New Constructions of Complete Permutation Polynomials over Finite Fields of Even Characteristic

Abstract

Complete permutation polynomials have many applications in mathematics and cryptography. In this paper, we study the complete permutation property of polynomial $xh{\left(x^{q-1}\right)}^{q+1}$ over ${\mathbb{F}}_{q^2}$, where $h\left(x\right)=h_1(x+x^q)+h_2(x+x^q)x^k$. Based on the trace functions and Dickson polynomials, we present several new constructions of such complete permutations by choosing suitable $h_1\left(x\right)$, $h_2\left(x\right)$, and k.

1. Introduction

Let p be a prime, n be a positive integer, and ${\mathbb{F}}_{p^n}$ be a finite field with $p^n$ elements. A polynomial $f\left(x\right)\in {\mathbb{F}}_{p^n}\left[x\right]$ is called a complete permutation polynomial (CPP) over ${\mathbb{F}}_{p^n}$ if both $f\left(x\right)$ and $f\left(x\right)+x$ are bijections from ${\mathbb{F}}_{p^n}$ to itself. The concept of CPP over finite fields was first introduced by Mann in 1942 while studying the construction of orthogonal Latin squares [1]. Subsequently, Niederreiter and Robinson conducted a detailed and systematic study on CPP over finite fields in 1982 and established their algebraic foundation [2]. With the development of modern cryptography, Mittenthal [3] first used CPPs with good cryptographic properties to design nonlinear dynamic substitution devices, and revealed their tremendous potential in the design of S-boxes for block ciphers. Since then, the construction and analysis of CPPs gradually became a hot topic in research on algebraic coding theory and cryptographic function for their extensive applications in combinatorial design and communication theory [4,5,6].

As we know, it is a challenge to construct sparse CPPs with controllable algebraic degrees over finite fields. In 2011, Akbary, Ghioca, and Wang [7] proposed the AGW criterion to provide a general characterization for permutation polynomials over finite fields. This criterion effectively reduces the permutation property of a polynomial over ${\mathbb{F}}_{p^n}$ to the permutation property of mapping on some subsets of ${\mathbb{F}}_{p^n}$. Applying the AGW criterion, three classes of monomials and one class of trinomials are proved to be CPPs over ${\mathbb{F}}_{2^n}$ [8]. Bassalygo and Zinoviev [9] generalized the definition of CPP with b-CPP and presented some constructions of b-CPPs. By determining the solutions of equations in a certain unit circle, Li, Zeng, and Cao [10] further investigated the permutation properties of $xh{\left(x^{q-1}\right)}^{q+1}$ over ${\mathbb{F}}_{q^2}$ and derived several explicit constructions of CPPs. In ref. [11], Chen et al. studied permutation polynomials of the form ${(x^{p^m}-x+\delta )}^s+L\left(x\right)$ and obtained some new CPPs. Given a subset S of ${\mathbb{F}}_{p^n}$, Coulter and Hearding [12] presented a definition of s-complete mapping over ${\mathbb{F}}_{p^n}$ and introduced a new method for constructing permutation polynomials. Recently, Chan et al. [13] completely characterized a class of complete permutation quadrinomials using the linear equivalence. Further developments on CPPs can be found in [14,15,16,17]. More recent results include [5,6,11,12,13,18,19,20,21] and the references therein.

Some of the known constructions of CPPs focus on specific polynomials or scattered results obtained through computer searches. However, the intrinsic connection between polynomial structures and permutation properties remains unclear. To address the aforementioned issues, we continue the work of [10] and explore CPPs of the form $xh{\left(x^{q-1}\right)}^{q+1}$ over ${\mathbb{F}}_{q^2}$. Inspired by the idea of [10], we study such polynomial by choosing $h\left(x\right)=h_1(x+x^q)+h_2(x+x^q)x^k$ for a general positive integer k. We define the unite circle of ${\mathbb{F}}_{q^2}$ as $U=\{z\in {\mathbb{F}}_{q^2}:z^{q+1}=1\}$. If $z\in U$ and $y=z+z^q$, then $z^2+yz+1=0$ and $h\left(z\right)=h_1\left(y\right)+h_2\left(y\right)z^k$. By recursively utilizing the equation $z^2+yz+1=0$, we can determine the values of $h\left(z\right)$ on U and hence find new CPPs. We transform the problem of constructing CPPs into the problem of avoiding “forbidden value sets” for the coefficients of polynomials.

The main contributions of this paper include the following two points: one is the phenomenon of “trace degeneration” and forbidden value set degeneration, and the other is the establishment of a general construction strategy based on linear operators. More precisely, the criterion of CPPs with the form $xh{\left(x^{q-1}\right)}^{q+1}$ is reduced to the nonvanishing of two explicit functions on $S\cup \left\{0\right\}$, which gives a concrete description of the admissible coefficients. Furthermore, this paper introduces Dickson polynomials to generalize the theory to the universal form and presents an in-depth analysis of the Gold exponent.

2. Preliminaries

Throughout this paper, let m, n, k, and q be positive integers satisfying $q=2^m$ and $m\mid n$. Let $\alpha \in {\mathbb{F}}_q$ and ${\mathbb{F}}_q^{\ast }$ denote the multiplicative group of ${\mathbb{F}}_q$. The trace function from ${\mathbb{F}}_{2^n}$ to ${\mathbb{F}}_{2^m}$ is defined as

$$\begin{array}{c}\hfill {Tr}_m^n\left(x\right)=x+x^{2^m}+x^{2^{2m}}+\cdots +x^{2^{(n/m-1)m}}.\end{array}$$

Let ℓ be a non-negative integer. In ref. [22], the Dickson polynomial of the first kind is defined by $D_0(y,\alpha )=2$, and

$$D_{\mathsf{\ell }}(y,\alpha )=\sum _{i=0}^{\lfloor \mathsf{\ell }/2\rfloor }{\displaystyle \frac{\mathsf{\ell }}{\mathsf{\ell }-i}}\left({\displaystyle \genfrac{}{}{0pt}{}{\mathsf{\ell }-i}{i}}\right){(-\alpha )}^i{y}^{\mathsf{\ell }-2i}$$

for $\mathsf{\ell }\ge 1$. Correspondingly, the ℓ-th Dickson polynomial of the second kind $E_{\mathsf{\ell }}(y,\alpha )\in {\mathbb{F}}_q\left[y\right]$ is defined by

$$E_{\mathsf{\ell }}(y,\alpha )=\sum _{i=0}^{\lfloor \mathsf{\ell }/2\rfloor }\left({\displaystyle \genfrac{}{}{0pt}{}{\mathsf{\ell }-i}{i}}\right){(-\alpha )}^i{y}^{\mathsf{\ell }-2i}.$$

For $\alpha =1$ and $\mathsf{\ell }\ge 2$, in characteristic 2, they satisfy the following recurrence relations:

$$D_{\mathsf{\ell }}(y,1)=y{D}_{\mathsf{\ell }-1}(y,1)+D_{\mathsf{\ell }-2}(y,1),\mathrm{with}{D}_0(y,1)=0,D_1(y,1)=y;$$

$$E_{\mathsf{\ell }}(y,1)=y{E}_{\mathsf{\ell }-1}(y,1)+E_{\mathsf{\ell }-2}(y,1),\mathrm{with}{E}_0(y,1)=1,E_1(y,1)=y.$$

When $y=x+x^{-1}$, it is known that

$$D_{\mathsf{\ell }}(x+x^{-1},1)=x^{\mathsf{\ell }}+x^{-\mathsf{\ell }},$$

$$E_{\mathsf{\ell }}(x+x^{-1},1)={\displaystyle \frac{x^{\mathsf{\ell }+1}+x^{-(\mathsf{\ell }+1)}}{x+x^{-1}}}.$$

Then,

$$y{E}_{\mathsf{\ell }-1}(y,1)=(x+x^{-1})·{\displaystyle \frac{x^{(\mathsf{\ell }-1)+1}+x^{-\left(\right(\mathsf{\ell }-1)+1)}}{x+x^{-1}}}=x^{\mathsf{\ell }}+x^{-\mathsf{\ell }}=D_{\mathsf{\ell }}(y,1)$$

(1)

and

$$\begin{array}{cc}\hfill E_{\mathsf{\ell }}(y,1)+E_{\mathsf{\ell }-2}(y,1)& ={\displaystyle \frac{x^{\mathsf{\ell }+1}+x^{-(\mathsf{\ell }+1)}+x^{\mathsf{\ell }-1}+x^{-(\mathsf{\ell }-1)}}{x+x^{-1}}}\hfill \\ \hfill & ={\displaystyle \frac{(x^{\mathsf{\ell }}+x^{-\mathsf{\ell }})(x+x^{-1})}{x+x^{-1}}}=x^{\mathsf{\ell }}+x^{-\mathsf{\ell }}\hfill \\ \hfill & =D_{\mathsf{\ell }}(y,1).\hfill \end{array}$$

Therefore, we have

$$D_{\mathsf{\ell }}(y,1)=y{E}_{\mathsf{\ell }-1}(y,1)=E_{\mathsf{\ell }}(y,1)+E_{\mathsf{\ell }-2}(y,1).$$

(2)

Let U denote the unit circle over the finite field ${\mathbb{F}}_{q^2}$, defined by

$$\begin{array}{c}\hfill U=\{x\in {\mathbb{F}}_{q^2}\mid x^{q+1}=1\}.\end{array}$$

For $z\in U$, we always let $y=z+z^q$. Then, one can obtain $y=z+z^{-1}\in {\mathbb{F}}_q$, and hence $y\in S\cup \left\{0\right\}$, where

$$\begin{array}{c}\hfill S=\{y\in {\mathbb{F}}_q^{\ast }\mid {Tr}_1^m(1/y)=1\}.\end{array}$$

The following lemma provides a sufficient and necessary condition for the polynomial $xh{\left(x^{q-1}\right)}^{q+1}$ being a CPP over ${\mathbb{F}}_{q^2}$.

