© Wuhan University 2025

0 Introduction

Let ${\mathrm{F}}_q$ be the finite field with $q$ elements. A polynomial $f(x)\in {\mathrm{F}}_q[x]$ is called a permutation polynomial if $f$ induces a bijection from ${\mathrm{F}}_q$ to itself. Permutation polynomials have wide applications in coding theory, cryptography and combinatorial designs. We refer the readers to Refs. [1-3] for more details of the recent advances.

Permutation polynomials with a few terms have attracted more attention in recent years for their simple algebraic forms and some special properties. There are several classes of permutation trinomials of the form $x^{r}h(x^{q-\mathrm{1}})$ over ${\mathrm{F}}_{q^{\mathrm{2}}}$ constructed in recent years. Kyureghyan and Zieve^[4] described a class of permutation trinomials having the form $x+\gamma \mathrm{T}{\mathrm{r}}_{q^{\mathrm{2}}/q}(x^{(q^{\mathrm{2}}+\mathrm{1})/\mathrm{4}})$ of ${\mathrm{F}}_{q^{\mathrm{2}}}$, where $\mathrm{T}{\mathrm{r}}_{q^{\mathrm{2}}/q}$ is the trace function from ${\mathrm{F}}_{q^{\mathrm{2}}}$ to ${\mathrm{F}}_q$. We note that this kind of permutation trinomials actually has the form $x(\mathrm{1}+\gamma x^{\frac{q+\mathrm{3}}{\mathrm{4}}(q-\mathrm{1})}+\gamma x^{(\frac{q^{\mathrm{2}}+\mathrm{3}q}{\mathrm{4}}+\mathrm{1})(q-\mathrm{1})})$. Zheng et al^[5] showed a class of permutation trinomials of the form $cx-x^s+x^{sq}$, where $s=\frac{\mathrm{3}{q}^{\mathrm{2}}+\mathrm{2}q-\mathrm{1}}{\mathrm{4}}$, which can be rewritten as $x(c-x^{\frac{\mathrm{3}q+\mathrm{5}}{\mathrm{4}}(q-\mathrm{1})}+x^{(\frac{\mathrm{3}{q}^{\mathrm{2}}+\mathrm{5}q}{\mathrm{4}}+\mathrm{1})(q-\mathrm{1})})$. In Ref. [6], the authors got several classes of more generalized permutation trinomials having similar forms to $x^r(c+x^{(\frac{q+\mathrm{3}}{\mathrm{4}}+k)(q-\mathrm{1})}+x^{(\frac{q^{\mathrm{2}}+\mathrm{3}q}{\mathrm{4}}++k+\mathrm{1})(q-\mathrm{1})})$

and $x^r(c-x^{(\frac{q+\mathrm{3}}{\mathrm{4}}+k)(q-\mathrm{1})}+x^{(\frac{q^{\mathrm{2}}+\mathrm{3}q}{\mathrm{4}}++k+\mathrm{1})(q-\mathrm{1})})$. By using a similar idea to Ref. [6], Lavorante^[7] constructed a few new families of permutation trinomials with the form $x^r(c+x^{s(q-\mathrm{1})}+x^{t(q-\mathrm{1})})$. By using monomial functions on the cosets of a subgroup of ${\mu }_{q+\mathrm{1}}$, Hou and Lavorante^[8] gave a general method to construct permutation polynomials over ${\mathrm{F}}_{q^{\mathrm{2}}}$. Specially, they presented several classes of permutation binomials and trinomials.

