FOR a positive integer n and a prime p, let F p n be

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1 1 Several new classes of Boolean functions with few Walsh transform values Guangkui Xu, Xiwang Cao, Shangding Xu arxiv: v1 [cs.it] Jun 2015 Abstract In this paper, several new classes of Boolean functions with few Walsh transform values, including bent, semi-bent five-valued functions, are obtained by adding the product of two or three linear functions to some known bent functions. Numerical results show that the proposed class contains cubic bent functions that are affinely inequivalent to all known quadratic ones. Meanwhile, we determine the distribution of the Walsh spectrum of five-valued functions constructed in this paper. Index Terms Boolean function; Bent function; Semi-bent function; five-valued function; Walsh transform I. INTRODUCTION FOR a positive integer n a prime p, let F p n be the finite field with p n elements, F p F n p n \ 0}. A Boolean function is a mapping from F 2 n to F 2. The Walsh transform is a powerful tool to investigate cryptographic properties of Boolean functions which have wide applications in cryptography coding theory. An interesting problem is to find Boolean functions with few Walsh transform values determine their distributions. Bent functions, introduced by [1] Rothaus, are Boolean functions with two Walsh transform values achieve the maximum Hamming distance to all affine Boolean functions. Such functions have been extensively studied because of their important applications in coding theory [2], [3], cryptography [], sequence designs [5] graph theory [6], [7]. Notice that bent functions exist only for an even number of variables can not be balanced. In 1985, Kumar, Scholtz Welch extended Rothaus definition to the case of an arbitrary prime p [8]. Complete classification of bent functions seems elusive even in the binary case. However, a number of recent interesting results on bent functions have been found through primary constructions secondary constructions see [9], [10], [11], [12], [13], [1], [15], [], [17], [18], [19], [20], [21], [22], references therein. As a particular case of the so-called plateaued Boolean functions [23], semi-bent functions are an important kind of Boolean functions with three Walsh transform values. The term of semi-bent function introduced by Chee et al. [2]. G. Xu is with the Department of Mathematics, Nanjing University of Aeronautics Astronautics, Nanjing 2100, China, also with the Department of Mathematics, Huainan Normal University, Huainan , China xuguangkuiy@3.com. X. Cao is with the Department of Mathematics, Nanjing University of Aeronautics Astronautics, Nanjing 2100, China, also with the State Key Laboratory of Information Security, Institute of Information Engineering, Chinese Academy of Sciences, Beijing , China xwcao@nuaa.edu.cn. S. Xu is with the Department of Mathematics, Nanjing University of Aeronautics Astronautics, Nanjing 2100, China, also with Department of Mathematics Physics, Nanjing Institute of Technology, Nanjing 2100, China sdxzx11@3.com. Semi-bent functions investigated under the name of threevalued almost optimal Boolean functions in [2], i.e., they have the highest possible nonlinearity in three-valued functions. They are also nice combinatorial objects have wide applications in cryptography coding theory. A lot of research has been devoted to finding new families of semibent functions see [25], [12], [26], [], [27], [28], [29] the references therein. However, there is only a few known constructions of semi-bent functions. In general, it is difficult to characterize all functions with few Walsh transform values. For any positive integers n, k dividing n, the trace function from F p n to F p k, denoted by Tr n k, is the mapping defined as: For k 1, Tr n 1 Tr n kx x+x pk +x p2k + +x pn k. n 1 x i0 x pi is called the absolute trace function. Recently, using a