Two Problems about Monomial Bent Functions
aa r X i v : . [ c s . I T ] F e b Two Problems about Monomial Bent Functions
Honggang Hu , Bei Wang , Xianhong Xie , and Yiyuan Luo School of Information Science and TechnologyUniversity of Science and Technology of ChinaHefei, China, 230027Email. [email protected], { wangbei,xianhxie } @mail.ustc.edu.cn School of Computer Science and EngineeringHuizhou UniversityHuizhou, China, 516007Email. [email protected]
Abstract
In 2008, Langevin and Leander determined the dual function of three classes of monomial bentfunctions with the help of Stickelberger’s theorem: Dillon, Gold and Kasami. In their paper, theyproposed one very strong condition such that their method works, and showed that both Goldexponent and Kasami exponent satisfy this condition. In 2018, Pott et al. investigated the issueof vectorial functions with maximal number of bent components. They found one class of binomialfunctions which attains the upper bound. They also proposed an open problem regarding monomialfunction with maximal number of bent components.In this paper, we obtain an interesting result about the condition of Langevin and Leander,and solve the open problem of Pott et al. . Specifically, we show that: 1) for a monomial bentfunction over F k , if the exponent satisfies the first part of the condition of Langevin and Leander,then it satisfies the entire condition; 2) x k +1 is the only monomial function over F k which hasmaximal number of bent components. Besides, as a byproduct, we dig out a very nice propertyabout Kloosterman sums. Key Words.
Bent function, finite field, Hamming weight, Walsh transform, vectorial function.
Bent functions have several areas of application such as cryptography, coding theory and communica-tions because they have the maximum nonlinearity [11, 15]. The nonlinearity measures the distanceof a Boolean function to the set of affine functions. Since Dillon and Rothaus introduced bent func-tions firstly [6, 35], this special class of Boolean functions has been an interesting research issue formore than 40 years [2, 3, 4, 9, 22, 23, 24, 26, 28, 32, 34, 37]. For more history on bent functions, thereader is referred to [27]. There are also some works about generalized bent functions over finite fields[12, 13, 14, 18, 19]. 1et F n be the finite field of 2 n elements, and F ∗ n be the set of nonzero elements in F n . If we fixone basis of F n over F , then any Boolean function from F n to F can be viewed as a function from F n to F . Let T r n ( · ) be the trace function from F n to F , and f ( x ) be a function from F n to F .Then the Hadamard transform of f ( x ) is defined by b f ( λ ) = X x ∈ F n ( − f ( x )+ T r n ( λx ) , where λ ∈ F n . f ( x ) is a bent function if and only if b f ( λ ) = ± n/ for any λ ∈ F n .A function f ( x ) from F n to F is called a monomial function if f ( x ) = T r n ( αx d ), where α ∈ F ∗ n and 0 < d < n −
1. Among all kinds of bent functions, monomial bent functions are especiallyinteresting. For any integer r , let ( F n ) r denote the set { y r | y ∈ F n } . So far, there are five classes ofknown monomial bent functions over F n with n = 2 k , and they are listed in [31]:1) d = 2 t + 1 with n/ gcd( n, t ) being even and α ( F n ) d [10];2) d = r (2 k −