On the other hand, permutation quadrinomials also have attracted attention in recent years. Especially, constructing permutation quadrinomials of the form

$f_{r,a,b,c,s,t,u}(x)=x^r(\mathrm{1}+a{x}^{s(q-\mathrm{1})}+b{x}^{t(q-\mathrm{1})}+c{x}^{u(q-\mathrm{1})})\in {\mathrm{F}}_{q^{\mathrm{2}}}[x]$(1)

where $r,s,t$ are integers, attracted great interest recently. Gupa^[9] studied several classes of permutation quadrinomials of the form (1) over ${\mathrm{F}}_{q^{\mathrm{2}}}$ with $\mathrm{C}\mathrm{h}\mathrm{a}\mathrm{r}(F_q)=\mathrm{3,5}$. Tu et al^[10] proposed a class of permutation quadrinomials having the form $x^{\mathrm{3}}(\mathrm{1}+a{x}^{q-\mathrm{1}}+b{x}^{\mathrm{2}(q-\mathrm{1})}+c{x}^{\mathrm{3}(q-\mathrm{1})})$ of ${\mathrm{F}}_{{\mathrm{2}}^{\mathrm{2}m}}$. In Ref. [11], the authors investigated some permutation quadrinomials of ${\mathrm{F}}_{{\mathrm{2}}^{\mathrm{2}m}}$ with the case of $(r,s,t,u)=(\mathrm{1},-\mathrm{1,1},\mathrm{2})$ in (1) under some restrictive conditions. In Ref. [12], the authors provided more classes of permutation quadrinomials of the form (1) in characteristic two. Lavorante^[13] used the Hasse-Weil type theorems to prove the necessary conditions for a polynomial in Ref. [12] to be a permutation polynomial. Ding and Zieve^[14] determined all permutation polynomials over ${\mathrm{F}}_{q^{\mathrm{2}}}$ having the form $x^{r}h(x^{q-\mathrm{1}})$, where, for some $Q$ which is the power of the character of ${\mathrm{F}}_q$, the terms of $h(x)$ have degrees $\{\mathrm{0,1},Q,Q+\mathrm{1}\}$ and $r\equiv Q+\mathrm{1}(\mathrm{m}\mathrm{o}\mathrm{d}q+\mathrm{1})$. The authors in Ref. [15] characterized two classes of permutation quadrinomials over ${\mathrm{F}}_{{\mathrm{2}}^n}$ by using self-reciprocal polynomials. In this paper, motivated by the method in Ref. [6], we continue to construct a new class of permutation quadrinomials of ${\mathrm{F}}_{q^{\mathrm{2}}}$.

This paper is organized as follows: In Section 1, we list some results, which will be used in our paper. In Section 2, by using monomials of ${\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$ and $-{\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$, we construct a class of permutation quadrinomials over ${\mathrm{F}}_{q^{\mathrm{2}}}$ of the form $x^r(\mathrm{1}+a{x}^{s(q-\mathrm{1})}+b{x}^{t(q-\mathrm{1})}+c{x}^{u(q-\mathrm{1})})$ for some integers $r,s,t,u$.

1 Preliminary

The following result was discovered independently by several authors.

Lemma 1^[16-17] Let $r$ be a positive integer. Then $f(x)=x^{r}h(x^{q-\mathrm{1}})\in {\mathrm{F}}_{q^{\mathrm{2}}}[x]$ is a permutation polynomial of ${\mathrm{F}}_{q^{\mathrm{2}}}$ if and only if each of the following is true:

(1) $\mathrm{g}\mathrm{c}\mathrm{d}(r,q-\mathrm{1})=\mathrm{1}$,

(2) $x^{r}h{(x)}^{q-\mathrm{1}}$ permutes $q+\mathrm{1}$-th roots of unity ${\mu }_{q+\mathrm{1}}$.

