Factorization of Dickson polynomials over Finite Fields

TL;DR

This paper explicitly factors Dickson polynomials of the first and second kind over finite fields when every prime dividing n also divides q-1, extending previous results in the literature.

Contribution

It provides explicit irreducible factorization formulas for Dickson polynomials under specific divisibility conditions, generalizing prior work.

Findings

Explicit factorization formulas derived

Generalizes previous factorization results

Applicable when prime divisors of n divide q-1

Abstract

Let $D_n(x;a)$ and $E_n(x;a)\in\mathbb F_q[x]$ be Dickson polynomials of first and second kind respectively, where $\mathbb F_q$ is a finite field with $q$ elements. In this article we show explicitly the irreducible factors these polynomials in the case that every prime divisor of $n$ divides $q-1$. This result generalizes the results find in Chou, W.S., The Factorization of Dickson polynomials over finite fields. Finite Fields Appl. {\bf3} (1997) 84-96, Fitzgerald R. W., Yucas J. L., Explicit factorization of cyclotomic and Dickson polynomials over finite fields. Arithmetic of Finite Fields. Lecture Notes in Computer Science, vol. {\bf 4547}, pp. 1-10. Springer, Berlin (2007), Tosun, S., Explicit factorizations of generalized Dickson polynomials of order $2^{m}$ via generalized cyclotomic polynomials over finite fields. Finite Fields Appl. {\bf 38} (2016) 40-56 and Tosun, S., Explicit factors of generalized cyclotomic polynomials and generalized Dickson polynomials of order $2^m3$ over finite fields. Discrete Math. 342 (2019) DOI: j.disc.2019.111618

Equations

$$D_n(x,a)=\sum _{i=0}^{\lfloor n/2\rfloor }\frac{n}{n-i}\left(\genfrac{}{}{0ex}{}{n-i}{i}\right)(-a{)}^i{x}^{n-2i}$$

$$E_n(x,a)=\sum _{i=0}^{\lfloor n/2\rfloor }\left(\genfrac{}{}{0ex}{}{n-i}{i}\right)(-a{)}^i{x}^{n-2i}$$

$$D_n\left(y+\frac{1}{y};a\right)=y^n+\frac{a^n}{y^n}\text{and}{E}_n\left(y+\frac{1}{y};a\right)=\frac{y^n-\frac{a^n}{y^n}}{y-\frac{a}{y}},$$

$$f_a^{\ast }(x)=\frac{x^n}{f(0)}f\left(\frac{a}{x}\right).$$

$$(f\cdot g{)}_a^{\ast }(x)=f_a^{\ast }(x)\cdot g_a^{\ast }(x)\text{and}(f_a^{\ast }{)}_a^{\ast }(x)=f(x),$$

$$f(x)=b_m{x}^m+\sum _{i=0}^{m-1}{b}_{2m-i}(x^{2m-i}+a^{m-i}{x}^i),$$

$${\Phi }_a:P_m\to S_{2m,a}\text{and}{\Psi }_a:S_{2m,a}\to P_m$$

$${\Phi }_a(f(x))=x^{m}f\left(x+\frac{a}{x}\right)$$

$${\Psi }_a(g(x))$$

$${\Phi }_a(D_n(x,a))=x^{2n}+a^n.$$

$${\Phi }_a(D_n(x,a))=x^n{D}_n\left(x+\frac{a}{x},a\right)=x^n\left(x^n+{\left(\frac{a}{x}\right)}^n\right)=x^{2n}+a^n.$$

$$\left(y^{t}-\alpha\right)\left(y^{t}-\alpha^{-1}\right)=y^{t}\Bigl{(}y^{t}+\frac{1}{y^{t}}-(\alpha+\alpha^{-1})\Bigr{)}=y^{t}\Bigl{(}D_{t}\Bigl{(}y+\frac{1}{y}\Bigr{)}-(\alpha+\alpha^{-1})\Bigr{)},$$

$$b^{t}f(b^{-1}x)=D_t\left(x,a\right)-b^t\left(\alpha +{\alpha }^{-1}\right)$$

$$\tau :{\mathbb{F}}_{q^2}$$

$$\beta$$

$$\left(D_t(x,a)-b^t(\alpha +{\alpha }^{-1})\right)\left(D_t(x,a)-b^t({\alpha }^q+{\alpha }^{-q})\right)\in {\mathbb{F}}_q[x]$$

$$\left(D_t(x,a)-b^t(\alpha +{\alpha }^{-1})\right)\left(D_t(x,a)-b^{qt}({\alpha }^q+{\alpha }^{-q})\right)$$

$${\Phi }_a(E_n(x,a))=\frac{x^{2(n+1)}-a^{n+1}}{x^2-a}.$$

$${\Phi }_a(E_n(x,a))=x^n{E}_n\left(x+\frac{a}{x},a\right)=x^n\left(\frac{x^{n+1}-\frac{a^{n+1}}{x^{n+1}}}{x-\frac{a}{x}}\right)=\frac{x^{2(n+1)}-a^{n+1}}{x^2-a},$$

