Observe that each
such $\theta$ satisfies $|\theta|<\alpha$. In view of this, we have the
following inequalities

$\displaystyle|n-\theta|\geq n-|\theta|>n-\alpha\geq d,$

using which we arrive at the following:

$\displaystyle d\geq|f_{1}(n)|=|c|\prod_{\theta}|n-\theta|$ $\displaystyle>$
$\displaystyle|c|d^{\deg f_{1}}\geq|c|d\geq d,$

leading to a contradiction.

Now assume that $p$ divides $|f_{1}(n)|$. Since $p$ divides $|f_{2}(n)|$, we
must have $k\geq 2$. Consequently, $p$ divides
$|{f_{1}}^{\prime}(n)f_{2}(n)+f_{1}(n){f_{2}}^{\prime}(n)|=|f^{\prime}(n)|$.
This contradicts the hypothesis. ∎

###### Proof of Theorem 2.

Let $|f(n)|/d=p^{k}$ for some prime $p$ and positive integer $k$. If
$z\in\mathbb{C}$ is such that $\alpha\leq|z|\leq\beta$, then using the
hypothesis we have

$\displaystyle|f(z)|\geq|a_{j}||z|^{j}-\sum_{i\neq
j,~{}i=0}^{m}|a_{i}||z|^{i}\geq|a_{j}|\alpha^{j}-\sum_{i\neq
j,~{}i=0}^{m}|a_{i}|\beta^{i}>0,$

which shows that each zero $\theta$ of $f$ satisfies $|\theta|<\alpha$ or
$\beta<|\theta|$. Now suppose on the contrary that $f=f_{1}(z)f_{2}(z)$ for
nonconstant polynomials $f_{1}$ and $f_{2}$ in $\mathbb{Z}[z]$. By the
hypothesis, we have $f(n)=\pm p^{k}d$ for some positive integer $k$. So,
$|f(n)|=p^{k}d=|f_{1}(n)||f_{2}(n)|$, which shows that $p$ divides at least
one of $|f_{1}(n)|$ or $|f_{2}(n)|$. Assume that $p$ divides $|f_{2}(n)|$. If
$p$ does not divide $|f_{1}(n)|$, then $p^{k}$ divides $|f_{2}(n)|$ so that
$p^{k}\leq|f_{2}(n)|$. We then have $|f_{1}(n)|\leq d$. We can write
$f_{1}(z)=c\prod_{\theta}(z-\theta)$, where $c\neq 0$ is the leading
coefficient of $f_{1}$, and the product runs over all zeros $\theta$ of
$f_{1}$. Further, for any zero $\theta$ of $f_{1}$, we have either
$|\theta|<\alpha$, or $\beta<|\theta|$, and so, we arrive at the following
inequalities.

$\displaystyle|n-\theta|$ $\displaystyle\geq$ $\displaystyle
n-|\theta|>n-\alpha\geq d,~{}\text{whenever}~{}|\theta|<\alpha,$
$\displaystyle|n-\theta|$ $\displaystyle\geq$
$\displaystyle|\theta|-n>\beta-n\geq d,~{}\text{whenever}~{}\beta<|\theta|.$

Consequently, each zero $\theta$ of $f_{1}$ satisfies $|n-\theta|>d$, and
therefore we have

$\displaystyle d\geq|f_{1}(n)|$ $\displaystyle=$
$\displaystyle|c|\prod_{\theta}|n-\theta|>|c|d^{\deg f_{1}}\geq|c|d\geq d,$

leading to a contradiction. ∎

###### Proof of Theorem 3.

If $z\in\mathbb{C}$ is such that $\alpha\leq|z|\leq\beta$, then we have
$\frac{1}{|z|^{m-j}}\geq\frac{1}{\beta^{m-j}}$ and
$\frac{-1}{|z|^{m-i}}\geq\frac{-1}{\alpha^{m-i}}$. Using these inequalities
along with the hypothesis we arrive at the following:

$\displaystyle|f(z)|\geq|z|^{m}\Bigl{(}\frac{|a_{j}|}{|z|^{m-j}}-\sum_{i\neq
j,~{}i=0}^{m}\frac{|a_{i}|}{|z|^{m-i}}\Bigr{)}$ $\displaystyle\geq$
$\displaystyle\alpha^{m}\Bigl{(}\frac{|a_{j}|}{\beta^{m-j}}-\sum_{i\neq
j,~{}i=0}^{m}\frac{|a_{i}|}{\alpha^{m-i}}\Bigr{)},$ $\displaystyle=$
$\displaystyle\Bigl{(}|a_{j}|\alpha^{j}\Bigl{(}\frac{\alpha}{\beta}\Bigr{)}^{m-j}-\sum_{i\neq
j,~{}i=0}^{m}|a_{i}|\alpha^{i}\Bigr{)}>0,$

which shows that each zero $\theta$ of $f$ satisfies $|\theta|<\alpha$ or
$\beta<|\theta|$. Now the irreducibility of $f$ can be deduced in exactly the
same way as in the proof of Theorem 2. ∎