Lemma 1 

([10]). Let q be a prime power, and let the polynomial $h\left(x\right)\in {\mathbb{F}}_{q^2}\left[x\right]$. Then, $f\left(x\right)=xh{\left(x^{q-1}\right)}^{q+1}$ is a CPP over ${\mathbb{F}}_{q^2}$ if and only if

$$h\left(z\right)(h{\left(z\right)}^{q+1}+1)\ne 0$$

(3)

for any $z\in U$.

Remark 1. 

The condition $h\left(z\right)(h{\left(z\right)}^{q+1}+1)\ne 0$ is equivalent to the joint requirement that $h\left(z\right)\ne 0$ and $h{\left(z\right)}^{q+1}\ne 1$. The condition $h\left(z\right)\ne 0$ ensures that the term $xh{\left(x^{q-1}\right)}^{q+1}$ does not vanish on the fibers determined by the unit circle U. Similarly, $h{\left(z\right)}^{q+1}\ne 1$ ensures the non-degeneracy of $f\left(x\right)+x=x(h{(x^{q-1})}^{q+1}+1)$. Thus, this condition serves to simultaneously characterize the permutation properties of both $f\left(x\right)$ and $f\left(x\right)+x$.

In the following sections, we will construct new CPPs by choosing $h\left(x\right)$ satisfying inequality (3).

3. Constructing CPPs with $\mathit{h}\left(\mathit{x}\right)={\mathit{h}}_{\mathbf{1}}(\mathit{x}+{\mathit{x}}^{\mathit{q}})+{\mathit{h}}_{\mathbf{2}}(\mathit{x}+{\mathit{x}}^{\mathit{q}}){\mathit{x}}^{{\mathbf{2}}^{\mathit{i}}}$

In this section, using the trace function, we primarily investigate the permutation properties of $xh{\left(x^{q-1}\right)}^{q+1}$ when $h\left(x\right)=h_1(x+x^q)+h_2(x+x^q)x^{2^i}$ as described in Lemma 1.

Lemma 2. 

Let $z\in U$ and $y=z+z^q$. Then,

$$\begin{array}{c}\hfill z^{2^i}=u_i\left(y\right)z+v_i\left(y\right),\end{array}$$

for every positive integer i, where $u_i\left(y\right)=y^{2^i-1}$, $v_i\left(y\right)={\sum }_{t=1}^i{y}^{2^i-2^t}$.

Proof. 

(i) Since $z\in U$ and $y=z+z^q$, one can directly get $z^2=yz+1$ in characteristic 2. Let $u_1\left(y\right)=y$ and $v_1\left(y\right)=1$. Clearly, the conclusion holds for $i=1$.

(ii) Assume the conclusion holds for $i=r$, that is,

$$\begin{array}{c}\hfill z^{2^r}=u_r\left(y\right)z+v_r\left(y\right).\end{array}$$

where $u_r\left(y\right)=y^{2^r-1}$ and $v_r\left(y\right)={\sum }_{t=1}^r{y}^{2^r-2^t}$. This leads to

$$\begin{array}{cc}\hfill z^{2^{r+1}}& ={\left(u_r\left(y\right)z+v_r\left(y\right)\right)}^2=u_r{\left(y\right)}^2{z}^2+v_r{\left(y\right)}^2\hfill \\ & =u_r{\left(y\right)}^2(yz+1)+v_r{\left(y\right)}^2\hfill \\ & =\underset{u_{r+1}\left(y\right)}{\underbrace{[y·u_r{\left(y\right)}^2]}}z+\underset{v_{r+1}\left(y\right)}{\underbrace{[u_r{\left(y\right)}^2+v_r{\left(y\right)}^2]}},\hfill \end{array}$$

where

$$u_{r+1}\left(y\right)=y·u_r{\left(y\right)}^2=y·{\left(y^{2^r-1}\right)}^2=y^{2^{r+1}-1},$$

$$v_{r+1}\left(y\right)=v_r{\left(y\right)}^2+u_r{\left(y\right)}^2={\left(\sum _{t=1}^r{y}^{2^r-2^t}\right)}^2+{\left(y^{2^r-1}\right)}^2=\sum _{t=1}^{r+1}{y}^{2^{r+1}-2^t}.$$

The conclusion also holds for $i=r+1$.

Therefore, the conclusion in Lemma 2 holds for any integer $i\ge 1$. □

For later use, we define

$$K_i\left(y\right):=u_i\left(y\right)+v_i\left(y\right)=\sum _{r=0}^i{y}^{2^i-2^r}.$$

(4)

It is easy to verify that

$$K_i{\left(y\right)}^2+y^{2^i}{K}_i\left(y\right)+1=y^{2^{i+1}-1}.$$

(5)

Moreover, for $y\in S$, it is clear that

$$K_{m-1}\left(y\right)=y^{2^{m-1}},$$

(6)

and

$$K_{m-2}\left(y\right)=y^{2^{m-2}}+y^{-2^{m-2}}.$$

(7)

In the following, we set

$$T\left(x\right):=x^{q-1}+x^{q^2-q}.$$

(8)

It is clear that $T\left(x\right)\in {\mathbb{F}}_q\left[x\right]$.

By Lemmas 1 and 2, we have the following theorem.

The following theorem reformulates the complete permutation condition in Lemma 1 in terms of two auxiliary polynomials on $S\cup \left\{0\right\}$.

Theorem 1. 

Let i be a positive integer, let $h_1\left(x\right),h_2\left(x\right)\in {\mathbb{F}}_q\left[x\right]$, and define $h\left(x\right)=h_1(x+x^q)+h_2(x+x^q)x^{2^i}$. Then, the polynomial

$$f\left(x\right)=x\left(h_1{\left(T\left(x\right)\right)}^2+h_2{\left(T\left(x\right)\right)}^2+T{\left(x\right)}^{2^i}{h}_1\left(T\left(x\right)\right)h_2\left(T\left(x\right)\right)\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if $F_1\left(y\right)F_2\left(y\right)\ne 0$ for all $y\in S\cup \left\{0\right\}$, where $F_1\left(y\right)=h_1\left(y\right)+K_i\left(y\right)h_2\left(y\right)$ and $F_2\left(y\right)=h_1{\left(y\right)}^2+h_2{\left(y\right)}^2+y^{2^i}{h}_1\left(y\right)h_2\left(y\right)+1$.

Proof. 

For any $z\in U$, obviously $y=z+z^q\in S\cup \left\{0\right\}$. It follows from Lemma 2 that $z^{2^i}=u_i\left(y\right)z+v_i\left(y\right)$, and hence,

$$h\left(z\right)=h_1\left(y\right)+h_2\left(y\right)v_i\left(y\right)+h_2\left(y\right)u_i\left(y\right)z.$$

Clearly, $h\left(1\right)=h_1\left(0\right)+h_2\left(0\right)=F_1\left(0\right)$. Suppose $h\left(z_0\right)=0$ for some $z_0\in U\setminus \left\{1\right\}$. Since $y=z+z^q\in {\mathbb{F}}_q$ and $h_1\left(y\right),h_2\left(y\right)\in {\mathbb{F}}_q\left[y\right]$, it is easy to check that

$$0=h\left(z_0\right)+z_{0}h{\left(z_0\right)}^q=(h_1\left(y_0\right)+h_2\left(y_0\right)v_i\left(y_0\right)+h_2\left(y_0\right)u_i\left(y_0\right))(z_0+1),$$

where $y_0=z_0+z_0^q\in S$. This leads to

$$h_1\left(y_0\right)+h_2\left(y_0\right)v_i\left(y_0\right)+h_2\left(y_0\right)u_i\left(y_0\right)=F_1\left(y_0\right)=0.$$

Therefore, $h\left(z\right)\ne 0$ for all $z\in U$ if $F_1\left(y\right)\ne 0$ for each $y\in S\cup \left\{0\right\}$.

