1. Introduction

Permutation polynomials and Dickson polynomials are two of the most important topics in the area of finite fields. Let ${\mathbb F}_{q}$ be the finite field of characteristic $p$ with $q$ elements. Let ${\mathbb F}_q[x]$ be the ring of polynomials over ${\mathbb F}_q$ in the indeterminate $x$. If the polynomial $f(x)\in {\mathbb F}_q[x]$ induces a bijective map from ${\mathbb F}_q$ to itself, then $f(x)\in {\mathbb F}_q[x]$ is called a permutation polynomial (denoted as PP for convenience) of ${\mathbb F}_q$. Properties, constructions and applications of permutation polynomials may be found in ^[5], ^[6] and ^[7]. The reversed Dickson polynomial of the first kind, denoted by $D_n(a, x)$, was introduced in ^[4] and defined as follows

$D_n(a, x):=\sum\limits_{i=0}^{[\frac{n}{2}]}\frac{n}{n-i}\binom{n-i}{i}(-x)^ia^{n-2i}$

if $n\ge 1$ and $D_0(a, x)=2$, where $[\frac{n}{2}]$ means the largest integer no more than $\frac{n}{2}$. Wang and Yucas ^[8] extended this concept to that of the $n$-th reversed Dickson polynomial of $(k+1)$-th kind $D_{n, k}(a, x)\in {\mathbb F}_{q}[x]$, which is defined for $n\ge 1$ by

$D_{n, k}(a, x):=\sum\limits_{i=0}^{[\frac{n}{2}]}\frac{n-ki}{n-i}\binom{n-i}{i}(-x)^ia^{n-2i}$

(1.1)

and $D_{0, k}(a, x)=2-k$. Some families of permutation polynomials from the revered Dickson polynomials of the first kind were obtained in ^[4]. Hong, Qin and Zhao ^[3] studied the revered Dickson polynomial $E_n(a, x)$ of the second kind. Very recently, the author ^[1] investigated the reversed Dickson polynomial $D_{n, k}(a, x)$ of the $(k+1)$-th kind and obtained some properties and permutational behaviors of them.

In this paper, we study the special reversed Dickson polynomial of the form $D_{p^{e_1}+...+p^{e_s}+\ell, k}(1, x)$, where $s, e_1, ..., e_s$ are positive integers, $\ell$ is an integer with $0\le \ell < p$. In fact, by using Hermite criterion we first give an answer to the question that the reversed Dickson polynomials of the forms $D_{p^{s}+1, k}(1, x)$, $D_{p^{s}+2, k}(1, x)$, $D_{p^{s}+3, k}(1, x)$, $D_{p^{s}+4, k}(1, x)$, $D_{p^{s}+p^{t}, k}(1, x)$ and $D_{p^{s}+p^{t}+1, k}(1, x)$ are permutation polynomials of ${\mathbb F}_{q}$ or not. Finally, utilizing the recursive formula of the reversed Dickson polynomials, we represent $D_{p^{e_1}+...+p^{e_s}+\ell, k}(1, x)$ as the linear combination of the elementary symmetric polynomials with the power of $1-4x$ being the variables. From this, we present a necessary and sufficient condition for $D_{p^{e_1}+...+p^{e_s}+\ell, k}(1, x)$ to be a permutation polynomial of ${\mathbb F}_{q}$.

Throughout this paper, as usual, for any given prime number $p$, we let $v _p(x)$ denote the $p$-adic valuation of any positive integer $x$, i.e., $v _p(x)$ is the largest nonnegative integer $k$ such that $p^k$ divides $x$. We also assume $p={\rm char}({\mathbb F}_{q})\ge 3$ and restrict $0\le k < p$.

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2. Preliminary lemmas

In this section, we list several properties of the reversed Dickson polynomials $D_{n, k}(a, x)$ of the $(k+1)$-th kind and some useful lemmas.

Lemma 2.1. ^[5] Let $f(x)\in {\mathbb F_{q}}[x]$. Then $f(x)$ is a PP of ${\mathbb F_{q}}$ if and only if $cf(dx)+b$ is a PP of ${\mathbb F_{q}}$ for any given $c, d\in {\mathbb F_{q}^*}$ and $b\in {\mathbb F_{q}}$.

Lemma 2.2. Let $s\ge 0$ be an integer and $a, b$ be in ${\mathbb F_{q}}^*$. Then the binomial $ax^{\frac{p^s-1}{2}}+bx^{\frac{p^s+1}{2}}$ is a PP of ${\mathbb F_{q}}$ if and only if $s=0$.

Proof. First we assume that the binomial $ax^{\frac{p^s-1}{2}}+bx^{\frac{p^s+1}{2}}$ is a PP of ${\mathbb F_{q}}$. If $s > 0$, then the equation $ax^{\frac{p^s-1}{2}}+bx^{\frac{p^s+1}{2}} =x^{\frac{p^s-1}{2}}(a+bx)=0$ has two distinct roots $0, -\frac{b}{a}$ which are in ${\mathbb F_{q}}$. This is a contradiction. So the integer $s$ must be zero. Conversely, if $s=0$, then it is easy to check that $ax^{\frac{p^s-1}{2}}+bx^{\frac{p^s+1}{2}}$ is a PP of ${\mathbb F_{q}}$. Therefore Lemma 2.2 is proved.

Lemma 2.3. ^[1] For any integer $n\ge 0$, we have

$D_{n, k}\Big(1, \frac{1}{4}\Big)=\frac{kn-k+2}{2^n}$

and

$D_{n, k}(1, x)=\frac{\big(k-1-(k-2)y\big)y^n-\big(1+(k-2)y\big)(1-y)^n}{2y-1}$

if $x=y(1-y)\ne \frac{1}{4}$.

Lemma 2.4. ^[1] Let $n\ge 2$ be an integer. Then the recursion

$D_{n, k}(1, x)=D_{n-1, k}(1, x)-xD_{n-2, k}(1, x)$

holds for any $x\in {\mathbb F_{q}}$.

Lemma 2.5. ^[1] Let $p={\rm char}({\mathbb F_{q}})\ge 3$ and $s$ be a positive integer. Then

$2D_{p^s, k}(1, x)+k-2=k(1-4x)^{\frac{p^s-1}{2}}.$

Lemma 2.6. ^[2] Let $\alpha$ and $e$ be positive integers. Let $d=\gcd(\alpha, e)$ and p be an odd prime. Then

$\begin{align*} \gcd(p^{\alpha}+1, p^e-1)={\left\{ \begin{array}{ll} 2, & if\ \frac{e}{d}\ \ is\ \ odd, \\ p^d+1, &if\ \ \frac{e}{d}\ \ is\ \ even. \end{array} \right.} \end{align*}$

Lemma 2.7. ^[5] Let $f(x)\in {\mathbb F_{q}}[x]$. Then $f(x)$ is permutation polynomial of ${\mathbb F_{q}}$ if and only if the following conditions hold:

($i$) $f(x)$ has exactly one root in ${\mathbb F_{q}}$;

($ii$) For each integer $t$ with $0 < t < q-1$ and $t\not\equiv 0\pmod{p}$, the reduction of $f(x)^t \pmod{x^q-x}$ has degree less than $q-1$.

Lemma 2.8. Let $p$ be a prime with $p > 3$ and $a$ be a nonzero element in ${\mathbb F_{p}}$. Then the binomial $x^{\frac{p^s-1}{2}}+ax$ is a PP of ${\mathbb F_{p^e}}$ if and only if $s=0$.

Proof. Let $p > 3, a\in {\mathbb F_{p}^*}$. Clearly, if $s=0$, then $w(x):=x^{\frac{p^s-1}{2}}+ax=1+ax$ is a PP of ${\mathbb F_{p^e}}$. In what follows, we show that $w(x)=x^{\frac{p^s-1}{2}}+ax$ is not a PP of ${\mathbb F_{p^e}}$ when $s > 0$. Let $s > 0$ and $s\equiv s_0\pmod{2e}$ with $0\le s_0\le 2e-1$. Then

$w(x)\equiv x^{\frac{p^{s_0}-1}{2}}+ax \pmod{x^{p^e}-x}$

for any $x\in {\mathbb F_{p^e}^*}$ since $\frac{p^s-1}{2}\equiv \frac{p^{s_0}-1}{2}\pmod{p^e-1}$, i.e.,

$w(x)=x^{\frac{p^{s_0}-1}{2}}+ax$

(2.1)

for any $x\in {\mathbb F_{p^e}^*}$. We consider the following three cases.

CASE 1. $s > 0$ and $s_0=0$. Then by (2.1) one has $w(x)=1+ax$ for any $x\in {\mathbb F_{p^e}^*}$. So $f_2(\frac{1}{a})=0$. It then follows from $f_2(0)=0$ that $w(x)=x^{\frac{p^s-1}{2}}+ax$ is not a PP of ${\mathbb F_{p^e}}$.

CASE 2. $s > 0$ and $s_0$ is a positive even number. Then $x^{\frac{p^{s_0}-1}{2}}=1$ for each $x\in {\mathbb F_{p}^*}$. By (2.1) one get $w(x)=1+ax$ for any $x\in {\mathbb F_{p}^*}$. Therefore $w(x)=0$ has one nonzero root $-\frac{1}{a}\in{\mathbb F_{p}^*}$. Hence $w(x)=x^{\frac{p^s-1}{2}}+ax$ does not permute ${\mathbb F_{p}}$ since $f_2(0)=0$. Note that $f_2({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. So one has that $w(x)=x^{\frac{p^s-1}{2}}+ax$ does not permute ${\mathbb F_{p^e}}$.

CASE 3. $s > 0$ and $s_0$ is an odd number. Then $x^{\frac{p^{s_0}-1}{2}}=x^{\frac{p-1}{2}}$ for each $x\in {\mathbb F_{p}^*}$. It follows from (2.1) that

$w(x)=x^{\frac{p-1}{2}}+ax$

for any $x\in {\mathbb F_{p}^*}$. Then we have

$(w(x))^2=x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1 \pmod{x^p-x}.$

Then by Lemma 2.7, we know that $w(x)$ is not a PP of ${\mathbb F_{p}}$. Also note that $f_2({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. So $w(x)$ is not a PP of ${\mathbb F_{p^e}}$.

The above three cases tell us that $w(x)=x^{\frac{p^s-1}{2}}+ax$ is not a PP of ${\mathbb F_{p^e}}$ when $s > 0$. This finishes the proof of Lemma 2.8.

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3. Reversed Dickson polynomials $D_{p^s+\ell, k}(1, x)$

In this section, we present an explicit formula for $D_{n, k}(1, x)$ when $n=p^s+\ell$ with $s\ge 0$ and $0\le \ell < p$. Then we characterize $D_{n, k}(1, x)$ to be a PP of ${\mathbb F}_q$ in this case.

Theorem 3.1. Let $p={\rm char}({\mathbb F_{q}})\ge 3$ and $s$ be a positive integer. Then

$D_{p^s+1, k}(1, x)=\frac{2-k}{4}(1-4x)^{\frac{p^s+1}{2}}+ \frac{k}{4}(1-4x)^{\frac{p^s-1}{2}}+\frac{1}{2}.$

(3.1)

Furthermore, we have

$D_{p^s+2\ell, k}(1, x)=\sum\limits_{i=0}^{\ell}A_{2\ell, p^s+2i-1}(1-4x)^{\frac{p^s+2i-1}{2}}+ \sum\limits_{j=0}^{\ell}A_{2\ell, 2j}(1-4x)^{j}, \ell\ge 0$

and

$D_{p^s+2\ell+1, k}(1, x)=\sum\limits_{i=0}^{\ell+1}A_{2\ell+1, p^s+2i-1}(1-4x)^{\frac{p^s+2i-1}{2}}+ \sum\limits_{j=0}^{\ell}A_{2\ell+1, 2j}(1-4x)^{j}, \ell\ge 0,$

where all the coefficients $A_{i, j}$ are given as follows:

$A_{0, p^s-1}=\frac{k}{2}, A_{0, 0}=\frac{2-k}{2}, A_{1, p^s+1}=\frac{2-k}{4}, A_{1, p^s-1}=\frac{k}{4}, A_{1, 0}=\frac{1}{2},$

and

$\begin{align}\label{ckm2} {\left\{ \begin{array}{ll} A_{2m+2, p^s+2m+1}=A_{2m+1, p^s+2m+1}+\frac{1}{4}A_{2m, p^s+2m-1}, & {\rm if} \ m\ge 0\\ A_{2m+2, p^s+2i-1}=A_{2m+1, p^s+2i-1}-\frac{1}{4}A_{2m, p^s+2i-1}+ \frac{1}{4}A_{2m, p^s+2i-3}, & {\rm if} \ 1\le i\le m\\ A_{2m+2, p^s-1}=A_{2m+1, p^s-1}-\frac{1}{4}A_{2m, p^s-1}, & {\rm if} \ m\ge 0\\ A_{2m+2, 0}=A_{2m+1, 0}-\frac{1}{4}A_{2m, 0}, & {\rm if} \ m\ge 0\\ A_{2m+2, 2j}=A_{2m+1, 2j}-\frac{1}{4}A_{2m, 2j}+\frac{1}{4}A_{2m, 2j-2}, & {\rm if} \ 1\le j\le m\\ A_{2m+2, 2m+2}=\frac{1}{4}A_{2m, 2m}, & {\rm if} \ m\ge 0\\ \end{array} \right.} \end{align}$

(3.2)

as well as

$\begin{align}\label{ckm3} {\left\{ \begin{array}{ll} A_{2m+1, p^s+2m+1}=\frac{1}{4}A_{2m-1, p^s+2m-1}, & {\rm if} \ m\ge 0\\ A_{2m+1, p^s+2i-1}=A_{2m, p^s+2i-1}-\frac{1}{4}A_{2m-1, p^s+2i-1}+\frac{1}{4}A_{2m-1, p^s+2i-3}, & {\rm if} \ 1\le i\le m\\ A_{2m+1, p^s-1}=A_{2m, p^s-1}-\frac{1}{4}A_{2m-1, p^s-1}, & {\rm if} \ m\ge 0\\ A_{2m+1, 0}=A_{2m, 0}-\frac{1}{4}A_{2m-1, 0}, & {\rm if} \ m\ge 0\\ A_{2m+1, 2j}=A_{2m, 2j}-\frac{1}{4}A_{2m-1, 2j}+\frac{1}{4}A_{2m-1, 2j-2}, & {\rm if} \ 1\le j\le m-1\\ A_{2m+1, 2m}=A_{2m, 2m}+\frac{1}{4}A_{2m-1, 2m-2}, & {\rm if}\ m\ge 0. \\ \end{array} \right.} \end{align}$

(3.3)

Proof. First of all, we show (3.1) is true. We consider the following two cases.

