Regular p-ary bent functions with five terms and Kloosterman sums

Abstract

Kloosterman sums are vital in the study of bent functions, including regular p-ary bent functions. In this paper, a congruence property for Kloosterman sums is presented first and is used to prove the nonexistence of a class of p-ary bent functions. Further, this paper considers p-ary functions of the form $f(x)= \text {Tr}^{n}_{1}(a_{1}x^{r_{1}(q-1)})+\text {Tr}^{n}_{1}\left (c_{1}x^{r_{1}(q-1)+\frac {q^{2}-1}{2}}\right ) +\text {Tr}^{n}_{1}\left (a_{2}x^{r_{2}(q-1)}\right )+\text {Tr}^{n}_{1}\left (c_{2}x^{r_{2}(q-1)+\frac {q^{2}-1}{2}}\right ) +bx^{\frac {q^{2}-1}{2}}$. We use Kloosterman sums in the characterization of this class of p-ary bent functions. Finally, an open problem of Jia et al. (IEEE Trans Inf. Theory 58(9): 6054–6063, 2012) is solved and we prove the nonexistence for a class of regular p-ary bent functions.

1 Introduction

Introduced by Rothaus [17], Boolean bent functions from $\mathbb {F}_{2}^{n}$ or $\mathbb {F}_{2^{n}}$ to $\mathbb {F}_{2}$ have important applications in cryptography, coding theory, and sequences. 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 present characterization and construction of monomial bent functions, binomial bent functions and quadratic bent functions [1,2,3,4,5, 12, 15, 16, 19, 20]. Boolean bent functions were generalized to the notation of functions over an arbitrary finite field in [11]. It is elusive to completely classify bent functions. The characterization of bent functions over finite fields of odd characteristic is more complicate than that of Boolean bent functions. Several work can be found in [7, 8].

Let p be an odd prime and m be an integer. Let n = 2m and q = p^m. Let $\text {Tr}^{n}_{1}(\cdot )$ be the trace function from $\mathbb {F}_{q^{2}}$ to $\mathbb {F}_{p}$. Helleseth and Kholosha [6] studied monomial functions with Dillon type of the form $f_{a,r}(x)=\text {Tr}^{n}_{1}(ax^{r(q-1)})$, 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. [9] considered binomial functions of the form $f_{a,b,r}(x)=\text {Tr}^{n}_{1}(ax^{r(q-1)})+ bx^{\frac {q^{2}-1}{2}}$, where $a\in \mathbb {F}_{q^{2}}$, $b\in \mathbb {F}_{p}$ and $\gcd (r,q + 1)= 1$. By Kloosterman sums, they presented the characterization of bentness for f_a,b,r. For p = 3 or q ≡ 3 (mod 4), they proved that f_a,b,r is bent if and only if $K_{m}(a)= 1-\sec \frac {2\pi b}{p}$. Zheng et al. [21] generalized Jia et al.’s result to the case q ≡ 1 (mod 4), i.e., f_a,b,r is bent if and only if $K_{m}(a)= 1-\sec \frac {2\pi b}{p}$. Further, when q ≡ 7 (mod 8), r is even and $\gcd (\frac {r}{2},q + 1)= 1$, Zheng et al. proved that $f_{a,b,r}(x)=\text {Tr}^{n}_{1}(ax^{r(q-1)})+bx^{\frac {q^{2}-1}{2}}(a\in \mathbb {F}_{q^{2}},b\in \mathbb {F}_{p})$ is not bent. This paper generalizes Zheng et al.’s results, presents the characterization of more regular p-ary bent functions and proves the nonexistence of a class of bent functions. Further, this paper also solves an open problem in the case q ≡ 3 (mod 8) presented by Jia et al. [9] and proves that f_a,b,r is not bent.

Li et al. [13] considered trinomial functions of the form $f_{a,c,b,r}(x)=\text {Tr}^{n}_{1}(ax^{r(q-1)})+\text {Tr}^{n}_{1}\left (cx^{r(q-1)+\frac {q^{2}-1}{2}}\right ) +bx^{\frac {q^{2}-1}{2}}$, where $a,c\in \mathbb {F}_{q^{2}}, b\in \mathbb {F}_{p}$, and $\gcd (r,q + 1)= 1$. They presented the relation between the bentness of f_a,c,b,r and Kloosterman sums K_m((a + c)^q+ 1),K_m((a − c)^q+ 1).

With similar methods in [9, 13, 21], this paper generalizes their results and considers functions with five terms of the form

$$\begin{array}{@{}rcl@{}} f(x)&=&\text{Tr}^{n}_{1}(a_{1}x^{r_{1}(q-1)})+\text{Tr}^{n}_{1}\left( c_{1}x^{r_{1}(q-1)+ \frac{q^{2}-1}{2}}\right)\\ &&+ \text{Tr}^{n}_{1}(a_{2}x^{r_{2}(q-1)})+ \text{Tr}^{n}_{1}\left( c_{2}x^{r_{2}(q-1)+\frac{q^{2}-1}{2}}\right)+ bx^{\frac{q^{2}-1}{2}}, \end{array} $$

where $a_{1},a_{2},c_{1},c_{2}\in \mathbb {F}_{q^{2}}$ and $b\in \mathbb {F}_{p}$. With the help of Kloosterman sums, we characterize the bentness of this class of p-ary functions. A congruence property of Kloosterman sums is deduced first, which is used to prove the nonexistence of some Dillon type bent functions.

This paper is organized as follows: Section 2 introduces some notations and results on character sums. Section 3 presents a congruence property and proves that some Dillon type functions are not bent. Section 4 presents the characterization of bentness for functions with five terms and solves an open problem proposed by Jia et al. [9]. Section 5 makes a conclusion for this paper.

