Explicit characterization of two classes of regular bent functions

Abstract

This paper considers two classes of p-ary functions studied by Li et al. (IEEE Trans Inf Theory 59(3):1818–1831, 2013). The first class of p-ary functions is of the form

$$\begin{aligned} f(x)=Tr^n_1\left( a x^{l(q-1)}+b x^{\left( l+\frac{q+1}{2}\right) (q-1)}\right) +\epsilon x^{\frac{q^2-1}{2}}. \end{aligned}$$

Another class of p-ary functions is of the form

$$\begin{aligned} f(x)={\left\{ \begin{array}{ll} \sum ^{q-1}_{i=0} Tr^n_1(a x^{(ri+s)(q-1)})+\epsilon x^{\frac{q^2-1}{2}},&{} x\ne 0,\\ f(0),&{} x=0. \end{array}\right. } \end{aligned}$$

We generalize Li et al.’s results, give necessary conditions for two classes of bent functions, and present more explicit characterization of these regular bent functions for different cases.

1 Introduction

Introduced by Rothaus, Boolean bent functions as functions from ${\mathbb {F}}_{2}^n$ or ${\mathbb {F}}_{2^n}$ to ${\mathbb {F}}_2$ have important applications in cryptography [2], coding theory [3, 6, 8] and sequences [21]. As a class of Boolean functions with maximal Hamming distance to the set of all affine functions, bent functions can be used to construct highly nonlinear cryptographic functions and attract much attention. Many research papers focus on the characterization and construction of monomial bent functions, binomial bent functions and quadratic bent functions [1, 4, 5, 7, 9, 16, 19, 20, 22,23,24]. Boolean bent functions were generalized to the notation of functions over an arbitrary finite field in [15]. It is elusive to completely classify bent functions. The characterization of bent functions over finite fields of odd characteristic is more complicated than that of Boolean bent functions. Several results can be found in [11, 12].

Let p be an odd prime and m be an integer. Let $n=2m$ and $q=p^m$. Let $Tr^n_1(\cdot )$ be the trace function from ${\mathbb {F}}_{q^2}$ to ${\mathbb {F}}_{p}$. Helleseth and Kholosha [10] studied monomial functions of the form

$$\begin{aligned} f_{a,r}(x)=Tr^n_1(ax^{r(q-1)}), \end{aligned}$$

where $a\in {\mathbb {F}}_{q^2}$ and $gcd(r,q+1)=1$. They proved that $f_{a,r}(x)$ is bent if and only if the Kloosterman sum $K_m(a^{q+1})$ on ${\mathbb {F}}_{p^m}$ is zero.

Jia et al. [13] considered binomial functions of the form

$$\begin{aligned} f_{a,b,r}(x)=Tr^n_1(ax^{r(q-1)})+bx^{\frac{q^2-1}{2}},a\in {\mathbb {F}}_{q^2},b\in {\mathbb {F}}_{p}, \end{aligned}$$

where $gcd(r,q+1)=1$. By Kloosterman sums, they presented the characterization of bentness for $f_{a,b,r}$. For $p=3$ or $q\equiv 3 \mod 4$, they proved that $f_{a,b,r}$ is bent if and only if $K_m(a)=1-\frac{1}{\cos (\frac{2\pi b}{p})}$. Zheng et al. [25] generalized Jia et al.’s result to the case $q\equiv 1 \mod 4$, i.e., $f_{a,b,r}$ is bent if and only if $K_m(a)=1-\frac{1}{\cos (\frac{2\pi b}{p})}$. Further, when $q\equiv 7 \mod 8$, r is even and $gcd(\frac{r}{2},q+1)=1$, Zheng et al. proved that $f_{a,b,r}(x)=Tr^n_1(ax^{r(q-1)})+bx^{\frac{q^2-1}{2}}(a\in {\mathbb {F}}_{q^2},b\in {\mathbb {F}}_{p})$ is not bent.

Li et al. [17] considered trinomial functions of the form

$$\begin{aligned} f(x)=Tr^n_1\left( a x^{l(q-1)}+b x^{\left( l+\frac{q+1}{2}\right) (q-1)}\right) +\epsilon x^{\frac{q^2-1}{2}}, \end{aligned}$$

(1)

where $a,b\in {\mathbb {F}}_{q^2}$ and $\epsilon \in {\mathbb {F}}_{p}$. When $gcd(l,q+1)=1$, they presented the relation between the bentness of f(x) and Kloosterman sums $K_m((a+b)^{q+1})$,$K_m((a-b)^{q+1})$ for different $a+b$ and $a-b$. Further, they considered another class of functions with multiple terms of the form

$$\begin{aligned} f(x)={\left\{ \begin{array}{ll} \sum ^{q-1}_{i=0} Tr^n_1(a x^{(ri+s)(q-1)})+\epsilon x^{\frac{q^2-1}{2}},&{} x\ne 0,\\ f(0),&{} x=0, \end{array}\right. } \end{aligned}$$

(2)

where $a\in {\mathbb {F}}_{q^2}$,$\epsilon \in {\mathbb {F}}_{p}$. When $gcd(r,q+1)=1$ and $gcd(s-r,q+1)=1$, they used Kloosterman sums to characterize regular bent function f(x) for different $-a$.

Based on results of Li et al. [17], this paper considers bent functions defined in (1) and (2), studies parameters of these bent functions, and presents more results of the characterization of bent functions in (1) and (2) for more explicit parameters. For bent functions defined in (1), Li et al. [17] gave the characterization for the case $gcd(l,q+1)=1$. We have some necessary conditions for these bent functions: $gcd(l,q+1)=1$, or, $gcd(l,q+1)=2$ and $q\equiv 1\mod 4$. When $p\ge 5$, $gcd(l,q+1)=1$ and $a-b, a+b$ are quadratic residues, p-ary functions defined in (1) are not bent. Further, we present explicit characterization of bentness for these functions defined in (1) for the case $gcd(l,q+1)=1$ and the case $gcd(l,q+1)=2$, $q\equiv 1\mod 4$. For bent functions defined in (2), Li et al. [17] gave the characterization for the case $gcd(r,q+1)=1$. We study their bentness for cases $gcd(r,q+1)=1$ and $gcd(r,q+1)=2$. When $p\ge 5$ and $\epsilon =0$, f(x) is not bent. When $\epsilon \ne 0$, we give necessary conditions for regular bent functions and present the characterization of these bent functions for the case $gcd(s-r,q+1)=1$ and the case $gcd(s-r,q+1)=2, q\equiv 1\mod 4$. Our work generalizes results of Li et al. [17] and Zheng et al. [25].