CASE 1. $x\ne \frac{1}{4}$. For this case, putting $x=y(1-y)$ in second identity of Lemma 2.3 gives us that

$\begin{align*} &D_{p^s+1, k}(1, x)=D_{p^s+1, k}(1, y(1-y))\\ =&\frac{\frac{k+(2-k)u}{2}\big(\frac{u+1}{2}\big)^{p^s+1} -\frac{k+(k-2)u}{2}\big(\frac{1-u}{2}\big)^{p^s+1}}{u}\\ =&\frac{2-k}{8}\Big((u+1)^{p^s}(u+1)+(1-u)^{p^s}(1-u)\Big)+ \frac{k}{8u}\Big((u+1)^{p^s}(u+1)-(1-u)^{p^s}(1-u)\Big)\\ =&\frac{2-k}{4}(u^{p^s+1}+1)+\frac{k}{4}(u^{p^s-1}+1)\\ =&\frac{2-k}{4}\big((u^2)^{\frac{p^s+1}{2}}\big)+ \frac{k}{4}\big((u^2)^{\frac{p^s-1}{2}}\big)+\frac{1}{2}, \end{align*}$

where $u=2y-1$ and $u^2=1-4x$. So (3.1) follows if $x\ne \frac{1}{4}$.

CASE 2. $x=\frac{1}{4}$. By the first identity of Lemma 2.3, one has

$D_{p^s+1, k}\big(1, \frac{1}{4}\big)=\frac{k(p^s+1)-k+2}{2^{p^s+1}}= \frac{2-k}{4}(1-4\times\frac{1}{4})^{\frac{p^s+1}{2}}+ \frac{k}{4}(1-4\times\frac{1}{4})^{\frac{p^s-1}{2}}+\frac{1}{2}$

as required.

Thus (3.1) is true for any $x\in {\mathbb F_{q}}$.

Now we give the the remainder proof of Theorem 3.1. By Lemmas 2.4-2.5 and (3.1), we readily find that there exists coefficients $A_{i, j}\in {\mathbb F_{q}}$ such that

$D_{p^s+2\ell, k}(1, x)=\sum\limits_{i=0}^{\ell}A_{2\ell, p^s+2i-1}(1-4x)^{\frac{p^s+2i-1}{2}}+ \sum\limits_{j=0}^{\ell}A_{2\ell, 2j}(1-4x)^{j}$

(3.4)

with $0\le\ell\le\frac{p-1}{2}$ and

$D_{p^s+2\ell+1, k}(1, x)=\sum\limits_{i=0}^{\ell+1}A_{2\ell+1, p^s+2i-1}(1-4x)^{\frac{p^s+2i-1}{2}}+ \sum\limits_{j=0}^{\ell}A_{2\ell+1, 2j}(1-4x)^{j}$

(3.5)

with $0\le\ell < \frac{p-1}{2}$. Therefore we now only need to determine all the coefficients $A_{i, j}$. Let $u^2=1-4x$. On the one hand, by (3.4) and (3.5), one then has

$\begin{align}\label{ckm6} &D_{p^s+2\ell, k}(1, x)-xD_{p^s+2\ell-1, k}(1, x)=D_{p^s+2\ell, k}(1, x)- \frac{1-u^2}{4}D_{p^s+2\ell-1, k}(1, x)\nonumber\\ &=\sum\limits_{i=0}^{\ell+1}A_{2\ell, p^s+2i-1}u^{p^s+2i-1}+ \sum\limits_{j=0}^{\ell}A_{2\ell, 2j}u^{2j}-\frac{1}{4}\sum\limits_{i=0}^{\ell}A_{2\ell-1, p^s+2i-1}u^{p^s+2i-1}\nonumber\\ &\ \ \ -\frac{1}{4}\sum\limits_{j=0}^{\ell-1}A_{2\ell-1, 2j}u^{2j}+\frac{1}{4}\sum\limits_{i=0}^{\ell}A_{2\ell-1, p^s+2i-1} u^{p^s+2i+1}+\frac{1}{4}\sum\limits_{j=0}^{\ell-1}A_{2\ell-1, 2j}u^{2j+2}\nonumber\\ &=\frac{1}{4}A_{2\ell-1, p^s+2\ell-1}u^{p^s+2\ell+1}+\sum\limits_{i=1}^{\ell}\big(A_{2\ell, p^s+2i-1}- \frac{1}{4}A_{2\ell-1, p^s+2i-1}+\frac{1}{4}A_{2\ell-1, p^s+2i-3}\big)u^{p^s+2i-1}\nonumber\\ &\ \ \ +\big(A_{2\ell, p^s-1}-\frac{1}{4}A_{2\ell-1, p^s-1}\big)u^{p^s-1}+\big(A_{2\ell, 2\ell}+ \frac{1}{4}A_{2\ell-1, 2\ell-2}\big)u^{2\ell}\nonumber\\ &\ \ \ +\sum\limits_{j=1}^{\ell-1}\big(A_{2\ell, 2j}-\frac{1}{4}A_{2\ell-1, 2j} +\frac{1}{4}A_{2\ell-1, 2j-2}\big)u^{2j}+A_{2\ell, 0}-\frac{1}{4}A_{2\ell-1, 0}. \end{align}$

(3.6)

On the other hand, Lemma 2.4 tells us that

$D_{p^s+2\ell+1, k}(1, x)=D_{p^s+2\ell, k}(1, x)-xD_{p^s+2\ell-1, k}(1, x).$

So by comparing the coefficient of the term $u^i$ in the right hand side of (3.6) and (3.5), one can get the desired results as (3.3). Following the similar way, one also obtain the recursions of $A_{i, j}$ as (3.2). So the proof Theorem 3.1 is complete.

For any nonzero integer $x$, let $v_2(x)$ be the 2-adic valuation of $x$. By Theorem 3.1, the following results are established.

Theorem 3.2. Let $q=p^e$ with $p$ being an odd prime and $e$ being a positive integer. Let $s$ be a nonnegative integer. Then each of following is true.

(ⅰ). If $k=0$, then $D_{p^s+1, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if either $p\equiv 1\pmod{4}$ and $v_2(s)\ge v_2(e)$, or $p\equiv 3\pmod{4}$ and $v_2(s)\ge \max\{v_2(e), 1\}$.

(ⅱ). If $k=2$, then $D_{p^s+1, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $p=3$, $v_2(s)=0$ and $\gcd(s, e)=1$.

(ⅲ). If $k\ne 2$ and $k\ne 0$, then $D_{p^s+1, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $s=0$.

Proof. By (3.1) of Theorem 3.1, we have that $D_{p^s+1, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if the polynomial

$(2-k)x^{\frac{p^s+1}{2}}+kx^{\frac{p^s-1}{2}}$

is a PP of ${\mathbb F_{q}}$.

(ⅰ). Let $k=0$. Then $D_{p^s+1, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if the monomial $x^{\frac{p^s+1}{2}}$ is a PP of ${\mathbb F_{q}}$, namely,

$\gcd\Big(\frac{p^s+1}{2}, p^e-1\Big)=1.$

So we consider the following two cases on the odd prime $p$.

CASE 1. $p\equiv 1\pmod{4}$. Then $\frac{p^s+1}{2}$ must be odd. It then follows that

$\gcd\Big(\frac{p^s+1}{2}, p^e-1\Big)=\gcd\Big(\frac{p^s+1}{2}, \frac{p^e-1}{2}\Big)= \frac{1}{2}\gcd(p^s+1, p^e-1).$

So in this case, by Lemma 2.6 we get that $\gcd(\frac{p^s+1}{2}, p^e-1)=1$ if and only if $\frac{e}{\gcd(s, e)}$ is odd which is equivalent to $v_2(e)\le v_2(s)$.

CASE 2. $p\equiv 3\pmod{4}$. Then $v_2(\frac{p^s+1}{2})\ge 1$ when $s$ is odd. In this case we have $2|\gcd(\frac{p^s+1}{2}, p^e-1)$ which is not allowed. So in the case of $p\equiv 3\pmod{4}$, $s$ must be even. Then $\frac{p^s+1}{2}$ is an odd number. It follows from Lemma 2.6 that $\gcd(\frac{p^s+1}{2}, p^e-1)=1$ if and only if $\frac{e}{\gcd(s, e)}$ is odd which is equivalent to $v_2(e)\le v_2(s)$ and $v_2(s)\ge 1$, i.e., $v_2(s)\ge \max\{1, v_2(e)\}$ as desired. Part (ⅰ) is proved.

(ⅱ). Let $k=2$. Assume that $D_{p^s+1, k}(1, x)$ is a PP of ${\mathbb F_{p^e}}$. Then $D_{p^s+1, k}(1, x)$ is a PP of ${\mathbb F_{p^e}}$ if and only if $x^{\frac{p^s-1}{2}}$ is a PP of ${\mathbb F_{p^e}}$. Clearly, $s > 0$ in this case. Suppose $p > 3$, then $x^{\frac{p^s-1}{2}}$ is a PP of ${\mathbb F_{p^e}}$ if and only if

$\gcd\Big(\frac{p^s-1}{2}, p^e-1\Big)=1.$

This is impossible since $\frac{p-1}{2}|\gcd\big(\frac{p^s-1}{2}, q-1\big)$ implies that

$\gcd\Big(\frac{p^s-1}{2}, q-1\Big)\ge\frac{p-1}{2}>1.$

So $p=3$ and $s > 0$ in what follows. Now Suppose $s > 0$ is even, then it is easy to see that $2|\gcd(\frac{3^s-1}{2}, 3^e-1)$ which is a contradiction. This means that $s$ must be an odd number and then so is $\frac{3^s-1}{2}$. Thus we have that $x^{\frac{3^s-1}{2}}$ is a PP of ${\mathbb F_{3^e}}$ if and only if

$\gcd\Big(\frac{3^s-1}{2}, 3^e-1\Big) =\frac{1}{2}\gcd\big(3^s-1, 3^e-1\big)=\frac{1}{2}(3^{\gcd(s, e)}-1)=1,$

which is equivalent to that $s$ is odd and $\gcd(s, e)=1$. So Part (ⅱ) is proved.

(ⅲ). $k\ne 0$ and $k\ne 2$. Then the desired result follows from Lemma 2.2 that $(2-k)x^{\frac{p^s+1}{2}}+kx^{\frac{p^s-1}{2}}$ is a PP of ${\mathbb F_{q}}$ if and only if $s=0$. Part (ⅲ) is proved.

Theorem 3.3. Let $q=p^e$ with $p$ being an odd prime and $e$ being a positive integer. Let $s$ be a nonnegative integer and $s_0$ be the least nonnegative residue of $s$ modulo $2e$. Then each of following is true.

(ⅰ). If $k=0, p=3$, then $D_{p^s+2, k}(1, x)$ is not a PP of ${\mathbb F_{3^e}}$.

(ⅱ). If $k=0, p > 3, s_0=0$, then $D_{p^s+2, k}(1, x)$ is a PP of ${\mathbb F_{p^e}}$.

(ⅲ). If $k=0, p > 3, s=e$, then $D_{p^e+2, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $q=p^e\equiv 1\pmod{3}$.

(ⅳ). If $k=2$, then $D_{p^s+2, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $s=0$.

(ⅴ). Let $k=4, p=3$. If $s=0$ or $s_0=1$, then the binomial $D_{p^s+2, k}(1, x)$ is a PP of ${\mathbb F_{3^e}}$. If $s > 0$ and $s_0$ is even, then $D_{p^s+2, k}(1, x)$ is not a PP of ${\mathbb F_{3^e}}$.

(ⅵ). Let $k=4, p > 3$. Then $D_{p^s+2, k}(1, x)$ is a PP of ${\mathbb F_{p^e}}$ if and only if $s=0$.

(ⅶ). If $k\ne 0, 2, 4$ and $p\nmid (4-k)$, then $D_{p^s+2, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $s=0$ and $k\ne 3$.

Proof. By Theorem 3.1, we have that $D_{p^s+2, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if the polynomial

$(4-k)x^{\frac{p^s+1}{2}}+kx^{\frac{p^s-1}{2}}+(2-k)x$

is a PP of ${\mathbb F_{q}}$.