2 Preliminaries

2.1 Regular bent functions

Throughout this paper, let p be an odd prime and m,n be positive integers. Let q = p^m, $\mathbb {F}_{q}$ be a finite field with q elements and $\mathbb {F}_{q}^{*}$ the multiplicative group composed of all nonzero elements in $\mathbb {F}_{q}$. Let k|m and $\text {Tr}^{m}_{k}(x) ={\sum }_{i = 0}^{m/k-1}x^{p^{ki}}$ be the trace function from $\mathbb {F}_{p^{m}}$ to $\mathbb {F}_{p^{k}}$. For any $x\in \mathbb {F}_{q^{2}}^{*}$, there exists a unique factorization x = y ∗ ξⁱ, where $y\in \mathbb {F}_{q}^{*}, 0\le i\le q$, and ξ is a primitive element of $\mathbb {F}_{q^{2}}$. Let U = {ξ⁰,ξ^(q− 1),…,ξ^(q− 1)q}, U₀ = U² = {u² : u ∈ U}, and U₁ = U ∖ U₀. Sets of squares and nonsquares 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 $\mathcal {C}_{0}^{+}=\{x\in \mathcal {C}_{0}:\text {Tr}^{m}_{1}(x^{\frac {p^{m}+ 1}{2}})\neq 0\}$.

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 $W_{f}(\lambda )={\sum }_{x\in \mathbb {F}_{p^{n}}}w^{f(x)-\text {Tr}^{n}_{1}(\lambda x)},$ 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(λ)|² = pⁿ 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^∗(λ) satisfying $W_{f}(\lambda )=p^{\frac {n}{2}}w^{f^{*}(\lambda )}$ for any $\lambda \in \mathbb {F}_{p^{n}}$. The function f^∗(λ) is called the dual of f(x). And the dual of a regular p-ary bent function is also bent. Let n = 2m for the rest of the paper.

2.2 Exponential sums

For $a\in \mathbb {F}_{p^{n}}$, the Kloosterman sum K_n(a) [14] of a is defined by $K_{n}(a)={\sum }_{x\in \mathbb {F}_{p^{n}}} w^{\text {Tr}^{n}_{1}(ax+\frac {1}{x})}$, where $\frac {1}{0}= 0$ for x = 0. Since $\overline {K_{n}(a)}={\sum }_{x\in \mathbb {F}_{p^{n}}}w^{-\text {Tr}^{n}_{1}(ax+\frac {1}{x})}=K_{n}(a)$, then K_n(a) is a real number.

Some notations are defined below.

$$\begin{array}{@{}rcl@{}} I&=&\left\{\begin{array}{llll} \frac{(-1)^{\frac{3m}{2}}p^{\frac{m}{2}}}{2}, &\text{\(p\equiv 3 \pmod 4\);}\\ \frac{(-1)^{m}p^{\frac{m}{2}}}{2}, &\text{otherwise.} \end{array}\right.\\ Q(a)&=&2\text{Tr}^{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{array} $$

Obviously, when q ≡ 1 (mod 4), I is a real number.

The following result on exponential sums is useful [9].

Proposition 1

Let$a\in \mathbb {F}_{q^{2}}^{*}$,then

$$\sum\limits_{u\in U_{0}}w^{\text{Tr}^{n}_{1}(au)}= \left\{\begin{array}{llll} R(a)+I(w^{Q(a)}-w^{-Q(a)}), & a\in \mathcal{C}_{0}^{+},\\ R(a), &otherwise, \end{array}\right. $$

and

$$\sum\limits_{u\in U_{1}}w^{\text{Tr}^{n}_{1}(au)}= \left\{\begin{array}{lll} R(a)-I(w^{Q(a)}-w^{-Q(a)}), & a\in \mathcal{C}_{0}^{+},\\ R(a), &otherwise. \end{array}\right. $$

3 A congruence property of Kloosterman sums and its application

Lemma 1

Let $a,x\in \mathbb {F}_{q}^{*}$ , and $y\in \mathbb {F}_{q}$ .

• (1) Ify² − 4ais not a quadratic residue in$\mathbb {F}_{q}$,thenax + x^− 1 = yhas no solution.

• (2) Ify² − 4a = 0,thenax + x^− 1 = yhas only a solution.

• (3) Ify² − 4ais a quadratic residue in$\mathbb {F}_{q}$,thenax + x^− 1 = yhas two solutions.

Proof

The equation ax + x^− 1 = y can be transformed into ax² − yx + 1 = 0. And Δ = y² − 4a is the discriminant for ax² − yx + 1 = 0. Hence, Results (1), (2), and (3) are obviously obtained. □

Proposition 2

Let w be a primitive p-th root of unity andQ(w) be the p-th cyclotomic field over rational field Q.Let$\mathfrak {R}$be a prime ideallying above 2 inQ(w) and$a\in \mathbb {F}_{q}^{*}$,then

• (1) $K_{m}(a)\equiv 1 \pmod {\mathfrak {R}}$ifand only if a is a nonsquare or a is a squaresatisfying$\text {Tr}_{1}^{m}(\sqrt {a})= 0$.

• (2) $K_{m}(a)\equiv 1+w^{t}+w^{-t}\pmod {\mathfrak {R}} (1\leq t\leq p-1)$ifand only if a is a square and$\text {Tr}_{1}^{m}(2\sqrt {a})=\pm t$.

