AAECC DOI 10.1007/s00200-015-0258-3 ORIGINAL PAPER

On more bent functions from Dillon exponents Long Yu1 · Hongwei Liu1 · Dabin Zheng2

Received: 14 April 2014 / Revised: 14 March 2015 / Accepted: 24 March 2015 © Springer-Verlag Berlin Heidelberg 2015

Abstract In this paper, we obtain several new classes of p-ary bent functions, where p is a prime. The bentness of all these functions is characterized by some exponential sums, which have close relations with Kloosterman sums. Moreover, we obtain some concise criterions on the bentness of p-ary functions in some special cases. In addition, our work generalizes some main results obtained by Li et al. (IEEE Trans Inf Theory 59(3):1818–1831, 2013). Keywords

Dickson polynomial · Kloosterman sum · p-ary bent function

1 Introduction In 1976, Rothaus [13] introduced boolean bent functions which are maximally nonlinear boolean functions with even number of variables. It means that boolean bent functions achieve the maximal Hamming distance between boolean functions and affine functions. Since then, boolean bent functions have attracted much attention due to their application in coding theory, cryptography and sequence design. Later, Kumar et al. [8] generalized the notion of boolean bent functions to the case of functions over

B

Hongwei Liu [email protected] Long Yu [email protected] Dabin Zheng [email protected]

1

School of Mathematics and Statistics, Central China Normal University, Wuhan 430079, Hubei, China

2

School of Mathematics and Statistics, Hubei University, Wuhan 430062, Hubei, China

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an arbitrary finite field F pn , where p is a prime and n is a positive integer. Some results on constructions of bent functions on monomial, binomial and quadratic functions can be found in [1–7,9,10,12,14–16]. Let p be a prime and m be a positive integer, n = 2m and F pn be the finite field with p n elements, and F∗pn = F pn \{0}. Let Tr n1 (·) be the trace function from F pn to n−1 pi x for all x ∈ F pn . The bentness of boolean monomials with F p , i.e. Tr n1 (x) = i=0 Dillon exponents was characterized by Dillon in [4] and Charpin et al. in [2]. The corrosponding p-ary case was investigated by Helleseth et al. [5]. Some multinomial bent functions with Dillon exponents were investigated in [7,12,14,15]. Recently, Li et al. [10] investigated the bentness of several special classes of p-ary functions of the following form f (x) =

m −1 p

Tr n1

    pn −1 o(d) i( p m −1) d bx , ai x + Tr 1

(1.1)

i=1

where ai ∈ F pn , b ∈ F po(d) , d is a positive integer with d | ( p m + 1) and o(d) is the smallest positive integer satisfying o(d) | n and d | ( p o(d) − 1). The bentness of m some special classes of p-ary functions with Dillon exponents ( p d+1 ± l)( p m − 1) is determined by some exponential sums, most of which have close relations with Kloosterman sums. This kind of Dillon exponents leads to some detailed new characterizations for the bentness of p-ary functions, and some new p-ary bent functions are obtained. In this paper, we first investigate the bentness of several classes of p-ary functions m of the form (1.1) with Dillon exponents (i p d+1 +l)( p m −1), where i ∈ {0, . . . , d −1}. m This kind of Dillon exponents is a generalization of exponents ( p d+1 ±l)( p m −1) and it also leads us to detailed new characterizations for the bentness of p-ary functions. Following the section 2 in [7] we further investigate a relationship between some partial exponential sums and Kloosterman sums. Based on this relationship, we get a new characterization for the bentness of a class of binomial p-ary functions (see Theorem 3.15). Last, the bentness of other classes of bent functions with Dillon exponents of the form (1.1) is also characterized by some exponential sums. In some special cases, we can characterize the bentness of these functions from simple approaches (see Corollaries 3.11, 3.21, 3.22). The remainder of this paper is organized as follows. Section 2 gives some preliminaries. In Sect. 3, the bentness of several classes of functions is characterized by some exponential sums. The concluding remarks are given in Sect. 4.

2 Preliminaries Throughout this paper, we fix some notations. • p is a prime, and m is a positive integer, n = 2m and d is a divisor of p m + 1. √ 2π −1 p

• ω=e of F pn .

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is a complex primitive p-th root of unity, and α is a primitive element

On more bent functions from Dillon exponents

In the following, we give some basic definitions and results. Definition 2.1 Let f : F pn → F p be a p-ary function. The Walsh transform of f is defined by 

W f (λ) =

ω f (x)−Tr1 (λx) , λ ∈ F pn . n

x∈F pn

Definition 2.2 Let f : F pn → F p be a p-ary function. Then f (x) is called a bent function if |W f (λ)|2 = p n for all λ ∈ F pn . A p-ary bent function f (x) is said to be n ∗ regular if for all λ ∈ F pn , W f (λ) = p 2 ω f (λ) for some function f ∗ from F pn to F p . The function f ∗ (x) is called the dual of f (x). Remark 2.3 In particular, for p = 2, a boolean bent function is always regular. Definition 2.4 The Dickson polynomial Dr (x) ∈ F2 [x] of degree r is defined by Dr (x) =

r/2  i=0

  k where = s

s−1 j=0 (k− j)  s j=1 j



r r −i

r −i i

and

 r2 

=



x r −2i , r = 2, 3, . . . ,

r/2, (r − 1)/2,

if r is even; otherwise.