This paper is organized as follows: Sect. 2 introduces some notations and results on exponential sums. Section 3 considers two classes of regular bent functions, gives necessary conditions for these regular bent functions, and characterizes these regular bent functions for different cases. Section 4 makes a conclusion.

2 Preliminaries

2.1 Regular bent functions

Throughout this paper, let p be an odd prime, m, n be positive integers and $n=2m$. Let ${\mathbb {F}}_{p^n}$ be a finite field with $p^n$ elements and ${\mathbb {F}}_{p^n}^*$ be the multiplicative group composed of all nonzero elements in ${\mathbb {F}}_{p^n}$. Let k|n and $Tr^n_k$ be the trace function from ${\mathbb {F}}_{p^n}$ to ${\mathbb {F}}_{p^k}$

$$\begin{aligned} Tr^n_k(x)=x+x^{p^k}+\cdots +x^{p^{n-k}}. \end{aligned}$$

Let $q=p^m$. For any $x\in {\mathbb {F}}_{q^2}^*$, there exists a unique factorization $x=y*\xi ^{i}$, where $y\in {\mathbb {F}}_{q}^*$, $\xi$ is a primitive element of ${\mathbb {F}}_{q^2}$ and $0\le i\le q$. Let $U=\{\xi ^0,\xi ^{(q-1)},\ldots ,\xi ^{(q-1)q}\}$, $U_0=U^2=\{u^2:u\in U\}$ and $U_1=U{\setminus } U_0$. Sets of quadratic residues and quadratic non-residues in ${\mathbb {F}}_{q^2}^*$ are defined as $\mathcal {C}_0=\{x^2:x\in {\mathbb {F}}_{q^2}^*\}$, $\mathcal {C}_1=\{\xi x^2:x\in {\mathbb {F}}_{q^2}^*\}$ respectively. Then ${\mathbb {F}}_{q^2}^*=\mathcal {C}_0\bigcup \mathcal {C}_1$ and $\mathcal {C}_0\bigcap \mathcal {C}_1=\emptyset$. Define $\mathcal {C}_0^{+}=\{x\in \mathcal {C}_0:Tr^m_1(x^{\frac{p^{m}+1}{2}})\ne 0\}$.

A p-ary function is a map from ${\mathbb {F}}_{p^n}$ to ${\mathbb {F}}_{p}$. The Walsh transform of a p-ary function f(x) over ${\mathbb {F}}_{p^n}$ is defined by

$$\begin{aligned} W_f(\lambda )=\sum _{x\in {\mathbb {F}}_{p^n}}w^{f(x)-Tr^n_1(\lambda x)}, \end{aligned}$$

where $w=e^{2\pi \sqrt{-1}/p}$ and $\lambda \in {\mathbb {F}}_{p^n}.$

A p-ary function f(x) is called a p-ary bent function if $|W_f(\lambda )|^2=p^n$ for any $\lambda \in {\mathbb {F}}_{p^n}$. A p-ary bent function f(x) is regular if there exists some p-ary function $f^*(\lambda )$ satisfying $W_f(\lambda )=p^{\frac{n}{2}}w^{f^*(\lambda )}$ for any $\lambda \in {\mathbb {F}}_{p^n}$. The function $f^*(\lambda )$ is called the dual of f(x). The dual of a p-ary bent function is also bent.

Let f(x) be a p-ary function with Dillon exponents of the form

$$\begin{aligned} f(x)={\left\{ \begin{array}{ll}\sum _{i=0}^{q-1} Tr^n_1(a_ix^{i(q-1)})+bx^{\frac{q^2-1}{2}},&{} x\ne 0,\\ f(0),&{} x=0, \end{array}\right. } \end{aligned}$$

(3)

where $n=2m$, $q=p^m$, $a_i\in {\mathbb {F}}_{p^n}$, and $b\in {\mathbb {F}}_p$. The characterization of bentness for f(x) is given in the following lemma [17, Theorem 1].

Lemma 1

Let f(x) be a p-ary function defined in (3). Then f(x) is bent if and only if $\varLambda _f=w^{f(0)}$, where

$$\begin{aligned} \varLambda _f=\sum _{u\in U}w^{\sum _{i=0}^{q-1} Tr^n_1(a_iu^{q-1})+bu^{\frac{q+1}{2}}}. \end{aligned}$$

Further, if f(x) is bent, then it is also regular bent.

2.2 Exponential sums

For $a\in {\mathbb {F}}_q$, the Kloosterman sum $K_m(a)$ of a [14, 18] is defined by

$$\begin{aligned} K_m(a)=\sum _{x\in {\mathbb {F}}_q}w^{Tr^m_1\left( ax+\frac{1}{x}\right) }, \end{aligned}$$

where $\frac{1}{0}=0$ for $x=0$. Since $\overline{K_m(a)}=\sum _{x\in {\mathbb {F}}_q}w^{-Tr^m_1(ax+\frac{1}{x})}=K_m(a)$, then $K_m(a)$ is a real number. Some notations are defined below.

$$\begin{aligned} I&={\left\{ \begin{array}{ll} \frac{(-1)^{\frac{3m}{2}}p^{\frac{m}{2}}}{2}, &{}p\equiv 3 \mod 4\text{; }\\ \frac{(-1)^{m}p^{\frac{m}{2}}}{2}, &{}\text{ otherwise. } \end{array}\right. }\\ Q(a)&=2Tr^m_1\left( a^{\frac{p^m+1}{2}}\right) ,~~a\in \mathcal {C}_0;\\ R(a)&=\frac{1-K_m(a^{p^m+1})}{2},~~a\in {\mathbb {F}}_{q^2}. \end{aligned}$$

Obviously, if $q\equiv 1 \mod 4$, then I is a real number. If $q\equiv 3 \mod 4$, then I is a pure imaginary number.

The following result on exponential sums is useful [13, Lemma 7].

Proposition 1

Let $a\in {\mathbb {F}}_{q^2}^*$. Let $S_i(a)=\sum _{u\in U_i}w^{Tr^n_1(au)}(i=0, 1)$. Then,

$$\begin{aligned} S_0(a)=\sum _{u\in U_0}w^{Tr^n_1(au)}= {\left\{ \begin{array}{ll} R(a)+I(w^{Q(a)}-w^{-Q(a)}), &{} a\in \mathcal {C}_0^{+},\\ R(a), &{}otherwise, \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} S_1(a)=\sum _{u\in U_1}w^{Tr^n_1(au)}= {\left\{ \begin{array}{ll} R(a)-I(w^{Q(a)}-w^{-Q(a)}), &{} a\in \mathcal {C}_0^{+},\\ R(a), &{}otherwise. \end{array}\right. } \end{aligned}$$

Remark

If $a\in \mathcal {C}_0{\setminus } \mathcal {C}_0^+$, then we have $Q(a)=0$. Hence, Proposition 1 still holds if $\mathcal {C}_0^+$ is replaced by $\mathcal {C}_0$. From Proposition 1, we have $S_i(a)=\sum _{u\in U_i}w^{Tr^n_1(au)}$ is an imaginary number if and only if $q\equiv 1 \mod 4$ and $a\in \mathcal {C}_0^{+}$.