Proof

We first prove that when 1 ≤ t ≤ p − 1, $1+w^{t}+w^{-t}\not \equiv 1\pmod {\mathfrak {R}}$, and when $1\leq t_{2}< t_{1}\leq \frac {p-1}{2}$, $1+w^{t_{1}}+w^{-t_{1}}\not \equiv 1+w^{t_{2}}+w^{-t_{2}}\pmod {\mathfrak {R}}$. Note that $w\not \equiv 1\pmod {\mathfrak {R}}$, i,e, $w\pmod {\mathfrak {R}}$ is also a primitive p-th root of unity. If $1+w^{t}+w^{-t}\equiv 1\pmod {\mathfrak {R}}$, then $w^{t}+w^{-t} \equiv 0 \pmod {\mathfrak {R}}$, i,e, $w^{2t}\equiv 1 \pmod {\mathfrak {R}}$. Then t ≡ 0 (mod p), which makes a contradiction with the supposition of t. Hence, $1+w^{t}+w^{-t}\not \equiv 1\pmod {\mathfrak {R}}$. If $1+w^{t_{1}}+w^{-t_{1}}\equiv 1+w^{t_{2}}+w^{-t_{2}}\pmod {\mathfrak {R}}$, then $(w^{t_{1}+t_{2}}+ 1)(w^{t_{1}-t_{2}}+ 1)\equiv 0\pmod {\mathfrak {R}}$. From the supposition of t₁,t₂, $w^{t_{1}+t_{2}}\not \equiv 1 \pmod {\mathfrak {R}}$, $w^{t_{1}-t_{2}}\not \equiv 1\pmod {\mathfrak {R}}$. Then it makes a contradiction. Hence, $1+w^{t_{1}}+w^{-t_{1}} \not \equiv 1+w^{t_{2}}+w^{-t_{2}}\pmod {\mathfrak {R}}$.

From the definition of Kloosterman sums, we have

$$\begin{array}{@{}rcl@{}} K_{m}(a)&=&1+\sum\limits_{x\in\mathbb{F}_{q}^{*}} w^{\text{Tr}_{1}^{m}(ax+x^{-1})}\\ &=&1+\sum\limits_{y\in\mathbb{F}_{q}} w^{\text{Tr}_{1}^{m}(y)}\#\{x\in \mathbb{F}_{q}^{*}: ax+x^{-1}=y \}. \end{array} $$

From Lemma 1,

$$\begin{array}{@{}rcl@{}} K_{m}(a) &=&1+\sum\limits_{y\in\mathbb{F}_{q}} w^{\text{Tr}_{1}^{m}(y)}\#\{x\in \mathbb{F}_{q}^{*}: ax+x^{-1}=y \}\\ &\equiv& 1+ \sum\limits_{y\in\mathbb{F}_{q}, y^{2}-4a = 0}w^{\text{Tr}_{1}^{m}(y)} \pmod{\mathfrak{R}}. \end{array} $$

If a is a nonsquare, then $K_{m}(a)\equiv 1\pmod {\mathfrak {R}}$.

If a is a square and $\text {Tr}_{1}^{m}(\sqrt {a})= 0$, then $K_{m}(a)\equiv 1\pmod {\mathfrak {R}}$.

If a is a square, and $\text {Tr}_{1}^{m}(2\sqrt {a})=t$, then $K_{m}(a)\equiv 1+w^{t}+w^{-t}\pmod {\mathfrak {R}}$.

Hence, this proposition follows. □

Remark 1

From Proposition 2, $K_{m}(a) \pmod {\mathfrak {R}} \in \{1+w^{t}+w^{-t} \pmod {\mathfrak {R}}:0\leq t\leq \frac {p-1}{2}\}$.

Proposition 2 can be used to discuss the nonexistence of some regular p-ary bent functions. The following theorem demonstrates that some regular p-ary bent functions in Theorem 10 in [13] do not exist.

Theorem 1

Let p be a prime bigger than 7, and 2 be a primitive root modulo p.Let$a\in \mathbb {F}_{q^{2}}$,andr,sbe two integers satisfying$\gcd (s-r,q + 1)= 1$.Then the function$f(x)={\sum }_{i = 0}^{q-1}\text {Tr}^{n}_{1}(ax^{(ri+s)(q-1)})$isnot bent.

Proof

From Theorem 10 in [13], f(x) is regular bent if and only if $\text {Tr}^{n}_{1}(a)=f(0)$ and K_m(a^q+ 1) = 2 − w^f(0) − w^−f(0). Denote f(0) = i. From Proposition 2, we just need to prove that

$$2-w^{i}-w^{-i}\not\equiv 1+w^{t}+w^{-t}\pmod{\mathfrak{R}}, $$

where $\mathfrak {R}$ is a prime ideal lying above 2 in Q(w) and $0\leq t\leq \frac {p-1}{2}$. Hence, we just prove that

$$w^{t}+w^{p-1-t}+w^{i}+w^{p-1-i}+ 1\not\equiv 0 \pmod{\mathfrak{R}}. $$

Suppose that $w^{t}+w^{p-1-t}+w^{i}+w^{p-1-i}+ 1\equiv 0 \pmod {\mathfrak {R}}$. Then $w^{t}+w^{p-1-t}+w^{i}+w^{p-1-i}+ 1\pmod {\mathfrak {R}}$ is an annihilating polynomial with no more than 5 terms of no more than p − 1 degree over $\mathbb {F}_{2}$. Since 2 is a primitive root modulo p, there is only an annihilating polynomial $w^{p-1}+w^{p-2}+\cdots +w + 1\pmod {\mathfrak {R}}$ of no more than p − 1 degree over $\mathbb {F}_{2}$. Since p ≥ 7, $w^{p-1}+w^{p-2}+\cdots +w + 1\pmod {\mathfrak {R}}$ has more than 5 terms, which makes a contradiction. Hence, $w^{t}+w^{p-1-t}+w^{i}+w^{p-1-i}+ 1\not \equiv 0 \pmod {\mathfrak {R}}$, and this theorem follows. □

Remark 2

The prime required in the above theorem is just an Artin prime for 2. Let S(2) be the set of primes p such that 2 is a primitive root modulo p. Then S(2) has a positive asymptotic density inside the set of primes. Let P_i and AP_i be the numbers of primes and primes in S(2) between 3 and 10ⁱ. Artin conjecture claims that S(2) has the density C_artin ≈ 0.3739558136…. Table 1 lists some values for $\frac {AP_{i}}{P_{i}}$. And all the primes in S(2) less than 100 are 3, 5, 11, 13, 19, 29, 37, 53, 59, 61, 67, 83.