1) with gcd( r, k + 1) = 1 and α ∈ F ∗ k satisfying P x ∈ F ∗ k ( − T r k ( αx +1 /x ) = − d = 2 t − t + 1 with gcd( t, n ) = 1 and α ( F n ) [8, 22];4) d = (2 t + 1) with t odd, n = 4 t , and α ∈ γ F t satisfying γ ∈ F \ F [5, 23];5) d = 2 t + 2 t + 1 with t > n = 6 t , and α ∈ F ∗ t satisfying T r tt ( α ) = 0 [2].In 2008, Langevin and Leander investigated the dual function of monomial bent functions, and solvedthree cases: Dillon, Gold and Kasami [22]. With the help of Stickelberger’s theorem and Teichm¨ullercharacter, they found a general way to compute the dual function of monomial bent functions overfinite fields with characteristic 2. They need a very strong condition such that their approach works.Fortunately, they proved that both Gold exponent and Kasami exponent satisfy this condition. Inparticular, for Kasami exponent, via this approach Langevin and Leander obtained a more generalresult than that of Dillon and Dobbertin in [8].For a vectorial function from F n to F m , if all nonzero linear combinations of the coordinate functionsof this function are also bent, it is called a vectorial bent function. Similar to the case of Booleanfunctions, any vectorial function from F n to F m can be viewed as a function from F n to F m . Let F ( x ) be a vectorial function from F n to F m . Then F ( x ) is a vectorial bent function if and only if T r m ( αF ( x )) is a bent function for any α ∈ F ∗ m . Nyberg showed that such vectorial bent functionscan only exist for the case of n ≥ m [33]. Two different constructions of vectorial bent functions fromsome known classes of bent functions have also been proposed in [33]. In 2014, vectorial bent functionsfrom multiple terms trace functions have been studied [32]. In 2018, Pott et al. investigated vectorial2unctions from F n to F n with maximal number of bent components [34]. They proved that the numberof bent components of any vectorial function is at most 2 k − k , where n = 2 k . They found one class ofbinomial functions which attains this upper bound. Furthermore, they also presented an open problemto show that x k +1 is the only monomial function over F k with maximal number of bent components.In this paper, we investigate the strong condition of Langevin and Leander in 2008, and solve theopen problem of Pott et al. in 2018 [22, 34]. Let n = 2 k . Our main contributions are as follows.1) We dig out an interesting result about the condition of Langevin and Leander. Suppose that d isthe exponent of a monomial bent function over F n . For any integer j , let wt( j ) be the Hammingweight of j modulo 2 n − 1, then we show that e jd ≡ n − d satisfies the first part of the condition, then it satisfies the entire condition.2) For the open problem of Pott et al. , based on one result of Pott et al. and our new observations,we find nice conditions under which a monomial function over F n has maximal number of bentcomponents. Furthermore, we show that if a monomial function over F n satisfies these conditions,then it must be x k +1 .3) As a byproduct, we obtain a very nice property about Kloosterman sums. Suppose that k ≥ α ∈ F k , if P x ∈ F ∗ k ( − T r k ( αx +1 /x ) = − 1, then we show that T r k ( α ) = 0.This paper is organized as follows. In Section 2, we provide some necessary notation and background.The strong condition of Langevin and Leander is studied in Section 3, and the open problem of Pott etal. is solved in Section 4. Finally, Section 5 concludes this paper. For any 0 ≤ t < n − 1, let n t > t ≡ t n t (mod 2 n − n t | n . The set C t = { t, t, ..., n t − t } is defined to be the cyclotomic coset containing t modulo 2 n − 1, and the smallest integer in C t is calledthe coset leader of C t . For simplicity, we may assume that t is the coset leader of C t . Proposition 1 ([11]) Let f ( x ) be a nonzero function from F n to F . Then f ( x ) can be representedas f ( x ) = X j ∈ Γ( n ) T r n j ( F j x j ) + F n − x n − , F j ∈ F nj , F n − ∈ F here Γ( n ) is the set of all coset leaders modulo n − , n j | n is the size of the coset C j , and T r n j ( x ) is the trace function from F nj to F . F k From now on, let n = 2 k .For any 0 ≤ j < n − 1, let j = P n − i =0 