Definition 2.5 Let β ∈ F pm , the Kloosterman sum K m (β) over F pm is defined as 

K m (β) =

ωTr1 (βx+x m

pm −2 )

.

x∈F pm

= 1, x ∈ F pn } be a cyclic subgroup of F∗pn . It is easy to see

m di | 0 ≤ i < p +1 }, that U can be decomposed into U = d−1 k=0 Vk , where V0 = {ξ d Vk = ξ k V0 for 1 ≤ k ≤ d − 1, and ξ is a generator of the cyclic group U . For i = 0, 1, . . . , d − 1 and a ∈ F pn , we define Let U = {x | x p

m +1

Si (a) =



ωTr1 (aξ n

i x)

.

x∈V0

It is well known that if p = 2, then 

ωTr1 (ax) = n

d−1 

x∈U

Si (a) = 1 − K m (a), a ∈ F2m .

i=0

If p > 2, then  x∈U

ωTr1 (ax) = n

d−1 

 m  Si (a) = 1 − K m a p +1 , a ∈ F pn ,

i=0

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which is given in [5]. In particular, if p = 2 and d = 3, Mesnager [11] gave a relationship between Si (a) and some well-known exponential sums, and then obtained a new class of binomial bent functions. Furthermore, if p = 2 and d = 5, Si (a) was determined by some well-known exponential sums in [14]. By using these results, Tang et al. [14] characterized the bentness of a new class of binomial functions. For p > 2 and d = 2, the only known results on Si (a), where i = 0, 1, were obtained by Jia et al. [7], which were used to characterize a new class of binomial bent functions. Following this line, the bentness of more p-ary functions can be concisely characterized if Si (a) is obtained for some p and d, where 0 ≤ i ≤ d − 1. When p = 2, Li et al. [10] obtained a relation between S0 (a) and some exponential sums as follows. Lemma 2.6 [10] Let p = 2, a = aξ k ∈ F2n , where 0 ≤ k ≤ 2m , and a ∈ F∗2m . If k ≡ 0 (mod d), then S0 (a) = where E m,d (a) =



x∈F2m (−1)

1 + 2E m,d (a) − K m (a) , d

Tr m 1 (a Dd (x)) ,

ξ is a generator of the cyclic group U .

Let p be an odd prime, Ct = {α 2i+t | i = 0, 1, . . . , p 2−3 } ⊆ F∗pn , where t = 0, 1. For a ∈ F pn and b ∈ C0 , we define R(a) and Q(b) as follows: n

 m 1 − K m a p +1 , R(a) = 2

Q(b) = 2Tr m 1 (b

pm +1 2

).

When d = 2, the values of S0 (a) and S1 (a) can be obtained as follows. Lemma 2.7 [7] Let the notations be given as above. We have S0 (a) =

R(a) + I (ω Q(a) − ω−Q(a) ), R(a),

a ∈ C0+ ; otherwise,

S1 (a) =

R(a) − I (ω Q(a) − ω−Q(a) ), R(a),

a ∈ C0+ ; otherwise,

and

where I =

⎧ m ⎨(−1) 3m 2 p2 2

m

⎩(−1)m p 2 2

,

,

p ≡ 3 (mod 4); otherwise,

and C0+ = {a ∈ C0 | Q(a) = 0}.

If p m ≡ 3 (mod 4), d = 4 and a = α j ( p +1) ∈ C0 , where j is a positive integer with 0 ≤ j ≤ p m − 2, we get the following relationship between Si (a) and Kloosterman sum, where i = 1, 3. m

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On more bent functions from Dillon exponents

Proposition 2.8 Let the notations be given as above. We have S1 (a) = S3 (a) =

R(a) − I (ω Q(a) − ω−Q(a) ) , 2

where R(a), Q(a), I are given in Lemma 2.7. Proof Note that p m ≡ 3 (mod 4), we have 4 | ( p m + 1), hence S1 (a) =





ωTr1 (aξ x) = n

x∈V0

=



ωTr1 (aξ n

pm x pm )

=

x∈V0

ω

Tr n1 (aξ 3 x)



ωTr1 (aξ n

3 ξ pm −3 x)

x∈V0

= S3 (a).

x∈V0

Since S1 (a) + S3 (a) = Lemma 2.7, then



ωTr1 (aξ x) , where H = {ξ 2i | 0 ≤ i ≤ n

x∈H

 S1 (a) = S3 (a) =

R(a)−I (ω Q(a) −ω−Q(a) ) , 2 R(a) 2 ,

p m −1 2 },

by

a ∈ C0+ ; otherwise.

We have Q(a) = 0 when a ∈ C0 \C0+ , then the result follows.



As we known, for an odd prime p, every x ∈ F∗pn has a unique representation as m x = uy, where u ∈ U = {1, α, . . . , α p }, y ∈ F∗pm . The following result can be easily obtained. Proposition 2.9 For λ ∈ F∗pn , there exists only one solution in U such that Tr nm (λu) = 0. To investigate the bentness of f (x) defined by (1.1), we need the following lemma. Lemma 2.10 [5] Let p be an odd prime, f : F pn → F p be a regular bent function such that f (x) = f (−x) and f (0) = 0, then f ∗ (0) = 0, where f ∗ is the dual function of f . A necessary and sufficient condition such that f (x) defined by (1.1) is regular bent, which was given in [10]. We restate this result and give a different proof. Lemma 2.11 [10] Assume the notations given as above. Then the function f (x) defined by (1.1) is regular bent if and only if S(a1 , a2 , . . . , a pm −1 , b) = 1, where S(a1 , . . . , a pm −1 , b) =

 x∈U

ω

 pm −1 i=1

o(d)

Tr n1 (ai x i )+Tr 1 (bx

pm +1 d )

.

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Proof We first compute the Walsh transform of f (x). If λ = 0, then 

W f (0) =

ω f (x) = 1 +

 

 = 1 + pm − 1 ω 

pm −1 i=1

u∈U

 = 1 + pm − 1 ω 

ω

i=1

   m o(d) bu Tr n1 ai u i ( p −1) +Tr 1

pn −1 d



y∈F∗pm

u∈U

x∈F pn

pm −1

pm −1 i=1

   m o(d) Tr n1 ai u i ( p −1) +Tr 1 bu

  o(d) Tr n1 ai x i +Tr 1 bx

pm +1 d

pn −1 d





x∈U

  = 1 + p m − 1 S a1 , a2 , . . . , a pm −1 , b .