Table 1 The density of Artin primes for 2

4 Regular bent functions with five terms

In this section, we consider functions of the form

$$\begin{array}{@{}rcl@{}} f(x)&=&\text{Tr}_{1}^{n}(a_{1}x^{r_{1}(q-1)})+\text{Tr}_{1}^{n}\left( c_{1}x^{r_{1}(q-1)+\frac{q^{2}-1}{2}}\right) +\text{Tr}_{1}^{n}\left( a_{2}x^{r_{2}(q-1)}\right)\\ &&+ \text{Tr}_{1}^{n}\left( c_{2}x^{r_{2}(q-1)+\frac{q^{2}-1}{2}}\right) +bx^{\frac{q^{2}-1}{2}}, \end{array} $$

(1)

where $a_{1}, c_{1}, a_{2}, c_{2}\in \mathbb {F}_{q^{2}}$, and b ∈ F_p. If b = 0, and $a_{1}, c_{1}, a_{2}, c_{2}\in \mathbb {F}_{q^{2}}^{*}$, f(x) has four terms.

In convenience, we denote the function over U induced by f(x)

$$\widetilde{f}(u) = \text{Tr}_{1}^{n}(a_{1}u^{r_{1}})+\text{Tr}_{1}^{n}\left( c_{1}u^{r_{1}+\frac{q + 1}{2}}\right) +\text{Tr}_{1}^{n}(a_{2}u^{r_{2}})+\text{Tr}_{1}^{n}\left( c_{2}u^{r_{2}+\frac{q + 1}{2}}\right) +bu^{\frac{q + 1}{2}}. $$

Define an exponential sum

$$ S_{f}=\sum\limits_{u\in U}w^{\widetilde{f}(u)}. $$

(2)

Then the following lemma determines regular bent function f(x).

Lemma 2

Letf(x) be a p-ary function defined in (1) andS_fbe the exponential sum in (2). Thenf(x) is bent if and only ifS_f = 1.Further, iff(x) is bent, thenf(x) is regular bent.

Proof

Suppose f(x) is bent. For $\lambda \in \mathbb {F}_{q^{2}}^{*}$,

$$\begin{array}{@{}rcl@{}} W_{f}(\lambda)&=&\sum\limits_{x\in\mathbb{F}_{q^{2}}}w^{f(x)- \text{Tr}_{1}^{n}(\lambda x)}\\ &=&1+\sum\limits_{i = 0}^{q}w^{f(\xi^{i})}\sum\limits_{y\in\mathbb{F}_{q}^{*}} w^{-\text{Tr}_{1}^{n}(\lambda \xi^{i} y)}\\ &=&1+\sum\limits_{i = 0}^{q}w^{f(\xi^{i})}\sum\limits_{y\in\mathbb{F}_{q}^{*}} w^{-\text{Tr}_{1}^{m}((\lambda \xi^{i}+\lambda^{q} (\xi^{i})^{q})y)}\\ &=&1-\sum\limits_{i = 0}^{q}w^{f(\xi^{i})} +\sum\limits_{i = 0}^{q}w^{f(\xi^{i})}\sum\limits_{y\in\mathbb{F}_{q}} w^{-\text{Tr}_{1}^{m}((\lambda \xi^{i}+\lambda^{q} (\xi^{i})^{q})y)}\\ &=&1-\sum\limits_{i = 0}^{q}w^{f(\xi^{i})} +q\sum\limits_{0\le i\le q, \lambda \xi^{i}+\lambda^{q} (\xi^{i})^{q}= 0} w^{f(\xi^{i})}\\ &=&1-\sum\limits_{u\in U}w^{\widetilde{f}(u)} +qw^{f(\xi^{i_{\lambda}})}, \end{array} $$

where i_λ is the unique number such that $0\le i_{\lambda }\le q, \lambda \xi ^{i_{\lambda }}+\lambda ^{q} (\xi ^{i_{\lambda }})^{q}= 0$. From the definition of S_f,

$$ W_{f}(\lambda)= 1-S_{f}+qw^{f(\xi^{i_{\lambda}})}. $$

(3)

Since f(x) is bent, from Property 8 in [11], there exists 0 ≤ j ≤ p − 1 satisfying W_f(λ) = ±qw^j. From (3), we have $S_{f}-1-qw^{f(\xi ^{i_{\lambda }})}\pm qw^{j}= 0.$ Suppose that $S_{f}-1-qw^{f(\xi ^{i_{\lambda }})}- qw^{j}= 0$. Then we have

$$ \sum\limits_{k = 0}^{p-1}N_{k}w^{k}-1-qw^{f(\xi^{i_{\lambda}})}- qw^{j}= 0 $$

(4)

where $N_{i}=\#\{u \in U:\widetilde {f}(u)=i\}$. Obviously, N₀ + N₁ + ⋯ + N_p− 1 = q + 1. Since f(x) is bent, then 1 ≤ N_i ≤ q. Since the minimal polynomial of w is w^p− 1 + w^p− 2 + ⋯ + w + 1 = 0, (4) does not hold. Hence, $S_{f}-1-qw^{f(\xi ^{i_{\lambda }})}+qw^{j}= 0$, i.e., $j=f(\xi ^{i_{\lambda }})$. We have S_f = 1.