(ⅰ). Let $k=0, p=3$. Then $D_{p^s+2, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if the monomial $x^{\frac{p^s+1}{2}}+\frac{1}{2}x$ is a PP of ${\mathbb F_{q}}$. Let

$f_1(x):=x^{\frac{p^s+1}{2}}+\frac{1}{2}x.$

It is easy to see that $f_1(x)$ is not a PP of ${\mathbb F_{3^e}}$ since $f_1(0)=f_1(1)=0$. So in this case $D_{p^s+2, k}(1, x)$ is not a PP of ${\mathbb F_{3^e}}$.

(ⅱ). Let $k=0, p > 3, s_0=0$. Then $\frac{p^s+1}{2}\equiv 1\pmod{p^e-1}$ which implies that

$f_1(x)\equiv \frac{3}{2}x \pmod{x^{p^e}-x}$

for any $x\in {\mathbb F_{q}^*}$. Note that $f_1(0)=\frac{3}{2}\times 0=0$ and the monomial $\frac{3}{2}x$ is a PP of ${\mathbb F_{q}}$. So $f_1(x)$ is a PP of ${\mathbb F_{q}}$. That is to say $D_{p^s+2, k}(1, x)$ is not a PP of ${\mathbb F_{p^e}}$.

(ⅲ). Let $k=0, p > 3, s=e$. Then by Theorem 7.11 in ^[5] we have $f_1(x)$ is a PP of ${\mathbb F_{q}}$ if and only if $\eta\big(\big(\frac{1}{2}\big)^2-1\big)=1$, i.e., $\eta(-3)=1$, where $\eta(\cdot)$ denotes the quadratic character of ${\mathbb F_{q}}$. One can also find that $\eta(-3)=1$ if and only if $q=p^e\equiv 1\pmod{3}$, as desired.

(ⅳ). If $k=2$, then the desired result follows from Lemma 2.2 that the binomial $2x^{\frac{p^s+1}{2}}+2x^{\frac{p^s-1}{2}}$ is a PP of ${\mathbb F_{q}}$ if and only if $s=0$. So $D_{p^s+2, k}(1, x)$ is not a PP of ${\mathbb F_{p^e}}$ if and only if $s=0$.

(ⅴ). Let $k=4, p=3$. Then $D_{p^s+2, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $2x^{\frac{p^s-1}{2}}-x$ is a PP of ${\mathbb F_{q}}$. Let

$f_2(x):=x^{\frac{p^s-1}{2}}+\frac{1}{2}x.$

Obviously $f_2(x)=x^{\frac{3^s-1}{2}}+x=1+x$ is a PP of ${\mathbb F_{3^e}}$ when $s=0$. Now let $0 < s\equiv s_0\pmod{2e}$ with $0\le {s_0}\le 2e-1$. Then $\frac{p^s-1}{2}\equiv \frac{p^{s_0}-1}{2}\pmod{p^e-1}$. Therefore

$f_2(x)\equiv x^{\frac{3^{s_0}-1}{2}}+x \pmod{x^{3^e}-x}$

for any $x\in {\mathbb F_{q}^*}$. If $s_0=1$, $f_2(x)\equiv 2x \pmod{x^{3^e}-x}$ and $f_2(0)=2\times 0=0$. This means that $f_2(x)=2x$ for any $x\in {\mathbb F_{q}}$. So $f_2(x)$ is a PP of ${\mathbb F_{q}}$ when $s_0=1$. If $s > 0$ and $s_0$ is even, then $x^{\frac{3^{s_0}-1}{2}}+x=1+x$ for any $x\in {\mathbb F_{3}^*}$. Note that $f_2({\mathbb F_{3}})\subseteq {\mathbb F_{3}}$ and $f_2(0)=f_2(-1)=0$, which tells us that $f_2(x)$ is not a PP of ${\mathbb F_{3}}$. Thus the desired results follows. Unfortunately, following the similar way, we cannot say anything for the case of $s > 0$ and $s_0$ being odd with $s_0\ge 3$.

(ⅵ). Let $k=4, p > 3$. Then $D_{p^s+2, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $2x^{\frac{p^s-1}{2}}-x$ is a PP of ${\mathbb F_{q}}$. It then follows from Lemma 2.8 that $D_{p^s+2, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $s=0$, as required.

(ⅶ). Let $p\ge 3$, $k\ne 0, 2, 4$ and $p\nmid (k-4)$. Denote

$f_3(x):=(4-k)x^{\frac{p^s+1}{2}}+kx^{\frac{p^s-1}{2}}+(2-k)x.$

First, if $s=0$, then $f_3(x)=(6-2k)x+k$ which is a PP of ${\mathbb F_{p^e}}$ if and only if $k\ne 3$. In what follows we will show that $f_3(x)$ is not a PP of ${\mathbb F_{p^e}}$ when $s > 0$. Let $0 < s\equiv s_0\pmod{2e}$ with $0\le s_0\le 2e-1$. Then

$f_3(x)\equiv (4-k)x^{\frac{p^{s_0}+1}{2}}+kx^{\frac{p^{s_0}-1}{2}}+(2-k)x\pmod{x^{p^e}-x}.$

(3.7)

We consider the following cases.

CASE 1. $s > 0, s_0=0$. By (3.7) we have $f_3(x)\equiv (6-2k)x+k\pmod{x^{p^e}-x}$, which means that $f_3(x)=(6-2k)x+k$ for any $x\in {\mathbb F_{p^e}^*}$. If $k=3$, then $\forall x\in {\mathbb F_{p^e}^*}$ $f_3(x)=k$. Obviously, $f_3(x)$ is not a PP of ${\mathbb F_{p^e}}$. If $k\ne 3$, then $f_3(x)=0$ has one nonzero root $\frac{k}{2k-6}\in {\mathbb F_{p^e}^*}$ since $k\ne 0$. But $f_3(0)=0$. So $f_3(x)$ is not a PP of ${\mathbb F_{p^e}}$ in this case.

CASE 2. $s > 0$ and $s_0$ is a positive even number. Then $x^{\frac{p^{s_0}+1}{2}}=x$ and $x^{\frac{p^{s_0}-1}{2}}=1$ for each $x\in {\mathbb F_{p}^*}$, which together with (3.7) imply that $f_3(x)=(6-2k)x+k$ for any $x\in {\mathbb F_{p}^*}$. If $k=3$, then $\forall x\in {\mathbb F_{p}^*}$, $f_3(x)=k$. Obviously, $f_3(x)$ is not a PP of ${\mathbb F_{p}}$. If $k\ne 3$, then $f_3(x)=0$ has one nonzero root $\frac{k}{2k-6}\in {\mathbb F_{p}^*}$ since $k\ne 0$. But $f_3(0)=0$. Therefore $f_3(x)$ is not a PP of ${\mathbb F_{p}}$ in this case. So is $f_3(x)$ of ${\mathbb F_{p^e}}$ since $f_3({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$.

CASE 3. $s > 0$ and $s_0$ is odd. Then $x^{\frac{p^{s_0}+1}{2}}=x^{\frac{p+1}{2}}$ and $x^{\frac{p^{s_0}-1}{2}}=x^{\frac{p-1}{2}}$ for each $x\in {\mathbb F_{p}^*}$, which together with (3.7) imply that $f_3(x)=(4-k)x^{\frac{p+1}{2}}+kx^{\frac{p-1}{2}}+(2-k)x$ for any $x\in {\mathbb F_{p}^*}$. If $p=3$, then $k$ must equal $1$ since $0\le k < p$ and $k\ne 0, 2, 4$, which contradicts to the condition $p\nmid (4-k)$. So one has $p > 3$. Then

$[f_3(x)]^2\equiv k^2x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

Then by Lemma 2.7, we know that $f_3(x)$ is not a PP of ${\mathbb F_{p}}$. Also note that $f_3({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. So $f_3(x)$ is not a PP of ${\mathbb F_{p^e}}$.

Combining the above cases, we verify that $f_3(x)$ is not a PP of ${\mathbb F_{p^e}}$ when $s > 0$. Thus Part (vii) is proved. So the proof of Theorem 3.3 is complete.

Theorem 3.4. Let $q=p^e$ with $p$ being an odd prime and $e$ being a positive integer. Let $s$ be a nonnegative integer and $s_0$ be the least nonnegative residue of $s$ modulo $2e$. Then each of following is true.

(ⅰ). If $k=0, p=3$, then $D_{3^s+3, 0}(1, x)$ is a PP of ${\mathbb F_{3^e}}$ if and only if $v_2(s-1)\ge \max\{1, v_2(e)\}$.

(ⅱ). Let $k=0, p > 3$. If $s_0$ is an even number, then $D_{p^s+3, 0}(1, x)$ is not a PP of ${\mathbb F_{p^e}}$.

(ⅲ). If $k=2, s=0$, then $D_{p^s+3, k}(1, x)$ is a PP of ${\mathbb F_{q}}$

(ⅳ). Let $k=2, s > 0, p=3$. If $s_0=1$, then $D_{3^s+3, 2}(1, x)$ is a PP of ${\mathbb F_{3^e}}$. If $s_0$ is even, then $D_{3^s+3, 2}(1, x)$ is not a PP of ${\mathbb F_{3^e}}$.

(ⅴ). Let $k=2, p > 3$. Then $D_{p^s+3, 2}(1, x)$ is a PP of ${\mathbb F_{p^e}}$ if and if $s=0$.

(ⅵ). If $k=3$, then $D_{p^s+3, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $p=3$ and $v_2(s-1)\ge \max\{1, v_2(e)\}$.

(ⅶ). If $k\ne 0, 2, 3$, then $D_{p^s+3, k}(1, x)$ is not a PP of ${\mathbb F_{p^e}}$.

Proof. By Theorem 3.1, we have that $D_{p^s+3, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if the polynomial

$(2-k)x^{\frac{p^s+3}{2}}+6x^{\frac{p^s+1}{2}}+ kx^{\frac{p^s-1}{2}}+(6-2k)x$

(3.8)

is a PP of ${\mathbb F_{q}}$.

(ⅰ). Letting $k=0$, we have $D_{p^s+3, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if (3.8) is a PP of ${\mathbb F_{q}}$, i.e., the trinomial $x^{\frac{p^s+3}{2}}+3x^{\frac{p^s+1}{2}}+3x$ is a PP of ${\mathbb F_{q}}$. Let

$f_4(x):=x^{\frac{p^s+3}{2}}+3x^{\frac{p^s+1}{2}}+3x.$

Now let $p=3$. Then $f_4(x)$ is a PP of ${\mathbb F_{q}}$ if and only if $x^{\frac{3^s+3}{2}}$ is a PP of ${\mathbb F_{q}}$. The latter is equivalent to $\gcd(\frac{3^s+3}{2}, 3^e-1)=1$. Now we let $\gcd(\frac{3^s+3}{2}, 3^e-1)=1$. If $s$ is even, then one has $v_2(\frac{3^s+3}{2})\ge 1$. It follows that $2|\gcd(\frac{3^s+3}{2}, 3^e-1)$, which is a contradiction. So $s$ must be odd. Then $\frac{3^s+3}{2}$ is an odd integer. It follows from Lemma 2.6 that $\gcd(\frac{3^s+3}{2}, 3^e-1)=\gcd(\frac{3^{s-1}+1}{2}, \frac{3^e-1}{2})= \frac{1}{2}\gcd(3^{s-1}+1, 3^e-1)=1$ if and only if $\frac{e}{\gcd(s-1, e)}$ is odd. This means that $\gcd(\frac{3^s+3}{2}, 3^e-1)=1$ if and only if $v_2(e)\le v_2(s-1)$ and $v_2(s-1)\ge 1$, namely, $v_2(s-1)\ge \max\{1, v_2(e)\}$, as desired.

(ⅱ). Let $k=0, p > 3$. Then $\frac{p^s+3}{2}\equiv \frac{p^{s_0}+3}{2}\pmod{p^e-1}$ and $\frac{p^s+1}{2}\equiv \frac{p^{s_0}+1}{2}\pmod{p^e-1}$. So

$f_4(x)\equiv x^{\frac{p^{s_0}+3}{2}}+3x^{\frac{p^{s_0}+1}{2}}+3x\pmod{x^{p^e}-x}.$

(3.9)

Clearly, if $s_0$ is even, then $x^{\frac{p^{s_0}+3}{2}}=x^2$ and $x^{\frac{p^{s_0}+1}{2}}=x$ for any $x\in {\mathbb F_{p}^*}$. Then by (3.9) we have $f_4(x)=x^2+6x=x(x+6)$ for any $x\in {\mathbb F_{p}^*}$. Hence $f_4(x)=0$ has one nonzero root $-6$ in ${\mathbb F_{p}^*}$. But $f_4(0)=0$. It then follows that $f_4(x)$ is not a PP of ${\mathbb F_{p}}$. One notes that $f_4({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Therefore $f_4(x)$ is not a PP of ${\mathbb F_{p^e}}$. It follows that $D_{p^s+3, k}(1, x)$ is not a PP of ${\mathbb F_{q}}$ when $s_0$ is even.

(ⅲ). Letting $k=2$, we have $D_{p^s+3, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if (3.8) is a PP of ${\mathbb F_{q}}$, i.e., $3x^{\frac{p^s+1}{2}}+x^{\frac{p^s-1}{2}}+x$ is a PP of ${\mathbb F_{q}}$. Let

$f_5(x):=3x^{\frac{p^s+1}{2}}+x^{\frac{p^s-1}{2}}+x.$

Let $s=0$. Then $f_5=4x+1$, which clearly is a PP of ${\mathbb F_{q}}$. So $D_{p^s+3, k}(1, x)$ is a PP of ${\mathbb F_{q}}$.