Specially, by using Lemma 1, constructing permutation polynomials of the form $x^{r}h(x^{q-\mathrm{1}})$ over ${\mathrm{F}}_{q^{\mathrm{2}}}$ translates to finding permutations having the form $x^{r}h{(x)}^{q-\mathrm{1}}$ on the set of $q+\mathrm{1}$-th roots of unity ${\mu }_{q+\mathrm{1}}$. For $x\in {\mu }_{q+\mathrm{1}}$, one has

$x^{r}h{(x)}^{q-\mathrm{1}}=x^r\frac{h{(x)}^q}{h(x)}=x^r\frac{h^q(x^{-\mathrm{1}})}{h(x)}$

where $h^q(x)$ denotes the polynomial obtained $h(x)$ by raising every coefficient to the $q$-th power. Thus to show that $x^{r}h(x^{q-\mathrm{1}})$ permutes ${\mathrm{F}}_{q^{\mathrm{2}}}$, the point is to prove that the rational function $x^r\frac{h^q(x^{-\mathrm{1}})}{h(x)}$ permutes ${\mu }_{q+\mathrm{1}}$.

Let $d|q+\mathrm{1}$ with $d\ge \mathrm{2}$ be a positive integer and $\xi$ be a primitive $d$-th root of unity. We make some denotations: $S_{\mathrm{0}}={\mu }_{\frac{q+\mathrm{1}}{d}}$ and $S_i={\xi }^i{S}_{\mathrm{0}}$ for $\mathrm{1}\le i\le d-\mathrm{1}$. It is easy to imply that ${\mu }_{q+\mathrm{1}}={\displaystyle \underset{i=\mathrm{0}}{\overset{d-\mathrm{1}}{\cup }}}{S}_i$ and $S_i\bigcap S_j=\mathrm{\varnothing }$ for $\mathrm{0}\le i\ne j\le d-\mathrm{1}$.

For $g(x)\in F_{q^{\mathrm{2}}}[x]$, if $g(x)$ is a monomial on each subset of ${\mu }_{q+\mathrm{1}}$, then by using the piecewise method, we can easily determine the permutational property of $g(x)$ on ${\mu }_{q+\mathrm{1}}$ in the following lemma.

Lemma 2^[6] Let $\frac{q+\mathrm{1}}{d}$ be a positive integer and $A_i\in {\mu }_{q+\mathrm{1}}$ for $\mathrm{0}\le i\le d-\mathrm{1}$. For $g(x)\in F_{q^{\mathrm{2}}}[x]$, if

$g(x)=A_i{x}^{r_i},forx\in S_i.$

Then $g(x)$ permutes ${\mu }_{q+\mathrm{1}}$ if and only if each of the following is true:

$(\mathrm{1})\mathrm{g}\mathrm{c}\mathrm{d}(r_i,\frac{q+\mathrm{1}}{d})=\mathrm{1},for\mathrm{0}\le i\le d-\mathrm{1};$

$(\mathrm{2})A_i{x_i}^{r_i}\ne A_j{x_j}^{r_j}for{x}_i\in S_{i}and{x}_j\in S_j.$

Lemma 2   provides an approach to study the permutational property of $x^r\frac{h^q(x^{-\mathrm{1}})}{h(x)}$ on ${\mu }_{q+\mathrm{1}}$ via monomials on the subsets $S_i$. In Refs. [4,5,7], the authors used the case $d=\mathrm{2}$ in Lemma 2 to construct a few classes of permutation trinomials of ${\mathrm{F}}_{q^{\mathrm{2}}}$. By using the cases $d=\mathrm{3}$, the authors obtained several kinds of permutation trinomials of ${\mathrm{F}}_{q^{\mathrm{2}}}$ in Refs. [16,18].

2 Main Results

Motivated by the method in Ref. [6], we characterize several classes of permutation quadrinomials over ${\mathrm{F}}_{q^{\mathrm{2}}}$ in this section.