On the other hand,

$$\begin{array}{@{}rcl@{}} W_{f}(0)&=&\sum\limits_{x\in\mathbb{F}_{q^{2}}}w^{f(x)}\\ &=& 1+\sum\limits_{i = 0}^{q}\sum\limits_{y\in \mathbb{F}_{q}^{*}}w^{f(y\xi^{i})}\\ &=&1+\sum\limits_{i = 0}^{q}\sum\limits_{y\in \mathbb{F}_{q}^{*}}w^{f(\xi^{i})}\\ &=&1+(q-1)\sum\limits_{i = 0}^{q}w^{f(\xi^{i})}\\ &=&1+(q-1)\sum\limits_{u\in U}w^{\widetilde{f}(u)}. \end{array} $$

From the definition of $\widetilde {f}(u)$ and S_f, we have

$$ W_{f}(0)= 1+(q-1)S_{f}. $$

(5)

If S_f = 1, from (3) and (5), f(x) is bent.

If f(x) is bent, from (3) and (5), f(x) is regular bent.

Hence, this lemma follows. □

The following lemma gives a simpler expression for S_f.

Lemma 3

Letf(x) be a p-ary function defined in (1) andS_fbe the exponential sum in (2). Then

$$S_{f}=w^{b} \sum\limits_{u\in U_{0}} w^{\text{Tr}_{1}^{n}((a_{1}+c_{1})u^{r_{1}})+\text{Tr}_{1}^{n} ((a_{2}+c_{2})u^{r_{2}})} +w^{-b}\sum\limits_{u\in U_{1}}w^{ \text{Tr}_{1}^{n}((a_{1}-c_{1})u^{r_{1}})+\text{Tr}_{1}^{n}((a_{2}-c_{2})u^{r_{2}})}. $$

Proof

We have

$$\begin{array}{@{}rcl@{}} S_{f}&=&\sum\limits_{u\in U}w^{\widetilde{f}(u)}\\ &=&\sum\limits_{u\in U_{0}}w^{\widetilde{f}(u)}+ \sum\limits_{u\in U_{1}} w^{\widetilde{f}(u)}\\ &=&\sum\limits_{u\in U_{0}}w^{\text{Tr}_{1}^{n}((a_{1}+c_{1})u^{r_{1}})+\text{Tr}_{1}^{n}((a_{2}+c_{2})u^{r_{2}})+b} +\sum\limits_{u\in U_{1}}w^{\text{Tr}_{1}^{n}((a_{1}-c_{1})u^{r_{1}})+\text{Tr}_{1}^{n}((a_{2}-c_{2})u^{r_{2}})-b}\\ &=&w^{b}\sum\limits_{u\in U_{0}}w^{\text{Tr}_{1}^{n}((a_{1}+c_{1})u^{r_{1}})+ \text{Tr}_{1}^{n}((a_{2}+c_{2})u^{r_{2}})} +w^{-b}\sum\limits_{u\in U_{1}}w^{\text{Tr}_{1}^{n}((a_{1}-c_{1})u^{r_{1}})+ \text{Tr}_{1}^{n}((a_{2}-c_{2})u^{r_{2}})}, \end{array} $$

which completes the proof. □

For general f(x), S_f is difficult to compute. We consider a subclass of functions in (1) defined by

$$\begin{array}{@{}rcl@{}} f(x)&=& \text{Tr}_{1}^{n}\left( a_{1}x^{r_{1}(q-1)}\right)+ \text{Tr}_{1}^{n}\left( a_{1}x^{r_{1}(q-1)+\frac{q^{2}-1}{2}}\right) +\text{Tr}_{1}^{n}(a_{2}x^{r_{2}(q-1)})\\ &&-\text{Tr}_{1}^{n}\left( a_{2}x^{r_{2}(q-1)+\frac{q^{2}-1}{2}}\right) +bx^{\frac{q^{2}-1}{2}}, \end{array} $$

(6)

where $a_{1}, a_{2}\in \mathbb {F}_{q^{2}}$ and b ∈ F_p.

Lemma 4

Letf(x) be a p-ary function defined in (6) andS_fbe the exponential sum in (2). Then$S_{f}=w^{b} {\sum }_{u\in U_{0}} w^{\text {Tr}_{1}^{n}(2a_{1}u^{r_{1}})} +w^{-b}{\sum }_{u\in U_{1}}w^{\text {Tr}_{1}^{n}(2a_{2}u^{r_{2}})}.$

Proof

From Lemma 3, this lemma can be obviously obtained. □

Theorem 2

Letf(x) be a p-ary function defined in (6). Let$\gcd \left (r_{1}, \frac {q + 1}{2}\right )=\gcd \left (r_{2}, \frac {q + 1}{2}\right )= 1$andr₂be odd. Thenf(x) is regular bent if and only if

$$w^{b}K_{m}\left( 4a_{1}^{q + 1}\right) +w^{-b}K_{m}\left( 4a_{2}^{q + 1}\right)= \left\{ \begin{array}{ll} M(A-B)+C-2, & {2a_{1}, 2a_{2} \in \mathcal{C}_{0}^{+};} \\ MA+C-2, & {2a_{1}\in \mathcal{C}_{0}^{+}, 2a_{2}\not \in \mathcal{C}_{0}^{+};} \\ -MB+C-2, & {2a_{1}\not\in \mathcal{C}_{0}^{+}, 2a_{2}\in \mathcal{C}_{0}^{+};} \\ C-2, & {2a_{1}, 2a_{2}\not \in\mathcal{C}_{0}^{+}.} \end{array} \right. $$

where $M = 4I\sqrt {-1}$, $A=w^{b} \sin \frac {2\pi Q(2a_{1})}{p}$, $B=w^{-b}\sin \frac {2\pi Q(2a_{2})}{p}$, and $C = 2\cos \frac {2\pi b}{p}$.