theorem proved by Carlet [30, Theorem 3], Mesnager [31] provided some primary secondary constructions of bent functions gave corresponding dual functions of these constructions. By means of the second order derivative of the duals of known bent functions, she obtained two new infinite families of bent functions with the forms fx Tr m Tr n λx2m 1 uxtrn 1 vx 1 fx Tr m Tr n x2m 1 2 r 1 1 i1 x 2m 1 i 2 r +1 +Tr n 1 uxtrn 1 vx 2 over F 2 n, where n 2m, λ F 2 u,v m F 2 n. She showed that the function defined by 1 is bent when Tr n 1 λ 1 u 2m v 0 the function defined by 2 is bent when u,v F 2 m. The aim of this paper is to present several classes of functions with few Walsh transform values. Inspired by the work of [31], we present several new classes of bent functions by adding the product of three or two linear functions to some known bent functions. Computer experiments show that we can obtain some cubic bent functions. Meanwhile, several new classes of semi-bent five-valued functions are also obtained. The proofs of our main results are based on the study of the Walsh transform, which are different from the ones used in [31]. The paper is organized as follows. In Section II, we give some notation recall the necessary background. In Section III, we present some new Boolean functions with few Walsh

2 2 transform values from Kasami function Gold function. A new family of bent functions via Niho exponents is presented in Section IV two new families of functions with few Walsh transform values via Maiorana-McFarls class are provided in Section V. II. PRELIMINARIES By viewing each x x 1 ξ 1 +x 2 ξ 2 + +x n ξ n F 2 n as a vector x 1,x 2,,x n F n 2 where ξ 1,,ξ n } is a basis of F 2 n over F 2, we identify F n 2 the n-dimensional vector space over F 2 with F 2 n, then every function f : F 2 n F 2 is equivalent to a Boolean function. For x,y F 2 n, the inner product is defined as x y Tr n 1xy. It is well known that every nonzero Boolean function defined on F 2 n can be written in the form of fx j Γ n Tr oj 1 a j x j +ǫ1 + x 2n 1, whereγ n is a set of integers obtained by choosing one element in each cyclotomic coset of 2 modulo 2 n 1, oj is the size of the cyclotomic coset containing j, a j F 2 oj ǫ wtfmod 2, where wtf is the cardinality of its support supp : x F 2 n fx 1}. The algebraic degree of f is equal to the maximum 2-weight of an exponent j for which a j 0 if ǫ 0 to n if ǫ 1. The Walsh transform of a Boolean function f : F 2 n F 2 is the function χ f : F 2 n Z defined by χ f a 1 fx+trn 1 ax,a F 2 n. x F 2 n The values χ f a,a F 2 n are called the Walsh coefficients of f. The Walsh spectrum of a Boolean function f is the multiset χ f a,a F 2 n}. A Boolean functionf is said to be balanced if χ f 0 0. Definition 1: [1] A Boolean function f is said to be bent if χ f a 2 n/2 for all a F 2 n. In view of Parseval s equation this definition implies that bent functions exist only for an even number of variables. For a bent function with n variables, its dual is the Boolean function f defined by χ f a 2 n/2 1 fa. It is easy to verify that the dual of f is again bent. Thus, Boolean bent functions occur in pair. However, determining the dual of a given bent function is not an easy thing. A bent function is said to be self-dual resp. anti-self-dual if f f resp. f f+1. For more study on self-dual anti-self-dual bent functions can be founded in [30], [32], [31], [33]. Definition 2: [2] A Boolean function f is said to be semibent if n+1 0,±2 2 }, if n is odd χ f a 0,±2 n 2 +1 }, if n is even for all a F 2 n. Our constructions can be derived from some known bent functions. The following result will be used in the sequel. Lemma 1: Let n be a positive integer u,v,r F 2 n. Let gx be a Boolean function over F 2 n. Define the Boolean function fx by fx gx+tr n 1 uxtrn 1 vxtrn 1 rx. Then, for every a F 2 n, χ f a 1 [3 χ ga+ χ g a+v+ χ g a+u χ g a+u+v + χ g a+r χ g a+r+v χ g a+r+u + χ g a+r+u+v]. In