(2.1)

If λ ∈ F∗pn , then W f (λ) =



ω f (x)−Tr1 (λx) n

x∈F pn

= 1+

pm −1

 

ω

i=1

   m o(d) Tr n1 ai u i ( p −1) +Tr 1 bu

pn −1 d

   m o(d) bu Tr n1 ai u i ( p −1) +Tr 1

pn −1 d

 −Tr n1 (λuy)

u∈U y∈F∗pm

= 1+

pm −1

 

ω

i=1

 −Tr n1 (λuy)

u∈U y∈F pm

−



pm −1

ω

i=1

   m o(d) Tr n1 ai u i ( p −1) +Tr 1 bu

u∈U

= 1+



pm −1

ω

i=1

pn −1 d

   m o(d) Tr n1 ai u i ( p −1) +Tr 1 bu



pn −1 d



u∈U

−



ω−Tr1 ( yTrm (λu)) m

n

y∈F pm pm −1

ω



i=1

 o(d) Tr n1 ai x i +Tr 1 bx

x∈U pm −1

= 1 + pm ω

i=1



pm +1 d



  pn −1   i ( pm −1) o(d) bu λ d Tr n1 ai u λ +Tr 1

 − S a1 , a2 , . . . , a pm −1 , b , (2.2)

where u λ satisfies Tr nm (λu) = 0 and the last equality in (2.2) is obtained by Proposition 2.9. Now, we finish the proof by two cases.

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On more bent functions from Dillon exponents

Case I, p = 2. If S(a1 , a2 , . . . , a2m −1 , b) = 1, we can get f (x) is bent from Eqs. (2.1) and (2.2). Conversely, if f (x) is bent, then W f (0) = 1 + (2m − 1)S(a1 , a2 , . . . , a2m −1 , b) ∈ {±2m }. Since S(a1 , a2 , . . . , a2m −1 , b) is an integer, then S(a1 , a2 , . . . , a2m −1 , b) = 1. ∗ Case II, p > 2. If f (x) is regular bent, then W f (0) = p m ω f (0) by Definition 2.2. By Lemma 2.10, we have W f (0) = 1+( p m −1)S(a1 , a2 , . . . , a pm −1 , b) = p m . Therefore, we get S(a1 , a2 , . . . , a pm −1 , b) = 1. Conversely, if S(a1 , a2 , . . . , a pm −1 , b) = 1, it is easy to find that f (x) is regular bent from Eqs. (2.1) and (2.2).



3 Binary and p-ary bent functions In this section, we study several classes of bent functions with Dillon exponents of the form (1.1) for p = 2 and p > 2, respectively. 3.1 Binary bent functions In the following, we investigate two classes of bent functions of the form (1.1). 3.1.1 First class of binary bent functions We assume that d and l are positive integers such that gcd(l, 2 the bentness of the following functions f a0 ,...,ad−1 ,b (x) =

d−1 

m +1

d

) = 1, and study

  2m +1    2n −1  l+i d (2m −1) o(d) + Tr 1 bx d , Tr n1 ai x

(3.1)

i=0

where ai ∈ F2n , 0 ≤ i ≤ d − 1, b ∈ F2o(d) . In what follows, we give a general characterization on the bentness of function f a0 ,...,ad−1 ,b (x) defined by (3.1). Based on this characterization, we investigate two special classes of binary bent functions of the form (3.1). Moreover, from these two special classes of binary bent functions, we can obtain some binary bent functions, which had been discussed by Li et al. [10]. Proposition 3.1 Let the notations be given as above. Then the function fa0 ,...,ad−1 ,b (x) defined by (3.1) is bent if and only if d−1 

o(d)

(−1)

Tr 1

 bξ

j (2m +1) d

j=0





d−1 

(−1) i=0

  m j (i 2 d+1 ) jl Tr n1 ai ξ ξ x

= 1.

x∈V0

Proof By Lemma 2.11, f a0 ,...,ad−1 ,b (x) is bent if and only if 

d−1 

(−1) i=0

Tr n1 (ai x

m 2m +1 l+i 2 d+1 o(d) )+Tr 1 (bx d )

= 1.

x∈U

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Note that d−1 



(−1) i=0

    m 2m +1 l+i 2 d+1 o(d) +Tr 1 bx d Tr n1 ai x

x∈U

=

d−1 



(−1) i=0

 o(d) Tr n1 ai x l +Tr 1 (b)

x∈V0

+··· +



d−1 

(−1) i=0

+

d−1 



(−1) i=0

x∈V0

 Tr n1 ai ξ

     m (d−1)(2m +1) (d−1) l+i 2 d+1 o(d) d bξ x l +Tr 1

x∈V0

=

d−1 



(−1) i=0

o(d)

Tr n1 (ai x)+Tr 1 (b)

x∈V0

+··· +



d−1 

(−1) i=0

x∈V0

=

d−1 

o(d)

(−1)

Tr 1

 bξ

+



d−1 

(−1) i=0

    m 2m +1 i 2 +1 o(d) bξ d Tr n1 ai ξ d ξ l x +Tr 1

x∈V0      m (d−1)(2m +1) (d−1) i 2 d+1 o(d) d ai ξ bξ ξ (d−1)l x +Tr 1

 Tr n1

j (2m +1) d

j=0

    m 2m +1 l+i 2 d+1 l o(d) bξ d Tr n1 ai ξ x +Tr 1





d−1 

(−1) i=0

 Tr n1 ai ξ

  m  j i 2 d+1 ξ jl x

,

(3.2)

x∈V0



then we finish the proof.