(ⅳ). Let $k=2, p=3, s > 0$. Then the desired result follows from the proof of Part (v) of Theorem 3.3.

(ⅴ). Let $k=2, p > 3$. By Part (iii) we only need to show that $D_{p^s+3, k}(1, x)$ is not a PP of ${\mathbb F_{q}}$ when $s > 0$. Let $0 < s\equiv s_0\pmod{2e}$ with $0\le s_0\le 2e-1$. Then one has

$f_5(x)\equiv 3x^{\frac{p^s+1}{2}}+x^{\frac{p^s-1}{2}}+x\pmod{x^{p^e}-x}.$

If $s_0$ is even, then $f_5(x)=4x+1$ for any $x\in {\mathbb F_{p}^*}$. In this situation, $f_5(x)=0$ has one nonzero root $-\frac{1}{4}\in {\mathbb F_{p}^*}$. So $f_5(x)$ is not a PP of ${\mathbb F_{p}}$ since $f_5(0)=0$. Also note that $f_5({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Thus $f_5(x)$ is not a PP of ${\mathbb F_{p^e}}$ in this case. If $s_0$ is odd, then $f_5(x)=3x^{\frac{p+1}{2}}+x^{\frac{p-1}{2}}+x$ for any $x\in {\mathbb F_{p}^*}$. So in ${\mathbb F_{p}^*}$, we have

$(f_5(x))^2\equiv x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

Then by Lemma 2.7, we know that $f_5(x)$ is not a PP of ${\mathbb F_{p}}$. We note that $f_5({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Therefore $f_5(x)$ is not a PP of ${\mathbb F_{p^e}}$. It infers that $D_{p^s+3, k}(1, x)$ is not a PP of ${\mathbb F_{q}}$ when $s > 0$. Part (v) is proved.

(ⅵ). Let $k=3$. Then $D_{p^s+3, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if (3.8) is a PP of ${\mathbb F_{q}}$, i.e., the trinomial

$f_6(x):=-x^{\frac{p^s+3}{2}}+6x^{\frac{p^s+1}{2}}+3x^{\frac{p^s-1}{2}}$

is a PP of ${\mathbb F_{q}}$. By the result of Part (ⅰ), we then have from the fact $D_{3^s+3, 3}(1, x)=D_{3^s+3, 0}(1, x)$ that $f_6(x)$ is a PP of of ${\mathbb F_{3^e}}$ if and only if $v_2(s-1)\ge \max\{1, v_2(e)\}$. Then we only need to show that $f_6(x)$ is not a PP of ${\mathbb F_{p^e}}$ when $p > 3$. Let $s\equiv s_0\pmod{2e}$ with $0\le s_0\le 2e-1$. Then one has

$f_6(x)\equiv -x^{\frac{p^{s_0}+3}{2}}+6x^{\frac{p^{s_0}+1}{2}}+3x^{\frac{p^{s_0}-1}{2}} \pmod{x^{p^e}-x}.$

If $s_0$ is even, then $f_6(x)=-x^2+6x+3$ for any $x\in {\mathbb F_{p}^*}$. Then $f_6(x)$ is not a PP of ${\mathbb F_{p}}$ since $f_6(2)=f_6(4)=11$. Also note that $f_6({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Thus $f_6(x)$ is not a PP of ${\mathbb F_{p^e}}$ for $p > 3$ and $s_0$ being even. If $s_0$ is odd, then $f_6(x)=-x^{\frac{p+3}{2}}+6x^{\frac{p+1}{2}}+3x^{\frac{p-1}{2}}$ for any $x\in {\mathbb F_{p}^*}$. So in ${\mathbb F_{p}^*}$, we have

$(f_6(x))^2\equiv 9x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

Then by Lemma 2.7, we know that $f_6(x)$ is not a PP of ${\mathbb F_{p}}$. We note that $f_6({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Therefore $f_6(x)$ is not a PP of ${\mathbb F_{p^e}}$ when $p > 3$ and $s_0$ is odd. So $f_6(x)$ is a PP of ${\mathbb F_{q}}$ if and only if $p=3$ and $v_2(s-1)\ge \max\{1, v_2(e)\}$, that is, $D_{p^s+3, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $p=3$ and $v_2(s-1)\ge \max\{1, v_2(e)\}$. Part (ⅵ) is proved.

(ⅳ). Let $k\ne 0, 2, 3$ and $0\le k < p$. Then $D_{p^s+3, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if (3.8) is a PP of ${\mathbb F_{q}}$, i.e., if and only if

$f_7(x):=(2-k)x^{\frac{p^s+3}{2}}+6x^{\frac{p^s+1}{2}} +kx^{\frac{p^s-1}{2}}+(6-2k)x$

is a PP of ${\mathbb F_{q}}$. Let $s\equiv s_0\pmod{2e}$ with $0\le s_0\le 2e-1$. Then one has

$f_7(x)\equiv (2-k)x^{\frac{p^{s_0}+3}{2}}+6x^{\frac{p^{s_0}+1}{2}} +kx^{\frac{p^{s_0}-1}{2}}+(6-2k)x \pmod{x^{p^e}-x}.$

If $s_0$ is even, then $f_7(x)=(2-k)x^2+(12-2k)x+k$ for any $x\in {\mathbb F_{p}^*}$. One then finds that $f_7(\frac{4}{k-2})=f_7(\frac{8-2k}{k-2})$ and $\frac{4}{k-2}\ne \frac{8-2k}{k-2}$ when $k\ne 4$. If $k=4$, then $p\ge 5$. In this case, $f_7(x)=-2x^2+4x+4$ for any $x\in {\mathbb F_{p}^*}$, which implies $f_7(-1)=f_7(3)$ when $k=4$. Therefore $f_7(x)$ is not a PP of ${\mathbb F_{p}}$. Also note that $f_7({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Thus $f_7(x)$ is not a PP of ${\mathbb F_{p^e}}$ when $s_0$ is even. If $s_0$ is odd, then

$f_7(x)=(2-k)x^{\frac{p+3}{2}}+6x^{\frac{p+1}{2}}+kx^{\frac{p-1}{2}}+(6-2k)x$

(3.10)

for any $x\in {\mathbb F_{p}^*}$. We consider the following two cases.

CASE 1. Let $p=3$. Then $k=1$ since $k < p$ and $k\ne 0, 2$. Hence $\forall x\in{\mathbb F_{3}^*}$, $f_7(x)=x^3+2x$. It then follows from $f_7(0)=f_7(1)=0$ that $f_7(x)$ is not a PP of ${\mathbb F_{3}}$. We note that $f_7({\mathbb F_{3}})\subseteq {\mathbb F_{3}}$. Therefore $f_7(x)$ is not a PP of ${\mathbb F_{3^e}}$.

CASE 2. Let $p > 3$. By (3.10), in ${\mathbb F_{p}}$ we have

$(f_7(x))^2\equiv k^2x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

Then by Lemma 2.7, we know that $f_7(x)$ is not a PP of ${\mathbb F_{p}}$. We note that $f_7({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Therefore $f_7(x)$ is not a PP of ${\mathbb F_{p^e}}$ when $p > 3$ and $s_0$ is odd.

Hence $f_7(x)$ is not a PP of ${\mathbb F_{p^e}}$ when $k\ne 0, 2, 3$, from which we deduce immediately that $D_{p^s+3, k}(1, x)$ is not a PP of ${\mathbb F_{p^e}}$ when $k\ne 0, 2, 3$. Part (vii) is proved. So we completes the proof of Theorem 3.4.

Theorem 3.5. Let $q=p^e$ with $p$ being an odd prime and $e$ being a positive integer. Let $s$ be a nonnegative integer and $s_0$ be the least nonnegative residue of $s$ modulo $2e$. If $D_{p^s+4, k}(1, x)$ is a PP of ${\mathbb F_{q}}$, then either $k=0$ and $s_0$ is odd, or $k > 0, k\ne 2$ and $s=0$.

Proof. It is sufficient to show that $D_{p^s+4, k}(1, x)$ is not a PP of ${\mathbb F_{q}}$ when $k=0$, $s_0$ is even, or $k > 0, s > 0$. By Theorem 3.1, we get

$\begin{align*} 32D_{p^s+4, k}(1, x)&=k(1-4x)^{\frac{p^s-1}{2}}+(8+2k)(1-4x)^{\frac{p^s+1}{2}}\\ &+(8-3k)(1-4x)^{\frac{p^s+3}{2}}+2+3k+(12-2k)(1-4x)+(2-k)(1-4x)^2. \end{align*}$

Then $D_{p^s+4, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $kx^{\frac{p^s-1}{2}}+(8+2k)x^{\frac{p^s+1}{2}} +(8-3k)x^{\frac{p^s+3}{2}}+(12-2k)x+(2-k)x^2$ is a PP of ${\mathbb F_{q}}$. Let

$f_8(x):=kx^{\frac{p^s-1}{2}}+(8+2k)x^{\frac{p^s+1}{2}} +(8-3k)x^{\frac{p^s+3}{2}}+(12-2k)x+(2-k)x^2.$

Now we show that $f_8(x)$ is not a PP of ${\mathbb F_{q}}$ when $k=0$, $s_0$ is even, or $k > 0, s > 0$. Then the following cases are considered.

CASE 1. $k=0$ and $s_0$ is an even. Then $f_8(x)=4x^{\frac{p^{s_0}+1}{2}} +4x^{\frac{p^{s_0}+3}{2}}+6x+x^2$. It infers that

$f_8(x)\equiv 4x^{\frac{p^{s_0}+1}{2}} +4x^{\frac{p^{s_0}+3}{2}}+6x+x^2\pmod{x^{q}-x}.$

Additionally, $\forall x\in{\mathbb F_{p}^*}$, $x^{\frac{p^{s_0}+1}{2}}=x$ and $x^{\frac{p^{s_0}+3}{2}}=x^2$ since $s_0$ is an even. Therefore

$f_8(x)=5x(x+2)$

for any $x\in{\mathbb F_{p}^*}$. Then $f_8(0)=f_8(-2)=0$. So $f_8(x)$ is not a PP of ${\mathbb F_{p}}$. Also $f_8(x)$ is not a PP of ${\mathbb F_{p^e}}$ since $f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$.

CASE 2. $k=2$. Then

$f_8(x)=2x^{\frac{p^{s}-1}{2}} +12x^{\frac{p^{s}+1}{2}}+2x^{\frac{p^{s}+3}{2}}+8x.$

SUBCASE 2-1. $p=3$. Then $f_8(x)=2x^{\frac{p^{s}-1}{2}} +2x^{\frac{p^{s}+3}{2}}+2x$. So $f_8(x)=2x^{2} +2x+2$ when $s=0$, which then follows that $f_8(0)=f_8(2)=2$. If $s > 0$, we have easily that $f_8(0)=f_8(1)=0$. Thus $f_8(x)$ is not a PP of ${\mathbb F_{3^e}}$ whenever.

SUBCASE 2-2. $p > 3$. Then

$f_8(x)\equiv 2x^{\frac{p^{s_0}-1}{2}} +12x^{\frac{p^{s_0}+1}{2}}+2x^{\frac{p^{s_0}+3}{2}} +8x\pmod{x^{p^e}-x}.$

If $s_0$ is even, then $f_8(x)=2x^2+20x+2$ for any $x\in {\mathbb F_{p}^*}$. This implies that $f_8(-4)=f_8(-6)$, which then follows that $f_8(x)$ is not a PP of ${\mathbb F_{p}}$. Note that $f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. So $f_8(x)$ is not a PP of ${\mathbb F_{q}}$ when $s_0$ is even. If $s_0$ is odd, then $f_8(x)=2x^{\frac{p-1}{2}}+12x^{\frac{p+1}{2}} +2x^{\frac{p+3}{2}}+8x$ for any $x\in {\mathbb F_{p}^*}$. We then deduces that

$(f_8(x))^2\equiv 4x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

Then by Lemma 2.7, we know that $f_8(x)$ is not a PP of ${\mathbb F_{p}}$. We note that $f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Therefore $f_8(x)$ is not a PP of ${\mathbb F_{p^e}}$ when $s_0$ is odd.

Thus $D_{p^s+4, 2}(1, x)$ is not a PP of ${\mathbb F_{p^e}}$ for any nonnegative integer $s$ and odd prime $p$.

CASE 3. $k=6, s > 0$. Then $p\ge 7$ and

$f_8(x)=6x^{\frac{p^{s}-1}{2}} +20x^{\frac{p^{s}+1}{2}}-10x^{\frac{p^{s}+3}{2}}-4x^2.$

If $s > 0$ and $s_0$ is even, then $f_8(x)=-14x^2+20x+6$ for any $x\in {\mathbb F_{p}^*}$. This implies that $f_8(0)=f_8(-1)=0$ if $p=7$, or $f_8(\frac{4}{7})=f_8(\frac{6}{7})$ if $p > 7$. This means that $f_8(x)$ is not a PP of ${\mathbb F_{p}}$. Note that $f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. So $f_8(x)$ is not a PP of ${\mathbb F_{q}}$ when $s_0$ is even. If $s > 0$ and $s_0$ is odd, then $f_8(x)=6x^{\frac{p-1}{2}}+20x^{\frac{p+1}{2}} -10x^{\frac{p+3}{2}}-4x^2$ for any $x\in {\mathbb F_{p}^*}$, which implies that

$(f_8(x))^2\equiv 36x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

Then by Lemma 2.7, we know that $f_8(x)$ is not a PP of ${\mathbb F_{p}}$. We note that $f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Therefore $f_8(x)$ is not a PP of ${\mathbb F_{p^e}}$ when $s_0$ is odd.