particular, if r v, then χ f a 1 2 [ χ ga+ χ g a+u+ χ g a+v χ g a+u+v]. Proof: For i,j 0,1} u,v F 2n, define denote T i,j x F 2 n Tr n 1 ux i,trn 1 vx j} S i,j a Q i,j a+r For each a F 2 n, we have χ f a x F 2 n x F 2 n 1 fx+trn 1 ax ω gx+tr n 1 ax x T i,j ω gx+tr n 1 a+rx. x T i,j 1 gx+trn 1 uxtrn 1 vxtrn 1 rx+trn 1 ax x T 0,0 x T 1,0 + 1 ax + 1 ax + 1 ax x T 0,1 1 a+rx x T 1,1 S 0,0 a+s 0,1 a+s 1,0 a+q 1,1 a+r χ g a S 1,1 a+q 1,1 a+r. 3 In the following, we will compute two sums S 1,1 a Q 1,1 a+r. Let T i,j be defined as above. Clearly, χ g a S 0,0 a+s 0,1 a+s 1,0 a+s 1,1 a. Furthermore, we have χ g a+v x T 0,0 x T 1,0 + 1 ax 1 ax x T 0,1 1 ax 1 ax x T 1,1 S 0,0 a S 0,1 a+s 1,0 a S 1,1 a. 5 Similarly, χ g a+u S 0,0 a+s 0,1 a S 1,0 a S 1,1 a 6 χ g a+u+v S 0,0 a S 0,1 a S 1,0 a+s 1,1 a. 7

3 3 From -7, we have S 0,0 a S 0,1 a S 1,0 a S 1,1 a χ g a χ g a+v χ g a+u χ g a+u+v. 8 Note that the coefficient matrix of 8 is a Hadamard matrix of order. Then we have S 1,1 a 1 [ χ ga χ g a+v χ g a+u+ χ g a+u+v]. 9 Substituting a by a+r in 9, we can get Q 1,1 a+r 1 [ χ ga+r χ g a+r+v χ g a+r+u+ χ g a+r+u+v]. 10 The desired conclusion follows from 3, In particular, if r v, it is easy to show that χ f a 1 2 [ χ ga+ χ g a+u+ χ g a+v χ g a+v +u]. The proof is completed. It must be pointed out that fx gx + Tr n 1 uxtrn 1 vxtrn 1 rx gx+trn 1 uxtrn 1 vxtrn 1 ux+ vx gx when u+v+r 0. In the following, we always assume that u+v +r 0. III. NEW INFINITE FAMILIES OF BENT, SEMI-BENT AND FIVE-VALUED FUNCTIONS FROM MONOMIAL BENT FUNCTIONS A. New infinite families of bent, semi-bent five-valued functions from Kasami function Let n 2m m is at least 2 be a positive even integer. The Kasami function gx Tr m 1 λx2m +1 is bent where λ F 2 m its dual g is given by gx Trm 1 λ 1 x 2m [31]. In other words, for each a F 2 n, the Walsh coefficient χ g a is χ g a 2 m 1 Trm 1 λ 1 a 2m In the following result, we will present some new bent five-valued functions by making use of the Kasami function. Theorem 1: Let n 2m be a positive even integer let u,v,r be three distinct pairwise elements in F 2n such that u+v +r 0. Define the Boolean function f on F 2 n as fx Tr m 1 λx2m +1 +Tr n 1 uxtrn 1 vxtrn 1 rx, where λ F 2 m. If Trn 1λ 1 u 2m v Tr n 1λ 1 r 2m u Tr n 1 λ 1 r 2m v 0, thenf is bent. Otherwise,f is five-valued the Walsh spectrum of f is 0,±2 m,±2 m+1 }. Moreover, if Tr n 1λ 1 r 2m v,tr n 1λ 1 r 2m u,tr n 1λ 1 u 2m v 0,0,1,1,0,0,0,1,0,1,1,1}, when a runs through all elements in F 2 n, the distribution of the Walsh spectrum of five-valued function f is given by 0, occurs 2 n 2 n 1 2 n 3 times 2 m, occurs 2 n 2 +2 m 1 times χ f a 2 m, occurs 2 n 2 2 m 1 times 2 m+1, occurs 2 n times 2 m+1, occurs 2 n times. If Tr n 1 λ 1 r 2m v,tr n 1 λ 1 r 2m u,tr n 1 λ 1 u 2m v 1, 1, 0,1, 0, 1,0, 1, 1}, when a runs through all elements in F 2 n, the distribution of the Walsh spectrum of five-valued function f is given by 0, occurs 2 n 2 n 1 2 n 3 times 2 m, occurs 2 n 2 times χ f a 2 m, occurs 2 n 2 times 2 m+1, occurs 2 n +2 m 2 times 2 m+1, occurs 2 n 2 m 2 times. Proof: Let gx Tr m 1 λx2m +1. For each a F 2 n, by Lemma 1, we have χ f a 1 [3 χ ga+ χ g a+v+ χ g a+u χ g a+u+v where + χ g a+r χ g a+r+v χ g a+r+u + χ g a+r+u+v] 1 + 2, 1 1 [3 χ ga+ χ g a+v+ χ g a+u χ g a+u+v] 2 1 [ χ ga+r χ g a+r+v χ g a+r+u + χ g a+r+u+v]. Now we use 11 to compute two sums 1 2 respectively m [ 3 1 Trm 1 λ 1 a 2m Trm 1 λ 1 a+v 2m Trm 1 λ 1 a+u 2m +1 1 Trm 1 λ 1 a+u+v 2m +1 ] 1 2m 1 Trm 1 λ 1 a 2m +1 [ 3+ 1 Trm 1 λ 1 a 2m v+av 2m 1 Trm 1 λ 1 v 2m Trm 1 λ 1 a 2m u+au 2m +u 2m +1 1 Trm 1 λ 1 a 2m v+av 2m +v 2m +1 +a 2m u+au 2m +u 2m +1 1 Trm 1 λ 1 u 2m v+uv 2m ]. Similarly, we have 2 1 2m 1 Trm 1 λ 1 a 2m +1 +a 2m r+ar 2m +r 2m +1 [ 1 1 Trm 1 λ 1 a 2m v+av 2m +v 2m +1 +r 2m v+rv 2m 1 Trm 1 λ 1 a 2m u+au 2m +u 2m +1 +r 2m u+ru 2m + 1 Trm 1 λ 1 a 2m v+av 2m +v 2m +1 +a 2m u+au 2m +u 2m +1 1 Trm 1 λ 1 r 2m v+rv 2m +r 2m u+ru 2m +u 2m v+uv 2m ].