From Proposition 3.1, the necessary and sufficient condition on the bentness of f a0 ,...,ad−1 ,b (x) defined by (3.1) is indeed complex. However, if we take some special values of ai and b, then the bentness of f a0 ,...,ad−1 ,b (x) defined by (3.1) can be concisely characterized as follows. Theorem 3.2 Let a0 ∈ F∗2m , a1 = a2 = · · · = ad−1 ∈ F2m and a0 = a1 , then f a0 ,...,ad−1 ,0 (x) defined by (3.1) is bent if and only if K m (a0 ) + (d − 1)K m (a0 + a1 )  2 E m,d (a0 ) + (d − 1)E m,d (a 0 + a1 ) , = 2 E m,d (a0 ) − E m,d (a0 + a1 ) ,

if d | l; if gcd(d, l) = 1,

where E m,d (a) is given in Lemma 2.6. Proof Since ξ j

2m +1 d

is a root of 1 + z + z 2 + · · · + z d−1 = 0 for each 1 ≤ j ≤ d − 1, d−1   n 2m +1 and a1 = a2 = · · · = ad−1 , we get Tr 1 (ai ξ i( j d ) ξ jl x) = Tr n1 (a0 + a1 )ξ jl x i=0

for each 1 ≤ j ≤ d − 1. Note that b = 0 and gcd(l, 2 equivalent to

123

m +1

d

) = 1, then Eq. (3.2) is

On more bent functions from Dillon exponents



d−1 

(−1) i=0

x∈U

=



  m l+i 2 d+1 Tr n1 ai x

(−1)Tr1 (a0 x) + n

x∈V0



(−1)Tr1 ((a0 +a1 )ξ n

l x)

+ ··· +

x∈V0



(−1)Tr1 ((a0 +a1 )ξ n

(d−1)l x)

.

x∈V0

(3.3) In the following, we discuss Eq. (3.3) in two cases. 1. If d | l, by Lemma 2.6, then 

d−1 

(−1) i=0

Tr n1 (ai x

m l+i 2 d+1 )

=

x∈U



(−1)Tr1 (a0 x) + (d − 1) n

x∈V0



(−1)Tr1 ((a0 +a1 )x) n

x∈V0

1 + 2E m,d (a0 + a1 ) − K m (a0 + a1 ) 1 + 2E m,d (a0 ) − K m (a0 ) + (d − 1) . = d d Hence, by Proposition 3.1, f a0 ,...,ad−1 ,b (x) is bent if and only if  K m (a0 ) + (d − 1)K m (a0 + a1 ) = 2 E m,d (a0 ) + (d − 1)E m,d (a0 + a1 ) . 2. If gcd(d, l) = 1, we have {l (mod d), 2l (mod d), . . . , (d − 1)l (mod d)} = {1, 2, . . . , d − 1}. By Lemma 2.6, 

d−1 

(−1) i=0

x∈U

=



Tr n1 (ai x

m l+i 2 d+1 )

(−1)Tr1 (a0 x) + n

x∈V0

=



=

(−1)Tr1 ((a0 +a1 )ξ x) + · · · + n

x∈V0

(−1)

Tr n1 (a0 x)

(−1)

Tr n1 (a0 x)

+

x∈V0







x∈V0

+

x∈U

(−1)Tr1 ((a0 +a1 )ξ n

d−1 x)

x∈V0

(−1)

Tr n1 ((a0 +a1 )x)

+ ··· +

x∈V1





(−1)

Tr n1 ((a0 +a1 )x)

−





(−1)Tr1 ((a0 +a1 )x) n

x∈Vd−1

(−1)Tr1 ((a0 +a1 )x) n

x∈V0

1 + 2E m,d (a0 ) − K m (a0 ) = 1 − K m (a0 + a1 ) + d 1 + 2E m,d (a0 + a1 ) − K m (a0 + a1 ) . − d Therefore, by Proposition 3.1, f a0 ,...,ad−1 ,b (x) is bent if and only if K m (a0 ) + (d − 1)K m (a0 + a1 ) = 2E m,d (a0 ) − 2E m,d (a0 + a1 ). This finishes the proof.



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Example 3.3 Let n = 2m = 6, d = 9 and l = 1, a0 ∈ F∗23 , a1 = a2 = · · · = a8 ∈ F∗23 , 8 Tr 61 (a1 x 7(1+i) ). By using Maple, there exist then f a0 ,...,a8 ,0 (x) = Tr 61 (a0 x 7 ) + i=1 9 pairs (a0 , a1 ) such that f a0 ,...,a8 ,0 (x) is bent. If we take d = 3 in Theorem 3.2, and combine the results on Si (a) in [11], i = 0, 1, 2, we obtain the following result, which is exact [10, Corollary 1]. Corollary 3.4 [10] Let d = 3, b = 0, a0 ∈ F∗2m , a1 = a2 ∈ F2m and a0 = a1 , then f a0 ,a1 ,a2 ,0 (x) defined by (3.1) is bent if and only if  2 (C (a ) + 2C (a + a1 )), if3 | l; K m (a0 ) + 2K m (a0 + a1 ) = 2 (Cm (a0 ) − C m(a 0+ a )), otherwise, m 0 m 0 1 where Cm (a) =



a∈F2m (−1)

3 Tr m 1 (ax +ax) .

For b = 0, by a similar proof as that in Theorem 3.2, we obtain the following result. Theorem 3.5 Let b = 0, d | l, a0 ∈ F∗2m , a1 = a2 = · · · = ad−1 ∈ F2m and a0 = a1 , then f a0 ,...,ad−1 ,b (x) defined by (3.1) is bent if and only if ρ K m (a0 ) + σ K m (a0 + a1 ) = 2(ρ E m,d (a0 ) + σ E m,d (a0 + a1 )) + ρ + σ − d, where ρ = Lemma 2.6.