Some results on $S_i(a)$ are given below.

Proposition 2

Let $q\equiv 1 \mod 4$, $a\in \mathcal {C}_0$, and $N^i_j(a)=\#\{v\in U_i:Tr^m_1\Big (a^{\frac{q+1}{2}}(v+\frac{1}{v})\Big )=j\}(i=0,1)$. Then,

$$\begin{aligned} N^i_j(a)\equiv {\left\{ \begin{array}{ll}0 \mod 2, &{}j\ne Tr^m_1\left( 2(-1)^ia^{\frac{q+1}{2}}\right) ,\\ 1 \mod 2, &{} j = Tr^m_1\left( 2(-1)^ia^{\frac{q+1}{2}}\right) . \end{array}\right. } \end{aligned}$$

Further, If $a^{\frac{q^2-1}{4}}=1$, then $S_i(a)=\sum _{i=0}^{p-1} N^i_{j}(a) w^j$. If $a^{\frac{q^2-1}{4}}=-1$, then $S_i(a)=\sum _{i=0}^{p-1} N^{i+1}_{j}(a) w^j$.

Proof

Note that $a=a^{\frac{q+1}{2}}\cdot a^{-\frac{q-1}{2}}$. Since $a\in \mathcal {C}_0$, then $a^{\frac{q+1}{2}} \in {\mathbb {F}}_{q}$ and $a^{-\frac{q-1}{2}} \in U$. We have

$$\begin{aligned} S_i(a)=\sum _{v\in U_i} w^{Tr^n_1(a v)} =\sum _{v\in U_i} w^{Tr^n_1\left( a^{\frac{q+1}{2}}\cdot a^{-\frac{q-1}{2}}v\right) }. \end{aligned}$$

If $a^{\frac{q^2-1}{4}}=1$, then $a^{-\frac{q-1}{2}}\in U_0$. Hence,

$$\begin{aligned} S_i(a)=\sum _{v\in U_i} w^{Tr^n_1\left( a^{\frac{q+1}{2}}v\right) } =\sum _{v\in U_i} w^{Tr^m_1\left( a^{\frac{q+1}{2}}\left( v+\frac{1}{v}\right) \right) } =\sum _{i=0}^{p-1} N^i_{j}(a) w^j, \end{aligned}$$

where $N^i_j(a)=\#\{v\in U_i:Tr^m_1(a^{\frac{q+1}{2}}(v+\frac{1}{v}))=j\}$. If $v\ne \pm 1$, then both v and $\frac{1}{v}$ lie in $\{v\in U_i:Tr^m_1(a^{\frac{q+1}{2}}(v+\frac{1}{v}))=j\}$ and $v\ne \frac{1}{v}$. Since $q \equiv 1 \mod 4$, then $-1\in U_1$, $(-1)^i\in U_i$, and $(-1)^{i+1}\not \in U_i$. Hence, we have $N^i_{Tr^m_1(2(-1)^ia^{(q+1)/2})}(a)$ is odd and $N^i_j(a)$ is even for $j\ne Tr^m_1(2(-1)^ia^{\frac{q+1}{2}})$.

If $a^{\frac{q^2-1}{4}}=-1$, then $a^{-\frac{q-1}{2}}\in U_1$ and

$$\begin{aligned} S_i(a)=\sum _{v\in U_{i+1}} w^{Tr^n_1\left( a^{\frac{q+1}{2}}v\right) } =\sum _{v\in U_{i+1}} w^{Tr^m_1\left( a^{\frac{q+1}{2}}\left( v+\frac{1}{v}\right) \right) } =\sum _{i=0}^{p-1} N^{i+1}_{j}(a) w^j, \end{aligned}$$

where $N^{i+1}_j(a)=\#\{v\in U_{i+1}:Tr^m_1(a^{\frac{q+1}{2}}(v+\frac{1}{v}))=j\}$.

Hence, this proposition follows. $\square$

Remark

From Proposition 2, for $q\equiv 1 \mod 4$ and any $a\in \mathcal {C}_0$, there exist non-negative integers $c_j$ satisfying $S_i(a)=c_0 +c_1 w+\cdots +c_{p-1}w^{p-1}$ and $\#\{i:2\not \mid c_i,i=0,1,\ldots , p-1\}=1$.

From a similar proof as that of Proposition 2, we have the following proposition.

Proposition 3

Let $q\equiv 3 \mod 4$, $a\in \mathcal {C}_0$, and $N^i_j(a)=\#\{v\in U_i:Tr^m_1(a^{\frac{q+1}{2}}(v+\frac{1}{v}))=j\}$. Then, $S_i(a)=\sum _{i=0}^{p-1} N^i_{j}(a) w^j$ and $N^1_j(a)\equiv 0 \mod 2$. Further, if $a\in \mathcal {C}_0^+$, then

$$\begin{aligned} N^0_j(a)\equiv {\left\{ \begin{array}{ll}0 \mod 2, &{}j\ne \pm Tr^m_1\left( 2a^{\frac{q+1}{2}}\right) ,\\ 1 \mod 2, &{} j = \pm Tr^m_1\left( 2a^{\frac{q+1}{2}}\right) .\end{array}\right. } \end{aligned}$$

If $a\in \mathcal {C}_0{\setminus } \mathcal {C}_0^+$, then $N^0_j(a)\equiv 0 \mod 2.$

Further, we can have the following lemma.

Lemma 2

Let $a\in {\mathbb {F}}_{q^2}^*$ and $N_j=\# \{u\in U:Tr^n_1(au)=j\}$, then $\sum _{u\in U} w^{Tr^n_1(a u)}= \sum _{i=0}^{p-1} N_i w^i$. Further, If $a\not \in \mathcal {C}_0^+$, then all $N_i$ are even. If $a\in \mathcal {C}_0^+$, then

$$\begin{aligned} N_i\equiv {\left\{ \begin{array}{ll}0 \mod 2, &{}i\ne \pm Tr^n_1\left( a^{\frac{q+1}{2}}\right) ,\\ 1 \mod 2,&{} i= \pm Tr^n_1\left( a^{\frac{q+1}{2}}\right) .\end{array}\right. } \end{aligned}$$

The following two lemmas are useful to obtain our results.