4 To simplify 1 2, we write t 1 Tr m 1 λ 1 r 2m v + rv 2m Tr n 1 λ 1 r 2m v, t 2 Tr m 1 λ 1 r 2m u + ru 2m Tr n 1λ 1 r 2m u t 3 Tr m 1 λ 1 u 2m v + uv 2m Tr n 1 λ 1 u 2m v due to the transitivity property of the trace function for every k dividing n, Tr n 1x Tr k 1Tr n kx. Meanwhile, denote c 1 Tr m 1 λ 1 a 2m v +av 2m +v 2m +1, c 2 Tr m 1 λ 1 a 2m u + au 2m + u 2m +1 c 3 Tr m 1 λ 1 a 2m r + ar 2m + r 2m +1. Then the sums 1 2 can be written as 1 1 2m 1 Trm 1 λ 1 a 2m +1 [ 3+ 1 c1 + 1 c2 1 c1+c2+t3] m 1 Trm 1 λ 1 a 2m +1 +c 3 [ 1 1 c1+t1 1 c2+t2 + 1 c1+c2+t1+t2+t3]. 13 Firstly, we prove that f is bent when t 1 t 2 t 3 0. If t 1 t 2 t 3 0, then 1 1 2m 1 Trm 1 λ 1 a 2m +1 [ 3+ 1 c1 + 1 c2 1 c1+c2] 2 1 2m 1 Trm 1 λ 1 a 2m +1 +c 3 [ 1 1 c 1 1 c2 + 1 c1+c2]. When c 3 0, we can get χ f a m 1 Trm 1 λ 1 a 2m +1. When c 3 1, we can get χ f a m 1 Trm 1 λ 1 a 2m +1 [ 1+ 1 c1 + 1 c2 1 c1+c2] 2 m 1 Trm 1 λ 1 a 2m +1, if c 1 c m 1 Trm 1 λ 1 a 2m +1, otherwise. Hence, f is bent if Tr n 1 λ 1 u 2m v Tr n 1 λ 1 r 2m u Tr n 1 λ 1 r 2m v 0. Secondly, we show that f is five-valued if at least one t i i 1,2,3} is equal to 1. We only give the proof of the case of t 1 t 2 0 t 3 1 since the others can be proven in a similar manner. In this case, become 1 1 2m 1 Trm 1 λ 1 a 2m +1 [ 3+ 1 c1 + 1 c2 + 1 c1+c2] 2 1 2m 1 Trm 1 λ 1 a 2m +1 +c 3 [ 1 1 c 1 1 c2 1 c1+c2]. When c 3 0, then we have χ f a m 1 Trm 1 λ 1 a 2m When c 3 1, then we have χ f a m 1 Trm 1 λ 1 a 2m +1 [ 1+ 1 c1 + 1 c2 + 1 c1+c2] 2 m+1 1 Trm 1 λ 1 a 2m +1, if c 1 c 2 0 0, otherwise. 15 It then follows from 1 15 that f is five-valued its Walsh spectrum is 0,±2 m,±2 m+1 }. Finally, we present the value distribution of the Walsh transform of five-valued function f. We only give the value distribution of f in the case of t 1 t 2 0 t 3 1 since the value distribution of others can be determined in a similar manner. Let N 2 m resp., N 2 m denote the number of a F 2 n such that χ f a 2 m resp., 2 m. For convenience of presentation, we denote Tr m 1 λ 1 a 2m +1 by c 0. The equality 1 implies that if c 3 0 c 0 Tr m 1 λ 1 a 2m +1 1, then χ f a 2 m. Hence we have N 2 m c c Trm 1 λ 1 a 2m Trn 1 λ 1 a 2m r+tr m 1 λ 1 r 2m +1 1 [ 2 n + 1 Trm 1 λ 1 r 2m +1 1 Trn 1 λ 1 a 2m r 1 Trm 1 λ 1 a 2m +1 1 Trm 1 λ 1 a+r 2m +1 ] 1 [ 2 n 21+2 m +1 1 Tr m 1 λ 1x ] x F 2 m 2 n 2 +2 m 1 where the fourth identity holds since raising elements of F 2 n to the power of 2 m +1 is a 2 m +1-to-1 mapping on to F 2 m. Similarly, it follows from 1 that N 2 m c c Trm 1 λ 1 a 2m Trn 1 λ 1 a 2m r+tr m 1 λ 1 r 2m +1 2 n 2 2 m 1. Let N 2 m+1 resp., N 2 m+1 denote the number of a F 2 n such that χ f a 2 m+1 resp., 2 m+1. From 15, we know that if c 3 1, c 1 c 2 0 c 0 1, then χ f a 2 m+1.