o(d)

(−1)Tr1 (b) ,

σ =

d−1 j=1

o(d)(bξ

(−1)Tr1

m j 2 d+1 )

and E m,d (a) is given in

Example 3.6 Let n = 2m = 4, d = 5, l = 5, and α be a primitive element of F24 , a0 ∈ F∗22 , a0 = a1 , a1 = a2 = a3 = a4 ∈ F∗22 , b ∈ F∗24 , then f a0 ,...,a4 ,b (x) defined by 4 Tr 41 (a1 x 3(5+i) ) + Tr 41 (bx 3 ). By using Maple, the (3.1) is equal to Tr 41 (a0 x 15 ) + i=1 number of (a0 , a1 , b) such that f a0 ,...,a4 ,b (x) is bent is 60. The following result is a corollary of Theorem 3.5, which is exact [10, Theorem 3]. Corollary 3.7 [10] Let d | l, a0 ∈ F∗2m and a1 = · · · = ad−1 = 0, then f a0 ,0,...,0,b (x) defined by (3.1) is bent if and only if m d−1  j 2 d+1 o(d) ) (−1)Tr1 (bξ =

j=0

d , 1 + 2E m,d (a0 ) − K m (a0 )

where E m,d (a) is given in Lemma 2.6. Remark 3.8 Some other special cases of Proposition 3.1 can be discussed as above. Li et al. [10] investigated the bentness of binary functions with Dillon exponent m ( 2 d+1 ± l)(2m − 1), and then obtained some new binomial, trinomial bent functions. In this subsection, we study the bentness of binary functions with Dillon exponent m (l + i 2 d+1 )(2m − 1), where i ∈ {0, . . . , d − 1}, and we give some new different

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On more bent functions from Dillon exponents

characterizations on their bentness. Since the bentness of the function defined by (3.1) for the case of a2 = · · · = ad−2 = 0 had been discussed in [10] and other cases are novel. From Theorem 3.2 and Theorem 3.5, we can obtain some new multinomial bent functions for suitable values of ai and b (see Example 3.3 and Example 3.6). 3.1.2 Second class of binary bent functions In this subsection, we assume that s, k, r are integers and g = gcd(r, 2m + 1). Let 2m +1 g −1



f a,r,s (x) =

  m Tr n1 ax (ri+s)(2 −1) ,

(3.4)

i=1

where a ∈ F∗2n and f (0) = 0. In the following, we give a necessary and sufficient condition on the bentness of f a,r,s (x) defined by (3.4) for arbitrary r . Theorem 3.9 Let the notations be given as above. 1. If gcd(s, 2m + 1) = 1, 0 ≤ k ≤ 2m and a = aξ k ∈ F2n with a ∈ F∗2m , then f a,r,s (x) defined by (3.4) is bent if and only if 

K m (a) = g −

(−1)Tr1 (ax) . n

x g =1,x∈U

2. If gcd(s, 2m + 1) = d, 0 ≤ k < 2 d+1 and a = aξ kd ∈ F2n with a ∈ F∗2m , then f a,r,s (x) defined by (3.4) is bent if and only if m



d S0 (a) =

(−1)Tr1 (ax ) + 1 − g, n

s

x g =1,x∈U

where S0 (a) is given in Lemma 2.6. Proof By Lemma 2.11, f a,r,s (x) is bent if and only if 

2m +1 −1 g

(−1)

i=1

Tr n1 (ax ri+s )

= 1.

x∈U

2

m +1 g −1

Tr n1 (ax (ri+s) ) = Tr n1 (ax s ) when x g = 1 and x ∈ U . Since m  2 g+1 −1 n  2m +1 Tr 1 (ax ri+s ) i=1 = g when x g = 1 and x ∈ U . x∈U (−1) g − 1 is even, then So we have Note that,

i=1

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L. Yu et al.



2m +1 −1 g

(−1)

i=1

Tr n1 (ax ri+s )

=



(−1)Tr1 (ax ) + g = n

s

x∈U \x g =1

x∈U

(−1)Tr1 (ax n

s)

x∈U



+g −



(−1)

Tr n1 (ax s )

.

x g =1,x∈U

  n n s If gcd(s, 2m + 1) = 1, then x∈U (−1)Tr1 (ax ) = x∈U (−1)Tr1 (ax) = 1 − K m (a) and gcd(s, g) = 1. Thus f a,r,s (x) is bent if and only if 

K m (a) = g −

(−1)Tr1 (ax) . n

x g =1,x∈U

If gcd(s, 2m + 1) = d and a = aξ kd , we have 

(−1)Tr1 (ax ) = d n

s

x∈U



(−1)Tr1 (ax) = d S0 (a). n

x∈V0

Hence, f a,r,s (x) is bent if and only if d S0 (a) =



(−1)Tr1 (ax ) + 1 − g. n

s

x g =1,x∈U

Let s1 be a positive integer, and g = gcd(r, 2m + 1) = 1, s = s1 − r . By (3.4), we have m −1 2 m f a,r,s1 −r (x) = Tr n1 (ax (ri+s1 )(2 −1) ). (3.5) i=0

According to Theorem 3.9, we have the following result, which is exact [10, Theorem 4]. Corollary 3.10 [10] Let the notations be given as above. 1. If gcd(s1 − r, 2m + 1) = 1, 0 ≤ k ≤ 2m and a = aξ k ∈ F2n with a ∈ F2m , then f a,r,s1 −r (x) defined by (3.5) is bent if and only if K m (a) = 1 − (−1)Tr1 (a) . n

2. If gcd(s1 − r, 2m + 1) = d, 0 ≤ k < 2 d+1 and a = aξ kd ∈ F2n with a ∈ F2m , then f a,r,s1 −r (x) defined by (3.5) is bent if and only if m

d S0 (a) = (−1)Tr1 (a) , n

where S0 (a) is given in Lemma 2.6.