Lemma 3

Let p be an odd prime, and $c_0', \ldots , c_{p-1}'$ be integers such that $p>\# \{i\in \{0,\ldots , p-1\}: c_i'\ne 0 \}$ and $\sum _{i=0}^{p-1} c_i' \in \{1,-1\}$. Then, for any integers $c_i ( i=0, \ldots , p-1)$ and positive integer $d>1$,

$$\begin{aligned} d\left( \sum _{i=0}^{p-1} c_i w^i\right) \ne \sum _{i=0}^{p-1} c_i' w^i. \end{aligned}$$

Proof

Suppose that $d(\sum _{i=0}^{p-1} c_i w^i) = \sum _{i=0}^{p-1} c_i' w^i$. Then, $\sum _{i=0}^{p-1} (d c_i -c_i') w^i=0$. The minimal polynomial of w over the rational field is $w^{p-1}+w^{p-2}+\cdots +w+1=0$. Thus, we have

$$\begin{aligned} dc_0-c_0'=\cdots =dc_{p-1} -c_{p-1}'. \end{aligned}$$

Since $p>\# \{i\in \{0,\ldots , p-1\}: c_i'\ne 0 \}$, there exists $i_0\in \{0, \ldots , p-1\}$ such that $c_{i_0}'=0$. Thus, $d | c_i'$ for any $i\in \{0, \ldots , p-1\}$. As a result, $d | \sum _{i=0}^{p-1} c_i' = \pm 1$, which conflicts with $d>1$. It completes the proof. $\square$

Lemma 4

Let p be an odd prime, and $c_0, c_0', \ldots , c_{p-1},c_{p-1}'$ be integers such that $\sum _{i=0}^{p-1} c_i \equiv 0 \pmod 2$ and $\sum _{i=0}^{p-1} c_i' \equiv 1 \pmod 2$. If $p> \# \{i\in \{0,\ldots , p-1\}: c_i\equiv 1 \pmod 2 \}+ \# \{i\in \{0,\ldots , p-1\}: c_i'\equiv 1 \pmod 2 \}$, then $\sum _{i=0}^{p-1} c_i w^i \ne \sum _{i=0}^{p-1} c'_i w^i$.

Proof

Suppose that $\sum _{i=0}^{p-1} c_i w^i = \sum _{i=0}^{p-1} c_i' w^i$. Then, $\sum _{i=0}^{p-1} ( c_i -c_i') w^i=0$. The minimal polynomial of w over the rational field is $w^{p-1}+w^{p-2}+\cdots +w+1=0$. Thus, we have

$$\begin{aligned} c_0-c_0'=\cdots =c_{p-1} -c_{p-1}'. \end{aligned}$$

Since $p> \# \{i\in \{0,\ldots , p-1\}: c_i\equiv 1 \pmod 2 \}+ \# \{i\in \{0,\ldots , p-1\}: c_i'\equiv 1 \pmod 2 \}$, there exists $i_0\in \{0, \ldots , p-1\}$ such that $c_{i_0}\equiv c_{i_0}' \equiv 0 \pmod 2$. Thus, $c_i-c_{i}' \equiv 0 \pmod 2$ for any $i\in \{0, \ldots , p-1\}$. As a result, $\sum _{i=0}^{p-1} c_i- \sum _{i=0}^{p-1} c_i' \equiv 0 \pmod 2$, which conflicts with $\sum _{i=0}^{p-1} c_i \equiv 0 \pmod 2$ and $\sum _{i=0}^{p-1} c_i' \equiv 1 \pmod 2$. It completes the proof. $\square$

Proposition 4

Let $p\ge 5$. Then for any $\epsilon \in {\mathbb {F}}_p$, and $\delta ,\theta \in \mathcal {C}_0$, $w^{\epsilon }S_0(\delta )+w^{-\epsilon }S_1(\theta )\ne 1$.

Proof

We first consider the case $q\equiv 1 \mod 4$. From Proposition 2, there exist non-negative integers $c_j(\delta ),c_j(\theta )$ $(j=0,1,\ldots , p-1)$ such that

$$\begin{aligned} w^{\epsilon }S_0(\delta )=\sum _{j=0}^{p-1} c_j(\delta ) w^j,~w^{-\epsilon }S_1(\theta )=\sum _{j=0}^{p-1} c_j(\theta ) w^j. \end{aligned}$$

Further, the number of odd $c_j(\delta )(j=0,1, \ldots , p-1)$ is one, and the number of odd $c_j(\theta )(j=0,1,\ldots , p-1)$ is also one. Then, $\sum _{j=0}^{p-1} ( c_j(\delta )+ c_j(\theta ))$ is even and $\# \{j \in \{0, \ldots , p-1\} : c_j(\delta )+ c_j(\theta ) \equiv 1 \pmod 2\} \le 2$. By Lemma 4, $w^{\epsilon }S_0(\delta )+w^{-\epsilon }S_1(\theta )\ne 1$.

If $q\equiv 3 \mod 4$, from Proposition 3, there exist non-negative integers $c_j(\delta ),c_j(\theta )$ $(j=0,1,\ldots , p-1)$ such that

$$\begin{aligned} w^{\epsilon }S_0(\delta )=\sum _{j=0}^{p-1} c_j(\delta ) w^j,~w^{-\epsilon }S_1(\theta )=\sum _{j=0}^{p-1} c_j(\theta ) w^j. \end{aligned}$$

Further, the number of odd $c_j(\delta )(j=0,1,\ldots , p-1)$ is 0 or 2, and $c_j(\theta )(j=0,1,\ldots , p-1)$ are even. Thus, we have $\sum _{j=0}^{p-1} ( c_j(\delta )+ c_j(\theta )) =0 \mod 2$ and $\# \{j \in \{0, \ldots , p-1\} : c_j(\delta )+ c_j(\theta ) \equiv 1 \pmod 2\} \le 2$. By Lemma 4, $w^{\epsilon }S_0(\delta )+w^{-\epsilon }S_1(\theta )\ne 1$.

Hence, this proposition follows. $\square$

Remark

Further, we have $w^{\epsilon }S_0(\delta )+w^{-\epsilon }S_0(\theta )\ne 1$. Proposition 4 can be used to prove the nonexistence of some class of bent functions.

3 Characterization of two classes of regular bent functions

In this section, we will consider two classes of p-ary functions in [17], generalize Li et al.’s work, and make more explicit characterization for bentness of these functions.

