On inverses of some permutation polynomials over finite fields of characteristic three

TL;DR

This paper derives explicit inverse formulas for certain permutation polynomials over finite fields of characteristic three using combinatorial and interpolation techniques.

Contribution

It provides explicit inverse expressions for a class of reversed Dickson and generalized cyclotomic permutation polynomials over characteristic three fields.

Findings

Explicit inverse formulas for specific permutation polynomials

Application of piecewise, Lagrange interpolation, and Lucas' theorem methods

Enhanced understanding of permutation polynomial inverses in characteristic three

Abstract

By using the piecewise method, Lagrange interpolation formula and Lucas' theorem, we determine explicit expressions of the inverses of a class of reversed Dickson permutation polynomials and some classes of generalized cyclotomic mapping permutation polynomials over finite fields of characteristic three.

Equations

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

$$D_n(a,x)=\sum _{i=0}^{\lfloor n/2\rfloor }\frac{n}{n-i}\left(\genfrac{}{}{0ex}{}{n-i}{i}\right)(-1{)}^i{a}^{n-2i}{x}^i,$$

$$f(x)=(x-x^2-x^3)x^{\frac{3^n-1}{2}}-x+x^2$$

$$f^{-1}(x)=x^{3^{n-1}}\big{(}x^{\frac{3^{n}-1}{2}}+1\big{)}+\Bigg{(}\sum_{j=0}^{n-1}\sum_{k=0}^{n-1}(-1)^{j+k}x^{\frac{3^{j}+3^{k}}{2}}\Bigg{)}\big{(}x^{\frac{3^{n}-1}{2}}-1\big{)}.$$

$$f(x)=\frac{1}{2}{f}_0(x)(1+x^{\frac{q-1}{2}})+\frac{1}{2}{f}_1(x)(1-x^{\frac{q-1}{2}})\text{with f(0)=0,}$$

$$f^{-1}(x)=\frac{1}{2}{f}_0^{-1}(x)(1+x^{\frac{q-1}{2}})+\frac{1}{2}{f}_1^{-1}(x)(1-x^{\frac{q-1}{2}}).$$

$$f^{-1}(x)=\frac{1}{2}{f}_0^{-1}(x)(1-x^{\frac{q-1}{2}})+\frac{1}{2}{f}_1^{-1}(x)(1+x^{\frac{q-1}{2}}).$$

$$\sum _{a\in C_s}{a}^k=\{\begin{array}{ll}-2^{-1}& \text{if k=q−1,}\\ (-1{)}^{s+1}{2}^{-1}& \text{if k=(q−1)/2,}\\ 0& \text{if 1≤k≤q−2 and k=(q−1)/2.}\end{array}$$

$$({\xi }^k+1)A=\sum _{a\in C_0}{a}^k+\sum _{a\in C_1}{a}^k=\sum _{a\in {\mathbb{F}}_q}{a}^k=\{\begin{array}{ll}-1& \text{if k=q−1,}\\ 0& \text{if k=1, 2, …, q−2.}\end{array}$$

$$f_s(x{)}^{q-1-i}\equiv \sum _{0\le k\le q-1}{b}_{ik}{x}^k(\mathrm{mod}{x}^q-x),$$

$$f_{s}^{-1}(x)=\sum_{(q+1)/2\leq i\leq q-2}(-1)^{s+t}\big{(}b_{i,(q-3)/2}+(-1)^{s}b_{i,q-2}\big{)}x^{i-\frac{q-1}{2}}$$

$$\begin{split}f_{s}^{-1}(x)&=\sum_{a\in C_{s}}a\big{(}1-(x-f_{s}(a))^{q-1}\big{)}\\ &=\sum_{a\in C_{s}}a\Big{(}-\sum_{1\leq i\leq q-1}(-1)^{i}(-f_{s}(a))^{q-1-i}x^{i}\Big{)}\\ &=\sum_{1\leq i\leq q-1}\Big{(}-\sum_{a\in C_{s}}af_{s}(a)^{q-1-i}\Big{)}x^{i}.\end{split}$$