5 5 Hence we have N 2 m c0 1 1 c c c1 1 [ 1+ 1 c c2 + 1 c1+c2 1 c3 1 c3+c1 1 c3+c2 1 c3+c2+c1 1 c0 1 c0+c1 1 c0+c2 1 c0+c2+c1 + 1 c0+c3 + 1 c0+c3+c1 + 1 c0+c3+c2 + 1 c0+c3+c1+c2]. On one h, 1 c1 1 Trm 1 λ 1 v 2m +1 1 Trn 1 λ 1 a 2m v 0. Similarly, 1c2 1c3 0. Note that u,v,r are pairwise distinct u + v + r 0. Then we have 1 c1+c2 1 Trm 1 λ 1 u 2m +1 +v 2m +1 1 Trn 1 λ 1 a 2m u+v 0. Similarly, a F 2 n 1c3+c1 a F 2 n 1c3+c2 1c3+c2+c1 0. On the other h, 1 c0+c1 1 Trm 1 λ 1 a+v 2m m +1 1 Tr m x F 2 m 2 m. 1 λ 1 x Similarly, 1 c0 1 c0+c2 1 c0+c3 2 m. By the condition t 1 t 2 0, we have 1 c0+c3+c1 1 c0+c3+c1+t1 1 Trm 1 λ 1 a+r+v 2m +1 2 m 1 c0+c3+c2 1 c0+c3+c1+t2 1 Trm 1 λ 1 a+r+u 2m +1 2 m. Since t 3 1, then 1 c0+c2+c1 1 c0+c2+c1+t3 1 Trm 1 λ 1 a+u+v 2m +1 2 m 1 c0+c3+c1+c2 1 c0+c3+c1+c2+t1+t2+t3 1 Trm 1 λ 1 a+r+u+v 2m +1 2 m. Therefore we have N 2 m+1 1 [2n 2 m 2 m 2 m + 2 m + 2 m + 2 m + 2 m 2 m ] 1 2n 2 n. Similarly, we have N 2 m c0 1 1 c c c1 2 n. Clearly, the number of a F 2 n such that χ f a 0 is equal to 2 n 2 n 1 2 n 3. This completes the proof. It is easily checked that Tr n 1λ 1 u 2m v Tr n 1λ 1 r 2m u Tr n 1λ 1 r 2m v 0 when u,v,r F 2 m. From Theorem 1, we get the following corollary. Corollary 1: Let n 2m be a positive even integer λ F 2 m. If u,v,r F 2m are three pairwise distinct elements such that u+v +r 0, then the Boolean function f fx Tr m 1 λx2m +1 +Tr n 1 uxtrn 1 vxtrn 1 rx is bent. Remark 1: If r v, the bent functions f presented in Theorem 1 become ones in [31, Theorem 9], i.e., if Tr n 1λ 1 u 2m v 0, then the Boolean function fx Tr m 1 +1 λx2m +Tr n 1 uxtrn 1 vx is bent. Now let us consider the algebraic degree of f in Theorem 1. Let i,j,k 0,1,, n 1} are pairwise distinct integers. Denote the set of all permutations on i,j,k by P. It is clear that the possible cubic term in the expression of f has the form i,j,k P u2i v 2j r 2k x 2i +2 j +2 k. If there exist three pairwise distinct integers i,j,k 0,1,,n 1} such that i,j,k P u2i v 2j r 2k 0, then the algebraic degree of f is 3. Next we will show that if Tr n 1 λ 1 u 2m v Tr n 1λ 1 r 2m u Tr n 1λ 1 r 2m v 0, the algebraic degree of f in Theorem 1 is not equal to 3 when m 2. Otherwise, this will contradict the fact that the algebraic degree of a bent function f is at most n/2. Let P 1 be the set of all all permutations on 0,1,3, P 2 be the set of all all permutations