$$F_n(x)=\sum _{i=0}^m\frac{n}{n-i}\left(\genfrac{}{}{0ex}{}{n-i}{i}\right)x^{m-i}.$$

$$x{F}_n(x{)}^2$$

$$x^{2n}+1={\Phi }_1(D_n(x,1))={\Phi }_1(x){\Phi }_1(F_n(x{)}^2)=(x^2+1){\Phi }_1(F_n(x){)}^2.$$

$$D_n(x,a)=b^n{D}_n(b^{-1}x,1)=b^{n-1}x(F_n(b^{-1}x){)}^2.$$

$$E_{n}(x,1)=E_{n}\Bigl{(}y+\dfrac{1}{y},1\Bigr{)}=\dfrac{y^{n+1}-\Bigl{(}\dfrac{1}{y}\Bigr{)}^{n+1}}{y-\dfrac{1}{y}}=\frac{D_{n+1}\Bigl{(}y+\dfrac{1}{y},1\Bigr{)}}{y+\dfrac{1}{y}}=\frac{D_{n+1}(x,1)}{x}.$$

Full text

1. Introduction

The Dickson polynomials $D_{n}(x;a)$ over finite fields was introduce in the nineteenth century by Leonard E. Dickson as part of his PhD thesis. These polynomials have several interesting applications and properties, being mainly examples of families of permutation polynomials and now $D_{n}(x,a)\in\mathbb{F}_{q}[x]$ are called Dickson polynomials of the first kind, to distinguish them from their variations introduced by Schur in 1923, which are actually called the Dickson polynomials of the second kind $E_{n}(x,a)\in\mathbb{F}_{q}[x]$.

When $a=0$, $D_{n}(x,0)=x^{n}$ is a permutation polynomial (PP) in $\mathbb{F}_{q}$, if and only if, $\gcd(n,q-1)=1$. In the case that $a\in\mathbb{F}_{q}^{*}$, it is known that the Dickson polynomial $D_{n}(x,a)$ induces a permutation of $\mathbb{F}_{q}$, if and only if $\gcd(n,q^{2}-1)=1$ (see [14, Theorem 7.16] or [15, Theorem 3.2]). This simple condition provides a very effective test for determining which polynomials $D_{n}(x,a)$ induce permutations of $\mathbb{F}_{q}$, and furthermore, once the condition is satisfied, we obtain $q-1$ different permutations, one for each of the elements $a\in\mathbb{F}_{q}^{*}$. These polynomials have several applications and interesting properties. In fact, with the digital advances, practical applications of permutation polynomials can be found in cryptography, combinatorial designs, error-correcting codes, as well as hardware implementation of turbo decoders, feedback shift-register, linear-feedback shift register, among other applications , as well as pure theoretical results.

In the field of complex numbers, Dickson polynomials are essentially equivalent to the classical Chebyshev polynomials $T_{n}(x)$ with a simple change of variable. In fact, Dickson polynomials are sometimes referred as Chebyshev Polynomials in mathematical literature. They were rediscovered by Brewer , who used certain Dickson polynomials of the first kind to calculate Brewer sums (see [1] and [4]).The Dickson polynomials are also related in some way to Kloosterman sum; in [17] Moisio proved that, if Kloosterman sum $K_{q^{n}}(a)$ is different to $1$ for some $a\in F_{q}^{*}$, then the minimal polynomial $f(x)$ of $K_{q}(a)$ over $\mathbb{Z}$ must be a factor of $D_{n}(x,q)+(-1)^{n-1}$.

In recent years, these polynomials have been object of intense studies; in [15], the authors give a comprehensive survey about Dickson polynomials, including applications such as Dickson cryptosystems, Dickson pseudoprimes analogous to Carmichael numbers, etc. In particular, the problem of factorization of these polynomials have been studied by several authors. For example, in [12], Gao and Mullen study the irreducibility of $D_{n}(x,a)+b$, where $n$ is odd (see also [20]). Their argument is based on a well-known irreducibility criterion for binomial $x^{t}-\alpha$ over finite fields. Chou [8] and later Bhargava and Zieve [2] study a factorization of the Dickson polynomials of the first kind on $\mathbb{F}_{q}$ using methods more complicated than those we use here, because even though the factorization found by Bhargava and Zieve is made in $\mathbb{F}_{q}$, the factors contain elements outside $\mathbb{F}_{q}$. Fitzgerald and Yucas [10] give factorizations of cyclotomic polynomials $Q_{3\cdot 2^{n}}(x)$ for all $\mathbb{F}_{q}$ of characteristic not igual to 2 or 3. They apply this result to get explicit factorizations of Dickson polynomials $D_{3\cdot 2^{n}}(x)$ and $E_{3\cdot 2^{n}-1}(x)$, respectively. Tosun [18] extends the previous one by obtaining explicit factors of Dickson polynomials of the first and second kind $D_{3\cdot 2^{m}}(x,a)$ and $E_{3\cdot 2^{n}-1}(x,a)$ over $\mathbb{F}_{q}$, where $a$ be an arbitrary element of $\mathbb{F}_{q}$ and odd characteristic.