$$c_i=-\sum _{a\in C_s}a{f}_s(a{)}^{q-1-i},\text{where 1≤i≤q−1.}$$

$$\begin{array}{rl}{c}_i& =-\sum _{a\in C_s}a\sum _{0\le k\le q-1}{b}_{ik}{a}^k=-\sum _{0\le k\le q-1}{b}_{ik}\sum _{a\in C_s}{a}^{k+1}\\ & =2^{-1}(b_{i,q-2}+(-1{)}^s{b}_{i,(q-3)/2}).\end{array}$$

$$f_s(a{)}^{q-1-i}=(-1{)}^t{f}_s(a{)}^{q-1-i}/f_s(a{)}^{\frac{q-1}{2}}=(-1{)}^t{f}_s(a{)}^{q-1-(i+\frac{q-1}{2})}.$$

$$c_i=(-1{)}^t{c}_{i+\frac{q-1}{2}},1\le i\le (q-1)/2.$$

$$c_{(q-1)/2}=(-1{)}^t{c}_{q-1}=(-1{)}^{t+1}\sum _{a\in C_s}a=0,$$

$$\begin{split}f_{s}^{-1}(x)&\overset{\eqref{bs=0}}{=}\sum_{1\leq i\leq(q-3)/2}\big{(}c_{i}x^{i}+c_{i+\frac{q-1}{2}}x^{i+\frac{q-1}{2}}\big{)}\\ &\equiv\sum_{1\leq i\leq(q-3)/2}\big{(}c_{i}+(-1)^{t}c_{i+\frac{q-1}{2}}\big{)}x^{i}\pmod{x^{\frac{q-1}{2}}+(-1)^{t+1}}\\ &\overset{\eqref{bi=-bi+s}}{=}\sum_{1\leq i\leq(q-3)/2}2(-1)^{t}c_{i+\frac{q-1}{2}}x^{i}\\ &=\sum_{(q+1)/2\leq i\leq q-2}2(-1)^{t}c_{i}x^{i-\frac{q-1}{2}}.\end{split}$$

$$f(x)=(x-x^2-x^3)x^{\frac{3^n-1}{2}}-x+x^2$$

$$f(x)=\{\begin{array}{ll}0& \text{if x=0,}\\ f_0(x):=-x^3& \text{if x∈C0,}\\ f_1(x):=x(x+1{)}^2& \text{if x∈C1.}\end{array}$$

$$f_0^{-1}(x)=-x^{3^{n-1}}.$$

$$f_1(x{)}^{w_i}=x^{w_i}(x+1{)}^{2w_i}=\sum _{0\le j\le 2w_i}\left(\genfrac{}{}{0ex}{}{2w_i}{j}\right)x^{w_i+j}.$$

$$\begin{array}{rl}{b}_{iu}& =\left(\genfrac{}{}{0ex}{}{2q-2-2i}{i-\frac{q+1}{2}}\right)\text{for (q+1)/2≤i≤(5q−3)/6,}\\ b_{iv}& =\left(\genfrac{}{}{0ex}{}{2q-2-2i}{i-1}\right)\text{for (q+1)/2≤i<2q/3.}\end{array}$$

$$\left(\genfrac{}{}{0ex}{}{q+m}{k}\right)\equiv \left(\genfrac{}{}{0ex}{}{m}{k}\right)(\mathrm{mod}p),$$

$$b_{iv}\equiv \left(\genfrac{}{}{0ex}{}{-2-2i}{i-1}\right)\equiv (-1{)}^{i-1}(\genfrac{}{}{0ex}{}{3i}{i-1})(\mathrm{mod}3),$$

$$b_{iu}\equiv \left(\genfrac{}{}{0ex}{}{-2-2i}{i-\frac{q+1}{2}}\right)\equiv (-1{)}^{i-\frac{q+1}{2}}(\genfrac{}{}{0ex}{}{3i-\frac{q-1}{2}}{i-\frac{q+1}{2}})(\mathrm{mod}3).$$

$$f^{-1}(x)=x^{3^{n-1}}\big{(}x^{\frac{3^{n}-1}{2}}+1\big{)}+f_{1}^{-1}(x)\big{(}x^{\frac{3^{n}-1}{2}}-1\big{)},$$