6 6 on 0,1,2, P 3 be the set of all all permutations on 1,2,3 P be the set of all all permutations on 0,2,3. The condition Tr 1 λ 1 r 22 v Tr 1 λ 1 r 22 u Tr 1 λ 1 u 22 v 0 can be written as r 22 v +r 23 v 2 +rv 22 +r 2 v 23 0 r 22 u+r 23 u 2 +ru 22 +r 2 u 23 0 u 22 v +u 23 v 2 +uv 22 +u 2 v Multiplying u, v r to the first, the second the third equation of respectively yields r 22 vu+r 23 v 2 u+rv 22 u+r 2 v 23 u 0 r 22 vu+r 23 vu 2 +rvu 22 +r 2 vu 23 0 ru 22 v +ru 23 v 2 +ruv 22 +ru 2 v Adding three equations of 17 gives i,j,k P 1 u 2i v 2j r 2k 0. Similarly, multiplying u 2, v 2 r 2 to the first, the second the third equation of respectively yields i,j,k P 2 u 2i v 2j r 2k 0. Multiplying u 22, v 22 r 22 to the first, the second the third equation of respectively yields i,j,k P 3 u 2i v 2j r 2k 0 multiplying u 23, v 23 r 23 to the first, the second the third equation of respectively yields i,j,k P u 2i v 2j r 2k 0. Therefore, there are no cubic terms in the expression of f when m 2, which implies that the algebraic degree of f in Theorem 1 is equal to 2. Remark 2: When m 2, the algebraic degree of the bent function f in Theorem 1 is equal to 2. When m 3, the bent functions f in Theorem 1 may be cubic according to our numerical results. Example 1: Let m 3, F 2 6 be generated by the primitive polynomial x 6 +x +x 3 +x+1 ξ be a primitive element of F 2 6. Take λ 1, u ξ, v ξ 9 r ξ 27. Let P be the set of all permutations on 0,1,2. By help of a computer, we can get Tr 6 1u 8 v Tr 6 1r 8 u Tr 6 1r 8 v 0, u+v+r 0, i,j,k P v u2i 2j v 2k ξ 5 0 the function fx Tr 3 1x 9 +Tr 6 1ξxTr 6 1ξ 9 xtr 6 1ξ 27 x is a cubic bent function, which coincides with the results in Theorem 1. Example 2: Let m, F 2 8 be generated by the primitive polynomial x 8 + x + x 3 + x ξ be a primitive element of F 2 8. Take λ ξ 17, u ξ 10, v ξ 9, r ξ 3. Then the function f in Theorem 1 is fx Tr 1 ξ17 x 17 + Tr 8 1ξ 10 xtr 8 1ξ 9 xtr 8 1ξ 3 x. By help of a computer, we can get Tr 8 1 u v 1, Tr 8 1 r u Tr 8 1 r v 0. Moreover, the function is a five-valued the distribution of the Walsh spectrum is 0, occurs 96 times, occurs 72 times ˆχ f a, occurs 56 times 32, occurs times 32, occurs times which is consistent with the results given in Theorem 1. As noted in Remark 1, if Tr n 1 λ 1 u 2m v 0, then the Boolean function fx Tr m 1 λx 2m +1 + Tr n 1uxTr n 1vx is bent. In the following result, we will prove that if Tr n 1 λ 1 u 2m v 1, the Boolean function fx Tr m 1 +1 λx2m + Tr n 1 uxtrn 1 vx is semi-bent by using Lemma 1. Theorem 2: Let n 2m be a positive even integer u,v F 2 n. Define a Boolean function f on F 2 n by fx Tr m 1 λx2m +1 +Tr n 1 uxtrn 1 vx, where λ F 2 m. If Trn 1λ 1 u 2m v 1, then f is semi-bent. Moreover, when Tr n 1 λ 1 u 2m v 1, if Tr n 1 λ 1 u 2m +1 1 