In this work, we study the problem of splitting $D_{n}(x,a)$ and $E_{n}(x,a)$ into irreducible factors over $\mathbb{F}_{q}$, where $\mathbb{F}_{q}$ is a finite field with $q$ elements, $n$ is a positive integer some that every prime divisor of $n$ divides $q-1$ for Dickson polynomials of the first kind and every prime divisor of $n+1$ divide $q-1$ for Dickson polynomials of the second kind. The result will be divide in several cases depending essentially on the class of $q$ modulo $4$ and also if $a$ is an square in $\mathbb{F}_{q}$ .

We note that when $n$ is a power of $2$, the condition $\mathop{\rm rad}(n)|(q-1)$ is trivial and then the results in [18] are particular case of our results.

The structure of this paper is as follows. In Section 2, we give a formal definition of Dickson polynomials with parameter $a$, some useful properties of that polynomial, and in particular, the $a$-self reciprocal property. In addition, we show, without proof, some results about the factorization of polynomial of the form $x^{m}\pm 1$. The factorization of Dickson polynomials of the first kind in odd characteristic is determined in Section 3. In Section 4, we give the factorization of Dickson polynomials of the second kind in odd characteristic. In Section 5, we provide a factorizations of Dickson polynomials in even characteristic.

2. Preliminaries

Throughout this paper, $\mathbb{F}_{q}$ will denote the finite field with $q$ elements, where $q$ is a power of a prime $p$. For any $a\in\mathbb{F}_{q}^{*}$, $\mathop{\rm ord}\nolimits_{q}(a)$ will denote the order of $a$ in the cyclic group $\mathbb{F}_{q}^{*}$ and for each positive integer $n\geq 2$, $\mathop{\rm rad}(n)$ denotes the product of every prime factor that divides $n$. The Dickson polynomials of first and second kind are defined as:

Definition 2.1**.**

Let $n\geq 1$ be an integer and $a\in\mathbb{F}_{q}$. The polynomial $D_{n}(x,a)\in\mathbb{F}_{q}[x]$ defined as

[TABLE]

is called the $n$-th Dickson polynomial of the first kind with parameter $a$.

Definition 2.2**.**

Let $n\geq 1$ be an integer and $a\in\mathbb{F}_{q}$. The polynomial $E_{n}(x,a)\in\mathbb{F}_{q}[x]$ defined as

[TABLE]

is called the $n$-th Dickson polynomial of the second kind with parameter $a$.

The $n$-th Dickson polynomials $D_{n}(x;a)$ and $E_{n}(x;a)$ are the unique degree $n$ polynomials satisfying the formal relations

[TABLE]

respectively. These equalities are known as Waring’s Identities.

In the case when $a=1$, we denote the polynomial $D_{n}(x;1)$ and $E_{n}(x;1)$ as $D_{n}(x)$ and $E_{n}(x)$, respectively. In this case, these polynomials can be see as polynomial with integer coefficients and then $D_{pn}(x)=(D_{n}(x))^{p}$ and $E_{pn}(x)=(E_{n}(x))^{p}$, where $p$ is the characteristic of the field.

The following lemma show some interesting properties of Dickson polynomials, that will be useful in the next sections.

Lemma 2.3** ([15], Lemma 2.6).**

The Dickson polynomials $D_{n}(x,a)$ and $E_{n}(x,a)$ satisfy the following properties

1. (i)

$D_{mn}(x,a)=D_{m}(D_{n}(x,a),a^{n})$* for $m\geq 0$ and $n\geq 0$,* 2. (ii)

$D_{np^{r}}(x,a)=[D_{n}(x,a)]^{p^{r}}$* for $n\geq 0$, $r\geq 0$, where $p$ is the characteristic of $\mathbb{F}_{q}$,* 3. (iii)

$b^{n}D_{n}(x,a)=D_{n}(bx,b^{2}a)$* for $n\geq 0$,* 4. (iv)

$b^{n}D_{n}(b^{-1}x,a)=D_{n}(x,b^{2}a)$* for $n\geq 0$ and $b\neq 0$.* 5. (v)

$E_{n}(x,a)=[E_{m}(x,a)]^{p^{r}}(x^{2}-4a)^{\frac{p^{r}-1}{2}}$* for $n\geq 0$ and $r\geq 0$, where $p$ is the characteristic of $\mathbb{F}_{q}$ and $n+1=(m+1)p^{r}$,* 6. (vi)

$b^{n}E_{n}(x,a)=E_{n}(bx,b^{2}a)$* for $n\geq 0$,* 7. (vii)

$b^{n}E_{n}(b^{-1}x,a)=E_{n}(x,b^{2}a)$* for $n\geq 0$ and $b\neq 0$.*

The following definition is borrowed from [9] and [11], that is also one possible generalization of the notion of reciprocal polynomial.

Definition 2.4**.**

Let $a$ be an element in $\mathbb{F}_{q}^{*}$. For each monic polynomial $f(x)\in\mathbb{F}_{q}[x]$ of degree $n$ with $f(0)\neq 0$, let define $f_{a}^{*}(x)$, the $a$-reciprocal of $f(x)$, by

[TABLE]

In the case when $f(x)=f_{a}^{*}(x)$ we say that $f(x)$ is $a$-self reciprocal.