Note that

$$\begin{array}{cc}\hfill & h{\left(z\right)}^{q+1}+1\hfill \\ \hfill =& (h_1\left(y\right)+h_2\left(y\right)v_i\left(y\right)+h_2\left(y\right)u_i\left(y\right)z)(h_1\left(y\right)+h_2\left(y\right)v_i\left(y\right)+h_2\left(y\right)u_i\left(y\right)z^{-1})+1\hfill \\ \hfill =& h_1{\left(y\right)}^2+h_2{\left(y\right)}^2(v_i{\left(y\right)}^2+u_i{\left(y\right)}^2+y{v}_i\left(y\right)u_i\left(y\right))+y{u}_i\left(y\right)h_1\left(y\right)h_2\left(y\right)\hfill \\ \hfill =& h_1{\left(y\right)}^2+h_2{\left(y\right)}^2+y^{2^i}{h}_1\left(y\right)h_2\left(y\right)+1=F_2\left(y\right)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill xh{\left(x^{q-1}\right)}^{q+1}& =x{\left(h_1\left(T\left(x\right)\right)+h_2\left(T\left(x\right)\right)x^{2^i(q-1)}\right)}^{q+1}\hfill \\ \hfill & =x(h_1{\left(T\left(x\right)\right)}^2+h_2{\left(T\left(x\right)\right)}^2\hfill \\ \hfill & +\left(x^{2^i(q-1)}+x^{2^i(q^2-q)}\right)h_1\left(T\left(x\right)\right)h_2\left(T\left(x\right)\right))\hfill \\ \hfill & =x\left(h_1{\left(T\left(x\right)\right)}^2+h_2{\left(T\left(x\right)\right)}^2+T{\left(x\right)}^{2^i}{h}_1\left(T\left(x\right)\right)h_2\left(T\left(x\right)\right)\right).\hfill \end{array}$$

If $F_1\left(y\right)·F_2\left(y\right)\ne 0$ for all $y\in S\cup \left\{0\right\}$, then $h\left(z\right)(h{\left(z\right)}^{q+1}+1)\ne 0$ for any $z\in U$. It follows from Lemma 1 that $f\left(x\right)=xh{\left(x^{q-1}\right)}^{q+1}$ is a CPP over ${\mathbb{F}}_{q^2}$. □

According to Theorem 1, we have the following results by choosing the appropriate terms $h_1\left(x\right)$ and $h_2\left(x\right)$. For later use, we introduce the following standing assumptions:

(A1)

$a+b\notin \{0,1\}$;

(A2)

${Tr}_1^m\left({\displaystyle \frac{ab}{a^2+b^2+1}}\right)=0$;

(A3)

${Tr}_1^m\left({\displaystyle \frac{b^2+ab}{a^2+1}}\right)=0$.

In the following proposition, we will characterize the case where $h_1$ is an affine function of $x^{2^i}$ and $h_2$ is a constant. In this setting, the two nonvanishing conditions in Theorem 1 reduce to a forbidden-value condition on the ratio $a/b$ and a trace condition formulated in terms of $ab/(a^2+b^2+1)$.

Proposition 1. 

Let $q=2^m$ and $a,b\in {\mathbb{F}}_q^{\ast }$. The polynomial

$$f\left(x\right)=x\left(a^2+b^2+ab{T\left(x\right)}^{2^i}\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if the following conditions are satisfied:

(i) 

Assumptions (A1) and (A2) hold;

(ii) 

${\displaystyle \frac{a}{b}}\notin \{y^{2^i}+K_i\left(y\right):y\in S\}$.

Proof. 

Choose $h_1\left(x\right)=a+b{x}^{2^i}$ and $h_2\left(x\right)=b$ in Theorem 1. Then, we have

$$\begin{array}{cc}\hfill f\left(x\right)& =x\left({\left(a+bT{\left(x\right)}^{2^i}\right)}^2+b^2+bT{\left(x\right)}^{2^i}\left(a+bT{\left(x\right)}^{2^i}\right)\right)\hfill \\ \hfill & =x\left(a^2+b^{2}T{\left(x\right)}^{2^{i+1}}+b^2+abT{\left(x\right)}^{2^i}+b^{2}T{\left(x\right)}^{2^{i+1}}\right)\hfill \\ \hfill & =x\left(a^2+b^2+abT{\left(x\right)}^{2^i}\right).\hfill \end{array}$$

$F_1\left(y\right)=a+b\left(y^{2^i}+K_i\left(y\right)\right)$ and $F_2\left(y\right)=a^2+b^2+1+ab{y}^{2^i}$. Clearly, $F_1\left(0\right)=a+b$ and $F_2\left(0\right)=a^2+b^2+1$. Since Assumption (A1) holds, we have $a+b\notin \{0,1\}$; hence, $F_1\left(0\right)F_2\left(0\right)\ne 0$.

Next we consider the case of $y\in S$. Since $a,b\in {\mathbb{F}}_q^{\ast }$, $F_1\left(y\right)=0$ is equivalent to $a+b\left(y^{2^i}+K_i\left(y\right)\right)=0$. That is to say, ${\displaystyle \frac{a}{b}}\in \{y^{2^i}+K_i\left(y\right):y\in S\}$. Similarly, $F_2\left(y\right)=0$ is equivalent to $y^{2^i}={\displaystyle \frac{a^2+b^2+1}{ab}}$. This implies that ${\displaystyle \frac{a^2+b^2+1}{ab}}\in S$, which contradicts Assumption (A2). Therefore, we can conclude that $F_1\left(y\right)·F_2\left(y\right)\ne 0$ for every $y\in S$ if conditions (i) and (ii) hold. This completes the proof. □

For $i\in \{1,2,m-2,m-1\}$, the forbidden-value set in Proposition 1 admits an explicit simplification. Specifically, the cases of $i=1$ and $i=2$ yield concrete polynomial constraints, whereas for $i=m-1$ and $i=m-2$, the conditions degenerate into trace-based characterizations. These results are summarized in the following corollaries.

Remark 2. 

In all computational examples below, the parameters are chosen by the following procedure. For a fixed value of m, we construct ${\mathbb{F}}_{2^m}$ and its quadratic extension ${\mathbb{F}}_{2^{2m}}$. We first compute the set $S=\{y\in {\mathbb{F}}_{2^m}^{\ast }:{Tr}_1^m(1/y)=1\}$ and the corresponding forbidden-value set appearing in the relevant corollary. Then, all candidate pairs $(a,b)\in {\mathbb{F}}_{2^m}^{\ast }\times {\mathbb{F}}_{2^m}^{\ast }$ are enumerated and only those satisfying the stated assumptions and forbidden-value conditions are retained. For each retained pair, we also verify directly over ${\mathbb{F}}_{2^{2m}}$ that both $f\left(x\right)$ and $f\left(x\right)+x$ are permutation polynomials. The pair displayed in each example is selected from this admissible list.

Corollary 1. 

Let $q=2^m$ with $m\ge 3$, and $a,b\in {\mathbb{F}}_q^{\ast }$. The polynomial

$$f\left(x\right)=(a^2+b^2)x+ab\left(x^{2q-1}+x^{q^2-2q+2}\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if the following conditions are satisfied:

(i) 

Assumptions (A1) and (A2) hold;

(ii) 

${\displaystyle \frac{a}{b}}\notin \{y^2+y+1:y\in S\}$.

Proof. 

Take $i=1$ in Proposition 1. By Equation (4), $K_1\left(y\right)=y+1$, and so ${\displaystyle \frac{a}{b}}\notin \{y^2+y+1:y\in S\}$. It follows from Proposition 1 that $f\left(x\right)$ is a CPP over ${\mathbb{F}}_{q^2}$. □

Example 1. 

Let u be a primitive element of ${\mathbb{F}}_{2^5}$. For $q=2^5$, we first compute $S=\{y\in {\mathbb{F}}_{2^5}^{\ast }:{Tr}_1^5(1/y)=1\}$ and the forbidden set $\{y^2+y+1:y\in S\}$. Then, all pairs $(a,b)\in {\mathbb{F}}_{2^5}^{\ast }\times {\mathbb{F}}_{2^5}^{\ast }$ satisfying the conditions in Corollary 1 are enumerated. By Magma computation, Corollary 1 produces exactly 280 pairs $(a,b)\in {\mathbb{F}}_{2^5}^{\ast }\times {\mathbb{F}}_{2^5}^{\ast }$ such that the corresponding polynomial is a CPP over ${\mathbb{F}}_{2^{10}}$. The pair $(a,b)=(1,u)$ is one of the admissible pairs. In particular, for $(a,b)=(1,u)$, we obtain that

$$f\left(x\right)=(1+u^2)x+u\left(x^{63}+x^{962}\right)$$

is a CPP over ${\mathbb{F}}_{2^{10}}$.