3.1 The first class of regular bent functions

The first class of p-ary functions is of the form

$$\begin{aligned} f(x)=Tr^n_1\left( a x^{l(q-1)}+b x^{\left( l+\frac{q+1}{2}\right) (q-1)}\right) +\epsilon x^{\frac{q^2-1}{2}}, \end{aligned}$$

(4)

where $n=2m$, $q=p^m$, $a,b\in {\mathbb {F}}_{q^2}$, and $\epsilon \in {\mathbb {F}}_p$.

The case $gcd(l,q+1)=1$ is studied in [17]. We will consider more general cases for these functions.

Theorem 1

Let f(x) be defined in (4). If f(x) is bent, then $gcd(l,q+1)=1$, or, $gcd(l,q+1)=2$ and $q\equiv 1 \mod 4$.

Proof

We first prove that if f(x) is bent, then $gcd(l,\frac{q+1}{2})=1$. Suppose that $gcd(l,\frac{q+1}{2})=d>1$. Then

$$\begin{aligned} \varLambda _f&=\sum _{u\in U} w^{Tr^n_1\left( \left( a+b u^{\frac{q+1}{2}}\right) u^{l}\right) +\epsilon u^{\frac{q+1}{2}}}\\&=w^{\epsilon }\sum _{v\in U_0} w^{Tr^n_1((a+b)v^{l})}+w^{-\epsilon }\sum _{v\in U_1} w^{Tr^n_1((a-b)v^{l})}\\&= d \left( w^{\epsilon }\sum _{v\in U_0^d} w^{Tr^n_1((a+b)v)}+w^{-\epsilon }\sum _{v\in U_1^d} w^{Tr^n_1((a-b)v)}\right) , \end{aligned}$$

where $U_i^d=\{v^d:v\in U_i\}$. From Lemma 1,

$$\begin{aligned} \varLambda _f=d \left( w^{\epsilon }\sum _{v\in U_0^d} w^{Tr^n_1((a+b)v)}+w^{-\epsilon }\sum _{v\in U_1^d} w^{Tr^n_1((a-b)v)}\right) =w^{f(0)}. \end{aligned}$$

From Lemma 3, $d=gcd(l,\frac{q+1}{2})=1$, i.e., $gcd(l,q+1)=1$ or 2. Further, from $gcd(l,q+1)=2$, $q\equiv 1\mod 4$. Thus, this theorem follows. $\square$

When $b=0$ and $\epsilon \ne 0$, Zheng et al. proved in Theorem 3 in [25] that when $q\equiv 3 \mod 8$ and $gcd(l,q+1)=2$, f(x) is not bent. They left the nonexistence of bent function f(x) as an open problem when $q\equiv 7\mod 8$. From Theorem 1, f(x) in the case $q\equiv 7\mod 8$ is not bent. Li et al. [17] studied the characterization of regular bentness of f(x) for the case $gcd(l,q+1)=1$.

Theorem 2

Let $p\ge 5$ and f(x) be a p-ary function defined in (4). Let $gcd(l,q+1)=1$, $\delta =a+b$, $\theta = a-b$, and $\delta ,\theta \in \mathcal {C}_0$. Then f(x) is not bent.

Proof

From [17], f(x) is bent if and only if $w^{\epsilon }S_0(\delta )+w^{-\epsilon }S_1(\theta )= 1$. From Proposition 4, this theorem follows. $\square$

From Theorem 2, if $\delta ,\theta \in \mathcal {C}_0$, then f(x) is not bent. Hence, we have the following corollary, which is a straightforward result of Theorem 9 of Li et al. [17].

Corollary 1

Let $p\ge 5$, and f(x) be a p-ary function defined in (4). Then f(x) is regular bent if and only if

$$\begin{aligned} w^{\epsilon }K_m(\delta ^{q+1})+w^{-\epsilon }K_m(\theta ^{q+1})= {\left\{ \begin{array}{ll} 4I\sqrt{-1}w^{\epsilon }\sin \frac{2\pi Q(\delta )}{p}+2\cos \frac{2\pi \epsilon }{p}-2,&{}\delta \in \mathcal {C}_0, \theta \not \in \mathcal {C}_0,\\ -4I\sqrt{-1}w^{-\epsilon }\sin \frac{2\pi Q(\theta )}{p}+2\cos \frac{2\pi \epsilon }{p}-2,&{}\delta \not \in \mathcal {C}_0, \theta \in \mathcal {C}_0,\\ 2\cos \frac{2\pi \epsilon }{p}-2,&{}\delta \not \in \mathcal {C}_0, \theta \not \in \mathcal {C}_0. \end{array}\right. } \end{aligned}$$

We will study the characterization of regular bent functions for the case $gcd(l,q+1)=2$ and $q\equiv 1 \mod 4$ in Theorem 1.

Theorem 3

Let $p\ge 5$, $q\equiv 1 \mod 4$, and f(x) be a p-ary function defined in (4). Let $gcd(l,q+1)=2$, $\delta =a+b$, and $\theta =a-b$. Then, f(x) is regular bent if and only if

$$\begin{aligned} w^{\epsilon }K_m(\delta ^{q+1})+w^{-\epsilon }K_m(\theta ^{q+1})= {\left\{ \begin{array}{ll} 4I\sqrt{-1}w^{\epsilon }\sin \frac{2\pi Q(\delta )}{p}+2\cos \frac{2\pi \epsilon }{p}-2,&{}\delta \in \mathcal {C}_0, \theta \not \in \mathcal {C}_0,\\ 4I\sqrt{-1}w^{-\epsilon }\sin \frac{2\pi Q(\theta )}{p}+2\cos \frac{2\pi \epsilon }{p}-2,&{}\delta \not \in \mathcal {C}_0, \theta \in \mathcal {C}_0,\\ 2\cos \frac{2\pi \epsilon }{p}-2,&{}\delta \not \in \mathcal {C}_0, \theta \not \in \mathcal {C}_0. \end{array}\right. } \end{aligned}$$

Proof

We have

$$\begin{aligned} \varLambda _f=w^{\epsilon }\sum _{v\in U_0} w^{Tr^n_1((a+b)v^{l})}+w^{-\epsilon }\sum _{v\in U_1} w^{Tr^n_1((a-b)v^{l})}. \end{aligned}$$

Since $gcd(l,q+1)=2$ and $q\equiv 1 \mod 4$, then $gcd(l,\frac{q+1}{2})=1$ and l is even. Further, $v\longmapsto v^l$ is a permutation of $U_0$ and also a bijection from $U_1$ to $U_0$. Hence,

$$\begin{aligned} \varLambda _f=w^{\epsilon }\sum _{v\in U_0} w^{Tr^n_1((a+b)v)}+w^{-\epsilon }\sum _{v\in U_0} w^{Tr^n_1((a-b)v)}. \end{aligned}$$