It is easy to prove that $f_{a}^{*}(x)$ is a monic polynomial and if $\alpha\in\overline{\mathbb{F}}_{q}$ is a root of $f(x)$, then $\frac{a}{\alpha}$ is a root of $f_{a}^{*}(x)$. In addition

[TABLE]

for any $f$ and $g$ monic polynomials with $(f\cdot g)(0)\neq 0$. In particular, the polynomial $f(x)$ is an irreducible polynomial in $\mathbb{F}_{q}[x]$ if and only if $f_{a}^{*}(x)$ is also irreducible.

Lemma 2.5**.**

Let $f(x)$ be a $a$-self reciprocal polynomial of degree $n=2m$ in $\mathbb{F}_{q}$. Then $f(x)$ can be write as

[TABLE]

where $b_{j}\in\mathbb{F}_{q}$ for every $j=0,1,\ldots,m$.

*Proof: * The result follows directly from comparing the coefficient of $f$ and $f_{a}^{*}$. $\square$

Definition 2.6**.**

Let $P_{m}$ be the family of monic polynomials of degree $n$ in $\mathbb{F}_{q}$ and $S_{2m,a}$ be the family of monic polynomials $a$-self reciprocal of degree $2m$ in $\mathbb{F}_{q}$. For each positive integer $m$, let

[TABLE]

be the applications defined as

[TABLE]

and

[TABLE]

Theorem 2.7** ([11] Theorem 3.1.).**

Let $a$ be an element in $\mathbb{F}_{q}^{*}$ and $\Phi_{a}$ and $\Psi_{a}$ be as in Definition 2.6. Then

1. a)

$\Phi_{a}\circ\Psi_{a}=Id_{S_{2m,a}}$* and $\Psi_{a}\circ\Phi_{a}=Id_{P_{m}}$* 2. b)

$\Phi_{a}$* and $\Psi_{a}$ are multiplicative functions.* 3. c)

If $f(x)$ is a monic irreducible non-trivial $a$-self reciprocal polynomial of degree $2m$, then $\Phi_{a}(f(x))$ is an irreducible polynomial. If $g(x)$ is an irreducible polynomial of degree $m$ and non a-self reciprocal, then $\Psi_{a}(g(x)g_{a}^{*}(x))$ is irreducible.

Corollary 2.8**.**

Let $\Phi_{a}$ be as in the definition 2.6. Then

[TABLE]

*Proof: *

[TABLE]

$\square$

From this corollary, in order to find the irreducible factors of $D_{n}(x,a)$, it is enough to split into irreducible factors the polynomial $x^{2n}+a^{n}$. The factorization of this last polynomial has been extensively studied (see [3], [5], [6], [7], [10] ), then we show, without proof, some results that will be useful in the following section.

Theorem 2.9** ([14] Theorem 3.35).**

Let $n$ be a positive integer and $f(x)\in\mathbb{F}_{q}[x]$ be an irreducible polynomial of degree $m$ and exponent $e$. Then the polynomial $f(x^{n})$ is irreducible over $\mathbb{F}_{q}$ if and only if the following conditions are satisfied:

1. (1)

$rad(n)$* divides $e$;* 2. (2)

gcd$(n,(q^{m}-1)/e)=1$* and* 3. (3)

if $4|n$ then $4|q^{m}-1$.

In addition, in the case when the polynomial $f(x^{n})$ is irreducible, then it has degree $mn$ and exponent $en$.

Theorem 2.10** ([5] Theorem 1).**

Let $\mathbb{F}_{q}$ be a finite field and $n\in\mathbb{N}$ such that

1. a)

$q\equiv 1\pmod{4}$* or $8\nmid n$,* 2. b)

$\mathop{\rm rad}(n)$* divides $q-1$.*

Then every irreducible factor of $x^{n}-1$ is of the form $x^{t}-a$, where $t$ divides $\frac{n}{\gcd(n,q-1)}$, $a\in\mathbb{F}_{q}$ and $ord_{q}(a)$ divides $\gcd(\frac{n}{t},q-1)$.

Theorem 2.11** ([5] Theorem 2).**

Let $q$ and $n$ such that $q\equiv 3\pmod{4}$, $8\mid n$ and $rad(n)\mid(q-1)$, then every irreducible factor of $x^{n}-1$ in $\mathbb{F}_{q}[x]$ is one of the following types

1. i)

$x^{t}-a$, where $a\in\mathbb{F}_{q}$, $t$ divides $\frac{n}{\gcd(n,q^{2}-1)}$ and $ord_{q}a$ divides $\gcd(\frac{n}{t},q-1)$, 2. ii)

$x^{2t}-(b+b^{q})x^{t}+b^{q+1}$, where $b\in\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$, $t$ divides $\frac{n}{\gcd(n,q^{2}-1)}$ and $ord_{q^{2}}b$ divides $\gcd(\frac{n}{t},q^{2}-1)$.