j i i be the binary representation of j , where j i ∈ { , } forany 0 ≤ i < n . Let wt( j ) = P n − i =0 j i . Moreover, for j < j ≥ n − 1, we use wt( j ) to denote wt( j ),where 0 ≤ j < n − j ≡ j (mod 2 n − < d < n − 2, let V d ( j ) = wt( j ) + wt( − jd ).In 2008, in order to study monomial bent functions over F k , Langevin and Leander proposed thefollowing condition about the exponent d . They proved that both Kasami exponent and Gold exponentsatisfy this condition. Condition 1 ([22]) min With notation as above, we have |S F | ≥ k . In particular, if |S F | = 2 k , then S F isa linear space of dimension k over F . If T r k ( αF ( x )) is a bent function, then it is called a bent component of F ( x ). By Theorem 1, thenumber of bent components of F ( x ) is at most 2 k − k . In addition to the known case of x k +1 , in[34], Pott et al. showed that x i ( x + x k ) also have 2 k − k bent components, where i ≥ et al. proposed the following open problem. Open Problem 1 ([34]) Show that x k +1 is the only monomial function over F k which has k − k bent components. For a finite field, there are two kinds of characters: additive character and multiplicative character. Let ψ be the mapping defined by ψ ( x ) = ( − T r n ( x ) . Then ψ is an additive character of F n . Let χ be a multiplicative character of F ∗ n . For simplicity, wedefine χ (0) = 0 which extends χ to F n . For the convenience, we denote the multiplicative characterset of F ∗ n by d F ∗ n . 4 efinition 1 ([25]) For any multiplicative character χ over F n , the Gauss sum G ( χ ) over F n isdefined by G ( χ ) = X x ∈ F n ψ ( x ) χ ( x ) . Lemma 1 ([25]) For any multiplicative character χ over F n , we have G ( χ ) = χ ( − G ( χ ) and G ( χ ) = G ( χ ) . If χ is trivial, then G ( χ ) = − . On the other hand, if χ is nontrivial, then G ( χ ) G ( χ ) = 2 n . Let ω n − = e πi/ (2 n − , which is the complex primitive (2 n − G ( χ ) ∈ Z [ ω n − ] for any multiplicative character χ . The algebraic integer ring Z [ ω n − ] is very useful for thestudy of Gauss sums over F n . In Z , (2) is a prime ideal. Let t = φ (2 n − /n . Then (2) can be factoredinto the product of t different prime ideals in Z [ ω n − ], i.e., (2) = P P · · · P t , where P i is a prime idealin Z [ ω n − ] for any 1 ≤ i ≤ t . For each P i , it holds that Z [ ω n − ] / P i ∼ = F n , because [ Z [ ω n − ] / P i : Z / (2)] = n . Henceforth, we fix one prime ideal P i and denote it by P forsimplicity.There is one special multiplicative character χ on F n satisfying χ ( x )(mod P ) = x. This character is called the Teichm¨uller character, and we denote it by χ p . The Teichm¨uller charactercan generate the group d F ∗ n . According to Stickelberger’s theorem [21], for any 0 < j < n − 1, we have G ( χ − j p ) ≡ wt( j ) mod 2 wt( j )+1 . Stickelberger’s theorem is very important for the proofs of three nice conjectures: Welch, Niho, and Linconjectures [29, 1, 16, 17].The lemma below is pretty useful and well known. Lemma 2 For any x ∈ F ∗ n , we have ( − T r n ( x ) = 12 n − X χ ∈ d F ∗ n G ( χ ) χ ( x ) . .5 The Binary Modular Add-With-Carry Algorithm For three integers 0 ≤ f, g, h < n − h ≡ f + g mod(2 n − f = P n − i =0 f i i , g = P n − i =0 g i i ,and h = P n − i =0 h i i , where f i , g i , h i ∈ { , } for any 0 ≤ i < n . Then there exists a unique integersequence −→ c = c , c , ..., c n − with c i ∈ { , } for any 0 ≤ i < n satisfying h i + 2 c i = f i + g i + c i − , ≤ i ≤ n − , where c i − = c n − if i = 0. Let wt( −→ c ) = c + c + ... + c n − . Then we havewt( f ) + wt( g ) = wt( h ) + wt( −→ c ) ≥ wt( f + g ) . The following lemma is known. Lemma 3 ([22]) For any < j ≤ k , we have wt( j (2 k − k . Suppose that f ( x ) = T r n ( αx d ) is a bent