$$\begin{array}{rl}{f}_1^{-1}(x)& =\sum _{(3^n+1)/2\le i\le (5\cdot 3^n-3)/6}(-1{)}^{i-\frac{3^n+1}{2}}(\genfrac{}{}{0ex}{}{3i-\frac{3^n-1}{2}}{i-\frac{3^n+1}{2}})x^{i-\frac{3^n-1}{2}}\\ & +\sum _{(3^n+1)/2\le i\le 2\cdot 3^n/3}(-1{)}^i(\genfrac{}{}{0ex}{}{3i}{i-1})x^{i-\frac{3^n-1}{2}}\end{array}$$

$$\begin{array}{rl}3i& =0+i_{0}3+i_1{3}^2+\cdots +i_{n-2}{3}^{n-1}+i_{n-1}{3}^n,\\ i-1& =i_0-1+i_{1}3+i_2{3}^2+\cdots +i_{n-1}{3}^{n-1}.\end{array}$$

$$\left(\genfrac{}{}{0ex}{}{3i}{i-1}\right)\equiv \left(\genfrac{}{}{0ex}{}{0}{i_0-1}\right)\left(\genfrac{}{}{0ex}{}{i_0}{i_1}\right)\left(\genfrac{}{}{0ex}{}{i_1}{i_2}\right)\cdots \left(\genfrac{}{}{0ex}{}{i_{n-2}}{i_{n-1}}\right)\left(\genfrac{}{}{0ex}{}{i_{n-1}}{0}\right)(\mathrm{mod}3).$$

Full text

1 Introduction

For $q$ a prime power, let $\mathbb{F}_{q}$ denote the finite field with $q$ elements, $\mathbb{F}_{q}^{*}=\mathbb{F}_{q}\setminus\{0\}$, and $\mathbb{F}_{q}[x]$ the ring of polynomials over $\mathbb{F}_{q}$. A polynomial $f(x)\in\mathbb{F}_{q}[x]$ is called a permutation polynomial (PP) of $\mathbb{F}_{q}$ if it induces a bijection from $\mathbb{F}_{q}$ to itself. For any PP $f(x)$ of $\mathbb{F}_{q}$, there exists a polynomial $f^{-1}(x)\in\mathbb{F}_{q}[x]$ such that $f^{-1}(f(c))=c$ for each $c\in\mathbb{F}_{q}$ or equivalently $f^{-1}(f(x))\equiv x\pmod{x^{q}-x}$, and the polynomial $f^{-1}(x)$ is unique in the sense of reduction modulo $x^{q}-x$. Hence $f^{-1}(x)$ is defined as the composition inverse of $f(x)$, and we simply call it the inverse of $f(x)$ on $\mathbb{F}_{q}$.

Recently, some classes of PPs are found; see for example [18, 32, 35, 21, 33] for PPs of the form $x^{r}h(x^{q-1})$ of $\mathbb{F}_{q^{2}}$, [24, 37] for PPs of the form $(x^{q}-x+c)^{s}+L(x)$ of $\mathbb{F}_{q^{2}}$, [17, 41, 34] for PPs of the form $(ax^{q}+bx+c)^{r}\phi((ax^{q}+bx+c)^{s})+ux^{q}+vx$ of $\mathbb{F}_{q^{2}}$, [15] for PPs with low boomerang uniformity, and [1, 12, 4] for PPs studied using the Hasse-Weil bound and Hermite’s criterion. For a detailed introduction to the developments on PPs, we refer the reader to [11, 28] and the references therein.

The construction of PPs of finite fields is not an easy subject. However, the problem of determining the inverse of a PP seems to be an even more complicated problem. In fact, there are a few known classes of PPs whose inverses have been obtained explicitly; see for example [16, 19, 25, 39] for PPs of the form $x^{r}h(x^{(q-1)/d})$, [30, 31] for linearized PPs, [27, 39] for generalized cyclotomic mapping PPs, [40] for general piecewise PPs, [3, 20, 36] for involutions over finite fields, and [22, 23] for more general classes of PPs. For a brief summary of the results concerning the inverses of PPs, we refer the reader to [38] and the references therein.