Corollary 2. 

Let $q=2^m$ and $a,b\in {\mathbb{F}}_q^{\ast }$. The polynomial

$$f\left(x\right)=(a^2+b^2)x+ab\left(x^{4q-3}+x^{q^2-4q+4}\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if the following conditions hold:

(i) 

Assumptions (A1) and (A2) hold;

(ii) 

${\displaystyle \frac{a}{b}}\notin \{y^4+y^3+y^2+1:y\in S\}$.

Proof. 

Take $i=2$ in Proposition 1. By Equation (4), $K_2\left(y\right)=y^3+y^2+1$, and therefore,

$${\displaystyle \frac{a}{b}}\notin \{y^4+y^3+y^2+1:y\in S\}.$$

Hence, by Proposition 1, $f\left(x\right)$ is a CPP over ${\mathbb{F}}_{q^2}$. □

Example 2. 

Let u be a primitive element of ${\mathbb{F}}_{2^5}$. For $q=2^5$, we compute $S=\{y\in {\mathbb{F}}_{2^5}^{\ast }:{Tr}_1^5(1/y)=1\}$ and the forbidden set $\{y^4+y^3+y^2+1:y\in S\}$. Then, all pairs $(a,b)\in {\mathbb{F}}_{2^5}^{\ast }\times {\mathbb{F}}_{2^5}^{\ast }$ satisfying the conditions in Corollary 2 are enumerated. Magma experiments show that Corollary 2 produces exactly 280 CPPs over ${\mathbb{F}}_{2^{10}}$. The pair $(a,b)=(1,u^3)$ is one of the admissible pairs. In particular, for $(a,b)=(1,u^3)$, we get that

$$f\left(x\right)=(1+u^6)x+u^3\left(x^{125}+x^{900}\right)$$

is a CPP over ${\mathbb{F}}_{2^{10}}$.

Corollary 3. 

Let $q=2^m$ and $a,b\in {\mathbb{F}}_q^{\ast }$. The polynomial

$$f\left(x\right)=(a^2+b^2)x+ab(x^{{\displaystyle \frac{q(q+1)}{2}}}+x^{{\displaystyle \frac{q^2-q+2}{2}}})$$

is a CPP over ${\mathbb{F}}_{q^2}$ if both Assumptions (A1) and (A2) hold.

Proof. 

Choose $i=m-1$ in Proposition 1. By Equation (6), we have $K_{m-1}\left(y\right)=y^{2^{m-1}}$ for every $y\in S$. This leads to

$$\{y^{2^{m-1}}+K_{m-1}\left(y\right):y\in S\}=\left\{0\right\}.$$

Since $a,b\in {\mathbb{F}}_q^{\ast }$, we have $a/b\ne 0$, and so condition (ii) in Proposition 1 is automatically satisfied. Then, we can get the desired result by Proposition 1. □

Example 3. 

Let u be a primitive element of ${\mathbb{F}}_{2^6}$. For $q=2^6$, the parameters are obtained by enumerating all pairs $(a,b)\in {\mathbb{F}}_{2^6}^{\ast }\times {\mathbb{F}}_{2^6}^{\ast }$ satisfying the two conditions in Corollary 3. Magma experiments show that there are 1860 pairs $(a,b)\in {\mathbb{F}}_{2^6}^{\ast }\times {\mathbb{F}}_{2^6}^{\ast }$ satisfying the two conditions of Corollary 3. The pair $(a,b)=(1,u^3)$ is one of the admissible pairs. In particular, for $(a,b)=(1,u^3)$, we obtain that

$$f\left(x\right)=(1+u^6)x+u^3\left(x^{2017}+x^{2080}\right)$$

is a CPP over ${\mathbb{F}}_{2^{12}}$.

Corollary 4. 

Let $a,b\in {\mathbb{F}}_q^{\ast }$. The polynomial

$$f\left(x\right)=(a^2+b^2)x+ab(x^{{\displaystyle \frac{3q^2+q}{4}}}+x^{{\displaystyle \frac{q^2-q+4}{4}}}),$$

is a CPP over ${\mathbb{F}}_{q^2}$ if all the following conditions hold:

(i) 

Assumptions (A1) and (A2) hold;

(ii) 

${Tr}_1^m\left({\displaystyle \frac{a}{b}}\right)=0$.

Proof. 

Let $i=m-2$ in Proposition 1. For every $y\in S$, it follows from Equation (7) that

$$\{y^{2^{m-2}}+K_{m-2}\left(y\right):y\in S\}=\{y^{-2^{m-2}}:y\in S\}.$$

Note that ${Tr}_1^m\left(y^{-2^{m-2}}\right)={Tr}_1^m\left({\displaystyle \frac{1}{y}}\right)=1$ for $y\in S$. Therefore, condition (ii) in Proposition 1 is equivalent to ${Tr}_1^m\left({\displaystyle \frac{a}{b}}\right)=0$. Then, we can get the desired result by Proposition 1. □

Example 4. 

Let u be a primitive element of ${\mathbb{F}}_{2^6}$. For $q=2^6$, we enumerate all pairs $(a,b)\in {\mathbb{F}}_{2^6}^{\ast }\times {\mathbb{F}}_{2^6}^{\ast }$ satisfying the conditions in Corollary 4, including the trace condition ${Tr}_1^6(a/b)=0$. Magma experiments show that Corollary 4 produces exactly 900 pairs $(a,b)\in {\mathbb{F}}_{2^6}^{\ast }\times {\mathbb{F}}_{2^6}^{\ast }$ such that the corresponding polynomial is a CPP over ${\mathbb{F}}_{2^{12}}$. The pair $(a,b)=(1,u^3)$ is one of the admissible pairs. Choosing $(a,b)=(1,u^3)$, we obtain that

$$f\left(x\right)=(1+u^6)x+u^3\left(x^{1009}+x^{3088}\right)$$

is a CPP over ${\mathbb{F}}_{2^{12}}$.

Next, we consider the case where $h_1$ is a constant and $h_2$ is a linear function. Under this setting, the obstruction from $F_1$ remains a forbidden-value condition, while the obstruction from $F_2$ is characterized by the trace condition in Assumption (A3).

Proposition 2. 

Let $a,b\in {\mathbb{F}}_q^{\ast }$ such that $a\ne 1$ and $b\ne a$. The polynomial

$$f\left(x\right)=x\left(a^2+(b^2+ab){T\left(x\right)}^{2^{i+1}}\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if both of the following conditions are satisfied:

(i) 

Assumption (A3) holds;

(ii) 

${\displaystyle \frac{a}{b}}\notin \{y^{2^i}{K}_i\left(y\right):y\in S\}$.

Proof. 

We apply Theorem 1 with $h_1\left(x\right)=a$ and $h_2\left(x\right)=b{x}^{2^i}$. Then,

$$\begin{array}{cc}\hfill f\left(x\right)& =x\left(a^2+b^{2}T{\left(x\right)}^{2^{i+1}}+abT{\left(x\right)}^{2^{i+1}}\right)\hfill \\ \hfill & =x\left(a^2+(b^2+ab)T{\left(x\right)}^{2^{i+1}}\right),\hfill \end{array}$$

while $F_1\left(y\right)=a+b{y}^{2^i}{K}_i\left(y\right)$ and $F_2\left(y\right)=a^2+1+(b^2+ab)y^{2^{i+1}}$.

By Theorem 1, it suffices to show that $F_1\left(y\right)·F_2\left(y\right)\ne 0$ for every $y\in S\cup \left\{0\right\}$. Assume to the contrary that there exists $y_0\in S\cup \left\{0\right\}$ such that $F_1\left(y_0\right)·F_2\left(y_0\right)=0$. Then, $F_1\left(y_0\right)=0$ or $F_2\left(y_0\right)=0$.

If $y_0=0$, then $F_1\left(0\right)=a=0$ or $F_2\left(0\right)=a^2+1=0$, which contradicts $a\notin \{0,1\}$.

If $y_0\ne 0$, then $y_0\in S$. Since $b\in {\mathbb{F}}_q^{\ast }$, the condition $F_1\left(y_0\right)=0$ is equivalent to ${\displaystyle \frac{a}{b}}=y_0^{2^i}{K}_i\left(y_0\right)$. Hence, ${\displaystyle \frac{a}{b}}\in \{y^{2^i}{K}_i\left(y\right):y\in S\}$, contrary to condition (ii).

Similarly, from $b\ne 0$ and $b\ne a$, we have $b^2+ab\ne 0$. Thus, the condition $F_2\left(y_0\right)=0$ holds if and only if $y_0^{2^{i+1}}={\displaystyle \frac{a^2+1}{b^2+ab}}$. It follows from $y_0\in S$ that ${\displaystyle \frac{a^2+1}{b^2+ab}}\in S$, i.e., ${Tr}_1^m\left({\displaystyle \frac{b^2+ab}{a^2+1}}\right)=1$, which contradicts Assumption (A3).