123

On more bent functions from Dillon exponents

In particular, if g = 3, gcd(s, 2m +1) = 1, then the following result can be obtained from Theorem 3.9. Corollary 3.11 Assume the notations given as above. Let a = aξ k ∈ F2n with a ∈ F2m and 0 ≤ k ≤ 2m and f (0) = 0, then f a,r,s (x) defined by (3.4) is bent if and only if K m (a) = 3 −

2 m  j 2 3+1 n ) (−1)Tr1 (aξ . j=0

Furthermore, if f a,r,s (x) is bent, then  K m (a) =

0, 4,

if Tr n1 (a) = Tr n1 (aξ otherwise.

2m +1 3

) = Tr n1 (aξ 2

2m +1 3

) = 0;

Proof By Theorem 3.9, f a,r,s (x) defined by (3.4) is bent if and only if K m (a) = 3 −

2 m  j 2 3+1 n ) (−1)Tr1 (aξ . j=0

Since ξ

2m +1 3

+ ξ2

2m +1 3

= 1, then Tr n1 (a) = Tr n1 (aξ Tr n1 (a)

Tr n1 (aξ

= to check that K m (a) = 0 if cases, K m (a) = 4. This completes the proof.

2m +1 3

2m +1 3

)=

) + Tr n1 (aξ 2

2 Tr n1 (aξ 2

m +1 3

2m +1 3

). It is easy

) = 0 and in other



Example 3.12 Let α be a primitive element of F26 , n = 2m = 6, r = 3, s = 1, a ∈ F∗26 , then f a,3,1 (x) = Tr 61 (ax 28 ) + Tr 61 (ax 49 ). By applying Maple, the number of this class of binomial bent functions is 36. Remark 3.13 Some other special cases of Theorem 3.9 can also be discussed as above. Theorem 3.9 is a generalization of Theorem 4 in [10], and derive a concise characterization on the bentness of f a,r,s (x) defined by (3.4) (see Corollary 3.11). 3.2 p-ary bent functions In this subsection, let p be an odd prime. The bentness of two classes of p-ary functions of the form (1.1) is characterized by some exponential sums, which have close relations with Kloosterman sums. 3.2.1 First class of p-ary bent functions Helleseth et al. [5] characterized the bentness of monomial Dillon functions f (x) = m Tr n1 (ax l( p −1) ) with gcd(l, p m + 1) = 1 by Kloosterman sum. Jia et al. in [7] studied the bentness of binomial functions f (x) = Tr n1 (ax l( p

m −1)

)+bx

pn −1 2

with gcd(l, p m +

123

L. Yu et al.

1) = 1. Later, Zheng et al. [15] further investigated the bentness of this class of binomial functions under the case of gcd( 2l , p m + 1) = 1. In [16], Zheng et al. proposed a class of binomial functions     pn −1 n l( p m −1) 2 4 , (3.6) + Tr 1 bx f a,b (x) = Tr 1 ax where p m ≡ 3 (mod 4), a ∈ F∗pn , b ∈ F∗p2 , and l is an integer with gcd(l, p 4+1 ) = 1, and determined the bentness of these functions by subsequences of two sequences. In this subsection, we further investigate the bentness of f a,b (x) defined by (3.6) and give a concise characterization on their bentness in some special case. The following result is a general characterization on the bentness of f a,b (x) defined by (3.6). m

Proposition 3.14 Assume the notations given as above. Then f a,b (x) defined by (3.6) is regular bent if and only if 3 

ωTr1 (bξ 2

j



pm +1 4 )

j=0

ωTr1 (aξ n

jl x)

= 1.

x∈V0

Proof By Lemma 2.11, f a,b (x) is regular bent if and only if 

ω

x∈U

Tr n1 (ax l )+Tr 21 (bx

pm +1 4 )

= 1.

Note that 

ω

  Tr n1 ax l +Tr 21 bx

pm +1 4



x∈U

=



ω

 Tr n1 ax l +Tr 21 (b)

+

x∈V0

+



ω

  Tr n1 aξ l x l +Tr 21 bξ

x∈V0



ω

Tr n1

 2l l aξ x −Tr 21 (b)

+

x∈V0



= ωTr1 (b) 2

ωTr1 (ax) + ω n



+ω

=

3 



ω

then the result follows.

ω

 Tr n1 aξ 2l x

x∈V0

Tr 21 bξ

j=0

123



pm +1 j 4



 x∈V0



+ω



  pm +1  Tr n1 aξ 3l x l +Tr 21 bξ 3 4

x∈V0  pm +1  Tr 21 bξ 4

x∈V0 Tr 21 (−b)

ω

pm +1 4

Tr 21 bξ

3

pm +1 4

n

ωTr1 

 l aξ x



n

ωTr1

 3l aξ x

x∈V0



n

ωTr1

 jl aξ x

,

x∈V0



On more bent functions from Dillon exponents

For given a and b, by Proposition 3.14, it is difficult to determine whether f a,b (x) defined by (3.6) is a regular bent function. So we consider some special cases of a and b to get a concise characterization for the bentness of f a,b (x), and we have the following result. Theorem 3.15 Assume the notations given as above. Let k be a positive integer with k ≡ 1 or 3 (mod 4), a = aξ k , where a ∈ F∗pm , and 4 | l, then f a,b (x) defined by (3.6) is regular bent if and only if √ 2π Q(a) K m (a 2 ) = 1 − 4I −1 sin − p