The Dickson polynomial $D_{n}(x,a)$ of degree $n$ in the indeterminate $x$ and with parameter $a\in\mathbb{F}_{q}$ is given as

[TABLE]

where $\lfloor n/2\rfloor$ denotes the largest integer $\leq n/2$, and the term $\frac{n}{n-i}\binom{n-i}{i}$ is an integer. By reversing the roles of the indeterminate $x$ and the parameter $a$ in $D_{n}(x,a)$, the $n$-th reversed Dickson polynomials $D_{n}(a,x)$ was defined in [14] by

[TABLE]

and their permutation properties were studied in [14]. Several families of reversed Dickson PPs over finite fields were given in [14, 10], which covered all the reversed Dickson PPs over $\mathbb{F}_{q}$ with $q<200$. Then, the notion of (reversed) Dickson polynomials of the $(k+1)$-th kind was introduced in [29], and the factorization and the permutation behavior of Dickson polynomials of the third kind were studied in [29]. Some necessary conditions for reversed Dickson polynomials of the first and second kinds to be PPs of finite fields were given in [13] and [9], respectively. The permutation behavior of reversed Dickson polynomials of the $(k+1)$-th kind was further investigated in [5, 6, 7]. In particular, Hou [10] proved the following result.

Lemma 1** ([10, Theorem1.1]).**

Let $n$ be a positive even integer. Then

[TABLE]

is a PP of $\mathbb{F}_{3^{n}}$.

Since the reversed Dickson polynomial $D_{3^{n}+5}(1,x)=f(1-x)-1$, Hou equivalently proved that $D_{3^{n}+5}(1,x)$ is a PP of $\mathbb{F}_{3^{n}}$ for even $n$.

The purpose of this paper is to find the inverse of $f(x)$ in Lemma 1. The main idea is the combination of the piecewise method in [39] and some techniques in [38].

The rest of the paper is organized as follows. Section 2 gives a formula for the inverse of a class of piecewise PPs $f(x)$, which converts the problem of determining the inverse of $f(x)$ on $\mathbb{F}_{q}$ to the problem of computing the inverse $f_{s}^{-1}(x)$ of piece function $f_{s}(x)$ when restricted to a subset $C_{s}$ of $\mathbb{F}_{q}$ for all $s$. Then an expression of $f_{s}^{-1}(x)$ is presented in Theorem 2, which provides all the coefficients of $f_{s}^{-1}(x)$ by computing the coefficients of $x^{(q-3)/2}$ and $x^{q-2}$ in $f_{s}(x)^{k}\pmod{x^{q}-x}$ for $1\leq k\leq(q-3)/2$. By applying the results in Section 2 to $f(x)$ in Lemma 1, the coefficients of $f^{-1}(x)$ are reduced into two classes of binomial coefficients in Section 3. Section 4 gives explicit values of these binomial coefficients by using a congruence of binomial coefficients and Lucas’ theorem. In other words, we determine the inverse of $f(x)$ as follows.

Theorem 1**.**

The inverse of $f(x)$ in Lemma 1 on $\mathbb{F}_{3^{n}}$ is

[TABLE]

Furthermore, the inverse of $D_{3^{n}+5}(1,x)$ on $\mathbb{F}_{3^{n}}$ is $D_{3^{n}+5}^{-1}(1,x)=1-f^{-1}(x+1)$.

In the last section, by an argument similar to the one used in Theorem 1, we also obtain explicit inverses of some generalized cyclotomic mapping PPs studied in [26].

2 The inverse of a class of piecewise PPs

The piecewise methods for constructing PPs and their inverses were summarized in [8, 2] and [39] respectively. Applying these methods, we can easily get the following result.