Theorem 1   Let $q$ be a prime power with $q\equiv \mathrm{1}(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{8})$, and $a,b,c\in {\mathrm{F}}_{q^{\mathrm{2}}}$ satisfy ${(a+b+c)}^{\frac{q+\mathrm{1}}{\mathrm{2}}}=\mathrm{1}$ and ${(b-c-a)}^{\frac{q+\mathrm{1}}{\mathrm{2}}}=\mathrm{1}$. Let $r$ be a positive integer and $k$ be an even integer. Then $f(x)=x^r(\mathrm{1}+a{x}^{(\frac{q+\mathrm{3}}{\mathrm{4}}+k)(q-\mathrm{1})}+b{x}^{(\frac{\mathrm{3}q+\mathrm{5}}{\mathrm{4}}+k)(q-\mathrm{1})}$

$+c{x}^{(\frac{\mathrm{3}{q}^{\mathrm{2}}+\mathrm{5}q}{\mathrm{4}}+k+\mathrm{1})(q-\mathrm{1})})$ permutes ${\mathrm{F}}_{q^{\mathrm{2}}}$ if and only if $\mathrm{g}\mathrm{c}\mathrm{d}(r,q-\mathrm{1})=\mathrm{1}$ and $\mathrm{g}\mathrm{c}\mathrm{d}(\mathrm{2}r-\mathrm{2}k-\mathrm{1},\frac{q+\mathrm{1}}{\mathrm{2}})=\mathrm{1}$.

Proof   It follows from Lemma 1 that $f(x)$ permutes ${\mathrm{F}}_{q^{\mathrm{2}}}$ if and only if $\mathrm{g}\mathrm{c}\mathrm{d}(r,q-\mathrm{1})=\mathrm{1}$ and $g(x)=x^r(\mathrm{1}+a{x}^{u+k}+b{x}^{v+k}+c{x}^{qv+k+\mathrm{1}}{)}^{q-\mathrm{1}}$ permutes ${\mu }_{q+\mathrm{1}}$, where $u=\frac{q+\mathrm{3}}{\mathrm{4}},v=\frac{\mathrm{3}q+\mathrm{5}}{\mathrm{4}}$.

In the following, we claim that if $\mathrm{g}\mathrm{c}\mathrm{d}(r,q-\mathrm{1})=\mathrm{1}$, then $g(x)$ permutes ${\mu }_{q+\mathrm{1}}$ if and only if $\mathrm{g}\mathrm{c}\mathrm{d}(\mathrm{2}r-\mathrm{2}k-\mathrm{1},\frac{q+\mathrm{1}}{\mathrm{2}})=\mathrm{1}$.

We divide ${\mu }_{q+\mathrm{1}}$ into two subsets ${\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$ and $-{\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$, and consider the following cases. For $x$ in ${\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$, it is easy to check that $x^u=x^v=x^{qv+\mathrm{1}}=x^{\mathrm{1}-v}$. One has $g(x)=x^r(\mathrm{1}+(a+b+c)x^{u+k}{)}^{q-\mathrm{1}}$.

Since $q\equiv \mathrm{1}(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{8})$, we have that $\frac{q+\mathrm{1}}{\mathrm{2}}$ is odd. Then by ${(a+b+c)}^{\frac{q+\mathrm{1}}{\mathrm{2}}}=\mathrm{1}$, we deduce that the equation $\mathrm{1}+(a+b+c)x^{u+k}=\mathrm{0}$ has no roots in ${\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$. Furthermore,

$g(x)=x^r\frac{\mathrm{1}+{(a+b+c)}^q{x}^{-u-k}}{\mathrm{1}+(a+b+c)x^{u+k}}={(a+b+c)}^q{x}^{r-k-u}\frac{\mathrm{1}+\frac{x^{k+u}}{{(a+b+c)}^q}}{\mathrm{1}+(a+b+c)x^{u+k}}.$

By using ${(a+b+c)}^{\frac{q+\mathrm{1}}{\mathrm{2}}}=\mathrm{1}$, $g(x)$ can be simplified as $\frac{\mathrm{1}}{a+b+c}{x}^{r-k-u}$. Since $x\in {\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$ can be written as $y^{\mathrm{2}}$ for $y\in {\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$ and $\mathrm{2}u\equiv \mathrm{1}(\mathrm{m}\mathrm{o}\mathrm{d}\frac{q+\mathrm{1}}{\mathrm{2}})$, thus $g(x)$ can be rewritten as $\frac{\mathrm{1}}{a+b+c}{y}^{\mathrm{2}r-\mathrm{2}k-\mathrm{1}}$.