2 cos

2π Tr 21 (b) p

+ cos

2π Tr 21 (bξ p

pm +1 4 )

,

where Q(a), I are given in Lemma 2.7. Proof By Proposition 3.14, f a,b (x) is regular bent if and only if 3 

ωTr1 (bξ 2

j



pm +1 4 )

j=0

ωTr1 (aξ n

jl x)

= 1.

x∈V0

Since 4 | l, then 3 

ωTr1 (bξ 2

j



pm +1 4 )

j=0

ωTr1 (aξ n

jl x)

=

x∈V0

3 

ωTr1 (bξ 2

j

pm +1 4 )

j=0



ωTr1 (ax) . n

x∈V0

Note that a = aξ k , a ∈ F∗pm , k ≡ 1 or 3 (mod 4), we have 



ωTr1 (ax) = n

x∈V0

ωTr1 (aξ x) = n

x∈V0



ωTr1 (aξ n

3 x)

= S1 (a) = S3 (a).

x∈V0

Hence, by Proposition 2.8, 

ωTr1 (ax) = n

x∈V0

R(a) − I (ω Q(a) − ω−Q(a) ) . 2

To sum up, f a,b (x) is regular bent if and only if 3  j=0

ωTr1 (bξ 2

j

pm +1 4 )

=

4 √ 1 − K m (a 2 ) − 4I −1 sin

2π Q(a) p

,

(3.7)

where Q(a), I are given in Lemma 2.7. Note that ⎛  pm +1  ⎞   pm +1 2 3 j 2 2 2π Tr 1 bξ 4  Tr1 bξ 4 2π Tr 1 (b) ⎟ ⎜ + cos ω = 2 ⎝cos ⎠ p p j=0

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L. Yu et al.



and simplify Eq. (3.7), we finish the proof. From Theorem 3.15, the following result can be obtained.

Corollary 3.16 If there exist (a, b) ∈ F∗3n × F∗32 such that f a,b (x) defined by (3.6) is a regular bent function, then the number of these regular bent functions is a multiple of 4. Proof Since b ∈ F∗32 , we get b ∈ {α i

3n −1 8

| 0 ≤ i ≤ 7}, where α is a primitive

element in F3n . Since ξ is a generator of U , then ξ n 3 3 4−1

−b, bα the proof.

have the same value of cos

2π Tr 21 (b) 3

3m +1 4

+ cos

3n −1 4 pm +1 2π Tr 21 (bξ 4 )

=α

. Hence, b, bα

3

3n −1 4

,

. This completes



Example 3.17 Let l = 4, a = aξ , a ∈ F∗33 , b ∈ F∗32 , ξ be a generator of cyclic

group U = {x ∈ F36 | x 3 +1 = 1}, then we have 33 ≡ 3 (mod 4) and f a,b (x) = Tr 61 (ax 144 ) + Tr 21 (bx 182 ). By using Maple, the number of this binomial regular bent functions is 48. 3

Remark 3.18 From Theorem 3.15, we obtain a new class of binomial p-ary regular bent functions, whose bentness is determined by Kloosterman sum. It is obvious that the characterization on the bentness of f a,b (x) in Theorem 3.15 is concise. 3.2.2 Second class of p-ary bent functions In the following, we assume s, r are integers with gcd(s, p m + 1) = 1, and g = gcd(r, p m + 1). Similar to the second class of binary bent functions, we discuss the bentness of the following functions pm +1 g −1

f a,b,r (x) =



Tr n1 (ax (ri+s)( p

m −1)

) + bx

pn −1 2

,

(3.8)

i=1

where a ∈ F∗pn , b ∈ F p and f (0) = 0. In what follows, we give a necessary and sufficient condition such that f a,b,r (x) defined by (3.8) is regular bent. Based on this characterization, we can obtain several classes of p-ary bent functions from simple approaches. Theorem 3.19 Let the notations be given as above. Then f a,b,r (x) defined by (3.8) is regular bent if and only if  m   2π b = 1 − K m a p +1 cos p where = 1.

123



ωTr1 (−ax n

x g =1,x∈U



s )+bx

4I sin

, pm +1 2

−

2π b p



sin

2π Q(−a) p

x g =1,x∈U

ω

(p

+ ,

−a ∈ C0+ ; otherwise,

pm +1 m +1 n s 2 g −1)Tr 1 (ax )+bx

+

On more bent functions from Dillon exponents

p

m +1 g −1

Tr n1 (x (ri+s) ) = −Tr n1 (x s ). m  p g+1 −1 n  Tr 1 (ax (ri+s) ) By Lemma 2.11, f a,b,r (x) is regular bent if and only if x∈U ω i=1

Proof Note that if

+bx

pm +1 2

xg

= 1 and x ∈ U , then

i=1

= 1, which is equivalent to



ω

(p

pm +1 m +1 n s 2 g −1)Tr 1 (ax )+bx

+

x g =1,x∈U



ωTr1 (−ax n

s )+bx

pm +1 2

= 1.

(3.9)

x g =1,x∈U

On the other hand, we have 

ωTr1 (−ax n

s )+bx

pm +1 2

=

x g =1,x∈U

ωTr1 (−ax n

s )+bx

pm +1 2

x∈U



−



ωTr1 (−ax n

s )+bx

pm +1 2

x∈U,x g =1

=ω

b



ω

Tr n1 (−ax s )

+ω



−b

x∈V0

ω

Tr n1 (−aξ s x s )



−

ω

Tr n1 (−ax s )+bx

pm +1 2

.

x g =1,x∈U

x∈V0

(3.10) Since gcd(s, p m + 1) = 1, then ωb

 x∈V0

=ω

b

=ω

b

ωTr1 (−ax ) + ω−b n



s

 x∈V0

ω

Tr n1 (−ax)

+ ω−b

ω

Tr n1 (−ax)

−b

x∈V0



x∈V0

ωTr1 (−aξ n



s xs)

ωTr1 (−aξ n

s x)

x∈V0

+ω



ωTr1 (−aξ(ξ n

s−1 x))

= ωb S0 (−a) + ω−b S1 (−a).

x∈V0

(3.11) From Eqs. (3.9), (3.10), (3.11) and Lemma 2.7, we complete this proof.