Lemma 2**.**

Let $q$ be odd and $f_{0}(x),f_{1}(x)\in\mathbb{F}_{q}[x]$. Define

[TABLE]

$C_{0}=\{e^{2}:e\in\mathbb{F}_{q}^{*}\}$, and $C_{1}=\mathbb{F}_{q}^{*}\setminus C_{0}$. Then $f(x)$ is a PP of $\mathbb{F}_{q}$ if and only if $f_{s}$ is injective on $C_{s}$ and $0\not\in f_{s}(C_{s})$ for $s\in\{0,1\}$, and $f_{0}(C_{0})\cap f_{1}(C_{1})=\emptyset$. Assume $f(x)$ is a PP of $\mathbb{F}_{q}$, and $f_{s}^{-1}(x)\in\mathbb{F}_{q}[x]$ satisfies that $f_{s}^{-1}(0)=0$ and $f_{s}^{-1}(f_{s}(c))=c$ for any $c\in C_{s}$ and $s\in\{0,1\}$.

1. (i)

If $f_{s}$ maps $C_{s}$ into $C_{s}$ for $s\in\{0,1\}$, then the inverse of $f(x)$ on $\mathbb{F}_{q}$ is

[TABLE] 2. (ii)

If $f_{s}$ maps $C_{s}$ into $C_{t}$ for $s\neq t\in\{0,1\}$, then the inverse of $f(x)$ on $\mathbb{F}_{q}$ is

[TABLE]

Lemma 2 converts the problem of determining $f^{-1}(x)$ into the problem of computing $f_{s}^{-1}(x)$, the inverse of piece function $f_{s}(x)$ when restricted to $C_{s}$. In Lemma 2, if $q=3$, then $f(x)=f^{-1}(x)=\pm x$ in the sense of reduction modulo $x^{3}-x$. We will give an expression of $f_{s}^{-1}(x)$ for $q>3$ after the following lemma.

Lemma 3**.**

For $q$ an odd prime power, let $C_{0}=\{e^{2}:e\in\mathbb{F}_{q}^{*}\}$ and $C_{1}=\mathbb{F}_{q}^{*}\setminus C_{0}$. Then

[TABLE]

Proof.

Let $A=\sum_{a\in C_{1}}a^{k}$ and $\xi$ a primitive element of $\mathbb{F}_{q}$. Obviously,

[TABLE]

If $k=q-1$, then $\xi^{k}=1$ and $2A=-1$, so $A=-1/2$. If $k=(q-1)/2$, then $a^{k}=-1$ for $a\in C_{1}$. Hence $A=(-1)(q-1)/2=1/2$. If $1\leq k\leq q-2$ and $k\neq(q-1)/2$, then $\xi^{k}\neq-1$ and $(\xi^{k}+1)A=0$. Thus $A=0$. Then the result follows from $\eqref{sum-all}$. ∎

Theorem 2**.**

For an odd prime power $q>3$, let $C_{0}=\{e^{2}:e\in\mathbb{F}_{q}^{*}\}$ and $C_{1}=\mathbb{F}_{q}^{*}\setminus C_{0}$. For $s,t\in\{0,1\}$, assume $f_{s}(x)\in\mathbb{F}_{q}[x]$ induces a bijection from $C_{s}$ to $C_{t}$, and

[TABLE]

where $(q+1)/2\leq i\leq q-2$. Then the inverse of $f_{s}(x)$ when restricted to $C_{s}$ is

[TABLE]

in the sense of reduction modulo $x^{\frac{q-1}{2}}+(-1)^{t+1}$.

Proof.

Let $f_{s}^{-1}(x)=\sum_{i=0}^{q-1}c_{i}x^{i}\in\mathbb{F}_{q}[x]$. Then by the Lagrange interpolation formula,

[TABLE]

Hence $c_{0}=0$ and

[TABLE]

Then by (3) and Lemma 3,

[TABLE]

We next reduce the degree of $f_{s}^{-1}(x)$. Since $f_{s}(x)$ induces a bijection from $C_{s}$ to $C_{t}$, we have $f_{s}(a)\in C_{t}$ and $f_{s}(a)^{(q-1)/2}=(-1)^{t}$ for any $a\in C_{s}$, and so

[TABLE]

Substituting it into (5) yields

[TABLE]