From Lemma 1, f(x) is regular bent if and only if

$$\begin{aligned} w^{\epsilon }S_0(\delta )+w^{-\epsilon }S_0(\theta ) =w^{f(0)}=1. \end{aligned}$$

If $\delta ,\theta \in \mathcal {C}_0$, then from the remark after Proposition 4, f(x) is not bent. From Proposition 1, this theorem can be obtained. $\square$

3.2 The second class of regular bent functions

The second class of p-ary functions is of the form

$$\begin{aligned} f(x)={\left\{ \begin{array}{ll} \sum ^{q-1}_{i=0} Tr^n_1(a x^{(ri+s)(q-1)})+\epsilon x^{\frac{q^2-1}{2}},&{} x\ne 0,\\ f(0),&{} x=0, \end{array}\right. } \end{aligned}$$

(5)

where $n=2m$, $q=p^m$, $a\in {\mathbb {F}}_{q^2}$, and $\epsilon \in {\mathbb {F}}_{p}$.

For simplicity, we first consider f(x) for the case $\epsilon =0$. When $gcd(r,q+1)=1$ or 2, the following theorem gives the nonexistence of bent f(x).

Theorem 4

Let $p\ge 7$ and f(x) be a p-ary function defined in (5). Let $\epsilon =0$ and $gcd(r,q+1)=1$ or 2. Then f(x) is not bent.

Proof

Suppose that f(x) is bent. We first prove that $gcd(s-r,q+1)=1$.

(1) $gcd(r,q+1)=1$: If $u\in U$, then

$$\begin{aligned} \sum _{i=0}^{q-1} u^{ri+s}={\left\{ \begin{array}{ll}-u^{s-r},&{}u\ne 1,\\ 0,&{}u=1. \end{array}\right. } \end{aligned}$$

We have

$$\begin{aligned} \varLambda _f&=\sum _{u\in U} w^{\sum _{i=0}^{q-1} Tr^n_1(a u^{ri+s})}\\&=1+\sum _{u\in U{\setminus } \{1\}} w^{Tr^n_1(-a u^{s-r})}\\&=1-w^{-Tr^n_1(a)}+\sum _{u\in U} w^{Tr^n_1(-a u^{s-r})}\\&=1-w^{-Tr^n_1(a)}+d\sum _{u\in U^d} w^{Tr^n_1(-a u)}, \end{aligned}$$

where $d=gcd(s-r,q+1)$ and $U^d=\{u^d:u\in U\}$. Since f(x) is bent, from Lemma 1, we have

$$\begin{aligned} d\sum _{u\in U^d} w^{Tr^n_1(-a u)}=-1+w^{Tr^n_1(-a)}+w^{f(0)}. \end{aligned}$$

There exist integers $c_i,c_i'(i=0,1,\ldots , p-1)$ such that

$$\begin{aligned} d\sum _{u\in U^d} w^{Tr^n_1(-a u)}=\sum _{i=0}^{p-1} c_iw^i, ~~~ -1+w^{Tr^n_1(a)}+w^{f(0)}=\sum _{i=0}^{p-1} c_i'w^i. \end{aligned}$$

By Lemma 3, we have $d=1= gcd(s-r,q+1)=1$.

(2) $gcd(r,q+1)=2$: From a similar discussion, we also have $gcd(s-r,q+1)=1$.

From Case (1) and Case (2), we have $gcd(s-r,q+1)=d=1$. Note that

$$\begin{aligned} \varLambda _f&=\sum _{u\in U} w^{\sum _{i=0}^{q-1} Tr^n_1(a u^{ri+s})}\\&=2+\sum _{u\in U{\setminus } \{\pm 1\}} w^{Tr^n_1(-a u^{s-r})}\\&=2-w^{-Tr^n_1(a)}-w^{-Tr^n_1(a(-1)^{s-r})}+ \sum _{u\in U} w^{Tr^n_1(-a u^{s-r})}\\&=2-w^{-Tr^n_1(a)}-w^{-Tr^n_1(a(-1)^{s-r})} +d\sum _{u\in U^d} w^{Tr^n_1(-a u)}. \end{aligned}$$

Note that $\sum _{u\in U} w^{Tr^n_1(-a (-u))}=\sum _{u\in U} w^{Tr^n_1(-a u)}$ and $\sum _{u\in U} w^{Tr^n_1(-a u)}$ is a real number. From Lemma 1, we have

$$\begin{aligned} \sum _{u\in U} w^{Tr^n_1(a u)}=-1+w^{Tr^n_1(a)}+w^{-Tr^n_1(a)}. \end{aligned}$$

From Lemma 2, there exist integers $c_i (i=0,1,\ldots ,p-1)$ such that

$$\begin{aligned} \sum _{u\in U} w^{Tr^n_1(a u)}=\sum _{i=0}^{p-1} c_i w^i, \end{aligned}$$

where $\sum _{i=0}^{p-1} c_i \equiv 0 \mod 2$ and $\# \{i \in \{0, \ldots , p-1\}: c_i \equiv 1 \pmod 2\} \le 2$. By Lemma 4, $\sum _{u\in U} w^{Tr^n_1(a u)} \ne -1+w^{Tr^n_1(a)}+w^{-Tr^n_1(a)}$. Hence, this theorem follows. $\square$

In Theorem 10, Li et al. [17] gave the necessary and sufficient conditions of bent f(x) for the case $\epsilon =0$ and $gcd(r,q+1)=1$. Theorem 4 demonstrates that these functions are not bent.

We will consider the case $\epsilon \ne 0$ and study the bentness of f(x). The following proposition gives necessary conditions of bent function f(x) in the case $gcd(r,q+1)=1$ or 2.

Proposition 5

Let $p\ge 7$ and f(x) be a p-ary bent function defined in (5). Let $\epsilon \ne 0$ and $gcd(r,q+1)=1$ or 2. Then, $gcd(s-r,q+1)=1$, or, $gcd(s-r,q+1)=2$ and $q\equiv 1\mod 4$.