3. Factorization of Dickson polynomials of the first kind

Throughout this section, $n$ represents a positive integer such that $\mathop{\rm rad}(n)$ divides $q-1$, i.e., every prime divisor of $n$ also divides $q-1$. In the case that $a$ is an square in $\mathbb{F}_{q}$, we can see that, by a linear change of the variable, the factorization of $D_{n}(x,a)$ and $D_{n}(x)$ are equivalent. In addition, the factorization of $D_{n}(x)$ also depend on the class of $q$ modulo $4$. Thus, our result will divide in three cases.

Theorem 3.1**.**

Let $a\in\mathbb{F}_{q}^{*}$ be a square in $\mathbb{F}_{q}$ and assume that either $q\equiv 1\pmod{4}$ or $n$ is an odd positive integer. If $\mathop{\rm rad}(n)|(q-1)$, then every irreducible factor of $D_{n}(x;a)$ in $\mathbb{F}_{q}$ is of the form $D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})$, where $b^{2}=a$, $\alpha\in\mathbb{F}_{q}^{*}$ and $t$ is a divisor of $\frac{4n}{\gcd(4n,q-1)}$, such that the following conditions are satisfied

1. (i)

$\alpha^{\frac{2n}{t}}=-1$, 2. (ii)

$\mathop{\rm rad}(t)\mid ord_{q}(\alpha)$, 3. (iii)

$\gcd\left(t,\frac{q-1}{ord_{q}(\alpha)}\right)=1$.

*Proof: * From Lemma 2.3 item (iv), we know that $D_{n}(x,a)=b^{n}D_{n}(b^{-1}x,1)$ for $n\geq 0$ and $b\neq 0$. Then, the factorization of $D_{n}(x,a)$ can be obtained from the factorization of $D_{n}(y)$ where $y=b^{-1}x$. In addition, by Corollary 2.8, we know that $\Phi_{1}(D_{n}(y))=y^{2n}+1$ and since $y^{4n}-1=(y^{2n}-1)(y^{2n}+1)$, if we find the irreducible factor of $y^{4n}-1$, in particular we obtain the irreducible factors of $y^{2n}+1$.

It follows from Theorem 2.10 that every irreducible factor of $y^{4n}-1$ is of the form $y^{t}-\alpha$, where $t$ divides $\frac{4n}{gcd(4n,q-1)}$ and $\alpha$ is an appropriate element of $\mathbb{F}_{q}$. In fact, $y^{t}-\alpha$ divides $y^{2n}+1$ is equivalent to $y^{2n}\equiv-1\pmod{y^{t}-\alpha}$, and therefore $y^{2n}\equiv\left(y^{t}\right)^{\frac{2n}{t}}\equiv-1\pmod{y^{t}-\alpha}$. Consequently $\alpha^{\frac{2n}{t}}=-1$.

Let suppose that $f(y)$ is an irreducible factor of $D_{n}(y)$ and $h(y)$ be the image of $f(y)$ by the map $\Phi_{1}$. Hence, $h(x)$ is a factor of $x^{2n}+1$, not necessarily irreducible in $\mathbb{F}_{q}[y]$. There exists an irreducible factor of the form $x^{t}-\alpha$ that divides $h(x)$. Since $h(x)$ is self-reciprocal polynomial, it follows that the reciprocal $x^{t}-\alpha^{-1}$ also divides $h(x)$.

At this point, we have two cases to consider

1. (a)

If $\alpha\neq\alpha^{-1}$, then $\left(y^{t}-\alpha\right)\left(y^{t}-\alpha^{-1}\right)\mid h(y)$. Since

[TABLE]

we get that $\Psi_{1}((y^{t}-\alpha)(y^{t}-\alpha^{-1}))=D_{t}(y)-(\alpha+\alpha^{-1})$ divides $f(y)$. Now, using the fact that $f(y)$ is a monic irreducible polynomial, we conclude that $f(y)=D_{t}\left(y\right)-\left(\alpha+\alpha^{-1}\right)$. Finally, rewrite this identity using the original variable we obtain $f(b^{-1}x)=D_{t}\left(b^{-1}x,1\right)-\left(\alpha+\alpha^{-1}\right)=b^{-t}D_{t}\left(x,a\right)-\left(\alpha+\alpha^{-1}\right)$ and therefore

[TABLE]

is a monic irreducible factor of $D_{n}\left(x,a\right)$ in $\mathbb{F}_{q}$. 2. (b)

In the case when $\alpha=\alpha^{-1}$, we have that $\alpha=\pm 1$ and $y^{t}-\alpha=y^{t}\pm 1$. It is clear that this polynomial is irreducible if $t=1$. In the case that $t\neq 1$, we have that $y^{t}-1$ is always reducible and $y^{t}+1$ is reducible if $t$ has an odd prime divisor. Lastly, if $t$ is a power of $2$, since $q\equiv 1\pmod{4}$, we have that $-1$ is an square in $\mathbb{F}_{q}$ and hence $y^{t}+1$ is also reducible. In any case we conclude that $y\pm 1$ divides $y^{2n}+1$, thus $y^{2n}+1\equiv(\mp 1)^{2n}+1\equiv 2\equiv 0\pmod{y\pm 1}$ implies that $char(\mathbb{F}_{q})=2$, which is a contradiction. $\square$

In the previous theorem, we consider the case when $a\in\mathbb{F}_{q}$ is an square in $\mathbb{F}_{q}$ and either $q\equiv 1\pmod{4}$ or $n$ is an odd integer. In the following theorems, we use this result in order to understand the complementary cases of this conditions.