Proof

Since $\gcd \left (r_{1}, \frac {q + 1}{2}\right )=\gcd \left (r_{2}, \frac {q + 1}{2}\right )= 1$ and r₂ is odd, the map $u\longmapsto u^{r_{1}}$ is a permutation from U₀ to U₀ and $u\longmapsto u^{r_{2}}$ is a permutation from U₁ to U₁. From Lemma 4, $S_{f}=w^{b}{\sum }_{u\in U_{0}} w^{\text {Tr}_{1}^{n}(2a_{1} u)} +w^{-b} {\sum }_{u\in U_{1}} w^{\text {Tr}_{1}^{n}(2a_{2} u)}.$ From Proposition 1 and Lemma 2, this theorem follows. □

Corollary 1

Letf(x) be a p-ary function defined in (6). Let$\gcd \left (r_{1}, \frac {q + 1}{2}\right )=\gcd \left (r_{2}, \frac {q + 1}{2}\right )= 1$,r₂be odd andb = 0.Thenf(x) is regular bent if and only if

$$K_{m}\left( 4a_{1}^{q + 1}\right) +K_{m}\left( 4a_{2}^{q + 1}\right)= \left\{ \begin{array}{ll} 4I\sqrt{-1}\left[\sin\frac{2\pi Q(2a_{1})}{p}- \sin\frac{2\pi Q(2a_{2})}{p}\right] , & {2a_{1}, 2a_{2} \in \mathcal{C}_{0}^{+};} \\ 4I\sqrt{-1} \sin\frac{2\pi Q(2a_{1})}{p} , & {2a_{1}\in \mathcal{C}_{0}^{+}, 2a_{2}\not \in \mathcal{C}_{0}^{+};} \\ -4I\sqrt{-1}\sin\frac{2\pi Q(2a_{2})}{p} , & {2a_{1}\not\in \mathcal{C}_{0}^{+}, 2a_{2}\in \mathcal{C}_{0}^{+};} \\ 0, & {2a_{1}, 2a_{2}\not \in\mathcal{C}_{0}^{+}.} \end{array} \right. $$

In particular, if q ≡ 1 (mod 4), then f(x) is regular bent if and only if $K_{m}(4a_{1}^{q + 1})+K_{m}(4a_{2}^{q + 1})= 0$.

Proof

From Theorem 2 , the first part of this corollary can be obviously obtained. Note that $K_{m}\left (4a_{1}^{q + 1}\right )$ and $K_{m}\left (4a_{2}^{q + 1}\right )$ are real. Since q ≡ 1 (mod 4), I is real. Hence, the rest part of this corollary also holds. □

Example: Let p = 7, m = 2, n = 2m, q = p^m ≡ 1 (mod 4), and $w=e^{\frac {2\pi \sqrt {-1}}{p}}$. Let $\mathbb {F}_{q^{2}} =\mathbb {F}_{p}(\xi )$, where the minimal polynomial of ξ is w⁴ + 5w² + 4w + 3 = 0. Then ξ is a primitive element of $\mathbb {F}_{q^{2}}$. Take r₁,r₂ satisfying $\gcd \left (r_{1}, \frac {q + 1}{2}\right )=\gcd \left (r_{2}, \frac {q + 1}{2}\right )= 1$ and r₂ is odd. Let b = 0, a₁ = ξ²⁸⁹, and a₂ = ξ⁸⁴¹. Then $K_{m}\left (4a_{1}^{q + 1}\right )=-6w^{5} - 4w^{4} - 4w^{3} - 6w^{2} - 1$ and $K_{m}\left (4a_{2}^{q + 1}\right )= 6w^{5} + 4w^{4} + 4w^{3} + 6w^{2} + 1$. And $K_{m}\left (4a_{1}^{q + 1}\right )+K_{m}(4a_{2}^{q + 1})= 0$. From Corollary 1, the function defined in (6) is a regular bent function with four terms.

Corollary 2

Letf(x) be a p-ary function defined in (6). Let$\gcd \left (r_{1}, \frac {q + 1}{2}\right )=\gcd \left (r_{2}, \frac {q + 1}{2}\right )= 1$,r₂be odd,b≠ 0 and$2a_{1}, 2a_{2}\not \in \mathcal {C}_{0}^{+}$.Thenf(x) is regular bent if and only if$K_{m}\left (4a_{1}^{q + 1}\right )= K_{m}\left (4a_{2}^{q + 1}\right )= 1-{\sec \frac {2\pi b}{p}}.$

Proof

From Theorem 2, if $2a_{1}, 2a_{2}\not \in \mathcal {C}_{0}^{+}$, f(x) is regular bent if and only if $w^{b} K_{m}\left (4a_{1}^{q + 1}\right )+w^{-b}K_{m}\left (4a_{2}^{q + 1}\right ) = 2\cos \frac {2\pi b}{p}-2.$ Take the complex conjugate of both sides. And we have $w^{-b}K_{m}\left (4a_{1}^{q + 1}\right )+w^{b}K_{m}\left (4a_{2}^{q + 1}\right ) = 2\cos \frac {2\pi b}{p}-2.$ Since b≠ 0, we have $K_{m}\left (4a_{1}^{q + 1}\right )=K_{m}\left (4a_{2}^{q + 1}\right )= 1-{\sec \frac {2\pi b}{p}}.$ Hence, this corollary follows. □

Remark 3

From Corollary 2 and Theorem 3.9 in [18], if p = 11, $\gcd \left (r_{1}, \frac {q + 1}{2}\right )=\gcd \left (r_{2}, \frac {q + 1}{2}\right )= 1$ and r₂ is odd, then for any $2a_{1}, 2a_{2}\not \in \mathcal {C}_{0}^{+}$ and b≠ 0, the function f(x) defined in (6) is not bent.