Proof

If $u\in U$, then

$$\begin{aligned} \sum _{i=0}^{q-1} u^{ri+s}={\left\{ \begin{array}{ll}-u^{s-r},&{} a^d\ne 1,\\ 0,&{} a^d=1, \end{array}\right. } \end{aligned}$$

where $d=gcd(r,q+1).$ We have

$$\begin{aligned} \varLambda _f&=\sum _{u\in U} w^{\sum _{i=0}^{q-1}Tr^n_1(a u^{ri+s})+\epsilon u^{\frac{q+1}{2}}}\\&=\sum _{u\in \{u\in U:u^d=1\}} w^{\epsilon u^{\frac{q+1}{2}}}+\sum _{u\in U{\setminus } \{u:u^d=1\}} w^{Tr^n_1(-a u^{s-r})+\epsilon u^{\frac{q+1}{2}}}\\&=\sum _{u\in \{u\in U:u^d=1\}} w^{\epsilon u^{\frac{q+1}{2}}}-\sum _{u\in \{u\in U:u^d=1\}} w^{Tr^n_1(-a u^{s-r})+\epsilon u^{\frac{q+1}{2}}}+\sum _{u\in U} w^{Tr^n_1(-a u^{s-r})+\epsilon u^{\frac{q+1}{2}}}\\&=\sum _{u\in \{u\in U:u^d=1\}} w^{\epsilon u^{\frac{q+1}{2}}}\!-\!\sum _{u\in \{u\in U:u^d=1\}} w^{Tr^n_1(-a u^{s-r})\!+\!\epsilon u^{\frac{q+1}{2}}}+e\sum _{u\in U^e} w^{Tr^n_1\left( -a u^{\frac{s-r}{e}}\right) +\epsilon u^{\frac{q+1}{2e}}}, \end{aligned}$$

where $e=gcd(s-r,\frac{q+1}{2})$. From Lemma 1, f(x) is bent if and only if

$$\begin{aligned} e\sum _{u\in U^e} w^{Tr^n_1\left( -a u^{\frac{s-r}{e}}\right) +\epsilon u^{\frac{q+1}{2e}}}=-\sum _{u\in \{u\in U:u^d=1\}} w^{\epsilon u^{\frac{q+1}{2}}}+\sum _{u\in \{u\in U:u^d=1\}} w^{Tr^n_1(-a u^{s-r})+\epsilon u^{\frac{q+1}{2}}}+w^{f(0)}. \end{aligned}$$

From $\# \{u\in U:u^d=1\} \le d \le 2$ and Lemma 3, $e=1$, i.e., $gcd(s-r,\frac{q+1}{2})=1$. Hence, $gcd(s-r,q+1)=1$ or $gcd(s-r,q+1)=2$, $q\equiv 1\mod 4$. $\square$

We will present the characterization of bentness of f(x) for different values of $gcd(r,q+1)$ and $gcd(s-r,q+1)$.

Theorem 5

Let f(x) be a p-ary function defined in (5) and $\epsilon \ne 0$.

(1) If $p\ge 7$, $gcd(r,q+1)=1$ and $gcd(r-s,q+1)=1$, then f(x) is regular bent if and only if $-a\not \in \mathcal {C}_0$ and $K_m(a^{q+1})=1-\frac{\rho }{\cos \frac{2\pi \epsilon }{p}}$, where $\rho =w^{f(0)}-w^{\epsilon }+ w^{-Tr^n_1(a)+\epsilon }$.

(2) If $p\ge 7$, $q\equiv 1\mod 4$, $gcd(r,q+1)=1$ and $gcd(s-r,q+1)=2$, then f(x) is regular bent if and only if $a\not \in \mathcal {C}_0$ and $K_m(a^{q+1})=1-\frac{\rho }{\cos \frac{2\pi \epsilon }{p}}$, where $\rho =-w^{\epsilon }+ w^{Tr^n_1(-a )+\epsilon }+w^{f(0)}$.

(3) If $p\ge 11$, $gcd(r,q+1)=2$, and $gcd(s-r,q+1)=1$, then f(x) is regular bent if and only if $-a\not \in \mathcal {C}_0$ and $K_m(a^{q+1})=1-\frac{\rho }{\cos \frac{2\pi \epsilon }{p}}$, where $\rho =-w^{\epsilon }-w^{\epsilon (-1)^{\frac{q+1}{2}}}+ w^{Tr^n_1(-a )+\epsilon }+w^{Tr^n_1(a )+\epsilon (-1)^{\frac{q+1}{2}} }+w^{f(0)}$.

(4) If $p\ge 11$, $q\equiv 1 \mod 4$, $gcd(r,q+1)=2$, and $gcd(s-r,q+1)=2$, then f(x) is regular bent if and only if $a\not \in \mathcal {C}_0$ and $K_m(a^{q+1})=1-\frac{\rho }{\cos \frac{2\pi \epsilon }{p}}$, where $\rho =-w^{\epsilon }-w^{-\epsilon }+ w^{Tr^n_1(-a )+\epsilon }+w^{Tr^n_1(-a )-\epsilon }+w^{f(0)}$.

Proof

(1) From Theorem 11 in [17], f(x) is regular bent if and only if

$$\begin{aligned} w^{\epsilon }S_0(-a)+w^{-\epsilon }S_1(-a)=w^{f(0)}-w^{\epsilon }+w^{-Tr^n_1(a)+\epsilon }. \end{aligned}$$

Suppose $-a\in \mathcal {C}_0$. From Propositions 2 and 3, there exist integers $c_i(i=0,1,\ldots , p-1)$ such that $w^{\epsilon }S_0(-a)+w^{-\epsilon }S_1(-a)= \sum _{i=0}^{p-1}c_iw^{i}$, where the number of odd $c_i$ is 0 or 2. By Lemma 4,

$$\begin{aligned} w^{\epsilon }S_0(-a)+w^{-\epsilon }S_1(-a)\ne w^{f(0)}-w^{\epsilon }+w^{-Tr^n_1(a)+\epsilon }, \end{aligned}$$

i.e., f(x) is not bent. From Theorem 11 in [17], this result can be obtained.

(2) From Proposition 5, f(x) is bent if and only if

$$\begin{aligned} \sum _{u\in U} w^{Tr^n_1(-a u^{s-r})+\epsilon u^{\frac{q+1}{2}}}=-w^{\epsilon }+ w^{Tr^n_1(-a )+\epsilon }+w^{f(0)}. \end{aligned}$$

Since $q\equiv 1\mod 4$ and $gcd(s-r,q+1)=2$, then $u\mapsto u^{s-r}$ is a permutation of $U_0$ and also a bijection from $U_1$ to $U_0$. We have

$$\begin{aligned} \sum _{u\in U} w^{Tr^n_1(-a u^{s-r})+\epsilon u^{\frac{q+1}{2}}}&=w^{\epsilon } \sum _{u\in U_0} w^{Tr^n_1(-a v)}+w^{-\epsilon } \sum _{u\in U_0} w^{Tr^n_1(-a v)}\\&=(w^{\epsilon }+w^{\epsilon })\sum _{u\in U_0} w^{Tr^n_1(-a v)}\\&=(w^{\epsilon }+w^{\epsilon })\sum _{u\in U_1} w^{Tr^n_1(a v)}~~~(-1\in U_1). \end{aligned}$$

Hence, f(x) is regular bent if and only if

$$\begin{aligned} (w^{\epsilon }+w^{-\epsilon })\sum _{u\in U_1} w^{Tr^n_1(a v)}=-w^{\epsilon }+ w^{Tr^n_1(-a )+\epsilon }+w^{f(0)}. \end{aligned}$$

From a similar discussion as (1), we have $a\not \in \mathcal {C}_0$. When $a\not \in \mathcal {C}_0$, from Proposition 1,

$$\begin{aligned} \sum _{u\in U_1} w^{Tr^n_1(a v)}=\frac{1-K_m(a^{q+1})}{2}. \end{aligned}$$

Hence, this result follows.