##  3\. Further generalizations

In this section, we generalize Theorems 1-3 to include the case when
$|f(n)|/d$ and $|f^{\prime}(n)|$ are not coprime. In the following, let
$f^{(0)}(z)=f(z)$ and for each natural number $i$, $f^{(i)}(z)$ denote the
$i$-th derivative of $f$ with respect to $z$.

######  Theorem 4.

Let $f=a_{0}+a_{1}z+\cdots+a_{m}z^{m}\in\mathbb{Z}[z]$ be primitive. Suppose
there exists a positive real number $\alpha$ such that

$\displaystyle|a_{m}|\alpha^{m}>|a_{0}|+|a_{1}|\alpha+\cdots+|a_{m-1}|\alpha^{m-1}.$

Further, if there exist natural numbers $n$, $d$, $k$, $\ell\leq m$, and a
prime $p\nmid d$ such that $n\geq\alpha+d$, $f(n)=\pm p^{k}d$,
$\gcd(k,\ell)=1$, $p^{k}\mid\frac{f^{(i)}(n)}{i!}$ for each index
$i=0,1,\ldots,\ell-1$, and for $k>1$, also
$p\nmid\frac{f^{(\ell)}(n)}{\ell!}$, then the polynomial $f$ is irreducible in
$\mathbb{Z}[z]$.

######  Theorem 5.

Let $f=a_{0}+a_{1}z+\cdots+a_{m}z^{m}\in\mathbb{Z}[z]$ be primitive. Suppose
there exist positive real numbers $\alpha$ and $\beta$ with $\alpha<\beta$ and
an index $j\in\\{0,1,\ldots,m\\}$ such that

$\displaystyle|a_{j}|\alpha^{j}>\sum_{i\neq j,~{}i=0}^{m}|a_{i}|\beta^{i}.$

Further, if there exist natural numbers $n$, $d$, $k$, $\ell\leq m$, and a
prime $p\nmid d$ such that $\beta-d\geq n\geq\alpha+d$, $f(n)=\pm p^{k}d$,
$\gcd(k,\ell)=1$, $p^{k}\mid\frac{f^{(i)}(n)}{i!}$ for each index
$i=0,1,\ldots,\ell-1$, and for $k>1$, also
$p\nmid\frac{f^{(\ell)}(n)}{\ell!}$, then the polynomial $f$ is irreducible in
$\mathbb{Z}[z]$.

######  Theorem 6.

Let $f=a_{0}+a_{1}z+\cdots+a_{m}z^{m}\in\mathbb{Z}[z]$ be primitive. Suppose
there exist positive real numbers $\alpha$ and $\beta$ with $\alpha<\beta$ and
an index $j\in\\{0,1,\ldots,m\\}$ such that

$\displaystyle|a_{j}|\alpha^{j}>\bigl{(}{\beta}/{\alpha}\bigr{)}^{m-j}\sum_{i\neq
j,~{}i=0}^{m}|a_{i}|\alpha^{i}.$

Further, if there exist natural numbers $n$, $d$, $k$, $\ell\leq m$, and a
prime $p\nmid d$ such that $\beta-d\geq n\geq\alpha+d$, $f(n)=\pm p^{k}d$,
$\gcd(k,\ell)=1$, $p^{k}\mid\frac{f^{(i)}(n)}{i!}$ for each index
$i=0,1,\ldots,\ell-1$, and for $k>1$, also
$p\nmid\frac{f^{(\ell)}(n)}{\ell!}$, then the polynomial $f$ is irreducible in
$\mathbb{Z}[z]$.

To prove Theorems 4-6, we need the following result of Singh and Kumar [11,
Lemma 3].

######  Lemma B (Singh and Kumar [11]).

Let $f=a_{0}+a_{1}z+\cdots+a_{m}z^{m}$; $f_{1}$ and $f_{2}$ be nonconstant
polynomials in $\mathbb{Z}[z]$ such that $f(z)=f_{1}(z)f_{2}(z)$. Suppose
there exists a prime $p,$ coprime natural numbers $k\geq 2$ and $\ell\leq m$
such that $p^{k}\mid\gcd(a_{0},a_{1},\ldots,a_{\ell-1})$ and $p^{k+1}\nmid
a_{0}$. If $p\mid f_{1}(0)$ and $p\mid f_{2}(0)$, then $p\mid a_{\ell}$.