Example. Let p = 5, m = 4, n = 2m, q = p^m ≡ 1 (mod 4), and $w=e^{\frac {2\pi \sqrt {-1}}{p}}$. Let $\mathbb {F}_{q^{2}} =\mathbb {F}_{p}(\xi )$, where the minimal polynomial of ξ is ξ⁸ + ξ⁴ + 3ξ² + 4ξ + 2 = 0. Then ξ is a primitive element of $\mathbb {F}_{q^{2}}$. Take r₁,r₂ satisfying $\gcd \left (r_{1}, \frac {q + 1}{2}\right )=\gcd \left (r_{2}, \frac {q + 1}{2}\right )= 1$ and r₂ is odd. Take b = 1, a₁ = ξ⁶⁴⁴⁰¹, and a₂ = ξ³⁷⁴⁹²⁵. Then $2a_{1}, 2a_{2}\not \in \mathcal {C}_{0}^{+}$ and $K_{m}(4a_{1}^{q + 1})=K_{m}\left (4a_{2}^{q + 1}\right )= 1-{\sec \frac {2\pi }{p}} =-\sqrt {5}$. From Corollary 2, the function defined in (6) is a regular bent function with five terms.

Corollary 3

Letf(x) be a p-ary function defined in (6). Let$\gcd \left (r_{1}, \frac {q + 1}{2}\right )=\gcd \left (r_{2}, \frac {q + 1}{2}\right )= 1$,r₂be odd,b = 0,anda₁ = a₂ = a.Thenf(x) is regular bent if and only ifK_m (4a^q+ 1) = 0.

Proof

From Corollary 1, this corollary can be obviously obtained. □

Remark 4

Kononen et al. [10] proved that if p ≥ 5, for any $a\in \mathbb {F}_{q}$, K_m(a)≠ 0. Hence, if p ≥ 5, a p-ary function in Corollary 3 is not bent.

Example. Let p = 3, m = 4, n = 2m, q = p^m ≡ 1 (mod 4), and $w=e^{\frac {2\pi \sqrt {-1}}{p}}$. Let $\mathbb {F}_{q^{2}} =\mathbb {F}_{p}(\xi )$, where the minimal polynomial of ξ is ξ⁸ + 2ξ⁵ + ξ⁴ + 2ξ² + 2ξ + 2 = 0. Then ξ is a primitive element of $\mathbb {F}_{q^{2}}$. Take r₁,r₂ satisfying $\gcd \left (r_{1}, \frac {q + 1}{2}\right )=\gcd \left (r_{2}, \frac {q + 1}{2}\right )= 1$ and r₂ is odd. Take b = 0, a₁ = a₂ = a = ξ⁴³⁴. Then K_m(4a^q+ 1) = 0. From Corollary 3, the function defined in (6) is a regular bent function with four terms.

Corollary 4

Letf(x) be a p-ary function defined in (6). Let$\gcd \left (r_{1}, \frac {q + 1}{2}\right )=\gcd \left (r_{2}, \frac {q + 1}{2}\right )= 1$,r₂be odd,a₁ = a₂ = aandb≠ 0.Thenf(x) is regular bent if and only if

$$K_{m}(4a^{q + 1})=\left\{ \begin{array}{ll} -4I\sin\frac{2\pi b}{p}\sin\frac{2 \pi Q(2a)}{p}{\sec\frac{2\pi b}{p}}+ 1- {\sec\frac{2\pi b}{p}}, & {2a\in \mathcal{C}_{0}^{+};} \\ 1- {\sec\frac{2\pi b}{p}} , & {2a\not \in \mathcal{C}_{0}^{+}.} \end{array} \right. $$

In particular, if q ≡ 3 (mod 4), f(x) is regular bent if and only if $K_{m}\left (4a^{q + 1}\right )= 1-{\sec \frac {2\pi b}{p}}$.

Proof

Note that if q ≡ 3 (mod 4), then I is not real. From Theorem 2, this corollary can be obviously obtained. □

Theorem 3

Letf(x) be a p-ary function defined in (6). Let$\gcd \left (r_{1}, \frac {q + 1}{2}\right )=\gcd \left (r_{2}, \frac {q + 1}{2}\right )= 1$andr₂be even. Thenf(x) is bent if and only if

$$w^{b}K_{m}\left( 4a_{1}^{q + 1}\right) +w^{-b}K_{m}(4a_{2}^{q + 1})= \left\{ \begin{array}{ll} M(A+B)+C-2, & {2a_{1}, 2a_{2} \in \mathcal{C}_{0}^{+};} \\ MA+C-2, & {2a_{1}\in \mathcal{C}_{0}^{+}, 2a_{2}\not \in \mathcal{C}_{0}^{+};} \\ MB+C-2, & {2a_{1}\not\in \mathcal{C}_{0}^{+}, 2a_{2}\in \mathcal{C}_{0}^{+};} \\ C-2, & {2a_{1}, 2a_{2}\not \in\mathcal{C}_{0}^{+}.} \end{array} \right. $$

where $M = 4I\sqrt {-1}$, $A=w^{b} \sin \frac {2\pi Q(2a_{1})}{p}$, $B=w^{-b}\sin \frac {2\pi Q(2a_{2})}{p}$, and $C = 2\cos \frac {2\pi b}{p}$.