Theorem 3.2**.**

Let $a$ be an square in $\mathbb{F}_{q}$, $q\equiv 3\pmod{4}$ and $n$ an even positive integer such that $\mathop{\rm rad}(n)|(q-1)$. Let $b\in\mathbb{F}_{q}$ such that $b^{2}=a$, $\alpha\in\mathbb{F}_{q^{2}}^{*}$ and $t$ be a divisor of $\frac{4n}{\gcd(4n,q^{2}-1)}$, such that $t$ and $\alpha$ satisfy the conditions (i), (ii) and (iii) of the Theorem 3.1 in the finite field $\mathbb{F}_{q^{2}}$. Then every irreducible factor of $D_{n}(x,a)$ in $\mathbb{F}_{q}$ is of the following types

1. (a)

$D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})$* in the case that $\alpha\in\mathbb{F}_{q}^{*}$ or $\alpha^{q+1}=1$,* 2. (b)

$\left(D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})\right)\left(D_{t}(x,a)-b^{t}(\alpha^{q}+\alpha^{-q})\right)$* otherwise.*

*Proof: * We observe that every irreducible factor of $D_{n}(x,a)$ in $\mathbb{F}_{q}[x]$ is also a factor, not necessarily irreducible, in $\mathbb{F}_{q^{2}}[x]$. Thus, at first time we consider the Dickson polynomial $D_{n}(x,a)$ as a polynomial in $\mathbb{F}_{q^{2}}[x]$. Since $q^{2}\equiv 1\pmod{4}$, by Theorem 3.1, every irreducible factor of $D_{n}(x,a)$ in $\mathbb{F}_{q^{2}}[x]$ is of the form $D_{t}\left(x,a\right)-b^{t}\left(\alpha+\alpha^{-1}\right)\in\mathbb{F}_{q^{2}}[x]$.

Some of these factors are also in $\mathbb{F}_{q}[x]$, but that happens when $\alpha+\alpha^{-1}\in\mathbb{F}_{q}$. This condition is equivalent to $\alpha+\alpha^{-1}=\left(\alpha+\alpha^{-1}\right)^{q}$ and that equation is equivalent to $(\alpha^{q+1}-1)(\alpha^{q-1}-1)=0$. Thus $\alpha\in\mathbb{F}_{q}^{*}$ or $\alpha^{q+1}=1$.

In the case when none of these conditions are satisfied, $D_{t}\left(x,a\right)-b^{t}\left(\alpha+\alpha^{-1}\right)\in\mathbb{F}_{q^{2}}[x]$ is a factor of $D_{n}(x,a)$, that is not in $\mathbb{F}_{q}[x]$. From the fact that the coefficient of $D_{n}(x,a)$ is invariant by the Frobenius Automorphism

[TABLE]

we conclude that $D_{t}\left(x,a\right)-b^{t}\left(\alpha^{q}+\alpha^{-q}\right)$ is also an irreducible factor of $D_{n}(x,a)$. In addition, from the fact that $\alpha+\alpha^{-1}\neq\alpha^{q}+\alpha^{-q}$, we concluded that $\left(D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})\right)\left(D_{t}(x,a)-b^{t}(\alpha^{q}+\alpha^{-q})\right)$ is a factor of $D_{n}(x,a)$.

Lastly, the coefficients of $\left(D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})\right)\left(D_{t}(x,a)-b^{t}(\alpha^{q}+\alpha^{-q})\right)$ are invariant by $\tau$, so

[TABLE]

is an irreducible factor of $D_{n}(x,a)$ in the polynomial ring $\mathbb{F}_{q}[x]$. $\square$

Theorem 3.3**.**

Let $a\in\mathbb{F}_{q}$ be a non-square in $\mathbb{F}_{q}$, $b\in\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$ such that $b^{2}=a$ and $n$ be a positive integer such that $\mathop{\rm rad}(n)|(q-1)$. Let $\alpha\in\mathbb{F}_{q^{2}}^{*}$ and $t$ be a divisor of $\frac{4n}{\gcd(4n,q^{2}-1)}$ satisfied the conditions (i), (ii) and (iii) of Theorem 3.1 in the field $\mathbb{F}_{q^{2}}$. Then every irreducible factor of $D_{n}(x,a)$ in $\mathbb{F}_{q}$ is one of the following forms

1. (a)

$D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})$* in the case that $t$ is even and either $\alpha\in\mathbb{F}_{q}^{*}$ or $\alpha^{q+1}=1$, or $t$ is odd and either $\alpha^{q-1}=-1$ or $\alpha^{q+1}=-1$,* 2. (b)

$\left(D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})\right)\left(D_{t}(x,a)-b^{qt}(\alpha^{q}+\alpha^{-q})\right)$* otherwise*

*Proof: * As in the previous theorem, we consider the Dickson polynomial $D_{n}(x,a)$ as a polynomial in $\mathbb{F}_{q^{2}}[x]$. In any case, we have that $q^{2}\equiv 1\pmod{4}$ and $a$ is an square in $\mathbb{F}_{q^{2}}$, i.e., $a=b^{2}$ where $b\in\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$. It follows from Theorem 3.1 that every irreducible factor of $D_{n}(x,a)$ in $\mathbb{F}_{q^{2}}[x]$ is of the form $D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})\in\mathbb{F}_{q^{2}}[x]$.