Therefore, $F_1\left(y\right)·F_2\left(y\right)\ne 0$ for every $y\in S\cup \left\{0\right\}$. It follows from Theorem 1 that $f\left(x\right)$ is a CPP over ${\mathbb{F}}_{q^2}$. □

Similarly, through the specialization of the index i in Proposition 2, we obtain explicit CPPs where the conditions on the admissible parameters are governed by either forbidden-value sets or trace identities. These instances demonstrate how the general criterion reduces to a more tractable form for certain significant exponents.

Corollary 5. 

Let $q=2^m$ with $m\ge 2$, and let $a,b\in {\mathbb{F}}_q^{\ast }$ satisfy $a\ne 1$ and $b\ne a$. Then, the polynomial

$$f\left(x\right)=a^{2}x+(b^2+ab)\left(x^{4q-3}+x^{q^2-4q+4}\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if both of the following conditions hold:

(i) 

Assumption (A3) holds;

(ii) 

${\displaystyle \frac{a}{b}}\notin \{y^3+y^2:y\in S\}$.

Proof. 

Take $i=1$ in Proposition 2. Since $K_1\left(y\right)=y+1$, we have $y^{2^i}{K}_1\left(y\right)=y^3+y^2$. Hence, condition (ii) in Proposition 2 is exactly ${\displaystyle \frac{a}{b}}\notin \{y^3+y^2:y\in S\}$. Then, by Proposition 2, $f\left(x\right)$ is a CPP over ${\mathbb{F}}_{q^2}$. □

Example 5. 

Let u be a primitive element of ${\mathbb{F}}_{2^5}$. For $q=2^5$, we compute $S=\{y\in {\mathbb{F}}_{2^5}^{\ast }:{Tr}_1^5(1/y)=1\}$ and the forbidden set $\{y^3+y^2:y\in S\}$. Then, all pairs $(a,b)$ with $a\in {\mathbb{F}}_{2^5}^{\ast }\setminus \left\{1\right\}$, $b\in {\mathbb{F}}_{2^5}^{\ast }$, and $a\ne b$ are tested against the conditions in Corollary 5. By Magma computation, there are exactly 280 pairs $(a,b)\in {\mathbb{F}}_{2^5}^{\ast }\setminus \left\{1\right\}\times {\mathbb{F}}_{2^5}^{\ast }$ with $a\ne b$ in Corollary 5 such that the corresponding polynomial is a CPP over ${\mathbb{F}}_{2^{10}}$. The pair $(a,b)=(u,u^2)$ is one of the admissible pairs. For example, given $(a,b)=(u,u^2)$, we have that

$$f\left(x\right)=u^{2}x+(u^4+u^3)\left(x^{125}+x^{900}\right)$$

is a CPP over ${\mathbb{F}}_{2^{10}}$.

Corollary 6. 

Let $q=2^m$ with $m\ge 3$, and let $a,b\in {\mathbb{F}}_q^{\ast }$ satisfy $a\ne 1$ and $b\ne a$. Then, the polynomial

$$f\left(x\right)=a^{2}x+(b^2+ab)\left(x^{8q-7}+x^{q^2-8q+8}\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if

(i) 

Assumption (A3) holds;

(ii) 

${\displaystyle \frac{a}{b}}\notin \{y^7+y^6+y^4:y\in S\}$.

Proof. 

Take $i=2$ in Proposition 2. Since $K_2\left(y\right)=y^3+y^2+1$, we obtain $y^{2^i}{K}_2\left(y\right)=y^7+y^6+y^4$. Hence, condition (ii) in Proposition 2 is exactly ${\displaystyle \frac{a}{b}}\notin \{y^7+y^6+y^4:y\in S\}$. It follows from Proposition 2 that $f\left(x\right)$ is a CPP over ${\mathbb{F}}_{q^2}$. □

Example 6. 

Let u be a primitive element of ${\mathbb{F}}_{2^5}$. For $q=2^5$, we compute $S=\{y\in {\mathbb{F}}_{2^5}^{\ast }:{Tr}_1^5(1/y)=1\}$ and the forbidden set $\{y^7+y^6+y^4:y\in S\}$. We then enumerate all pairs $(a,b)$ satisfying the assumptions and the forbidden-value condition in Corollary 6. By Magma computation, there are exactly 280 pairs $(a,b)$ in Corollary 6 such that the corresponding polynomial is a CPP over ${\mathbb{F}}_{2^{10}}$. The pair $(a,b)=(u,u^4)$ is one of the admissible pairs. In particular, for $(a,b)=(u,u^4)$, we obtain that

$$f\left(x\right)=u^{2}x+(u^8+u^5)\left(x^{249}+x^{776}\right)$$

is a CPP over ${\mathbb{F}}_{2^{10}}$.

Corollary 7. 

Let $q=2^m$ with $m\ge 3$, and let $a,b\in {\mathbb{F}}_q^{\ast }$ satisfy $a\ne 1$ and $b\ne a$. Then, the polynomial

$$f\left(x\right)=a^{2}x+(b^2+ab)\left(x^{{\displaystyle \frac{q(q+1)}{2}}}+x^{{\displaystyle \frac{q^2-q+2}{2}}}\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if both of the following conditions hold:

(i) 

Assumption (A3) holds;

(ii) 

${Tr}_1^m\left({\displaystyle \frac{b}{a+b}}\right)=0$.

Proof. 

We apply Proposition 2 with $i=m-2$. Then,

$$f\left(x\right)=a^{2}x+(b^2+ab)\left(x^{{\displaystyle \frac{q(q+1)}{2}}}+x^{{\displaystyle \frac{q^2-q+2}{2}}}\right).$$

For every $y\in S$, by Equation (7), we have $K_{m-2}\left(y\right)=y^{2^{m-2}}+y^{-2^{m-2}}$, and therefore,

$$y^{2^{m-2}}{K}_{m-2}\left(y\right)=y^{2^{m-1}}+1.$$

Note that $a,b\in {\mathbb{F}}_q^{\ast }$ with $a\ne 1$ and $b\ne a$. If ${\displaystyle \frac{a}{b}}=y_0^{2^{m-2}}{K}_{m-2}\left(y_0\right)=y_0^{2^{m-1}}+1$ for some $y_0\in S$, then $y_0^{2^{m-1}}={\displaystyle \frac{a+b}{b}}$. This leads to

$${Tr}_1^m\left({\displaystyle \frac{b}{a+b}}\right)={Tr}_1^m\left({\displaystyle \frac{1}{y_0^{2^{m-1}}}}\right)={Tr}_1^m\left({\displaystyle \frac{1}{y_0}}\right)=1,$$

which contradicts condition (ii). Therefore, ${\displaystyle \frac{a}{b}}\notin \{y^{2^i}{K}_i\left(y\right):y\in S\}$. Hence, the result follows from Proposition 2. □

Example 7. 

Let u be a primitive element of ${\mathbb{F}}_{2^5}$. For $q=2^5$, all pairs $(a,b)$ with $a\in {\mathbb{F}}_{2^5}^{\ast }\setminus \left\{1\right\}$, $b\in {\mathbb{F}}_{2^5}^{\ast }$, and $a\ne b$ are enumerated. We retain those satisfying the two trace conditions in Corollary 7. By Magma computation, there are exactly 210 pairs $(a,b)\in {\mathbb{F}}_{2^5}^{\ast }\setminus \left\{1\right\}\times {\mathbb{F}}_{2^5}^{\ast }$ with $a\ne b$ satisfying the two conditions of Corollary 7. The pair $(a,b)=(u,u^2)$ is one of the admissible pairs. In particular, for $(a,b)=(u,u^2)$, it can be checked that

$$f\left(x\right)=u^{2}x+(u^4+u^3)\left(x^{528}+x^{497}\right),$$

is a CPP over ${\mathbb{F}}_{2^{10}}$.

Corollary 8. 

Let $q=2^m$ and let $a,b\in {\mathbb{F}}_q^{\ast }$ satisfy $a\ne 1$ and $b\ne a$. The polynomial

$$f\left(x\right)=a^{2}x+(ab+b^2)(x^{q^2-q+1}+x^q)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if the following conditions hold:

(i) 

Assumption (A3) holds;

(ii) 

${Tr}_1^m\left({\displaystyle \frac{b}{a}}\right)=0$.

Proof. 