Thus $D_{p^s+4, 6}(1, x)$ is not a PP of ${\mathbb F_{p^e}}$ when $s > 0$.

CASE 4. $k=p-4, s > 0$. Then $p\ge 5$ and

$f_8(x)=(p-4)x^{\frac{p^{s}-1}{2}} +(20-3p)x^{\frac{p^{s}+3}{2}}+(20-2p)x+(6-p)x^2.$

If $s > 0, s_0$ is even, then $f_8(x)=26x^2+20x-4$ for any $x\in {\mathbb F_{p}^*}$. This implies that $f_8(0)=f_8(\frac{1}{5})=0$ if $p=13$, or $f_8(\frac{-4}{13})=f_8(\frac{-6}{13})$ if $p\ne 13$. This means that $f_8(x)$ is not a PP of ${\mathbb F_{p}}$. Note that $f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. So $f_8(x)$ is not a PP of ${\mathbb F_{q}}$ when $s_0$ is even. If $s > 0, s_0$ is odd, then $f_8(x)=-4x^{\frac{p-1}{2}}+20x^{\frac{p+3}{2}} +20x+6x^2$ for any $x\in {\mathbb F_{p}^*}$, which implies that

$(f_8(x))^2\equiv 16x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

Then by Lemma 2.7, we know that $f_8(x)$ is not a PP of ${\mathbb F_{p}}$. We note that $f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Therefore $f_8(x)$ is not a PP of ${\mathbb F_{p^e}}$ when $s_0$ is odd.

Thus $D_{p^s+4, p-4}(1, x)$ is not a PP of ${\mathbb F_{p^e}}$ when $s > 0$.

CASE 5. $p\mid (3k-8), s > 0$. Then $p\ge 5$, $p\nmid (2-k)$ and

$f_8(x)=kx^{\frac{p^{s}-1}{2}} +(8+2k)x^{\frac{p^{s}+1}{2}}+(12-2k)x+(2-k)x^2.$

If $s > 0, s_0$ is even, then $f_8(x)=(2-k)x^2+20x+k$ for any $x\in {\mathbb F_{p}^*}$. This implies that $f_8(\frac{-11}{2-k})=f_8(\frac{-9}{2-k})=0$. This means that $f_8(x)$ is not a PP of ${\mathbb F_{p}}$. Note that $f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. So $f_8(x)$ is not a PP of ${\mathbb F_{q}}$ when $s_0$ is even. If $s > 0, s_0$ is odd, then $f_8(x)=kx^{\frac{p-1}{2}}+(8+2k)x^{\frac{p+1}{2}} +(12-2k)x+(2-k)x^2$ for any $x\in {\mathbb F_{p}^*}$, which implies that

$(f_8(x))^2\equiv k^2x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

Then by Lemma 2.7, we know that $f_8(x)$ is not a PP of ${\mathbb F_{p}}$. We note that $f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Therefore $f_8(x)$ is not a PP of ${\mathbb F_{p^e}}$ when $s_0$ is odd.

Thus $D_{p^s+4, k}(1, x)$ is not a PP of ${\mathbb F_{p^e}}$ when $p\mid (3k-8)$ and $s > 0$.

CASE 6. $k\ne 0, 2, 6, p-4$, $s > 0$ and $p\nmid (3k-8)$. Then

$f_8(x)=kx^{\frac{p^{s}-1}{2}} +(8+2k)x^{\frac{p^{s}+1}{2}}+(8-3k)x^{\frac{p^{s}+3}{2}}+(12-2k)x+(2-k)x^2.$

If $s > 0, s_0$ is even, then $f_8(x)=(10-4k)x^2+20x+k$ for any $x\in {\mathbb F_{p}^*}$. If $p\mid (2k-5)$, then $p\ne 5$ and $f_8(x)=20x+k, \forall x\in\mathbb{F}_p^*$. It implies that $f_8(0)=f_8(\frac{-k}{20})=0$. So $f_8(x)$ is not a PP of ${\mathbb F_{p}}$ when $p\mid (2k-5)$. If $p\nmid (2k-5)$, then $f_8(\frac{4}{2k-5})=f_8(\frac{6}{2k-5})$, which means that $f_8(x)$ is not a PP of ${\mathbb F_{p}}$ when $p\nmid (2k-5)$. Thus $f_8(x)$ is not a PP of ${\mathbb F_{p}}$ when $s > 0, s_0$ is even. Note that $f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. So $f_8(x)$ is not a PP of ${\mathbb F_{q}}$ when $s_0$ is even. If $s > 0, s_0$ is odd, then $f_8(x)=kx^{\frac{p-1}{2}}+(8+2k)x^{\frac{p+1}{2}} +(8-3k)x^{\frac{p+3}{2}}+(12-2k)x+(2-k)x^2$ for any $x\in {\mathbb F_{p}^*}$. If $p=3$, then $k=1$. In this case $f_8(x)=2x+2x^2+2x^3$, which implies that $f_8(0)=f_8(1)=0$. It then follows that $f_8(x)$ is not a PP of ${\mathbb F_{3}}$. If $p > 3$, then

$(f_8(x))^2\equiv k^2x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

Then by Lemma 2.7, we know that $f_8(x)$ is not a PP of ${\mathbb F_{p}}$. Thus $f_8(x)$ is not a PP of ${\mathbb F_{p}}$ when $s_0$ is odd. We note that $f_8({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Therefore $f_8(x)$ is not a PP of ${\mathbb F_{p^e}}$ when $s_0$ is odd.

Thus $D_{p^s+4, k}(1, x)$ is not a PP of ${\mathbb F_{p^e}}$ when $k\ne 0, 2, 6, p-4$, $s > 0$ and $p\nmid (3k-8)$. Combining all of the above cases, we have the desired result. Therefore Theorem 3.5 is proved.

Corollary 3.6. Let $q=p^e$ with $p$ being an odd prime and $e$ being a positive integer. Let $s$ and $k$ be nonnegative integers with $0 < k < p$. Then $D_{p^s+4, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $s=0$ and $p\mid(2k-5)$.

Proof. The desired result follows immediately from the proof of Theorem 3.5.

---

4. Reversed Dickson polynomials $D_{p^s+p^t+\ell, k}(1, x)$

In this section, we present an explicit formula for $D_{n, k}(1, x)$ when $n=p^s+p^t+\ell$ with $\le s < t$ and $0\le \ell < p$. Then we characterize $D_{n, k}(1, x)$ to be a PP of ${\mathbb F}_q$ in this case.

Theorem 4.1. Let $p={\rm char}({\mathbb F_{q}})$ be an odd prime. Let $s$ and $t$ be integers such that $0\le s < t$. Then

$D_{p^s+p^{t}, k}(1, x)=\frac{k}{4}\big((1-4x)^{\frac{p^s-1}{2}}+ (1-4x)^{\frac{p^{t}-1}{2}}\big)-\frac{k-2}{4}\big(1 +(1-4x)^{\frac{p^{s}+p^{t}}{2}}\big).$

Proof. We consider the following two cases.

CASE 1. $x\ne \frac{1}{4}$. For this case, putting $x=y(1-y)$ in the second identity of Lemma 2.3 gives us that

$\begin{align*} D_{p^s+p^{t}, k}(1, x)=&D_{p^s+p^{t}, k}(1, y(1-y))\\ =&\frac{\big(k-1-(k-2)y\big)y^{p^s+p^{t}}-\big(1+(k-2)y\big)(1-y)^{p^s+p^{t}}}{2y-1}\\ =&\frac{\frac{k+(2-k)u}{2}\big(\frac{u+1}{2}\big)^{p^s+p^{t}} -\frac{k+(k-2)u}{2}\big(\frac{1-u}{2}\big)^{p^s+p^{t}}}{u}\\ =&\frac{k}{4}(u^{p^s-1}+u^{p^{t}-1})-\frac{k-2}{4}(1+u^{p^s+p^{t}})\\ =&\frac{k}{4}\big((u^2)^{\frac{p^s-1}{2}}+(u^2)^{\frac{p^{t}-1}{2}}\big)-\frac{k-2}{4}(1+(u^2)^{\frac{p^s+p^{t}}{2}}), \end{align*}$

where $u=2y-1$ and $u^2=1-4x$. So we obtain that

$D_{p^s+p^{t}, k}(1, x)=\frac{k}{4}\big((1-4x)^{\frac{p^s-1}{2}}+ (1-4x)^{\frac{p^{t}-1}{2}}\big)-\frac{k-2}{4}\big(1 +(1-4x)^{\frac{p^{s}+p^{t}}{2}}\big)$

as desired.

CASE 2. $x=\frac{1}{4}$. By the first identity of Lemma 2.3, one has

$D_{p^s+p^{t}, k}\big(1, \frac{1}{4}\big)=\frac{k(p^s+p^{t})-k+2}{2^{p^s+p^{t}}}=\frac{-k+2}{4}.$

Besides,

$\frac{k}{4}\big((1-4\times \frac{1}{4})^{\frac{p^s-1}{2}}+ (1-4\times \frac{1}{4})^{\frac{p^{t}-1}{2}}\big)-\frac{k-2}{4}\big(1 +(1-4\times \frac{1}{4})^{\frac{p^{s}+p^{t}}{2}}\big)=\frac{-k+2}{4}.$

Thus the required result follows. So Theorem 4.1 is proved.

Theorem 4.2. Let $q=p^e$ with $p$ being an odd prime and $e$ being a positive integer. Let $s$ and $t$ be positive integers with $s < t$. Then each of following is true.

(ⅰ). If $k=0$, then $D_{p^s+p^t, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if either $p\equiv 1\pmod{4}$ and $v_2(t-s)\ge v_2(e)$, or $p\equiv 3\pmod{4}$ and $v_2(t-s)\ge \max\{v_2(e), 1\}$.

(ⅱ). Let $k=2$. If $p > 3$, then $D_{p^s+p^t, k}(1, x)$ is not a PP of ${\mathbb F_{p^e}}$. If $p=3$ and $st$ is even, then $D_{p^s+p^t, k}(1, x)$ is not a PP of ${\mathbb F_{p^e}}$.

(ⅲ). If $k\ne 0, 2$, then $D_{p^s+p^t, k}(1, x)$ is not a PP of ${\mathbb F_{q}}$.

Proof. By Theorem 4.1, we have that $D_{p^s+p^{t}, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if

$k\big(x^{\frac{p^s-1}{2}}+ x^{\frac{p^{t}-1}{2}}\big)-(k-2)x^{\frac{p^{s}+p^{t}}{2}}$

is a PP of ${\mathbb F_{q}}$.

(ⅰ). Let $k=0$. Then $D_{p^s+p^{t}, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $x^{\frac{p^{s}+p^{t}}{2}}$ is a PP of ${\mathbb F_{q}}$ if and only if

$\gcd\Big(\frac{p^s+p^t}{2}, p^e-1\Big)=1.$

Additionally, $\gcd\big(\frac{p^s+p^t}{2}, p^e-1\big)= \gcd\big(\frac{p^{t-s}+1}{2}, p^e-1\big)$. Then the desired result follows from the same way as proving Part (ⅰ) of Theorem 3.2.

(ⅱ). Let $k=2$. Then $D_{p^s+p^{t}, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if $x^{\frac{p^{s}-1}{2}}+x^{\frac{p^{t}-1}{2}}$ is a PP of ${\mathbb F_{q}}$. Let

$g_1(x):=x^{\frac{p^{s}-1}{2}}+x^{\frac{p^{t}-1}{2}}.$

So

$g_1(x)\equiv x^{\frac{p^{s_0}-1}{2}}+x^{\frac{p^{t_0}-1}{2}}\pmod{x^{p^e}-x}.$

Then the following cases are considered.

CASE 1. $s > 0$ and both $s_0$ and $t_0$ are even. Then $g_1(x)=2$ for any $x\in {\mathbb F_{p}^*}$. So $g_1(x)$ is not a PP of ${\mathbb F_{p}}$. One also notices that $g_1({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Thus $g_1(x)$ is not a PP of ${\mathbb F_{q}}$.

CASE 2. $s > 0$ and one of $s_0$ and $t_0$ is even, the other is odd. Then $g_1(x)=x^{\frac{p-1}{2}}+1$ for any $x\in {\mathbb F_{p}^*}$. If $p=3$, then $\forall x\in{\mathbb F_{p}^*}$, $g_1(x)=x+1$, which implies $g_1(0)=g_1(-1)=0$. So $g_1(x)$ is not a PP of ${\mathbb F_{3}}$. If $p > 3$, then

$(g_1(x))^2\equiv x^{p-1}+2x^{\frac{p-1}{2}}+1\pmod{x^p-x}.$

It follows from Lemma 2.7 that $g_1(x)$ is not a PP of ${\mathbb F_{p}}$ when $p > 3$. Therefore $g_1(x)$ is not a PP of ${\mathbb F_{p}}$. Obviously, $g_1({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Hence $g_1(x)$ is not a PP of ${\mathbb F_{q}}$ in this case.

CASE 3. $s > 0, p > 3$ and both $s_0$ and $t_0$ are odd. Then $g_1(x)=2x^{\frac{p-1}{2}}$ for any $x\in {\mathbb F_{p}}$. But $\gcd(\frac{p-1}{2}, p-1)=\frac{p-1}{2} > 1$. Therefore $g_1(x)$ is not a PP of ${\mathbb F_{p}}$. Note that $g_1({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Hence $g_1(x)$ is not a PP of ${\mathbb F_{q}}$ in this case.