In particular, if $q>3$, then

[TABLE]

where the last identity follows from Lemma 3. Therefore,

[TABLE]

Substituting (6) into (9) gives the desired result. ∎

3 The inverse of the PP in Lemma 1

In this section, we will employ the results in Section 2 to compute the inverse of the PP $f(x)$ in Lemma 1. First, let $n>0$ be even, $C_{0}=\{e^{2}:e\in\mathbb{F}_{3^{n}}^{*}\}$ and $C_{1}=\mathbb{F}_{3^{n}}^{*}\setminus C_{0}$. Then $x^{(3^{n}-1)/2}=1$ if $x\in C_{0}$, and $x^{(3^{n}-1)/2}=-1$ if $x\in C_{1}$. Therefore,

[TABLE]

can be written as

[TABLE]

Lemma 1 stated that $f(x)$ is a PP of $\mathbb{F}_{3^{n}}$. It means that $f_{0}(x)$ (resp. $f_{1}(x)$) induces an injection on $C_{0}$ (resp. $C_{1}$), and $f_{0}(C_{0})\cap f_{1}(C_{1})=\emptyset$. Since $n$ is even, $3^{n}\equiv 1\pmod{4}$, and so $-1\in C_{0}$. Hence $f_{0}(x)$ (resp. $f_{1}(x)$) induces a permutation on $C_{0}$ (resp. $C_{1}$).

Since $(x^{3})^{3^{n-1}}=x^{3^{n}}=x$ for any $x\in C_{0}$, the inverse of $f_{0}(x)$ on $C_{0}$ is

[TABLE]

We next apply Theorem 2 to determine the inverse of $f_{1}(x)$ on $C_{1}$. Denote $q=3^{n}$, $u=(q-3)/2$, $v=q-2$ and $w_{i}=q-1-i$, where $(q+1)/2\leq i\leq q-2$. Then

[TABLE]

The degree of $f_{1}(x)^{w_{i}}$ is $3w_{i}$, and $3\leq 3w_{i}<q-1+u$, as shown in Figure 1.

Hence the coefficient of $x^{u}$ in (3) equals the coefficients of $x^{u}$ in (11), i.e., $b_{iu}=\binom{2w_{i}}{u-w_{i}}$. Similarly, $b_{iv}=\binom{2w_{i}}{v-w_{i}}$. If $3w_{i}<u$, i.e., $i>(5q-3)/6$, then $b_{iu}=0$. If $3w_{i}<v$, i.e., $i\geq 2q/3$, then $b_{iv}=0$. Hence we only need to consider the binomial coefficients

[TABLE]

Lemma 4** ([38, Lemma 9]).**

Let $q$ be a prime $p$ power, and let $m$, $k$ be integers with $0\leq k<q$. Then

[TABLE]

where $\binom{m}{k}=m(m-1)\cdots(m-k+1)/k!$.

Employing Lemma 4 and the fact $\binom{-m}{k}=(-1)^{k}\binom{m+k-1}{k}$ for $m$, $k>0$, we have

[TABLE]

where $(q+1)/2\leq i<2q/3$. Similarly, if $(q+1)/2\leq i\leq(5q-3)/6$, then

[TABLE]

By Lemma 2, Theorem 2, (10), (12) and (13), the inverse of $f(x)$ in Lemma 1 on $\mathbb{F}_{3^{n}}$ is

[TABLE]

where

[TABLE]

in the sense of reduction modulo $x^{\frac{3^{n}-1}{2}}+1$.

4 Explicit values of binomial coefficients

This section will give the explicit values of binomial coefficients in (15).

Theorem 3**.**

Let $1\leq i<3^{n}$ with $n\geq 1$. Then $\binom{3i}{i-1}$ $\not\equiv 0\pmod{3}$ if and only if $i=(3^{k}-1)/2$, where $k=1$, $2$, $\ldots$, $n$.

Proof.