or Tr n 1 λ 1 v 2m +1 1, then f is a balanced semi-bent function. Proof: Let gx Tr m 1 λx 2m +1. By Lemma 1 11, for each a F 2n, we have χ f a 1 2 [ χ ga+ χ g a+v+ χ g a+u χ g a+u+v] 1 2 χ ga [ 1+ 1 Trm 1 λ 1 a 2m v+av 2m 1 Trm 1 λ 1 v 2m Trm 1 λ 1 a 2m u+au 2m +u 2m +1 1 Trm 1 λ 1 a 2m v+av 2m +v 2m +1 +a 2m u+au 2m +u 2m +1 + ] 1 Trm 1 λ 1 u 2m v+uv 2m 1 2 χ ga [ 1+ 1 Trm 1 λ 1 a 2m v+av 2m +v 2m Trm 1 λ 1 a 2m u+au 2m +u 2m Trm 1 λ 1 a 2m v+av 2m +v 2m +1 +a 2m u+au 2m +u 2m +1 ] 18 where the last identity holds because Tr n 1λ 1 u 2m v Tr m 1 u2m v +uv 2m 1. Denote c 1 Tr m 1 λ 1 a 2m v +av 2m +v 2m +1 c 2 Tr m 1 λ 1 a 2m u+au 2m +u 2m +1. Then 18 can be written as χ f a 1 2 2m 1 Trm 1 λ 1 a 2m c1 + 1 c2 + 1 c1+c2 ] 2 m+1 1 Trm 1 λ 1 a 2m +1, if c 1 c 2 0 0, otherwise. It then follows from Definition 2 that f is semi-bent. Furthermore, from 18, if Tr n 1λ 1 u 2m v Tr m 1 λ 1 u 2m v + uv 2m 1 then the Walsh transform coefficient of the function f evaluated at 0 is equal to χ f m 1+ 1 Trm 1 λ 1 v 2m Trm 1 λ 1 u 2m Trm 1 λ 1 v 2m +1 +u 2m +1 ]. It is easy to check that χ f 0 0 if Tr m 1 λ 1 v 2m +1 1 or Tr m 1 λ 1 u 2m Therefore,fx is a balanced semi-bent function. Remark 3: For a given λ F 2m, the number of semibent functions f in Theorem 2 is equal to 1 2 u,v F n1 2 1 Trn 1 λ 1 u 2m v 2 n 1 2 n 1.

7 7 B. New infinite families of bent, semi-bent five-valued functions from Gold-like monomial function In [], Carlet et.al proved that the Gold-like monomial function gx Tr1 k +1 λx2k over F 2 k where k is at least 2 λ F 2, is self-dual or anti-self-dual bent if only if k λ 2 +λ 23k+1 1 λ 2k +1 +λ 23k +2 k 0. Recently, Mesnager showed that gx Tr k 1 +1 λx2k over F 2 k is self-dual bent when λ+λ 23k 1 in [31, Lemma 23], i.e., for each a F 2, k the Walsh coefficient χ f a is χ g a 2 2k 1 Trk 1 λa2k +1 when λ+λ 23k 1. Lemma 2: Let k > 1 be a positive integer let λ F 2 k such that λ+λ 23k 1. Then the linearized polynomial lx λx+λ 2k x 22k is a permutation polynomial over F 2 k. Proof: Firstly, we will prove that λ +λ 23k 1 implies that λ / x 2k +1 x F 2 k}. It follows fromλ+λ 23k 1 that λ+λ 2k 1, which leads toλ 23k +λ 2k 0, i.e.,λ 22k 1 1. If λ a 2k +1 for some a F 2 k, then a 2k +12 2k 1 1. Since gcd2 k k 1,2 k 1 2 2k 1, we know that a 22k 1 1, i.e., a 2k 12 k +1 λ 2k 1 1. Thus, λ F 2 k which is contradiction with λ+λ 23k 1. Secondly, we will show that lx is a permutation polynomial. Since lx is a linearized polynomial, we have to prove that the equationλx+λ 2k x 22k 0 has the only solutionx 0 in F 2 k. Assume that β 0 is a solution of λx+λ 2k x 22k 0. Then we have β 22k 1 λ 1 2k, i.e., β 