Combining the above cases, we know that part (ⅱ) is true.

(ⅲ). Let $k\ne 0$ and $k\ne 2$. Let

$g_2(x):=k\big(x^{\frac{p^s-1}{2}}+ x^{\frac{p^{t}-1}{2}}\big)-(k-2)x^{\frac{p^{s}+p^{t}}{2}}.$

Then

$g_2(x)\equiv k\big(x^{\frac{p^{s_0}-1}{2}}+ x^{\frac{p^{{t_0}}-1}{2}}\big)-(k-2)x^{\frac{p^{{s_0}}+p^{{t_0}}}{2}}\pmod{x^q-x}.$

Then we divide the proof into the following three cases.

CASE 1. Both $s_0$ and $t_0$ are even. Then $g_2(x)=2k-(k-2)x$ for any $x\in {\mathbb F_{p}^*}$. So $g_2(x)$ is not a PP of ${\mathbb F_{p}}$ since $g_2(0)=g_2(\frac{2k}{k-2})$. One also notices that $g_2({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Thus $g_2(x)$ is not a PP of ${\mathbb F_{q}}$.

CASE 2. One of $s_0$ and $t_0$ is even, the other is odd. Then $g_2(x)=k+kx^{\frac{p-1}{2}}-(k-2)x^{\frac{p+1}{2}}$ for any $x\in {\mathbb F_{p}^*}$. If $p=3$, then $k=1$ and so $g_2(0)=g_2(1)=0$, which implies $g_2(x)$ is not a PP of ${\mathbb F_{3}}$. If $p > 3$, then

$(g_2(x))^2\equiv k^2x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

It follows from Lemma 2.7 that $g_2(x)$ is not a PP of ${\mathbb F_{p}}$ when $p > 3$. Therefore $g_2(x)$ is not a PP of ${\mathbb F_{p}}$. Obviously, $g_2({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Hence $g_2(x)$ is not a PP of ${\mathbb F_{q}}$ in this case.

CASE 3. Both $s_0$ and $t_0$ are odd. Then $g_2(x)=2kx^{\frac{p-1}{2}}-(k-2)x$ for any $x\in {\mathbb F_{p}^*}$. If $p=3$, then $k=1$ and so $g_2(x)=0, \forall x\in\mathbb{F}_p$, which implies $g_2(x)$ is not a PP of ${\mathbb F_{3}}$. If $p > 3$, then

$(g_2(x))^2\equiv 4k^2x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

Since $4k^2\in \mathbb{F}_p^*$, it then follows from Lemma 2.7 that $g_2(x)$ is not a PP of ${\mathbb F_{p}}$ when $p > 3$. Therefore $g_2(x)$ is not a PP of ${\mathbb F_{p}}$. Obviously, $g_2({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. Hence $g_2(x)$ is not a PP of ${\mathbb F_{q}}$ in this case.

Combining the above cases, we deduce that $g_2(x)$ is not a PP of ${\mathbb F_{q}}$ in the condition of $k\ne 0, 2$. Thus $D_{p^s+p^t, k}(1, x)$ is not a PP of ${\mathbb F_{q}}$. The proof of Theorem 4.2 is completed.

Theorem 4.3. Let $q=p^e$ with $p$ being an odd prime and $e$ being a positive integer. Let $s$ and $t$ be positive integers with $s < t$. Then

$\begin{align} D_{p^s+p^{t}+1, k}(1, x)&=\frac{1}{4}(1-4x)^{\frac{p^s+p^{t}}{2}}+ \frac{1}{4}+\frac{1}{8}\big((1-4x)^{\frac{p^s-1}{2}}+ (1-4x)^{\frac{p^{t}-1}{2}}\big)\nonumber\\ &-\frac{k-2}{8}\big((1-4x)^{\frac{p^s+1}{2}}+ (1-4x)^{\frac{p^{t}+1}{2}}\big).\label{cheng5} \end{align}$

(4.1)

Furthermore, $D_{p^s+p^{t}+1, k}(1, x)$ is not a PP of ${\mathbb F_{q}}$.

Proof We consider the following two cases.

CASE 1. $x\ne \frac{1}{4}$. For this case, putting $x=y(1-y)$ in the second identity of Lemma 2.3 gives us that

$\begin{align*} &D_{p^s+p^{t}+1, k}(1, x)=D_{p^s+p^{t}+1, k}(1, y(1-y))\\ =&\frac{\big(k-1-(k-2)y\big)y^{p^s+p^{t}+1}-\big(1+(k-2)y\big)(1-y)^{p^s+p^{t}+1}}{2y-1}\\ =&\frac{\frac{k+(2-k)u}{2}\big(\frac{u+1}{2}\big)^{p^s+p^{t}+1} -\frac{k+(k-2)u}{2}\big(\frac{1-u}{2}\big)^{p^s+p^{t}+1}}{u}\\ =&\frac{k}{8}(1+u^{p^s-1}+u^{p^{t}-1}+u^{p^s+p^{t}})-\frac{k-2}{8}(1+u^{p^s+1}+u^{p^{t}+1}+u^{p^s+p^{t}})\\ =&\frac{1}{4}u^{p^s+p^{t}}+\frac{1}{4}+\frac{k}{8}(u^{p^s-1}+u^{p^{t}-1})-\frac{k-2}{8}(u^{p^s+1}+u^{p^{t}+1}), \end{align*}$

where $u=2y-1$ and $u^2=1-4x$. Then we have that

$\begin{align*} &D_{p^s+p^{t}+1, k}(1, x)=\frac{1}{4}(1-4x)^{\frac{p^s+p^{t}}{2}}+\frac{1}{4}\\ &+\frac{1}{8}\big((1-4x)^{\frac{p^s-1}{2}}+ (1-4x)^{\frac{p^{t}-1}{2}}\big)-\frac{k-2}{8}\big((1-4x)^{\frac{p^s+1}{2}}+ (1-4x)^{\frac{p^{t}+1}{2}}\big) \end{align*}$

as desired.

CASE 2. $x=\frac{1}{4}$. On the one hand, by the first identity of Lemma 2.3, one has

$D_{p^s+p^{t}+1, k}\big(1, \frac{1}{4}\big) =\frac{k(p^s+p^{t}+1)-k+2}{2^{p^s+p^{t}}}=\frac{1}{4}.$

On the other hand,

$\frac{1}{4}(1-4\times \frac{1}{4})^{\frac{p^s+p^{t}}{2}}+\frac{1}{4}+ \frac{1}{8}\big((1-4\times \frac{1}{4})^{\frac{p^s-1}{2}}+ (1-4\times \frac{1}{4})^{\frac{p^{t}-1}{2}}\big)- \frac{k-2}{8}\big((1-4\times \frac{1}{4})^{\frac{p^s+1}{2}}+ (1-4\times \frac{1}{4})^{\frac{p^{t}+1}{2}}\big)=\frac{1}{4}.$

Combing Case 1 and Case 2, we know that (4.1) always holds. So $D_{p^s+p^{t}+1, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if the polynomial

$g_3(x):=2x^{\frac{p^s+p^{t}}{2}}+2\big(x^{\frac{p^s-1}{2}}+ x^{\frac{p^{t}-1}{2}}\big)-(k-2)\big(x^{\frac{p^s+1}{2}}+ x^{\frac{p^{t}+1}{2}}\big)$

is a PP of ${\mathbb F_{q}}$.

In what follows, we show that $g_3(x)$ is not a PP of ${\mathbb F_{q}}$. Now let $s\equiv s_0\pmod{2e}$ and $t\equiv t_0\pmod{2e}$ with $0\le s_0\le 2e-1, 0\le t_0\le 2e-1$. Then

$g_3(x)\equiv 2x^{\frac{p^{s_0}+p^{{t_0}}}{2}}+2\big(x^{\frac{p^{s_0}-1}{2}}+ x^{\frac{p^{{t_0}}-1}{2}}\big)-(k-2)\big(x^{\frac{p^{s_0}+1}{2}}+ x^{\frac{p^{{t_0}}+1}{2}}\big)\pmod{x^q-x}.$

First we let $k=2$. In this case we have

$g_3(x)\equiv 2x^{\frac{p^{s_0}+p^{{t_0}}}{2}}+2x^{\frac{p^{s_0}-1}{2}}+ 2x^{\frac{p^{{t_0}}-1}{2}}\pmod{x^q-x}.$

If both $s_0$ and $t_0$ are even, then $\forall x\in{\mathbb F_{p}^*}$, $g_3(x)=2x+4$. It follows that $g_3(x)$ is not a PP of ${\mathbb F_{p}}$ since $g_3(0)=g_3(-2)=0$.

If exactly one of $s_0$ and $t_0$ is even, then $g_3(x)=2x^{\frac{p-1}{2}}+2x^{\frac{p+1}{2}}+2$ for any $x\in {\mathbb F_{p}^*}$. In this case if $p=3$, then $g_3(0)=g_3(1)=0$, which implies $g_3(x)$ is not a PP of ${\mathbb F_{3}}$. If $p > 3$, then

$(g_3(x))^2\equiv 4x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

It follows from Lemma 2.7 that $g_3(x)$ is not a PP of ${\mathbb F_{p}}$ when $p > 3$. Therefore $g_3(x)$ is not a PP of ${\mathbb F_{p}}$.

If both $s_0$ and $t_0$ are odd, then $g_3(x)=2x+4x^{\frac{p-1}{2}}$ for any $x\in {\mathbb F_{p}^*}$. If $p=3$, then $g_3(x)=0, \forall x\in {\mathbb F_{p}^*}$, which implies $g_3(x)$ is not a PP of ${\mathbb F_{3}}$. If $p > 3$, then

$(g_3(x))^2\equiv 4x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

It follows from Lemma 2.7 that $g_3(x)$ is not a PP of ${\mathbb F_{p}}$ when $p > 3$. Therefore $g_3(x)$ is not a PP of ${\mathbb F_{p}}$.

Combining the above discussions, we derive that $g_3(x)$ is not a PP of ${\mathbb F_{p}}$. Note that $g_3({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. So $g_3(x)$ is not a PP of ${\mathbb F_{q}}$ when $k=2$.

Now let $k\ne 2$. The following cases are considered.

If both $s_0$ and $t_0$ are even, then $\forall x\in{\mathbb F_{p}^*}$, $g_3(x)=(6-2k)x+4$. Clearly, if $k=3$, then $g_3(x)$ is not a PP of ${\mathbb F_{p}}$. If $k\ne 3$, then $g_3(0)=g_3(\frac{2}{k-3})=0$. This implies that $g_3(x)$ is not a PP of ${\mathbb F_{p}}$.

If exactly one of $s_0$ and $t_0$ is even then $g_3(x)=(4-k)x^{\frac{p+1}{2}}+2x^{\frac{p-1}{2}}-(k-2)x+2$ for any $x\in {\mathbb F_{p}^*}$. If $p=3$, then $k=0$ or $k=1$. And $g_3(0)=0, g_3(1)=k+2, g_3(-1)=2$. So in this case, either $g_3(1)=g_3(-1)=2$ if $k=0$, or $g_3(1)=g_3(0)=0$ if $k=1$, which implies $g_3(x)$ is not a PP of ${\mathbb F_{3}}$. If $p > 3$, then

$(g_3(x))^2\equiv 4x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

It follows from Lemma 2.7 that $g_3(x)$ is not a PP of ${\mathbb F_{p}}$ when $p > 3$. Therefore $g_3(x)$ is not a PP of ${\mathbb F_{p}}$.

If both $s_0$ and $t_0$ are odd then $g_3(x)=2x^p+4x^{\frac{p-1}{2}}-2(k-2)x^{\frac{p+1}{2}}$ for any $x\in {\mathbb F_{p}^*}$. If $p=3$, then $g_3(1)=g_3(-1)=k-2$, which implies $g_3(x)$ is not a PP of ${\mathbb F_{3}}$. If $p > 3$, then

$(g_3(x))^2\equiv 16x^{p-1}+{\rm the\ \ terms\ \ of} \ \ x\ \ {\rm with \ \ the\ \ degree\ \ less\ \ than }\ \ p-1\pmod{x^p-x}.$

It follows from Lemma 2.7 that $g_3(x)$ is not a PP of ${\mathbb F_{p}}$ when $p > 3$. Therefore $g_3(x)$ is not a PP of ${\mathbb F_{p}}$.

From them, we derive that $g_3(x)$ is not a PP of ${\mathbb F_{p}}$ when $k\ne 2$. Note that $g_3({\mathbb F_{p}})\subseteq {\mathbb F_{p}}$. So $g_3(x)$ is not a PP of ${\mathbb F_{q}}$ when $k\ne 2$. Hence $g_3(x)$ is not a PP of ${\mathbb F_{q}}$. Thus $D_{p^s+p^{t}+1, k}(1, x)$ is not a PP of ${\mathbb F_{q}}$.

By Lemma 2.4, Theorem 4.1 and Theorem 4.3, we have the following general result.