If $1\leq i<3^{n}$, then we can write $i=i_{0}+i_{1}3+i_{2}3^{2}+\cdots+i_{n-1}3^{n-1},$ where $i_{t}=0$, $1$, $2$ for $0\leq t\leq n-1$. Then

[TABLE]

By Lucas’ theorem, if $i_{0}=0$, then $\binom{3i}{i-1}\equiv\binom{0}{2}\equiv 0\pmod{3}$, and if $i_{0}=1,2$, then

[TABLE]

Hence $\binom{3i}{i-1}\not\equiv 0\pmod{3}$ if and only if one of the following conditions holds:

$i_{0}=1$ and $1\geq i_{1}\geq i_{2}\geq\cdots\geq i_{n-2}\geq i_{n-1}\geq 0$; 2. 2.

$i=1$, $1+3$, $1+3+3^{2}$, $\ldots$, $1+3+3^{2}+\cdots+3^{n-1}$; 3. 3.

$i=(3^{k}-1)/2$, where $k=1$, $2$, $\ldots$, $n$. ∎

Corollary 1**.**

Let $(3^{n}-1)/2<i<3^{n}$ with $n\geq 1$. Then $\binom{3i}{i-1}$ $\equiv 0\pmod{3}$.

It is easy to obtain the following $3$-adic expansion:

[TABLE]

Thus, when $(3^{n}+1)/2\leq i\leq(5\cdot 3^{n}-3)/6$, we can write

[TABLE]

where $i_{t}=0$, $1$, $2$ for $0\leq t\leq n-2$. Then by an argument similar to the one used in [38, Theorems 11 and 12], we obtain the following two results.

Theorem 4**.**

Let $(3^{n}+1)/2\leq i\leq(5\cdot 3^{n}-3)/6$ with $n\geq 1$. Then the following statements are equivalent:

1. $(i)$

$\displaystyle\binom{3i-\frac{3^{n}-1}{2}}{i-\frac{3^{n}+1}{2}}$* $\not\equiv 0\pmod{3}$;* 2. $(ii)$

$2\geq i_{0}\geq i_{1}\geq\cdots\geq i_{n-2}\geq 0$, where $i_{0},i_{1},\ldots,i_{n-2}$ are defined by (16); 3. $(iii)$

$i=\frac{3^{n}+1}{2}+\frac{3^{j}-1}{2}+\frac{3^{k}-1}{2}$, where $0\leq j\leq k\leq n-1$.

Theorem 5**.**

Let $i=\frac{3^{n}+1}{2}+\frac{3^{j}-1}{2}+\frac{3^{k}-1}{2}$, where $0\leq j\leq k\leq n-1$. Then

[TABLE]

According to Theorem 5, we obtain the following result.

Theorem 6**.**

Let $i=\frac{3^{n}+1}{2}+\frac{3^{j}-1}{2}+\frac{3^{k}-1}{2}$, where $0\leq j\leq k\leq n-1$. Then

[TABLE]

Proof.

Since $3^{m}=(1+2)^{m}=1+\binom{m}{1}2+\binom{m}{2}2^{2}+\cdots+\binom{m}{m}2^{m}$, we have $(-1)^{\frac{3^{m}-1}{2}}=(-1)^{m}$, and so $(-1)^{i-\frac{3^{n}+1}{2}}=(-1)^{\frac{3^{j}-1}{2}+\frac{3^{k}-1}{2}}=(-1)^{j+k}$. Substituting it into Theorem 5 gives the desired result. ∎

By Corollary 1, Theorems 4 and 6, we can write (15) as the following form:

[TABLE]

Substituting (17) into (14) completes the proof of Theorem 1.

5 Slight generalization

In this section, we also let $C_{0}=\{e^{2}:e\in\mathbb{F}_{3^{n}}^{*}\}$ and $C_{1}=\mathbb{F}_{3^{n}}^{*}\setminus C_{0}$. Let $\eta$ be the quadratic character. By an argument similar to that used in Theorem 1, we deduce the inverses of some generalized PPs studied in [26].

Lemma 5** ([26, Theorem 4.7]).**

Let $\alpha$, $\beta$, $\gamma$, $\theta\in\mathbb{F}_{3^{n}}^{*}$ with $n\geq 1$, and let

[TABLE]

Then $f(x)$ is a PP of $\mathbb{F}_{3^{n}}$ if and only if $\eta(\gamma)=-1$, $\eta(\theta)=1$ and $\eta(\alpha)=\eta(\beta)$.