Let s1 be a positive integer, and g = gcd(r, p m + 1) = 1, s = s1 − r . By (3.8), we have m −1 p pn −1 m Tr n1 (ax (ri+s1 )( p −1) ) + bx 2 . (3.12) f a,b,r (x) = i=0

From Theorem 3.19, we have the following result, which is exact [10, Theorems 10,11]. Corollary 3.20 [10] Let the notations be given as above.

123

L. Yu et al.

1. If b = 0 and gcd(s1 − r, p m + 1) = 1, then f a,0,r (x) defined by (3.12) is regular bent if and only if  m  n K m a p +1 = 1 − ωTr1 (−a) . 2. If b = 0 and gcd(s1 − r, p m + 1) = 1, then f a,b,r (x) defined by (3.12) is regular bent if and only if 

 m  2π b = 1 − K m a p +1 cos p

 4I sin

,

2π b p

sin

2π Q(−a) p

+ ,

−a ∈ C0+ ; otherwise,

where = ωTr1 (−a)+b − ωb + 1. n

Note that if p = 3, g = 2, then 3 2+1 − 1 = 1 + 3 + · · · + 3m−1 ≡ 1 (mod 3). Together with gcd(s, 3m + 1) = 1 and b = 0, we have m



 3m +1 n n s s ωTr1 (−ax ) − ω( 2 −1)Tr1 (ax ) + 1 2 x 2 =1,x∈U  Trn (−ax s )  x =1,x∈U  Trn (−ax)  Trn (ax) n s = ω 1 − ωTr1 (ax ) + 1 = ω 1 − ω 1 + 1 = 1.

=

x=±1

x=±1

x=±1

x=±1

Then, by Theorem 3.19, we have the following result. Corollary 3.21 Assume the notations given as above. Let p = 3, g = 2 and b = 0, then f a,0,r (x) defined by (3.8) is regular bent if and only if K m (a 3

m +1

) = 0.

Note that if p = 3, g = 2 and 3m ≡ 3 (mod 4), then 3 2+1 −1 = 1+3+· · ·+3m−1 ≡ m 1 (mod 3) and 3 2+1 is an even integer. Together with gcd(s, 3m + 1) = 1, we have m



=

ωTr1 (−ax n

s )+bx

3m +1 2

x 2 =1,x∈U



=

ωTr1 (−ax



x=±1

ω(

3m +1 n s 2 −1)Tr 1 (ax )+bx

3m +1 2

+1

x 2 =1,x∈U

n

s )+bx

3m +1 2

−

x=±1

= ωb



− 

ωTr1 (ax n

s )+bx

3m +1 2

+1

x=±1



ωTr1 (−ax) − ωb n

ωTr1 (ax) + 1 = 1. n

x=±1

Hence, the following result can be derived from Theorem 3.19. Corollary 3.22 Assume the notations given as above. Let p = 3, g = 2, 3m ≡ 3 (mod 4) and b = 0, then f a,b,r (x) defined by (3.8) is regular bent if and only if K m (a 3

123

m +1

)=1−

1 cos 2π3 b

.

On more bent functions from Dillon exponents

Proof Since = 1, by Theorem 3.19, we have that f a,b,r (x) is regular bent if and only if (1 − K m (a

3m +1

2π b )) cos = 3



4I sin 1,

2π b 3

sin

2π Q(−a) 3

+ 1, −a ∈ C0+ ; otherwise.

Note that p = 3, 3m ≡ 3 (mod 4), then I is a complex number in Lemma 2.7. Together m with real number (1 − K m (a 3 +1 )) cos 2π3 b , we have (1 − K m (a 3

m +1

)) cos

√ 2π Q(−a) 2π b 2π b = −4I −1 sin sin +1 3 3 3

if and only if Q(−a) = 0, which contradicts with −a ∈ C0+ . That is to say f a,b,r (x)

can not be bent if −a ∈ C0+ . This finishes the proof. Remark 3.23 Some other special cases of Theorem 3.19 can also be discussed as above. From Theorem 3.19, we not only obtain Corollary 3.20, which is exact Theorems 10, 11 in [10], but we also get some characterizations for the bentness of f a,b,r (x) defined by (3.8) from simple approaches (see Corollaries 3.21, 3.22).

4 Conclusion In this paper, we investigate the bentness of several special classes of p-ary functions of the following form f (x) =

m −1 p

    pn −1 m o(d) bx d . Tr n1 ai x i( p −1) + Tr 1

i=1

We obtain some new binary and p-ary bent functions with different kinds of Dillon exponents, which include binomials and functions with multiple trace terms. All of these bent functions are determined by some exponential sums, which have close relations with Kloosterman sums. In addition, Corollaries 3.4, 3.7, 3.10, 3.20 obtained in this paper are the corresponding results in [10]. Acknowledgments We sincerely thank the anonymous reviewers for their helpful comments and suggestions. The work of H. Liu was supported by NSFC (Grant No. 11171370) and self-determined research funds of CCNU from the colleges’ basic research and operation of MOE (Grant No. CCNU14F01004). The work of D. Zheng was supported by NSFC (Grant No. 11101131) and the Natural Science Foundation of Hubei Province under Grant 2014CFB537.

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