Theorem 4.4. Let $q=p^e$ with $p$ being an odd prime and $e$ being a positive integer. Let $s$ and $t$ be positive integers with $s < t$. Then

$\begin{align} D_{p^s+p^t+2, k}(1, x)&=\frac{2-k}{16}(1-4x)^{\frac{p^s+p^t+2}{2}}+\frac{2+k}{16}(1-4x)^{\frac{p^s+p^t}{2}} +\frac{4-k}{16}\Big((1-4x)^{\frac{p^s+1}{2}}+(1-4x)^{\frac{p^t+1}{2}}\Big)\nonumber\\ &\ \ \ +\frac{2-k}{16}\Big((1-4x)^{\frac{p^s-1}{2}}+(1-4x)^{\frac{p^t11}{2}}\Big) +\frac{2-k}{16}(1-4x)+\frac{2-k}{16}. \end{align}$

(4.2)

Consequently, $D_{p^s+p^{t}+2, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if the polynomial

$(2-k)(x^{\frac{p^s+p^{t}+2}{2}}+x^{\frac{p^s-1}{2}}+x^{\frac{p^t-1}{2}})+ (2+k)x^{\frac{p^s+p^{t}}{2}}+(4-k)(x^{\frac{p^s+1}{2}}+x^{\frac{p^t+1}{2}})$

is a PP of ${\mathbb F_{q}}$. Furthermore, let $\ell\ge 0$ be an integer. Then

$\begin{align*} &D_{p^s+p^t+2\ell, k}(1, x)=\sum\limits_{i=0}^{\ell}B_{2\ell, p^s+p^t+2i}(1-4x)^{\frac{p^s+p^t+2i}{2}} +\sum\limits_{j=0}^{\ell}B_{2\ell, 2j}(1-4x)^{j}\\ &\ \ \ +\sum\limits_{i=0}^{\ell}B_{2\ell, p^s+2i-1}\big((1-4x)^{\frac{p^s+2i-1}{2}}+ (1-4x)^{\frac{p^t+2i-1}{2}}\big), \ \ 0\le\ell\le\frac{p-1}{2}, \end{align*}$

and

$\begin{align*} &D_{p^s+p^t+2\ell+1, k}(1, x)=\sum\limits_{i=0}^{\ell}B_{2\ell+1, p^s+p^t+2i}(1-4x)^{\frac{p^s+p^t+2i}{2}} +\sum\limits_{j=0}^{\ell}B_{2\ell+1, 2j}(1-4x)^{j}\\ &\ \ \ +\sum\limits_{i=0}^{\ell+1}B_{2\ell+1, p^s+2i-1} \big((1-4x)^{\frac{p^s+2i-1}{2}}+(1-4x)^{\frac{p^t+2i-1}{2}}\big), \ \ 0\le\ell<\frac{p-1}{2}, \end{align*}$

where all the coefficients $B_{i, j}$ are given as follows:

$B_{0, p^s+p^t}=\frac{2-k}{4}, B_{0, p^s-1}=\frac{k}{4}, B_{0, 0}=\frac{2-k}{4},$

$B_{1, p^s+p^t}=\frac{1}{4}, B_{1, p^s+1}=\frac{2-k}{8}, B_{1, p^s-1}=\frac{1}{8}, B_{1, 0}=\frac{1}{4},$

and

$\begin{align}\label{ckm7} {\left\{ \begin{array}{ll} B_{2m+2, p^s+p^t+2m+2}=\frac{1}{4}B_{2m, p^s+p^t+2m}, & {\rm if} \ m\ge 0\\ B_{2m+2, p^s+p^t+2i}=B_{2m+1, p^s+2i}-\frac{1}{4}B_{2m, p^s+p^t+2i}+\frac{1}{4}B_{2m, p^s+2i-2}, & {\rm if} \ 1\le i\le m\\ B_{2m+2, p^s+p^t}=B_{2m+1, p^s+p^t}-\frac{1}{4}B_{2m, p^s+p^t}, & {\rm if} \ m\ge 0\\ B_{2m+2, p^s+2m+1}=B_{2m+1, p^s+2m+1}+\frac{1}{4}B_{2m, p^s+2m-1}, & {\rm if} \ m\ge 0\\ B_{2m+2, p^s+2i-1}=B_{2m+1, p^s+2i-1}-\frac{1}{4}B_{2m, p^s+2i-1}+\frac{1}{4}B_{2m, p^s+2i-3}, & {\rm if} \ 1\le i\le m\\ B_{2m+2, p^s-1}=B_{2m+1, p^s-1}-\frac{1}{4}B_{2m, p^s-1}, & {\rm if} \ m\ge 0\\ B_{2m+2, 0}=B_{2m+1, 0}-\frac{1}{4}B_{2m, 0}, & {\rm if} \ m\ge 0\\ B_{2m+2, 2j}=B_{2m+1, 2j}-\frac{1}{4}B_{2m, 2j}+\frac{1}{4}B_{2m, 2j-2}, & {\rm if} \ 1\le j\le m\\ B_{2m+2, 2m+2}=\frac{1}{4}B_{2m, 2m}, & {\rm if} \ m\ge 0\\ \end{array} \right.} \end{align}$

(4.3)

as well as

$\begin{align}\label{ckm8} {\left\{ \begin{array}{ll} B_{2m+1, p^s+p^t+2m}=B_{2m, p^s+p^t+2m}+\frac{1}{4}B_{2m-1, p^s+p^t+2m-2}, & {\rm if} \ m\ge 0\\ B_{2m+1, p^s+p^t+2i}=B_{2m, p^s+2i}-\frac{1}{4}B_{2m-1, p^s+p^t+2i}+\frac{1}{4}B_{2m-1, p^s+2i-2}, & {\rm if} \ 1\le i\le m-1\\ B_{2m+1, p^s+p^t}=B_{2m, p^s+p^t}-\frac{1}{4}B_{2m-1, p^s+p^t}, & {\rm if} \ m\ge 0\\ B_{2m+1, p^s+2m+1}=\frac{1}{4}B_{2m-1, p^s+2m-1}, & {\rm if} \ m\ge 0\\ B_{2m+1, p^s+2i-1}=B_{2m, p^s+2i-1}-\frac{1}{4}B_{2m-1, p^s+2i-1}+\frac{1}{4}B_{2m-1, p^s+2i-3}, & {\rm if} \ 1\le i\le m\\ B_{2m+1, p^s-1}=B_{2m, p^s-1}-\frac{1}{4}B_{2m-1, p^s-1}, & {\rm if} \ m\ge 0\\ B_{2m+1, 0}=B_{2m, 0}-\frac{1}{4}B_{2m-1, 0}, & {\rm if} \ m\ge 0\\ B_{2m+1, 2j}=B_{2m, 2j}-\frac{1}{4}B_{2m-1, 2j}+\frac{1}{4}B_{2m-1, 2j-2}, & {\rm if} \ 1\le j\le m-1\\ B_{2m+1, 2m}=B_{2m, 2m}+\frac{1}{4}B_{2m-1, 2m-2}, & {\rm if}\ m\ge 0\\ \end{array} \right.} \end{align}$

(4.4)

Proof. The identity immediately follows from Lemma 2.4, Theorem 4.1 and Theorem 4.3. Moreover we readily find that there exists coefficients $B_{i, j}\in {\mathbb F_{q}}$ such that

$\begin{align}\label{ckm9} &D_{p^s+p^t+2\ell, k}(1, x)=\sum\limits_{i=0}^{\ell}B_{2\ell, p^s+p^t+2i}u^{p^s+p^t+2i} +\sum\limits_{j=0}^{\ell}B_{2\ell, 2j}u^{2j}\nonumber\\ &\ \ \ +\sum\limits_{i=0}^{\ell+1}B_{2\ell, p^s+2i-1} \big(u^{p^s+2i-1}+u^{p^t+2i-1}\big), \ \ 0\le\ell<\frac{p-1}{2}, \end{align}$

(4.5)

and

$\begin{align}\label{ckm10} &D_{p^s+p^t+2\ell-1, k}(1, x)=\sum\limits_{i=0}^{\ell}B_{2\ell-1, p^s+p^t+2i}u^{p^s+p^t+2i} +\sum\limits_{j=0}^{\ell}B_{2\ell-1, 2j}u^{2j}\nonumber\\ &\ \ \ +\sum\limits_{i=0}^{\ell+1}B_{2\ell-1, p^s+2i-1} \big(u^{p^s+2i-1}+u^{p^t+2i-1}\big), \ \ 0\le\ell<\frac{p-1}{2}, \end{align}$

(4.6)

where $u^2=1-4x$. Now let's determine all the coefficients $B_{i, j}$. On the one hand, by (4.5) and (4.6), one then has

$\begin{align}\label{ckm11} &D_{p^s+p^t+2\ell, k}(1, x)-xD_{p^s+p^t+2\ell-1, k}(1, x)=D_{p^s+p^t+2\ell, k}(1, x)- \frac{1-u^2}{4}D_{p^s+p^t+2\ell-1, k}(1, x)\nonumber\\ &=\sum\limits_{i=0}^{\ell}B_{2\ell, p^s+p^t+2i}u^{p^s+p^t+2i} +\sum\limits_{i=0}^{\ell}B_{2\ell, p^s+2i-1}\big(u^{p^s+2i-1}+u^{p^t+2i-1}\big)+ \sum\limits_{j=0}^{\ell}B_{2\ell, 2j}u^{2j}\nonumber\\ &\ \ \ -\frac{1}{4}\sum\limits_{i=0}^{\ell-1}B_{2\ell-1, p^s+p^t+2i}u^{p^s+p^t+2i} -\frac{1}{4}\sum\limits_{j=0}^{\ell}B_{2\ell-1, p^s+2i-1}\big(u^{p^s+2i-1}+u^{p^t+2i-1}\big)\nonumber\\ &\ \ \ -\frac{1}{4}\sum\limits_{j=0}^{\ell-1}B_{2\ell-1, 2j}u^{2j}+ \frac{1}{4}\sum\limits_{j=0}^{\ell-1}B_{2\ell-1, p^s+p^t+2i}u^{p^s+p^t+2i+2}\nonumber\\ &\ \ \ +\frac{1}{4}\sum\limits_{j=0}^{\ell}B_{2\ell-1, p^s+2i-1}\big(u^{p^s+2i+1}+u^{p^t+2i+1}\big) +\frac{1}{4}\sum\limits_{j=0}^{\ell-1}B_{2\ell-1, 2j}u^{2j+2}\nonumber\\ &=\big(B_{2\ell, p^s+p^t+2l}+\frac{1}{4}\big)u^{p^s+p^t+2\ell}+\sum\limits_{i=1}^{\ell-1}\big( B_{2\ell, p^s+p^t+2i}-\frac{1}{4}B_{2\ell-1, p^s+p^t+2i}+\frac{1}{4}B_{2\ell-1, p^s +p^t+2i-2}\big)u^{p^s+p^t+2i}\nonumber\\ &\ \ \ +\big(B_{2\ell, p^s+p^t}-\frac{1}{4}B_{2\ell-1, p^s+p^t}\big)u^{p^s+p^t}+\frac{1}{4} B_{2\ell-1, p^s+2\ell-1}\big(u^{p^s+2\ell+1}+u^{p^t+2\ell+1}\big)\nonumber\\ &+\sum\limits_{i=1}^{\ell}\big(B_{2\ell, p^s+2i-1}-\frac{1}{4}B_{2\ell-1, p^s+2i-1}+ \frac{1}{4}B_{2\ell-1, p^s+2i-3}\big)\big(u^{p^s+2i-1}+u^{p^t+2i-1}\big)\nonumber\\ &+\big(B_{2\ell, p^s-1}-\frac{1}{4}B_{2\ell-1, p^s-1}\big)\big(u^{p^s-1}+u^{p^t-1}\big) +\big(B_{2\ell, 2\ell}+\frac{1}{4}B_{2\ell-1, 2\ell-2}\big)u^{2\ell}\nonumber\\ &+\sum\limits_{j=1}^{\ell-1}\big(B_{2\ell2j}-\frac{1}{4}B_{2\ell-1, 2j}+ \frac{1}{4}B_{2\ell-1, 2j-2}\big)u^{2j}+B_{2\ell, 0}-\frac{1}{4}B_{2\ell-1, 0}. \end{align}$

(4.7)

On the other hand, Lemma 2.4 tells us that

$D_{p^s+p^t+2\ell+1, k}(1, x)=D_{p^s+p^t+2\ell, k}(1, x)-xD_{p^s+p^t+2\ell-1, k}(1, x).$

So by comparing the coefficient of the term $u^i$ in the right hand side of (4.6) and (4.7), one can get the desired results as (4.4). Following the similar way, one also obtain the recursions of $B_{i, j}$ as (4.3). So the proof Theorem 4.4 is complete.

---

5. Reversed Dickson polynomials $D_{p^{e_1}+p^{e_2}+\cdots+p^{e_s}+\ell, k}(1, x)$

Let $s\ge 1$ be an integer. Let $e_1, e_2, \cdots, e_s, \ell$ be integers with $0\le e_1 < e_2 < \cdots < e_s$ and $0\le \ell < p$. In this section, we present an explicit formula for $D_{n, k}(1, x)$ presented by elementary symmetric polynomials in terms of the power of $(1-4x)$ when $n=p^{e_1}+p^{e_2}+\cdots+p^{e_s}+\ell$. Then we characterize $D_{n, k}(1, x)$ to be a PP of ${\mathbb F}_q$ in this case.

Let $\sigma_i(x_1, x_2, \cdots, x_s)$ be the elementary polynomials in $s$ variables $x_1, x_2, \cdots, x_s$ which are defined by

$\begin{align*} &\sigma_0(x_1, x_2, \cdots, x_s)=1, \\ &\sigma_1(x_1, x_2, \cdots, x_s)=\sum\limits_{1\le j\le n}x_j, \\ &\sigma_2(x_1, x_2, \cdots, x_s)=\sum\limits_{1\le j<k\le n}x_jx_k, \\ &\sigma_3(x_1, x_2, \cdots, x_s)=\sum\limits_{1\le j<k<\ell \le n}x_jx_kx_{\ell}, \\ \end{align*}$

and so forth, ending with

$\sigma_s(x_1, x_2, \cdots, x_s)=x_1x_2\cdots x_s.$

Now we give the first result of this section.