For each $i=0,\ldots,m$, let $s_{i}=\frac{f^{(i)}(n)}{i!}$, and define

(2) $g(z)=f(z+n)=s_{0}+s_{1}z+\cdots+s_{m}z^{m}.$

To prove Theorems 4 and 5, it will be enough to show that $g$ is irreducible.
In view of the hypothesis of each of Theorems 4 and 5, $p$ divides $s_{i}$ for
each $i=0,\ldots,\ell-1$. Suppose on the contrary that $g(z)=g_{1}(z)g_{2}(z)$
for nonconstant polynomials $g_{1}$ and $g_{2}$ in $\mathbb{Z}[z]$. We find
that

(3) $|g(0)|=|f(n)|=p^{k}d=|g_{1}(0)||g_{2}(0)|.$

So, $p$ divides at least one of the factors $|g_{1}(0)|$ or $|g_{2}(0)|$.
Assume that $p$ divides $|g_{2}(0)|$.

###### Proof of Theorem 4.

If $p$ divides $|g_{1}(0)|$, then $k>1$ and $p$ divides $|g_{i}(0)|$ for each
$i=1$, $2$. Since $p\nmid d$, it follows that $p^{k+1}\nmid|g(0)|$. Now
applying Lemma B on $g$, we get $p\mid s_{\ell}$. This contradicts the
hypothesis.

Now consider the case when $p\nmid|g_{1}(0)|$. From (3), we find that
$|g_{1}(0)|$ divides $d$ so that $|g_{1}(0)|\leq d$. If $c\neq 0$ is the
leading coefficient of $g_{1}$, then $g_{1}(z)=c\prod_{\theta}(z-\theta)$,
where the product runs over all zeros $\theta$ of $g_{1}$. If $g(\theta)=0$,
then from (2) we have $f(\theta+n)=0$, which in view of the hypothesis gives
$|\theta+n|<\alpha$. Consequently, we have

$\displaystyle|\theta|$ $\displaystyle=$ $\displaystyle|n-(n+\theta)|\geq
n-|n+\theta|>n-\alpha\geq d,$

and hence we have
$d\geq|g_{1}(0)|=|c|\prod_{\theta}|\theta|>|c|d^{\deg{g_{1}}}\geq|c|d\geq d$,
leading to a contradiction. ∎

###### Proof of Theorem 5.

As in the preceding proof, if $p$ divides $|g_{1}(0)|$, then $k>1$ and $p$
divides $|g_{i}(0)|$ for each $i=1$, $2$. Now applying Lemma B on $g$, we get
$p\mid s_{\ell}$, which contradicts the hypothesis.

Now consider the case when $p$ does not divide $|g_{1}(0)|$. From (3), we find
that $|g_{1}(0)|\leq d$. If $c\neq 0$ is the leading coefficient of $g_{1}$,
then $g_{1}(z)=c\prod_{\theta}(z-\theta)$, where the product runs over all
zeros $\theta$ of $g_{1}$. If $g(\theta)=0$, then from (2) we have
$f(\theta+n)=0$, which by the hypothesis gives either $|\theta+n|<\alpha$, or
$\beta<|\theta+n|$. Now arguing as before, we have the following inequalities.

$\displaystyle|\theta|$ $\displaystyle=$ $\displaystyle|n-(n+\theta)|\geq
n-|n+\theta|>n-\alpha\geq d,~{}\text{if}~{}|\theta+n|<\alpha,$
$\displaystyle|\theta|$ $\displaystyle=$
$\displaystyle|n+\theta-n|\geq|n+\theta|-n>\beta-n\geq
d,~{}\text{if}~{}\beta<|\theta+n|.$

Thus, $|\theta|>d$ for each zero $\theta$ of $g_{1}$, and so,
$d\geq|g_{1}(0)|=|c|\prod_{\theta}|\theta|>|c|d^{\deg{g_{1}}}\geq d$, which is
a contradiction. ∎

###### Proof of Theorem 6.

The inequality $|a_{j}|\alpha^{j}>(\beta/\alpha)^{m-j}\sum_{i\neq
j,~{}i=0}^{m}|a_{i}|\alpha^{i}$ shows that each zero $\theta$ of $f$ satisfies
$|\theta|<\alpha$, or $\beta<|\theta|$. So, the irreducibility of $f$ follows
as in the proof of Theorem 5. ∎