Take $i=m-1$ in Proposition 2. For every $y\in S$, by Equation (6), we have $K_{m-1}\left(y\right)=y^{2^{m-1}}$, and therefore,

$$y^{2^{m-1}}{K}_{m-1}\left(y\right)=y.$$

This leads to $\{y^{2^{m-1}}{K}_{m-1}\left(y\right):y\in S\}=S$. Hence, condition (ii) in Proposition 2 is exactly ${Tr}_1^m\left({\displaystyle \frac{b}{a}}\right)=0$. Hence, the result follows from Proposition 2. □

Example 8. 

Let u be a primitive element of ${\mathbb{F}}_{2^5}$. For $q=2^5$, the admissible parameters are obtained by enumerating all pairs $(a,b)$ with $a\in {\mathbb{F}}_{2^5}^{\ast }\setminus \left\{1\right\}$, $b\in {\mathbb{F}}_{2^5}^{\ast }$, and $a\ne b$, and then imposing the trace conditions in Corollary 8. By Magma computation, Corollary 8 produces exactly 210 pairs $(a,b)\in {\mathbb{F}}_{2^5}^{\ast }\setminus \left\{1\right\}\times {\mathbb{F}}_{2^5}^{\ast }$ and $a\ne b$ such that the corresponding polynomial is a CPP over ${\mathbb{F}}_{2^{10}}$. The pair $(a,b)=(u,u^2)$ is one of the admissible pairs. In particular, when $(a,b)=(u,u^2)$, we have that

$$f\left(x\right)=u^{2}x+(u^4+u^3)\left(x^{32}+x^{993}\right),$$

is a CPP over ${\mathbb{F}}_{2^{10}}$.

4. Constructing CPPs as $\mathit{h}\left(\mathit{x}\right)={\mathit{h}}_{\mathbf{1}}(\mathit{x}+{\mathit{x}}^{\mathit{q}})+{\mathit{h}}_{\mathbf{2}}(\mathit{x}+{\mathit{x}}^{\mathit{q}}){\mathit{x}}^{\mathit{k}}$

When $k\ne 2^i$, the recursive expression in Lemma 2 is not directly applicable to the general power $z^k$. To treat arbitrary integers $k\ge 2$, we use Dickson polynomials. Recall that Dickson polynomials are finite-field analogues of Chebyshev polynomials. In the present setting, their relevance comes from the identity $D_k(z+z^{-1},1)=z^k+z^{-k}$. Thus, for $z\in U$ and $y=z+z^q=z+z^{-1}$, the term $z^k+z^{-k}$ can be written as $D_k(y,1)\in {\mathbb{F}}_q\left[y\right]$. In addition, Dickson polynomials of the second kind provide the linear representation $z^k=E_{k-1}(y,1)z+E_{k-2}(y,1)$, which is used to reduce the condition in Lemma 1 to explicit nonvanishing conditions on $S\cup \left\{0\right\}$. In this section, we apply these identities to characterize the complete permutation property of $xh{\left(x^{q-1}\right)}^{q+1}$, where $h\left(x\right)=h_1(x+x^q)+h_2(x+x^q)x^k$. We first establish two lemmas needed for this general exponent case.

Lemma 3. 

Let $z\in U$ and $y=z+z^q$. For every integer $k\ge 2$, there exist polynomials ${\mathcal{U}}_k\left(y\right),{\mathcal{V}}_k\left(y\right)\in {\mathbb{F}}_q\left[y\right]$ such that

$$z^k={\mathcal{U}}_k\left(y\right)z+{\mathcal{V}}_k\left(y\right),$$

where ${\mathcal{U}}_k\left(y\right)=E_{k-1}(y,1)$ and ${\mathcal{V}}_k\left(y\right)=E_{k-2}(y,1)$.

Proof. 

Since $z\in U$ and $y=z+z^q$, we have $y\in {\mathbb{F}}_q$ and $z^2=yz+1$. For $k=2$, this gives

$$z^2=E_1(y,1)z+E_0(y,1).$$

Assume that, for some $k\ge 2$,

$$z^k=E_{k-1}(y,1)z+E_{k-2}(y,1).$$

Multiplying both sides by z and using $z^2=yz+1$, we obtain

$$\begin{array}{cc}\hfill z^{k+1}& =E_{k-1}(y,1)z^2+E_{k-2}(y,1)z\hfill \\ & =\left(y{E}_{k-1}(y,1)+E_{k-2}(y,1)\right)z+E_{k-1}(y,1)\hfill \\ & =E_k(y,1)z+E_{k-1}(y,1),\hfill \end{array}$$

where the last equality follows from the recurrence relation of Dickson polynomials of the second kind. Therefore the assertion follows by induction, with ${\mathcal{U}}_k\left(y\right)=E_{k-1}(y,1)$ and ${\mathcal{V}}_k\left(y\right)=E_{k-2}(y,1)$. □

Lemma 4. 

Let $z\in U$ and $y=z+z^q$. If $z^n={\mathcal{U}}_n\left(y\right)z+{\mathcal{V}}_n\left(y\right)$ for ${\mathcal{U}}_n\left(y\right),{\mathcal{V}}_n\left(y\right)\in {\mathbb{F}}_q\left[y\right]$, then

$${\mathcal{U}}_n{\left(y\right)}^2+{\mathcal{V}}_n{\left(y\right)}^2+y{\mathcal{U}}_n\left(y\right){\mathcal{V}}_n\left(y\right)=1.$$

(9)

Proof. 

Since $z\in U$, $z^n$ must also lie in the unit circle U, yielding ${\left(z^n\right)}^{q+1}=1$. Note that $y=z+z^q\in {\mathbb{F}}_q$ and ${\mathcal{U}}_n\left(y\right),{\mathcal{V}}_n\left(y\right)\in {\mathbb{F}}_q\left[y\right]$. It is easy to check that

$$\begin{array}{cc}\hfill 1& ={({\mathcal{U}}_n\left(y\right)z+{\mathcal{V}}_n\left(y\right))}^{q+1}=({\mathcal{U}}_n\left(y\right)z+{\mathcal{V}}_n\left(y\right))·({\mathcal{U}}_n\left(y\right)z^{-1}+{\mathcal{V}}_n\left(y\right))\hfill \\ \hfill & ={\mathcal{U}}_n{\left(y\right)}^2+y{\mathcal{U}}_n\left(y\right){\mathcal{V}}_n\left(y\right)+{\mathcal{V}}_n{\left(y\right)}^2.\hfill \end{array}$$

This completes the proof. □

Recall that $T\left(x\right)$ is defined by (8). Employing Lemma 3, we have the following theorem.

Theorem 2. 

Let $k\ge 2$ be an integer. Let $h_1\left(x\right),h_2\left(x\right)\in {\mathbb{F}}_q\left[x\right]$ and $h\left(x\right)=h_1(x+x^q)+h_2(x+x^q)x^k$. Then, the polynomial

$$f\left(x\right)=x\left(h_1{\left(T\left(x\right)\right)}^2+h_2{\left(T\left(x\right)\right)}^2+D_k(T\left(x\right),1)h_1\left(T\left(x\right)\right)h_2\left(T\left(x\right)\right)\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if $F_1\left(y\right)·F_2\left(y\right)\ne 0$ for all $y\in S\cup \left\{0\right\}$, where $F_1\left(y\right):=h_1\left(y\right)+(E_{k-2}(y,1)+E_{k-1}(y,1))h_2\left(y\right)$ and $F_2\left(y\right):=h_1{\left(y\right)}^2+h_2{\left(y\right)}^2+D_k(y,1)h_1\left(y\right)h_2\left(y\right)+1$.

Proof. 

Let $z\in U$ and $y=z+z^q$. Obviously, $y\in S\cup \left\{0\right\}$. By Lemma 3, we have $z^k=E_{k-1}(y,1)z+E_{k-2}(y,1)$, and hence,

$$h\left(z\right)=h_1\left(y\right)+h_2\left(y\right)(E_{k-2}(y,1)+E_{k-1}(y,1)z).$$

Clearly, $h\left(1\right)=h_1\left(0\right)+h_2\left(0\right)(E_{k-2}(0,1)+E_{k-1}(0,1))=F_1\left(0\right)$. Assume $h\left(z_0\right)=0$ for some $z_0\in U\setminus \left\{1\right\}$. Since $y=z+z^q\in {\mathbb{F}}_q$ and $h_1\left(y\right)$, $h_2\left(y\right)\in {\mathbb{F}}_q\left[y\right]$, it is easy to check that

$$0=h\left(z_0\right)+z_{0}h{\left(z_0\right)}^q=\left(h_1\left(y_0\right)+h_2\left(y_0\right)\left(E_{k-2}(y_0,1)+E_{k-1}(y_0,1)\right)\right)(z_0+1),$$

where $y_0=z_0+z_0^q\in S$. It leads to

$$h_1\left(y_0\right)+h_2\left(y_0\right)(E_{k-2}(y_0,1)+E_{k-1}(y_0,1))=F_1\left(y_0\right)=0.$$

Therefore, $h\left(z\right)\ne 0$ for all $z\in U$ if $F_1\left(y\right)\ne 0$ for each $y\in S\cup \left\{0\right\}$.