function, where 0 < d < n − α ∈ F ∗ n . Let J d = { e j | < e j < n − , V d ( e j ) = min With notation as above, if min 2. However, there are 2 n − g α ( λ ) = 0. Therefore, g α ( λ ) is a constant polynomial,and jd ≡ n − 1) for any j ∈ J d . Thus, Π d ( α ) = 1. (cid:3) On the other hand, if min Let t = min In the following, we study the case of Dillon exponent in more detail. Lemma 4 Let k ≥ , and d = 2 k − . Then it holds that min Let t = min 6∈ { , , , ..., k − } , then V d ( j ) = wt((2 k + 1) e j ) > 2. Thus, it follows that t = 2 and J d = { k + 1 , k + 1) , (2 k + 1) , ..., k − (2 k + 1) } . (cid:3) Theorem 4 Let k ≥ . For any α ∈ F k , if P x ∈ F ∗ k ( − T r k ( αx +1 /x ) = − , then T r k ( α ) = 0 . Proof. If α ∈ F ∗ k , let d = 2 k − 1. Then f ( x ) = T r n ( αx d ) is a bent function if P x ∈ F ∗ k ( − T r k ( αx +1 /x ) = − 1. By Lemma 4, J d = { k + 1 , k + 1) , (2 k + 1) , ..., k − (2 k + 1) } . By Theorem 3, Π d ( α ) = 0. Letus compute Π d ( α ) = X j ∈J d α j = k − X i =0 α i (2 k +1) = k − X i =0 α i +1 = T r k ( α ) . Therefore, we get T r k ( α ) = 0.If α = 0, the result holds too. (cid:3) emark 1 By computer program, we have checked Theorem 4 for ≤ k ≤ . Note that Theorem 4does not hold in the case of k = 2 . Remark 2 In 2008, if k is even and k > , Charpin and Gong showed that P x ∈ F ∗ k ( − T r k ( αx +1 /x ) = − for any α ∈ F ∗ k/ [4]. In 2009, Shparlinski generalized this result [36], and Moisio finished the proofof the subfield conjecture [30]. et al. Let F ( x ) = x d , where 0 < d < n − 1. In this section, we dig out more properties about S F and d , andsolve Open Problem 1. Lemma 5 With notation as above, if α ∈ S F , then α ∈ S F . Proof. Let f α ( x ) = T r n ( αx d ), and f α ( x ) = T r n ( α x d ). Then, for any λ , we have b f α ( λ ) = X x ∈ F n ( − T r n ( α x d + λx ) = X x ∈ F n ( − T r n ( α x d + λx ) = X x ∈ F n ( − T r n ( αx d + √ λx ) = b f α ( √ λ ) . Thus, if α ∈ S F , then α ∈ S F . (cid:3) For any α ∈ F n , let O ( α ) be the smallest integer t > α t = α . Then α ∈ F O ( α ) .Moreover, let L ( α ) be the linear space over F generated by α, α , ..., α O ( α ) . Then L ( α ) = F O ( α ) . Inparticular, if O ( α ) = n , then L ( α ) = F n . Lemma 6 Suppose that k = 2 e p e p e ...p e t t , where t ≥ , e ≥ , e i ≥ , and p i are prime numberssatisfying ≤ p < p < ... < p t . Then we have k/p + 2 k/p + ... + 2 k/p t + 2 k/ < k . Proof. There are two cases.1) p = 3. In this case, we have 2 k/p + 2 k/p + ... + 2 k/p t + 2 k/ < ( t + 1)2 k/ . Thus, weonly need to prove that ( t + 1)2 k/ ≤ k , which is equivalent to t + 1 ≤ k/ . One may check that2 k/ ≥ t − ≥ t + 1. The result follows.2) p > 3. In this case, we have 2 k/p + 2 k/p + ... + 2 k/p t + 2 k/ < ( t + 1)2 k/ . Thus, weonly need to prove that ( t + 1)2 k/ ≤ k , which is equivalent to t + 1 ≤ k/ . One may check that2 k/ > k/ ≥ t + 1. The result follows, too. (cid:3) Lemma 7 Suppose that k = p e p e ...p e t t , where t ≥ , e i ≥ , and p i are prime numbers satisfying ≤ p < p < ... < p t . The we have k/p + 2 k/p + ... + 2 k/p t < k . roof. The proof is similar to that of Lemma 6, so we omit the details. (cid:3) Lemma 8 With notation as above, if |S F | = 2 k , then S F = F k . Proof. Suppose that S F = F k . Let S = S F \ F k , and S = S F T F k . Then |S | ≥ 1, and |S | < k .For simplicity, let α ∈ S , and β ∈ S . By Theorem 1 and Lemma 5, we have L ( α ) , L ( β ) ⊆ S F . If O ( α ) = n , then F n ⊆ S F . We get a contradiction. If O ( β ) = k , then F k ⊆ S F . We get a contradictionagain. Hence, in the following, we can assume that O ( α ) < n , and O ( β ) < k .We divide the proof into three cases.1) k is a prime number. There are two subcases.