For $x$ in $-{\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$, one has $x^u=-x^v=x^{qv+\mathrm{1}}=x^{\mathrm{1}-v}$. Then

$\begin{array}{l}g(x)=x^r(\mathrm{1}+a{x}^{u+k}+b{x}^{v+k}+c{x}^{qv+\mathrm{1}+k}{)}^{q-\mathrm{1}}\\ =x^r(\mathrm{1}+(a-b+c)x^{u+k}{)}^{q-\mathrm{1}}.\end{array}$

Since $q\equiv \mathrm{1}(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{8})$, we have that $u$ and $\frac{q+\mathrm{1}}{\mathrm{2}}$ are odd, thus $u+k$ is odd. Then by ${(b-c-a)}^{\frac{q+\mathrm{1}}{\mathrm{2}}}=\mathrm{1}$, we know that $\mathrm{1}+(a-b+c)x^{u+k}\ne \mathrm{0}$ for $x\in -{\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$. Thus

$g(x)=x^r\frac{\mathrm{1}+{(a-b+c)}^q{x}^{-u-k}}{\mathrm{1}+(a-b+c)x^{u+k}}={(a-b+c)}^q{x}^{r-k-u}\frac{\mathrm{1}+\frac{x^{k+u}}{{(a-b+c)}^q}}{\mathrm{1}+(a-b+c)x^{u+k}}.$

Since ${(a-b+c)}^q=\frac{\mathrm{1}}{a-b+c}$ and $\frac{\mathrm{1}}{{(a-b+c)}^q}=a-b+c$, then $g(x)=\frac{\mathrm{1}}{a-b+c}{x}^{r-k-u}$. For $x\in -{\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$, there exists $y\in {\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$ such that $x$ can be presented by $-y^{\mathrm{2}}$. Then $g(x)=\frac{\mathrm{1}}{a-b-c}{y}^{\mathrm{2}r-\mathrm{2}k-\mathrm{1}}$.

Note that $\frac{\mathrm{1}}{a+b+c}\in {\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$ and $\frac{\mathrm{1}}{a-b+c}\in -{\mu }_{\frac{q+\mathrm{1}}{\mathrm{2}}}$. Then it follows from Lemma 2 that $g(x)$ permutes ${\mu }_{q+\mathrm{1}}$ if and only if $\mathrm{g}\mathrm{c}\mathrm{d}(\mathrm{2}r-\mathrm{2}k-\mathrm{1},\frac{q+\mathrm{1}}{\mathrm{2}})=\mathrm{1}$. Namely, the claim is true.

Therefore, we can conclude that $f(x)$ permutes ${\mathrm{F}}_{q^{\mathrm{2}}}$ if and only if $\mathrm{g}\mathrm{c}\mathrm{d}(r,q-\mathrm{1})=\mathrm{1}$ and $\mathrm{g}\mathrm{c}\mathrm{d}(\mathrm{2}r-\mathrm{2}k-\mathrm{1},\frac{q+\mathrm{1}}{\mathrm{2}})=\mathrm{1}.$

We complete the proof of Theorem 1.

Similarly, we can get the following results, and we omit their detailed proofs.

Theorem 2   Let $q$ be a prime power with $q\equiv \mathrm{1}(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{8})$, and $a,b,c\in {\mathrm{F}}_{q^{\mathrm{2}}}$ satisfy ${(a+b+c)}^{\frac{q+\mathrm{1}}{\mathrm{2}}}=\mathrm{1}$ and ${(b-a+c)}^{\frac{q+\mathrm{1}}{\mathrm{2}}}=\mathrm{1}$. Let $r$ be a positive integer and $k$ be an even integer. Then $f(x)=x^r(\mathrm{1}+a{x}^{(\frac{q+\mathrm{3}}{\mathrm{4}}+k)(q-\mathrm{1})}+b{x}^{(\frac{\mathrm{3}q+\mathrm{5}}{\mathrm{4}}+k)(q-\mathrm{1})}$