Theorem 5.1. Let $q=p^e$ with $p$ being an odd prime and $e$ being a positive integer. Let $s$ be a positive integer. Let $e_1, \cdots, e_s$ be nonnegative integers with $e_1 < \cdots < e_s$. Then

$\begin{align*} D_{p^{e_1}+...+p^{e_s}, k}(1, x)&=\frac{1}{2^s}\Big((2-k)\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}}{2}}\big)\\ &+k\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}-1/i}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}-1/i}{2}}\big)\Big). \end{align*}$

Consequently, $D_{p^{e_1}+...+p^{e_s}, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if the polynomial

$(2-k)\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big(x^{\frac{p^{e_1}}{2}}, \cdots, x^{\frac{p^{e_s}}{2}}\big) +k\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big(x^{\frac{p^{e_1}-1/i}{2}}, \cdots, x^{\frac{p^{e_s}-1/i}{2}}\big)$

is a PP of ${\mathbb F_{q}}$.

Proof. We divide the proof into the following two cases.

CASE 1. $x\ne \frac{1}{4}$. For this case, putting $x=y(1-y)$ in the second identity of Lemma 2.3 gives us that

$\begin{align*} &D_{p^{e_1}+...+p^{e_s}, k}(1, x)=D_{p^{e_1}+...+p^{e_s}, k}(1, y(1-y))\\ =&\frac{\big(k-1-(k-2)y\big)y^{p^{e_1}+...+p^{e_s}}-\big(1+(k-2)y\big)(1-y)^{p^{e_1}+...+p^{e_s}}}{2y-1}\\ =&\frac{\frac{k+(2-k)u}{2}\big(\frac{u+1}{2}\big)^{p^{e_1}+\cdots+p^{e_s}} -\frac{k+(k-2)u}{2}\big(\frac{1-u}{2}\big)^{p^{e_1}+\cdots+p^{e_s}}}{u}\\ =&\frac{k+(2-k)u}{2^{p^{e_1}+\cdots+p^{e_s}+1}u}\prod\limits_{i=1}^{s}(u^{p^{e_i}}+1) -\frac{k+(k-2)u}{2^{p^{e_1}+\cdots+p^{e_s}+1}u}\prod\limits_{i=1}^{s}(1-u^{p^{e_i}})\\ =&\frac{1}{2^{s+1}u}\Big((k+(2-k)u)\sum\limits_{0\le i\le s}\sigma_i(u^{p^{e_1}}, \cdots, u^{p^{e_s}}) -(k+(k-2)u)\sum\limits_{0\le i\le s}(-1)^i\sigma_i(u^{p^{e_1}}, \cdots, u^{p^{e_s}})\Big)\\ =&\frac{1}{2^s}\Big((2-k)\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big(u^{p^{e_1}}, \cdots, u^{p^{e_s}}\big)+k\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big(u^{p^{e_1}-1/i}, \cdots, u^{p^{e_s}-1/i}\big)\Big), \end{align*}$

where $u=2y-1$ and $u^2=1-4x$. Then we have that

$\begin{align*} D_{p^{e_1}+...+p^{e_s}, k}(1, x)&=\frac{1}{2^s}\Big((2-k)\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}}{2}}\big)\\ &+k\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}-1/i}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}-1/i}{2}}\big)\Big). \end{align*}$

as desired.

CASE 2. $x=\frac{1}{4}$. On the one hand, by the first identity of Lemma 2.3, one has

$D_{p^{e_1}+...+p^{e_s}, k}\big(1, \frac{1}{4}\big) =\frac{k(p^{e_1}+\cdots+p^{e_s})-k+2}{2^{p^{e_1}+\cdots+p^{e_s}}}=\frac{2-k}{2^s}.$

On the other hand,

$\begin{align*} &\frac{1}{2^s}\Big((2-k)\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}}{2}}\big)\\ &+k\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}-1/i}{2}}, \cdots, (1-4x)^ {\frac{p^{e_s}-1/i}{2}}\big)\Big)\Big |_{x=1/4}=\frac{2-k}{2^s}. \end{align*}$

Thus the required result follows. So Theorem 5.1 is proved.

heorem 5.2. Let $q=p^e$ with $p$ being an odd prime and $e$ being a positive integer. Let $s$ be a positive integer. Let $e_1, \cdots, e_s$ be nonnegative integers with $e_1 < \cdots < e_s$. Then

$\begin{align*} D_{p^{e_1}+...+p^{e_s}+1, k}(1, x)&=\frac{1}{2^{s+1}}\Big(2\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}}{2}}\big)\\ &+\big((2-k)(1-4x)+k\big)\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}-1/i}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}-1/i}{2}}\big)\Big). \end{align*}$

Consequently, $D_{p^{e_1}+...+p^{e_s}+1, k}(1, x)$ is a PP of ${\mathbb F_{q}}$ if and only if the polynomial

$2\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big(x^{\frac{p^{e_1}}{2}}, \cdots, x^{\frac{p^{e_s}}{2}}\big) +\big((2-k)x+k\big)\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big(x^{\frac{p^{e_1}-1/i}{2}}, \cdots, x^{\frac{p^{e_s}-1/i}{2}}\big)$

is a PP of ${\mathbb F_{q}}$.

Proof. We consider the following two cases.

CASE 1. $x\ne \frac{1}{4}$. For this case, putting $x=y(1-y)$ in the second identity of Lemma 2.3 gives us that

$\begin{align*} &D_{p^{e_1}+...+p^{e_s}+1, k}(1, x)=D_{p^{e_1}+...+p^{e_s}+1, k}(1, y(1-y))\\ =&\frac{\big(k-1-(k-2)y\big)y^{p^{e_1}+...+p^{e_s}+1}-\big(1+(k-2)y\big)(1-y)^{p^{e_1}+...+p^{e_s}+1}}{2y-1}\\ =&\frac{\frac{k+(2-k)u}{2}\big(\frac{u+1}{2}\big)^{p^{e_1}+\cdots+p^{e_s}+1} -\frac{k+(k-2)u}{2}\big(\frac{1-u}{2}\big)^{p^{e_1}+\cdots+p^{e_s}+1}}{u}\\ =&\frac{k+(2-k)u}{2^{p^{e_1}+\cdots+p^{e_s}+2}u}\prod\limits_{i=1}^{s}(u^{p^{e_i}}+1)(u+1) -\frac{k+(k-2)u}{2^{p^{e_1}+\cdots+p^{e_s}+2}u}\prod\limits_{i=1}^{s}(1-u^{p^{e_i}})(1-u)\\ =&\frac{1}{2^{s+2}u}\Big((k+2+(2-k)u^2)\sum\limits_{0\le i\le s}\sigma_i(u^{p^{e_1}}, \cdots, u^{p^{e_s}}) -(2k+(4-2k)u^2)\sum\limits_{0\le i\le s}(-1)^i\sigma_i(u^{p^{e_1}}, \cdots, u^{p^{e_s}})\Big)\\ =&\frac{1}{2^{s+1}}\Big(2\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big(u^{p^{e_1}}, \cdots, u^{p^{e_s}}\big)+\big((2-k)u^2+k\big)\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big(u^{p^{e_1}-1/i}, \cdots, u^{p^{e_s}-1/i}\big)\Big), \end{align*}$

where $u=2y-1$ and $u^2=1-4x$. Then we have

$\begin{align*} D_{p^{e_1}+...+p^{e_s}+1, k}(1, x)&=\frac{1}{2^{s+1}}\Big(2\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}}{2}}\big)\\ &+\big((2-k)(1-4x)+k\big)\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}-1/i}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}-1/i}{2}}\big)\Big). \end{align*}$

as desired.

CASE 2. $x=\frac{1}{4}$. On the one hand, by the first identity of Lemma 2.3, one has

$D_{p^{e_1}+...+p^{e_s}+1, k}\big(1, \frac{1}{4}\big) =\frac{k(p^{e_1}+\cdots+p^{e_s}+1)-k+2}{2^{p^{e_1}+\cdots+p^{e_s}+1}}=\frac{2}{2^{s+1}}.$

On the other hand,

$\begin{align*} &\frac{1}{2^{s+1}}\Big(2\sum\limits_{\mbox{$\begin{array}{c} 0\le i\le s\\ i\ {\rm even}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}}{2}}\big)\\ &+\big((2-k)(1-4x)+k\big)\sum\limits_{\mbox{$\begin{array}{c} 1\le i\le s\\ i\ {\rm odd}\end{array}$}}\sigma_i \big((1-4x)^{\frac{p^{e_1}-1/i}{2}}, \cdots, (1-4x)^{\frac{p^{e_s}-1/i}{2}}\big)\Big)\Big|_{x=1/4} =\frac{2}{2^{s+1}}. \end{align*}$

Thus the required result follows. So Theorem 5.3 is proved.

Then Theorems 5.1-5.2 together with Lemma 2.4 show that the general result is true.

Theorem 5.3. Let $q=p^e$ with $p$ being an odd prime and $e$ being a positive integer. Let $s$ be a positive integer. Let $e_1, \cdots, e_s$ be nonnegative integers with $e_1 < \cdots < e_s$. Then for any $\ell \ge 0$ each of the identities is true.

$\begin{align*} D_{p^{e_1}+...+p^{e_s}+2\ell, k}(1, x)=\frac{1}{2^{s+2\ell}}\Big(\sum\limits_{j=0}^{\ell} C_{2\ell, 2j}u^{2j}\sum\limits_{0\le i\le s \atop i\ {\rm even}} \sigma_i\big(u^{p^{e_1}}, \cdots, u^{p^{e_s}}\big) +\sum\limits_{j=0}^{\ell}Q_{2\ell, 2j}u^{2j}\sum\limits_{0\le i\le s \atop i\ {\rm odd}} \sigma_i \big(u^{p^{e_1}-\frac{1}{i}}, \cdots, u^{p^{e_s}-\frac{1}{i}}\big)\Big), \end{align*}$

$\begin{align*} D_{p^{e_1}+...+p^{e_s}+2\ell+1, k}(1, x)&=\frac{1}{2^{s+2\ell+1}}\Big(\sum\limits_{j=0}^{\ell} C_{2\ell+1, 2j}u^{2j}\sum\limits_{0\le i\le s \atop i\ {\rm even}} \sigma_i\big(u^{p^{e_1}}, \cdots, u^{p^{e_s}}\big) +\sum\limits_{j=0}^{\ell+1} Q_{2\ell+1, 2j}u^{2j}\sum\limits_{0\le i\le s \atop i\ {\rm odd}} \sigma_i \big(u^{p^{e_1}-\frac{1}{i}}, \cdots, u^{p^{e_s}-\frac{1}{i}}\big)\Big), \end{align*}$

where $u^2=1-4x$, and the coefficients $C_{a, 2b}$ and $Q_{a, 2b}$ can be determined as follows:

$C_{0, 0}=2-k, Q_{0, 0}=k, C_{1, 0}=k, Q_{1, 2}=2-k,$

$\begin{align}\label{ch4} {\left\{ \begin{array}{ll} C_{2m+2, 0}=2C_{2m+1, 0}-C_{2m, 0}, & {\rm if} \ m\ge 0\\ C_{2m+2, 2j}=2C_{2m+1, 2j}+C_{2m, 2j-2}-C_{2m, 2j}, & {\rm if} \ 1\le j\le m\\ C_{2m+2, 2m+2}=C_{2m, 2m}, & {\rm if} \ m\ge 0\\ Q_{2m+2, 0}=2Q_{2m+1, 0}-Q_{2m, 0}, & {\rm if} \ m\ge 0\\ Q_{2m+2, 2j}=2Q_{2m+1, 2j}+Q_{2m, 2j-2}-Q_{2m, 2j}, & {\rm if} \ 1\le j\le m\\ Q_{2m+2, 2m+2}=2Q_{2m+1, 2m+2}+Q_{2m, 2m}, & {\rm if} \ m\ge 0 \end{array} \right.} \end{align}$

(5.1)

as well as

$\begin{align}\label{ch5} {\left\{ \begin{array}{ll} C_{2m+1, 0}=2C_{2m, 0}-C_{2m-1, 0}, & {\rm if} \ m\ge 1\\ C_{2m+1, 2j}=2C_{2m, 2j}+C_{2m-1, 2j-2}-C_{2m-1, 2j}, & {\rm if} \ 1\le j\le m-1\\ C_{2m+1, 2m}=2C_{2m, 2m}+C_{2m-1, 2m-2}, & {\rm if} \ m\ge 1\\ Q_{2m+1, 0}=2Q_{2m, 2j}-Q_{2m-1, 0}, & {\rm if} \ m\ge 0\\ Q_{2m+1, 2j}=2Q_{2m, 2j}+Q_{2m-1, 2j-2}-Q_{2m-1, 2j}, & {\rm if} \ 1\le j\le m\\ Q_{2m+1, 2m+2}=2Q_{2m-1, 2m}, & {\rm if} \ m\ge 1 \end{array} \right.} \end{align}$

(5.2)

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Acknowledgement

Cheng was supported partially by the General Project of Department of Education of Sichuan Province 15ZB0434. [2000]Primary 11T06, 11T55, 11C08.

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Conflict of Interest

The author declares no conflicts of interest in this paper.

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