By Equations (2) and (9), one can verify that

$$\begin{array}{cc}\hfill h{\left(z\right)}^{q+1}+1& =\left(h_1\left(y\right)+h_2\left(y\right)(E_{k-2}(y,1)+E_{k-1}(y,1)z)\right)\hfill \\ \hfill & ·\left(h_1\left(y\right)+h_2\left(y\right)(E_{k-2}(y,1)+E_{k-1}(y,1)z^{-1})\right)+1\hfill \\ \hfill & =h_1{\left(y\right)}^2+h_2{\left(y\right)}^2(E_{k-2}{(y,1)}^2+E_{k-1}{(y,1)}^2\hfill \\ \hfill & +y{E}_{k-2}(y,1)E_{k-1}(y,1))\hfill \\ \hfill & +\left(y{E}_{k-1}(y,1)\right)h_1\left(y\right)h_2\left(y\right)+1\hfill \\ \hfill & =h_1{\left(y\right)}^2+h_2{\left(y\right)}^2+D_k(y,1)h_1\left(y\right)h_2\left(y\right)+1=F_2\left(y\right).\hfill \end{array}$$

For $x\ne 0$, let $z=x^{q-1}$. Then, $z\in U$ and $T\left(x\right)=x^{q-1}+x^{q^2-q}=z+z^{-1}$. Thus,

$$\begin{array}{c}\hfill D_k(T\left(x\right),1)=D_k(z+z^{-1},1)=z^k+z^{-k}=x^{k(q-1)}+x^{k(q^2-q)}.\end{array}$$

It follows that

$$\begin{array}{cc}\hfill xh{\left(x^{q-1}\right)}^{q+1}& =x{\left(h_1\left(T\left(x\right)\right)+h_2\left(T\left(x\right)\right)x^{k(q-1)}\right)}^{q+1}\hfill \\ \hfill & =x\left(h_1{\left(T\left(x\right)\right)}^2+h_2{\left(T\left(x\right)\right)}^2+D_k(T\left(x\right),1)h_1\left(T\left(x\right)\right)h_2\left(T\left(x\right)\right)\right).\hfill \end{array}$$

If $F_1\left(y\right)·F_2\left(y\right)\ne 0$ for all $y\in S\cup \left\{0\right\}$, then $h\left(z\right)(h{\left(z\right)}^{q+1}+1)\ne 0$ for each $z\in U$. The desired result follows from Lemma 1. □

The following proposition presents a specialization of Theorem 2 to the case of constant coefficients. In this setting, the Dickson polynomial $D_k(T\left(x\right),1)$ captures the term $x^{k(q-1)}+x^{k(q^2-q)}$, thereby enabling a compact representation of the resulting CPP.

Proposition 3. 

Let $q=2^m$ and $a,b\in {\mathbb{F}}_q^{\ast }$. The polynomial

$$f\left(x\right)=x\left(a^2+b^2+ab{D}_k(T\left(x\right),1)\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if the following conditions hold:

(i) 

Assumption (A1) holds;

(ii) 

${\displaystyle \frac{a}{b}}\notin \{E_{k-2}(y,1)+E_{k-1}(y,1):y\in S\}$;

(iii) 

${\displaystyle \frac{a^2+b^2+1}{ab}}\notin \{D_k(y,1):y\in S\}$.

Proof. 

Choose $h_1\left(x\right)=a$ and $h_2\left(x\right)=b$ in Theorem 2. Then, we have

$$\begin{array}{c}\hfill f\left(x\right)=x\left(a^2+b^2+ab{D}_k(T\left(x\right),1)\right).\end{array}$$

$F_1\left(y\right)=a+b(E_{k-2}(y,1)+E_{k-1}(y,1))$ and $F_2\left(y\right)=a^2+b^2+1+ab{D}_k(y,1)$. Clearly, $F_1\left(0\right)=a+b$ and $F_2\left(0\right)={(a+b+1)}^2$. Since Assumption (A1) holds, we have $a+b\notin \{0,1\}$; hence, $F_1\left(0\right)·F_2\left(0\right)\ne 0$.

Next, we consider the case of $y\in S$. Since $a,b\in {\mathbb{F}}_q^{\ast }$, $F_1\left(y\right)=0$ is equivalent to $a+b(E_{k-2}(y,1)+E_{k-1}(y,1))=0$. That is to say, ${\displaystyle \frac{a}{b}}\in \{E_{k-2}(y,1)+E_{k-1}(y,1):y\in S\}$. Similarly, $F_2\left(y\right)=0$ is equivalent to $a^2+b^2+1+ab{D}_k(y,1)=0$. This implies that ${\displaystyle \frac{a^2+b^2+1}{ab}}\in \{D_k(y,1):y\in S\}$. Therefore, we conclude that $F_1\left(y\right)·F_2\left(y\right)\ne 0$ for every $y\in S$ if conditions (ii) and (iii) hold. This completes the proof. □

By specializing k in Proposition 3, we obtain several explicit families. The choices $k=3,2^{m-2},2^{m-1}-2^j,$ and $2^j+1$ are of particular interest because the resulting Dickson polynomials and forbidden-value sets are amenable to significantly simpler descriptions.

Corollary 9. 

Let $q=2^m$ and $a,b\in {\mathbb{F}}_q^{\ast }$. Then, the polynomial

$$f\left(x\right)=(a^2+b^2)x+ab\left(x^{3q-2}+x^{q^2-3q+3}\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if the following conditions are satisfied:

(i) 

Assumption (A1) holds;

(ii) 

${\displaystyle \frac{a}{b}}\notin \{y^2+y+1:y\in S\}$;

(iii) 

${\displaystyle \frac{a^2+b^2+1}{ab}}\notin \{y^3+y:y\in S\}$.

Proof. 

Take $k=3$ in Proposition 3. Since $E_1(y,1)=y$ and $E_2(y,1)=y^2+1$, we have ${\displaystyle \frac{a}{b}}\notin \{y^2+y+1:y\in S\}$. It follows from $D_3(y,1)=y^3+y$ that ${\displaystyle \frac{a^2+b^2+1}{ab}}\notin \{y^3+y:y\in S\}$. Hence, by Proposition 3, $f\left(x\right)$ is a complete permutation polynomial over ${\mathbb{F}}_{q^2}$. □

Example 9. 

Let u be a primitive element of ${\mathbb{F}}_{2^5}$. For $q=2^5$, we compute $S=\{y\in {\mathbb{F}}_{2^5}^{\ast }:{Tr}_1^5(1/y)=1\}$ and the two forbidden sets $\{y^2+y+1:y\in S\}$ and $\{y^3+y:y\in S\}$. Then, all pairs $(a,b)\in {\mathbb{F}}_{2^5}^{\ast }\times {\mathbb{F}}_{2^5}^{\ast }$ satisfying the conditions in Corollary 9 are retained. It was checked via Magma calculation that Corollary 9 produces exactly 500 pairs $(a,b)\in {\mathbb{F}}_{2^5}^{\ast }\times {\mathbb{F}}_{2^5}^{\ast }$ such that $f\left(x\right)$ is a CPP over ${\mathbb{F}}_{2^{10}}$. The pair $(a,b)=(1,u)$ is one of the admissible pairs. In particular, for $(a,b)=(1,u)$, we have that

$$f\left(x\right)=(1+u^2)x+u\left(x^{94}+x^{931}\right)$$

is a CPP over ${\mathbb{F}}_{2^{10}}$.

Corollary 10. 

Let $q=2^m$ with $m\ge 3$, and let $a,b\in {\mathbb{F}}_q^{\ast }$. Then, the polynomial

$$f\left(x\right)=(a^2+b^2)x+ab\left(x^{{\displaystyle \frac{q^2-q+4}{4}}}+x^{{\displaystyle \frac{3q^2+q}{4}}}\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if all the following conditions are satisfied:

(i) 

Assumptions (A1) and (A2) hold;

(ii) 

${\displaystyle \frac{a}{b}}\notin \{u+u^{-1}:u\in S\}$.

Proof. 

Take $k=2^{m-2}$ in Proposition 3. For $z\in U$ and $y=z+z^q$, it follows from Lemmas 2 and 3 that $z^{2^{m-2}}$ has two representations as follows:

$$z^{2^{m-2}}=u_{m-2}\left(y\right)z+v_{m-2}\left(y\right)=E_{2^{m-2}-1}(y,1)z+E_{2^{m-2}-2}(y,1).$$

Hence,

$$E_{2^{m-2}-1}(y,1)=u_{m-2}\left(y\right),E_{2^{m-2}-2}(y,1)=v_{m-2}\left(y\right).$$

Therefore, by Equations (4) and (7), we have

$$E_{2^{m-2}-2}(y,1)+E_{2^{m-2}-1}(y,1)=u_{m-2}\left(y\right)+v_{m-2}\left(y\right)=y^{2^{m-2}}+y^{-2^{m-2}}.$$

Set $u=y^{2^{m-2}}$. Then, ${Tr}_1^m({\displaystyle \frac{1}{u}})={Tr}_1^m({\displaystyle \frac{1}{y}})=1$, i.e., $u\in S$. Hence, condition (ii) in Proposition 3 is equivalent to ${\displaystyle \frac{a}{b}}\notin \{u+u^{-1}:u\in S\}$.