At this point, we need to determine which factors are in $\mathbb{F}_{q}[x]$. Since $D_{t}(x,a)\in\mathbb{F}_{q}[x]$, we need find conditions in order that $b^{t}\left(\alpha+\alpha^{-1}\right)\in\mathbb{F}_{q}$, which is equivalent to $b^{t}\left(\alpha+\alpha^{-1}\right)=b^{qt}\left(\alpha+\alpha^{-1}\right)^{q}$. This equation can be rewritten as $b^{t(q-1)}\left(\alpha^{q}+\alpha^{-q}\right)=\alpha+\alpha^{-1}$.

It follows from $a$ not being an square in $\mathbb{F}_{q}$ that $b^{q-1}=\left(b^{2}\right)^{\frac{q-1}{2}}=a^{\frac{q-1}{2}}=-1$. At this point, we have two cases to consider

1. (1)

When $t$ is even: $b^{t(q-1)}\left(\alpha^{q}+\alpha^{-q}\right)=\alpha+\alpha^{-1}$ if and only if $\alpha^{q}+\alpha^{-q}=\alpha+\alpha^{-1}$, and so $\alpha\in\mathbb{F}_{q}^{*}$ or $\alpha^{q+1}=1$. 2. (2)

When $t$ is odd: $b^{t(q-1)}\left(\alpha^{q}+\alpha^{-q}\right)=\alpha+\alpha^{-1}$ if and only if $-\alpha^{q}-\alpha^{-q}=\alpha+\alpha^{-1}$, that is equivalent to $(\alpha^{q-1}+1)(\alpha^{q+1}+1)=0$.

Finally, in the case when none of the conditions are satisfied, the factor $D_{t}\left(x,a\right)-b^{t}\left(\alpha+\alpha^{-1}\right)\in\mathbb{F}_{q^{2}}[x]$ of $D_{n}(x,a)$ is not in $\mathbb{F}_{q}[x]$. Using the Frobenius Automorphism we also conclude that $D_{t}\left(x,a\right)-b^{qt}\left(\alpha^{q}+\alpha^{-q}\right)$ is an irreducible factor of $D_{n}(x,a)$ in $\mathbb{F}_{q^{2}}[x]$ and

[TABLE]

is an irreducible factor of $D_{n}(x,a)$ in $\mathbb{F}_{q}[x]$. $\square$

4. Dickson polynomials of the second kind

Throughout this section, $n$ represents a positive integer such that $\mathop{\rm rad}(n+1)$ divides $q-1$, i.e., every prime divisor of $n+1$ also divides $q-1$.

Lemma 4.1**.**

Let $\Phi_{a}$ be the function defined at 2.6. Then

[TABLE]

*Proof: * Using the Waring’s Identity for $E_{n}$, we have

[TABLE]

as we want to prove. $\square$

Using the previous lemma and essentially the same steps of the proof of Theorems 3.1, 3.2 and 3.3, we obtain the following result, enunciated without proof.

Theorem 4.2**.**

Let $a$ be an square in $\mathbb{F}_{q}^{*}$ with $q\equiv 1\pmod{4}$ or $n$ an even positive integer. In addition, $n$ satisfies that $\mathop{\rm rad}(n+1)$ divides $q-1$. Then every irreducible factor of $E_{n}(x,a)$ in $\mathbb{F}_{q}[x]$ is of the form $D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})$, where $b^{2}=a$, $\alpha\in\mathbb{F}_{q}^{*}$ and $t$ is a divisor of $\frac{2(n+1)}{\gcd(2(n+1),q-1)}$, satisfying the following conditions

1. (i)

$\alpha^{\frac{2(n+1)}{t}}=1$, 2. (ii)

$\mathop{\rm rad}(t)\mid ord_{q}(\alpha)$, 3. (iii)

$\gcd\left(t,\frac{q-1}{ord_{q}(\alpha)}\right)=1$, 4. (iv)

$(\alpha,t)\notin\{(1,1),(1,-1)\}$.

Theorem 4.3**.**

Let $a\in\mathbb{F}_{q}$ be a square in $\mathbb{F}_{q}^{*}$ with $q\equiv 3\pmod{4}$ and $n$ an odd positive integer. In addition, $n$ satisfies that $\mathop{\rm rad}(n+1)$ divides $q-1$. Let $b^{2}=a$, $\alpha\in\mathbb{F}_{q^{2}}^{*}$ and $t$ a divisor of $\frac{2(n+1)}{\gcd(2(n+1),q-1)}$, satisfying the conditions (i), (ii), (iii) and (iv) of the previous theorem in the field $\mathbb{F}_{q^{2}}$. Then every irreducible factor of $E_{n}(x,a)$ in $\mathbb{F}_{q}$ has one of the following forms