(1) k = 2. In this subcase, O ( α ) | k . Thus, α ∈ F k , which is a contradiction.(2) k > 2. In this subcase, O ( α ) | 2, and O ( β ) = 1. Thus, α ∈ F , and β ∈ F . It follows that S F = S ∪ S ⊆ F . Therefore, |S F | ≤ < k . This is a contradiction.2) k is an even composite number. There are three subcases.(1) k = 2 e , where e ≥ 2. In this subcase, O ( α ) | k . Thus, α ∈ F k , which is a contradiction.(2) k = 2 e p e , where e ≥ , e ≥ 1, and p ≥ O ( α ) | e +1 p e − , and O ( β ) | e − p e or O ( β ) | e p e − . Thus, α ∈ F k/p , and β ∈ F k/ or β ∈ F k/p . It follows that S F = S ∪ S ⊆ F k/p ∪ F k/ . Therefore, by Lemma 6, we get |S F | ≤ k/p + 2 k/ < k . This is a contradiction.(3) k = 2 e p e p e ...p e t t , where t ≥ , e ≥ , e i ≥ 1, and p i are prime numbers satisfying 3 ≤ p
2, and p ≥ α ∈ F k/p , and β ∈ F k/p .It follows that S F = S ∪ S ⊆ F k/p . Therefore, |S F | ≤ k/p < k . This is a contradiction.(2) k = p e p e ...p e t t , where t ≥ , e i ≥ 1, and p i are prime numbers satisfying 3 ≤ p < p < ... < p t .In this subcase, we have α ∈ F k/p ∪ F k/p ∪ ... ∪ F k/pt , and β ∈ F k/p ∪ ... ∪ F k/pt . It follows that S F = S ∪ S ⊆ F k/p ∪ F k/p ∪ ... ∪ F k/pt . |S F | ≤ k/p + 2 k/p + ... + 2 k/p t < k . This is a contradiction, too. (cid:3) Lemma 9 With notation as above, if |S F | = 2 k , then (2 k + 1) | d . Proof. Suppose that (2 k + 1) d . Let α be a primitive element of F n . Then α d F k . By Lemma 8,1 ∈ S F . It follows that α d ∈ S F . However, by Lemma 8, we know S F = F k . Hence, α d 6∈ S F . This isa contradiction. Hence, we get (2 k + 1) | d . (cid:3) Theorem 5 Let d = (2 k + 1) s with < s < k − . If gcd( s, k − 1) = 1 and s 6∈ { , , , ..., k − } ,then f ( x ) = T r n ( αx d ) is not a bent function for any α ∈ F n . Proof. Suppose that there exists α ∈ F n such that f ( x ) = T r n ( αx d ) is a bent function. Let0 < t < k − t ≡ s − (mod 2 k − s 6∈ { , , , ..., k − } , we have 2 ≤ wt( t ) < k . Let e t = 2 k − − t . Then 0 < wt( e t ) ≤ k − 2. Consequently, we compute V d ( e t ) = wt( e t ) + wt( − e td ) = wt( e t ) + wt((2 k + 1) st ) = wt( e t ) + wt(2 k + 1) ≤ k. There are two cases.1) min Let d = (2 k + 1) s with < s < k − . If gcd( s, k − > , then F ( x ) = x d can not have k − k bent components. Proof. Suppose that F ( x ) = x d has 2 k − k bent components. By Lemma 8, for any α ∈ F n \ F k , f ( x ) = T r n ( αx d ) is a bent function.Let t = gcd( s, k − j = (2 k − /t . Then 0 < j < k − 1. Thus, we have V d ( j ) = wt( j ) < k ,which means that min 1) + i (2 k − /t , where 0 ≤ i ≤ k ,and 0 ≤ i ≤ t − 1. In particular, if i = 2 k , then V d ( i (2 k − 1) + i (2 k − /t ) = wt(2 k (2 k − 1) + i (2 k − /t ) ≥ k ≤ i ≤ t − 1. Therefore, 2 k (2 k − i (2 k − /t 6∈ J d for any 0 ≤ i ≤ t − 1. As a consequence,the degree of Π d ( x ) = P j ∈J d x j is smaller than 2 k − k . However, by Theorem 3, Π d ( α ) = 0 for any α ∈ F n \ F k . This is a contradiction. Hence, F ( x ) = x d can not have 2 k − k bent components. (cid:3) Theorem 7 x k +1 is the only monomial function over F k which has maximal number of bent com-ponents. Proof. By Lemma 9, Theorem 5, and Theorem 6, the result follows. (cid:3) In this paper, we solve an open problem of Pott et al. regarding bent components in 2018, and showthat x k +1 is the only monomial function over F k with maximal number of bent components. Wefind that this open problem is closely related to the strong condition of Langevin and Leander in2008. Via new observations, we dig out interesting and very useful properties about this condition.Moreover, we also obtain a very nice result about Kloosterman sums: let k ≥ α ∈ F k , if P x ∈ F ∗ k ( − T r k ( αx +1 /x ) = − 1, then T r k ( α ) = 0. 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