Note that $D_{2^{m-2}}(y,1)=y^{2^{m-2}}$. It leads to $\{D_{2^{m-2}}(y,1):y\in S\}=S$. Thus, condition (iii) in Proposition 3 is equivalent to ${\displaystyle \frac{a^2+b^2+1}{ab}}\notin S$, namely, ${Tr}_1^m\left({\displaystyle \frac{ab}{a^2+b^2+1}}\right)=0$. Then, by Proposition 3, we have that $f\left(x\right)$ is a CPP over ${\mathbb{F}}_{q^2}$. □

Example 10. 

Let u be a primitive element of ${\mathbb{F}}_{2^5}$. For $q=2^5$, we compute $S=\{y\in {\mathbb{F}}_{2^5}^{\ast }:{Tr}_1^5(1/y)=1\}$ and the forbidden set $\{w+w^{-1}:w\in S\}$. We then enumerate all pairs $(a,b)$ satisfying the conditions in Corollary 10. It was checked via Magma calculation that there are exactly 280 pairs $(a,b)$ satisfying the conditions of Corollary 10. The pair $(a,b)=(1,u^3)$ is one of the admissible pairs. In particular, for $(a,b)=(1,u^3)$, we obtain that

$$f\left(x\right)=(1+u^6)x+u^3\left(x^{249}+x^{776}\right)$$

is a CPP over ${\mathbb{F}}_{2^{10}}$.

Corollary 11. 

Let $q=2^m$ with $m\ge 3$, let $0\le j\le m-3$, and set $k=2^{m-1}-2^j$. For $a,b\in {\mathbb{F}}_q^{\ast }$, the polynomial

$$f\left(x\right)=(a^2+b^2)x+ab\left(x^{1+k(q-1)}+x^{q^2-k(q-1)}\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if the following three conditions are satisfied:

(i) 

Assumption (A1) holds;

(ii) 

${\displaystyle \frac{a}{b}}\notin \{E_{2^{m-1}-2^j-2}(y,1)+E_{2^{m-1}-2^j-1}(y,1):y\in S\}$;

(iii) 

${\displaystyle \frac{a^2+b^2+1}{ab}}\notin \{D_{2^{j+1}+1}(y,1):y\in S\}$.

Proof. 

Take $k=2^{m-1}-2^j$ in Proposition 3. We will simplify condition (iii). It can be checked that

$$k=2^{m-1}-2^j\equiv 2^{m-1}(2^{j+1}+1)(\mathrm{mod}q+1).$$

For $y=z+z^q\in S$ (where $z\in U\setminus \left\{1\right\}$), we get

$$D_k(y,1)=z^k+z^{-k}={\left(z^{2^{j+1}+1}+z^{-2^{j+1}-1}\right)}^{2^{m-1}}=D_{2^{j+1}+1}{(y,1)}^{2^{m-1}}.$$

Note that

$$D_{2^{j+1}+1}(y,1)=\sum _{i=0}^{2^j}{\displaystyle \frac{2^{j+1}+1}{2^{j+1}+1-i}}\left({\displaystyle \genfrac{}{}{0pt}{}{2^{j+1}+1-i}{i}}\right)y^{2^{j+1}+1-2i}.$$

This leads to

$$D_k(y,1)=D_{2^{j+1}+1}(y^{2^{m-1}},1).$$

Clearly, $y\in S$ if and only if $y^{2^{m-1}}\in S$. Thus,

$$\{D_k(y,1):y\in S\}=\{D_{2^{j+1}+1}(y,1):y\in S\}.$$

Then, condition (iii) in Proposition 3 becomes exactly ${\displaystyle \frac{a^2+b^2+1}{ab}}\notin \{D_{2^{j+1}+1}(y,1):y\in S\}$. The desired result follows from Proposition 3. □

Example 11. 

Let $q=2^6$ and $j=1$ in Corollary 11. Then, $k=30$, and the corresponding polynomial is

$$f\left(x\right)=(a^2+b^2)x+ab\left(x^{1891}+x^{2206}\right).$$

For this choice of q and j, we compute the set S and the two forbidden sets determined by $\{E_{k-2}(y,1)+E_{k-1}(y,1):y\in S\}$ and $\{D_{2^{j+1}+1}(y,1):y\in S\}$. All pairs $(a,b)\in {\mathbb{F}}_{2^6}^{\ast }\times {\mathbb{F}}_{2^6}^{\ast }$ satisfying the conditions in Corollary 11 are then enumerated. By Magma computation, there are exactly 2296 pairs $(a,b)$ such that the polynomial defined in Corollary 11 is a CPP over ${\mathbb{F}}_{2^{12}}$. Let u denote the primitive element of ${\mathbb{F}}_{2^6}$. The pair $(a,b)=(1,u^3)$ is one of the admissible pairs. For $(a,b)=(1,u^3)$, we obtain that

$$f\left(x\right)=(1+u^6)x+u^3\left(x^{1891}+x^{2206}\right),$$

is a CPP over ${\mathbb{F}}_{2^{12}}$.

Let $k=2^j+1$ be a Gold exponent in Proposition 3. One can get the following result directly.

Corollary 12. 

Let $q=2^m$ and $a,b\in {\mathbb{F}}_q^{\ast }$. The polynomial

$$f\left(x\right)=x\left(a^2+b^2+ab{D}_{2^j+1}(T\left(x\right),1)\right)$$

is a CPP over ${\mathbb{F}}_{q^2}$ if the following conditions hold:

(i) 

Assumption (A1) holds;

(ii) 

${\displaystyle \frac{a}{b}}\notin \{E_{2^j-1}(y,1)+E_{2^j}(y,1):y\in S\}$;

(iii) 

${\displaystyle \frac{a^2+b^2+1}{ab}}\notin \{D_{2^j+1}(y,1):y\in S\}$.

Example 12. 

Let $q=2^6$ and $j=2$ in Corollary 12. Then, $k=2^j+1=5$. We compute the set S and the two forbidden sets $\{E_{2^j-1}(y,1)+E_{2^j}(y,1):y\in S\}$ and $\{D_{2^j+1}(y,1):y\in S\}$. The admissible pairs $(a,b)$ are obtained by enumerating ${\mathbb{F}}_{2^6}^{\ast }\times {\mathbb{F}}_{2^6}^{\ast }$ and retaining those satisfying the conditions in Corollary 12. It can be verified by Magma calculation that there are exactly 2128 pairs $(a,b)$ such that the polynomial

$$f\left(x\right)=(a^2+b^2)x+ab\left(x^{316}+x^{3781}\right)$$

is a CPP over ${\mathbb{F}}_{2^{12}}$. Let $(a,b)=(1,u)$, where u is a primitive element of ${\mathbb{F}}_{2^6}$. This pair is one of the admissible pairs. From Corollary 12, we get that

$$f\left(x\right)=(1+u^2)x+u\left(x^{316}+x^{3781}\right)$$

is a CPP over ${\mathbb{F}}_{2^{12}}$.

5. Conclusions

In this paper, we investigated CPPs of the form $xh{\left(x^{q-1}\right)}^{q+1}$ over ${\mathbb{F}}_{q^2}$, where $h\left(x\right)=h_1(x+x^q)+h_2(x+x^q)x^k$. Using the trace functions, we constructed several complete permutation trinomials for the case where $k=2^i$. Based on Dickson polynomials, we characterized the complete permutation properties of the polynomials for general $k\ge 2$, and presented some new CPPs.

The main contribution of this work is the reduction of the complete permutation property to explicit nonvanishing conditions on $S\cup \left\{0\right\}$. For $k=2^i$, these conditions are described in terms of the polynomials $K_i\left(y\right)$ and trace functions. For general $k\ge 2$, the same criterion is expressed by Dickson polynomials, through the polynomials $D_k(y,1)$ and $E_k(y,1)$. This gives a uniform approach to deriving sparse CPPs over finite fields of even characteristic.

The resulting constructions contribute to the study of permutation polynomials over finite fields and may be relevant to related problems in combinatorial designs and cryptographic functions. Possible directions for future work include extending the method to other exponent families and to finite fields of odd characteristic, classifying the obtained CPPs up to natural equivalence, determining their inverse polynomials, and studying cryptographic parameters such as differential uniformity and nonlinearity.