(3) Note that f(x) is bent if and only if

$$\begin{aligned} \sum _{u\in U} w^{Tr^n_1(-a u^{s-r})+\epsilon u^{\frac{q+1}{2}}}&=-w^{\epsilon }-w^{\epsilon (-1)^{\frac{q+1}{2}}}+ w^{Tr^n_1(-a )+\epsilon }\\&\qquad +w^{Tr^n_1(a )+\epsilon (-1)^{\frac{q+1}{2}} }+w^{f(0)}. \end{aligned}$$

Since $gcd(s-r,q+1)=1$, then $u\mapsto u^{s-r}$ is a permutation of $U_0$ and also a permutation of $U_1$. We have

$$\begin{aligned} \sum _{u\in U} w^{Tr^n_1(-a u^{s-r})+\epsilon u^{\frac{q+1}{2}}}=&w^{\epsilon } \sum _{u\in U_0} w^{Tr^n_1(-a v)}+w^{-\epsilon } \sum _{u\in U_1} w^{Tr^n_1(-a v)}. \end{aligned}$$

Hence, f(x) is regular bent if and only if

$$\begin{aligned}&w^{\epsilon } \sum _{u\in U_0} w^{Tr^n_1(-a v)}+w^{-\epsilon } \sum _{u\in U_1} w^{Tr^n_1(-a v)}=-w^{\epsilon }-w^{\epsilon (-1)^{\frac{q+1}{2}}}\\&\quad \qquad + w^{Tr^n_1(-a )+\epsilon }+w^{Tr^n_1(a )+\epsilon (-1)^{\frac{q+1}{2}} }+w^{f(0)}. \end{aligned}$$

From a similar discussion as (1), we have $-a\not \in \mathcal {C}_0$. When $-a\not \in \mathcal {C}_0$, from Proposition 1,

$$\begin{aligned} \sum _{u\in U_0} w^{Tr^n_1(-a v)}=\sum _{u\in U_1} w^{Tr^n_1(-a v)}=\frac{1-K_m(a^{q+1})}{2}. \end{aligned}$$

Hence, this result follows.

(4) Note that f(x) is bent if and only if

$$\begin{aligned} \sum _{u\in U} w^{Tr^n_1(-a u^{s-r})+\epsilon u^{\frac{q+1}{2}}}=-w^{\epsilon }-w^{-\epsilon }+ w^{Tr^n_1(-a )+\epsilon }+w^{Tr^n_1(-a )-\epsilon }+w^{f(0)}. \end{aligned}$$

Since $q\equiv 1 \mod 4$ and $gcd(s-r,q+1)=2$, then $u\mapsto u^{s-r}$ is a permutation of $U_0$ and also a bijection from $U_1$ to $U_0$. We have

$$\begin{aligned} \sum _{u\in U} w^{Tr^n_1(-a u^{s-r})+\epsilon u^{\frac{q+1}{2}}}&=w^{\epsilon } \sum _{u\in U_0} w^{Tr^n_1(-a v)}+w^{-\epsilon } \sum _{u\in U_0} w^{Tr^n_1(-a v)}\\&=(w^{\epsilon }+w^{-\epsilon }) \sum _{u\in U_1} w^{Tr^n_1(a v)}~~~(-1\in U_1). \end{aligned}$$

Hence, f(x) is regular bent if and only if

$$\begin{aligned} (w^{\epsilon }+w^{-\epsilon }) \sum _{u\in U_1} w^{Tr^n_1(a v)}=-w^{\epsilon }-w^{-\epsilon }+ w^{Tr^n_1(-a )+\epsilon }+w^{Tr^n_1(-a )-\epsilon }+w^{f(0)}. \end{aligned}$$

From a similar discussion as (1), we have $a\not \in \mathcal {C}_0$. When $a\not \in \mathcal {C}_0$, from Proposition 1,

$$\begin{aligned} \sum _{u\in U_1} w^{Tr^n_1(a v)}=\frac{1-K_m(a^{q+1})}{2}. \end{aligned}$$

Hence, this result follows. $\square$

Remark

Theorem 5 is a generalization of Theorem 11 in [17]. Note that there is a minor error in Theorem 11 in [17], where $\rho =w^{f(0)}-w^{\epsilon }+w^{-Tr^n_1(a)}$ should be $\rho =w^{f(0)}-w^{\epsilon }+w^{-Tr^n_1(a)+\epsilon }$.

4 Conclusion

This paper studies two classes of p-ary regular bent functions introduced by Li et al. [17]. The first class of p-ary functions is defined by

$$\begin{aligned} f(x)=Tr^n_1\left( a x^{l(q-1)}+b x^{\left( l+\frac{q+1}{2}\right) (q-1)}\right) +\epsilon x^{\frac{q^2-1}{2}}, \end{aligned}$$

where $a,b\in {\mathbb {F}}_{q^2}$. We prove that when $gcd(l,q+1)=1$ and both $a+b$ and $a-b$ are quadratic residues, f(x) is not bent. Further, we present the characterization of these regular bent functions for the case $gcd(l,q+1)=1$ and the case $gcd(l,q+1)=2$ and $q\equiv 1\mod 4$. The second class of p-ary functions is defined by

$$\begin{aligned} f(x)={\left\{ \begin{array}{ll} \sum ^{q-1}_{i=0} Tr^n_1\left( a x^{(ri+s)(q-1)}\right) +\epsilon x^{\frac{q^2-1}{2}},&{} x\ne 0,\\ f(0),&{} x=0, \end{array}\right. } \end{aligned}$$

where $a\in {\mathbb {F}}_{q^2}$ and $\epsilon \in {\mathbb {F}}_{p}$. When $p\ge 7$ and $\epsilon =0$, f(x) is not bent. Further, we present the concrete characterization for these functions for different values of $gcd(r,q+1)$ and $gcd(s-r,q+1)$. Our work generalizes results in [17].