Proof

Since $\gcd \left (r_{1}, \frac {q + 1}{2}\right )=\gcd \left (r_{2}, \frac {q + 1}{2}\right )= 1$ and r₂ is even, then the map $u\longmapsto u^{r_{1}}$ is a permutation from U₀ to U₀ and $u\longmapsto u^{r_{2}}$ is a bijection between U₁ and U₀. From Lemma 4, $S_{f}=w^{b} {\sum }_{u\in U_{0}} w^{\text {Tr}_{1}^{n}(2a_{1} u)}+ w^{-b}{\sum }_{u\in U_{0}}w^{\text {Tr}_{1}^{n}(2a_{2} u)}.$ From Proposition 1 and Lemma 2, this theorem follows. □

Example. Let p = 3, m = 6 and n = 2m. Let $\mathbb {F}_{q^{2}} =\mathbb {F}_{p}(\xi )$, where the minimal polynomial of ξ is ξ¹² + ξ⁶ + ξ⁵ + ξ⁴ + ξ² + 2 = 0. Then ξ is a primitive element of $\mathbb {F}_{q^{2}}$. Take r₁,r₂ satisfying $\gcd \left (r_{1}, \frac {q + 1}{2}\right )=\gcd \left (r_{2}, \frac {q + 1}{2}\right )= 1$ and r₂ is even. Take b = 1, a₁ = ξ⁸⁸⁹⁷⁶ and a₂ = ξ³²⁵¹⁸⁹. Then $2a_{1}, 2a_{2}\not \in \mathcal {C}_{0}^{+}$ and $K_{m}\left (4a_{1}^{q + 1}\right )=K_{m}\left (4a_{2}^{q + 1}\right )= 1-{\sec \left (\frac {2\pi }{p}\right )}= 3$. From Theorem 3, the function defined in (6) is a regular bent function with five terms.

Theorem 4

Letf(x) be a p-ary function defined in (1). If$\gcd \left (r_{1},r_{2}, \frac {q + 1}{2}\right )>1$,thenf(x) is not bent.

Proof

Let $d=\gcd \left (r_{1},r_{2}, \frac {q + 1}{2}\right )$. From Lemma 3,

$$S_{f}=dw^{b}\sum\limits_{v\in \mathcal{H}_{0}} w^{\text{Tr}_{1}^{n}((a_{1}+c_{1})u^{r_{1}/d})+ \text{Tr}_{1}^{n}((a_{2}+c_{2})u^{r_{2}/d})}+dw^{-b}\sum\limits_{v\in \mathcal{H}_{1}} w^{\text{Tr}_{1}^{n}((a_{1}-c_{1})u^{r_{1}/d})+\text{Tr}_{1}^{n}((a_{2}-c_{2})u^{r_{2}/d})}, $$

where $\mathcal {H}_{0}={U_{0}^{d}}$ and $\mathcal {H}_{1}= {U_{1}^{d}}$. Hence, S_f ≡ 0 (mod d). Since d > 1, then S_f≠ 1. From Lemma 2, f(x) is not bent. □

Corollary 5

Letq ≡ 3 (mod 4).Let$f(x)=\text {Tr}_{1}^{n}(ax^{r(p^{m}-1)})+bx^{\frac {q^{2}-1}{2}}$,where$a\in \mathbb {F}_{q}$,$b\in \mathbb {F}_{p}$, r is even, and$\gcd (\frac {r}{2}, q + 1)= 1$.Thenf(x) is not bent.

Proof

In Theorem 4, take a₁ = a, c₁ = 0, a₂ = c₂ = 0, r₁ = r, and r₂ = 0. Then $2|\gcd \left (r,0, \frac {q + 1}{2}\right )$. From Theorem 4, f(x) is not bent. □

Remark 5

Corollary 5 is a generalization of Theorem 3 in [21]. [21] just discussed the case q ≡ 7 (mod 8) and did not solve the case q ≡ 3 (mod 8).

5 Conclusion

This paper first presents a congruence property for Kloosterman sums and with it prove the nonexistence of some regular p-ary bent functions. Further, we study p-ary functions of the form $f(x)=\text {Tr}^{n}_{1}\left (a_{1}x^{r_{1}(q-1)}\right )+\text {Tr}^{n}_{1}\left (a_{1}x^{r_{1}(q-1)+\frac {q^{2}-1}{2}}\right ) +\text {Tr}^{n}_{1}\left (a_{2}x^{r_{2}(q-1)}\right )-\text {Tr}^{n}_{1}\left (a_{2}x^{r_{2}(q-1)+\frac {q^{2}-1}{2}}\right )+bx^{\frac {q^{2}-1}{2}}$ and characterize the bentness of these functions with Kloosterman sums. Finally, we solve an open problem in [9] and prove the nonexistence of some regular bent functions. A natural problem is to study general regular p-ary bent functions of the form $f(x)=\text {Tr}^{n}_{1}\left (a_{1}x^{r_{1}(q-1)}\right )+\text {Tr}^{n}_{1}\left (c_{1}x^{r_{1}(q-1) +\frac {q^{2}-1}{2}}\right )+\text {Tr}^{n}_{1}\left (a_{2}x^{r_{2}(q-1)}\right ) +\text {Tr}^{n}_{1}\left (c_{2}x^{r_{2}(q-1)+\frac {q^{2}-1}{2}}\right )+bx^{\frac {q^{2}-1}{2}}$, which is our further work.