$+c{x}^{(\frac{q^{\mathrm{2}}+\mathrm{3}q}{\mathrm{4}}+k+\mathrm{1})(q-\mathrm{1})})$ permutes ${\mathrm{F}}_{q^{\mathrm{2}}}$ if and only if $\mathrm{g}\mathrm{c}\mathrm{d}(r,q-\mathrm{1})=\mathrm{1}$ and $\mathrm{g}\mathrm{c}\mathrm{d}(\mathrm{2}r-\mathrm{2}k-\mathrm{1},\frac{q+\mathrm{1}}{\mathrm{2}})=\mathrm{1}$.

Theorem 3   Let $q$ be a prime power with $q\equiv \mathrm{1}(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{8})$, and $a,b,c\in {\mathrm{F}}_{q^{\mathrm{2}}}$ satisfy ${(a+b+c)}^{\frac{q+\mathrm{1}}{\mathrm{2}}}=\mathrm{1}$ and ${(b-a-c)}^{\frac{q+\mathrm{1}}{\mathrm{2}}}=\mathrm{1}$. Let $r$ be a positive integer and $k$ be an even integer. Then $f(x)=x^r(\mathrm{1}+a{x}^{(\frac{q+\mathrm{3}}{\mathrm{4}}+k)(q-\mathrm{1})}+b{x}^{(\frac{q^{\mathrm{2}}+\mathrm{3}q}{\mathrm{4}}+k)(q-\mathrm{1})}$

$+c{x}^{(\frac{\mathrm{3}{q}^{\mathrm{2}}+\mathrm{5}q}{\mathrm{4}}+k+\mathrm{1})(q-\mathrm{1})})$ permutes ${\mathrm{F}}_{q^{\mathrm{2}}}$ if and only if $\mathrm{g}\mathrm{c}\mathrm{d}(r,q-\mathrm{1})=\mathrm{1}$ and $\mathrm{g}\mathrm{c}\mathrm{d}(\mathrm{2}r-\mathrm{2}k-\mathrm{1},\frac{q+\mathrm{1}}{\mathrm{2}})=\mathrm{1}$.

Theorem 4   Let $q$ be a prime power with $q\equiv \mathrm{1}(\mathrm{m}\mathrm{o}\mathrm{d}\mathrm{8})$, and $a,b,c\in {\mathrm{F}}_{q^{\mathrm{2}}}$ satisfy ${(a+b+c)}^{\frac{q+\mathrm{1}}{\mathrm{2}}}=\mathrm{1}$ and ${(a+b-c)}^{\frac{q+\mathrm{1}}{\mathrm{2}}}=\mathrm{1}$. Let $r$ be a positive integer and $k$ be an even integer. Then $f(x)=x^r(\mathrm{1}+a{x}^{(\frac{\mathrm{3}q+\mathrm{5}}{\mathrm{4}}+k)(q-\mathrm{1})}+b{x}^{(\frac{q^{\mathrm{2}}+\mathrm{3}q}{\mathrm{4}}+k)(q-\mathrm{1})}$

$+c{x}^{(\frac{\mathrm{3}{q}^{\mathrm{2}}+\mathrm{5}q}{\mathrm{4}}+k+\mathrm{1})(q-\mathrm{1})})$ permutes ${\mathrm{F}}_{q^{\mathrm{2}}}$ if and only if $\mathrm{g}\mathrm{c}\mathrm{d}(r,q-\mathrm{1})$

$=\mathrm{1}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{g}\mathrm{c}\mathrm{d}(\mathrm{2}r-\mathrm{2}k-\mathrm{1},\frac{q+\mathrm{1}}{\mathrm{2}})=\mathrm{1}.$

References