1. (a)

$D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})$* in the case that $\alpha\in\mathbb{F}_{q}^{*}$ or $\alpha^{q+1}=1$,* 2. (b)

$\left(D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})\right)\left(D_{t}(x,a)-b^{t}(\alpha^{q}+\alpha^{-q})\right)$* in the case when $\alpha$ does not satisfy neither conditions of the item above.*

Theorem 4.4**.**

Let $a\in\mathbb{F}_{q}$ be a non-square in $\mathbb{F}_{q}$ and $n$ be an integer such that $\mathop{\rm rad}(n+1)$ divides $q-1$. Let $b\in\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$ such that $b^{2}=a$, $\alpha\in\mathbb{F}_{q^{2}}^{*}$ and $t$ be a divisor of $2(n+1)$, satisfying the conditions (i), (ii), (iii) and (iv) of Theorem 4.2 in $\mathbb{F}_{q^{2}}$. Then every irreducible factor of $E_{n}(x,a)$ in $\mathbb{F}_{q}$ has one of the following forms

1. (a)

$D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})$* in the case that $t$ is even and either $\alpha\in\mathbb{F}_{q}^{*}$ or $\alpha^{q+1}=1$, or $t$ is odd and either $\alpha^{q-1}=-1$ or $\alpha^{q+1}=-1$.* 2. (b)

$\left(D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})\right)\left(D_{t}(x,a)-b^{qt}(\alpha^{q}+\alpha^{-q})\right)$* in the case that $\alpha$ and $t$ do not satisfy neither conditions of the item above.*

5. Dickson polynomial in characteristic 2

Throughout this section, $\mathbb{F}_{q}$ is a finite field of characteristic $2$. We note that if $n=2^{r}s$, where $r\geq 0$ and $s$ odd, then $D_{n}(x)=[D_{s}(x)]^{2^{r}}$, therefore, in order to splits $D_{n}(x)$ into irreducible factors, it is enough to factorize $D_{s}(x,1)$. Hence, we assume that $n$ is positive integer such that $\mathop{\rm rad}(n)$ divides $q-1$.

Lemma 5.1**.**

Let $n=2m+1$ be a positive odd integer.

1. (a)

$D_{n}(x)=xF_{n}(x)^{2}$, for some polynomial $F_{n}(x)$ of degree $m$, with $F_{n}(0)\neq 0$, 2. (b)

$(x+1)\Phi_{1}(F_{n}(x))=x^{n}+1$.

*Proof: *

1. (a)

Let define

[TABLE]

Then

[TABLE] 2. (b)

By Corollary 2.8 item (i) and Theorem 2.7 follows that

[TABLE]

Therefore $(x^{n}+1)^{2}=[(x+1)\Phi_{1}(F_{n}(x))]^{2}$. $\square$

Theorem 5.2**.**

Let $\mathbb{F}_{q}$ be a finite field such that $char(\mathbb{F}_{q})=2$, $a\in\mathbb{F}_{q}^{*}$ and $n$ be a positive integer such that $\mathop{\rm rad}(n)|(q-1)$. Then every irreducible factor of $D_{n}(x,a)$ different than $x$, has multiplicity $2$ and is of the form $D_{t}(x,a)-b^{t}(\alpha+\alpha^{-1})$, where $b^{2}=a$, $\alpha\in\mathbb{F}_{q}^{*}$ and $t$ is a divisor of $\frac{n}{\gcd(n,q-1)}$, satisfying the following conditions

1. (i)

$\alpha^{\frac{n}{t}}=1$, 2. (ii)

$rad(t)\mid ord_{q}(\alpha)$, 3. (iii)

$gcd\left(t,\frac{q-1}{ord_{q}(\alpha)}\right)=1$, 4. (iv)

$(\alpha,t)\neq(1,1)$.

*Proof: * In characteristic 2, every element of $\mathbb{F}_{q}$ is an square. In addition, by Lemma 2.3 and Lemma 5.1, we have that

[TABLE]

At this point, in order to find the factorization of $D_{n}(x,a)$, it is enough to find the factorization of $F_{n}(x)$. On the other hand, $\Phi_{1}(F_{n}(x))=\dfrac{x^{n}+1}{x+1}$, so from here the proof is essentially the same proof of Theorem 3.1. $\square$

To conclude this section, we analyse the factorization of Dickson polynomials of the second kind in characteristic $2$, that, as we shall see, is almost the same factorization of Dickson polynomials of the first kind. In fact, if $n+1=2^{r}(s+1)$ with $r\geq 0$ and $s$ even, then $E_{n}(x,1)=[E_{s}(x,1)]^{2^{r}}x^{2^{r}-1}$. Therefore, in order to find the factors of $E_{n}(x,1)$, we need to factorize $E_{s}(x,1)$ with $s+1$ odd. So, we can assume that $n+1$ is odd. Then

[TABLE]

From Lemma 5.1 item (1), we conclude that $E_{n}(x,1)=F_{n+1}(x)^{2}$, and therefore the factorization of $E_{n}(x,a)$ with the condition that $\mathop{\rm rad}(n+1)|(q-1)$ can be found using Theorem 5.2.