Lemma 1 is a special case of Lemma 5 for $n$ is even, $\gamma=0$ and $\alpha=-\beta=\theta=-1$. The following result gives the inverse of $f(x)$ in Lemma 5.

Theorem 7**.**

If $f(x)$ in Lemma 5 is a PP of $\mathbb{F}_{3^{n}}$ and $\eta(\alpha)=(-1)^{m}$ with $m\in\{0,1\}$, then

[TABLE]

where

[TABLE]

[TABLE]

Proof.

Let $\tau,\lambda\in\mathbb{F}_{3^{n}}^{*}$, $s\in\{0,1\}$ and $f_{s}(x)=\tau(x^{3}+\lambda x^{2}+\lambda^{2}x)=\tau x(x-\lambda)^{2}.$ If $f(x)$ in Lemma 5 is a PP of $\mathbb{F}_{3^{n}}$ and $\eta(\tau)=(-1)^{m}$ with $m\in\{0,1\}$, then $f_{s}(x)$ induces a bijection from $C_{s}$ to $C_{t}$, where $s,t\in\{0,1\}$. Clearly, $s=t$ if $m=0$, and $s\neq t$ if $m=1$. Hence $s+t+m\equiv 0\pmod{2}$. Let $w_{i}=3^{n}-1-i$, where $(3^{n}+1)/2\leq i\leq 3^{n}-2$. Then

[TABLE]

By an argument similar to that used in the previous sections, we have

[TABLE]

Substituting it into Lemma 2 gives the desired result. ∎

Lemma 6** ([26, Theorem 4.2]).**

Let $\alpha$, $\beta$, $\theta\in\mathbb{F}_{3^{n}}^{*}$ with $n\geq 1$, and let

[TABLE]

Then $f(x)$ is a PP of $\mathbb{F}_{3^{n}}$ if and only if \big{(}t,\frac{3^{n}-1}{2}\big{)}=1, $\eta(\theta)=1$ and $\eta(\alpha)=\eta(\beta)$.

Theorem 8**.**

If $f(x)$ in Lemma 6 is a PP of $\mathbb{F}_{3^{n}}$ and $\eta(\alpha)=(-1)^{m}$ with $m\in\{0,1\}$, then

[TABLE]

where $v(x)$ is defined by (20), $u(x)=(\alpha^{-1}x)^{s}$, and $s$ is the inverse of $t$ modulo $(3^{n}-1)/2$.

Lemma 7** ([26, Corollary 4.5]).**

Let $n$, $t$ be positive integers. Let $\alpha$, $\beta$, $\gamma\in\mathbb{F}_{3^{n}}^{*}$ and

[TABLE]

Then $f(x)$ is a PP of $\mathbb{F}_{3^{n}}$ if and only if \big{(}t,\frac{3^{n}-1}{2}\big{)}=1, $\eta(\gamma)=-1$ and $\eta(\alpha)=\eta(\beta)(-1)^{t+1}$.

Theorem 9**.**

If $f(x)$ in Lemma 7 is a PP of $\mathbb{F}_{3^{n}}$ and $\eta(\alpha)=(-1)^{m}$ with $m\in\{0,1\}$, then

[TABLE]

where $u(x)$ is defined by (19), $v(x)=(-1)^{r}(\beta^{-1}x)^{s}$, $r$ and $s$ are integers such that $1\leq s<(3^{n}-1)/2$ and $st+r(3^{n}-1)/2=1$.

When $n=1$, it is easy to verify that $f(x)\equiv f^{-1}(x)\equiv\pm x\pmod{x^{3}-x}$ in all results of this section. Hence Theorems 7, 8 and 9 are all true for $n=1$. If $\gamma=\theta=0$, then Lemmas 5, 6 and 7 are the special cases of [26, Corollary 2.3], and their inverses are given in [27, 39].

Acknowledgments

We are grateful to the referees for many useful comments and suggestions.