######  Example 5.

Let $Y_{\ell,d}$ be the polynomial defined by

$Y_{\ell,d}=p^{\ell}+p^{\ell-1}z+p^{\ell-2}z^{2}+\ldots+pz^{\ell-1}-\ell
z^{\ell}\pm(p^{k}d)z^{m},$

where $\ell$, $k\geq m+\ell$, $m>\ell$, and $d$ are positive integers, and
$p\geq 1+d$ is a prime number with $\gcd(p,\ell)=1$, $\gcd(k,\ell)=1$. Here,
$a_{i}=p^{\ell-i}$ for $i=0,\ldots,\ell-1$, $a_{\ell}=-\ell$, $a_{i}=0$ for
$i=\ell+1,\ldots,m-1$, and $a_{m}=\pm p^{k}d$. Taking $\alpha=1$, $n=p$, we
have

$\displaystyle|a_{m}|\alpha^{m}=p^{k}d\geq p^{m+\ell+1}$ $\displaystyle=$
$\displaystyle\underbrace{p^{m}+p^{m}+\cdots+p^{m}}_{p^{\ell+1}\text{-times}}$
$\displaystyle>$
$\displaystyle(p^{\ell}+p^{\ell-1}+p^{\ell-2}+\cdots+p^{1})+\ell=\sum_{i=0}^{m-1}|a_{i}|\alpha^{i},$

where $n=p\geq 1+d=\alpha+d$. Since $\gcd(p,\ell)=1$, the polynomial
$Y_{\ell,d}$ is primitive, and $|Y_{\ell,d}(p)|/d=p^{k+m}$, where $p$ divides
${Y_{\ell,d}}^{(i)}(p)/i!$ for $i=0,\ldots,\ell-1$, and
${Y_{\ell,d}}^{(\ell)}(p)/\ell!=-\ell$, which is assumed to be coprime to $p$.
By Theorem 4, the polynomial $Y_{\ell,d}$ is irreducible in $\mathbb{Z}[z]$.

######  Example 6.

Consider the polynomial

$Z_{\ell,d}=p^{m}+(p^{\ell-1}z+p^{\ell-2}z^{2}+\ldots+pz^{\ell-1})-(\ell-1)z^{\ell}\pm(p^{k-j}d)z^{j}-z^{m},$

where $\ell\geq 2$, $j>\ell$, $k\geq 2(m+\ell)+j$, $m\geq j+1$ and $d$ are
positive integers, and $p\geq 1+d$ is a prime number with $\gcd(p,\ell-1)=1$,
$\gcd(k,\ell)=1$. Here, $a_{0}=p^{m}$, $a_{i}=p^{\ell-i}$ for
$i=1,\ldots,\ell-1$, $a_{\ell}=-(\ell-1)$, $a_{i}=0$ for $i\neq j$ and
$i\in\\{\ell+1,\ldots,m-1\\}$, $a_{j}=\pm p^{k-j}d$, and $a_{m}=-1$. Taking
$\alpha=1/p$, $\beta=p^{2}$, and $n=p$, we have

$\displaystyle|a_{j}|\alpha^{j}=p^{k}d\geq p^{2m+2\ell}$ $\displaystyle=$
$\displaystyle\underbrace{p^{2m}+p^{2m}+\cdots+p^{2m}}_{p^{2\ell}\text{-times}}$
$\displaystyle>$ $\displaystyle
p^{m}+(p^{\ell+1}+p^{\ell+2}+\cdots+p^{2\ell-1})+(\ell-1)p^{2\ell}+p^{2m}$
$\displaystyle=$ $\displaystyle\sum_{i\neq j,~{}i=0}^{m}|a_{i}|\beta^{i},$

where $\beta-d=p^{2}-d>p=n\geq 1+d>\alpha+d$. Further, the polynomial
$Z_{\ell,d}$ is primitive, and $|Z_{\ell,d}(p)|/d=p^{k}$, where $p$ divides
${Z_{\ell,d}}^{(i)}(p)/i!$ for $i=0,\ldots,\ell-1$, and
${Z_{\ell,d}}^{(\ell)}(p)/\ell!=-(\ell-1)$, which is assumed to be coprime to
$p$. So, by Theorem 5, the polynomial $Z_{\ell,d}$ is irreducible in
$\mathbb{Z}[z]$.

### Compliance with Ethical Standards

The authors declare to have no conflict of interest.

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