2k +1 2k 1 λ 1 2k. It is easy to see that the left-h side is a 2 k +1th power, while the right-h side is not a 2 k + 1th power because λ / x 2k +1 x F 2 k}. This gives a contradiction. Thus, λx + λ 2k x 22k is a permutation polynomial over λ F 2 k when λ+λ 23k 1. Theorem 3: Let k be a positive integer such that k > 1 let u,v,r be three pairwise distinct elements in F 2 such k that u + v + r 0. Let λ F 2 such that λ + λ 23k k 1. If Tr k 1 λu2k v + uv 2k Tr k 1 λr2k u + ru 2k Tr k 1 λr2k v +rv 2k 0, then the Boolean function fx Tr k 1 λx2k +1 +Tr k 1 uxtrk 1 vxtrk 1 rx over F 2 k is a bent function. Otherwise, fx is a fivevalued function. Moreover, if Tr k 1 λr 2k v + rv 2k, Tr k 1 λr 2k u + ru 2k, Tr k 1 λu 2k v + uv 2k 0,0,1,1,0,0,0,1,0,1,1,1}, when a runs through all elements in F 2 k, the distribution of the Walsh spectrum of five-valued function f is given by 0, occurs 2 k 2 k 1 2 k 3 times 2 2k, occurs 2 k k 1 times χ f a 2 2k, occurs 2 k 2 2 2k 1 times 2 2k+1, occurs 2 k times 2 2k+1, occurs 2 k times. If Tr k 1 λr 2k v + rv 2k, Tr k 1 λr2k u + ru 2k, Tr k 1 λu 2k v + uv 2k 1,1,0,1,0,1,0,1,1}, when a runs through all elements in F 2 n, the distribution of the Walsh spectrum of five-valued function f is given by 0, occurs 2 k 2 k 1 2 k 3 times 2 2k, occurs 2 k 2 times χ f a 2 2k, occurs 2 k 2 times 2 2k+1, occurs 2 k +2 2k 2 times 2 2k+1, occurs 2 k 2 2k 2 times. Proof: Let gx Tr k 1 λx 2k +1. We write Tr k 1 λr 2k v + rv 2k t 1, Tr k 1 λr 2k u + ru 2k t 2, Tr k 1 λu2k v + uv 2k t 3. Denote c 1 Tr k 1 λa2k v + av 2k + v 2k +1, c 2 Tr k 1 λa2k u + au 2k + u 2k +1 c 3 Tr k 1 λa2k r + ar 2k + r 2k +1. By analyses similar to those in Theorem 1, we have where χ f a 1 + 2, k 1 Trk 1 λa2k +1 [ 3+ 1 c1 + 1 c2 1 c1+c2+t3] k 1 Trk 1 λa2k +1 +c 3 [ 1 1 c1+t1 1 c2+t2 + 1 c1+c2+t1+t2+t3]. 20 Similar to Theorem 1, we can prove that fx is bent if t 1 t 2 t 3 0. Next we will prove that f is five-valued determine its distribution of the Walsh transform in the case of t 1 t 2 0 t 3 1 since the others can be proven in a similar manner. In this case, become k 1 Trk 1 λa2k +1 [ 3+ 1 c1 + 1 c2 + 1 c1+c2] k 1 Trk 1 λa2k +1 +c 3 [ 1 1 c1 1 c2 1 c1+c2]. When c 3 0, then we have χ f a k 1 Trk 1 λa2k When c 3 1, then we have χ f a k 1 Trk 1 λa2k +1 [ 1+ 1 c1 + 1 c2 + 1 c1+c2] 2 2k+1 1 Trk 1 λa2k +1, if c 1 c 2 0 0, otherwise. 22 Thus, f is a five-valued function if t 1 t 2 0 t 3 1.

8 8 Let c 0 Tr k 1 +1 λa2k. Let N 2 2k resp., N 2 2k denote the number of a F 2 k such that χ f a 2 2k resp., 2 2k. From 21, we know that χ f a 2 2k if c 3 0 c 0 0. Then we have N 2 2k c c3 1 a F 2 k a F 2 k 1+ 1 Trk 1 λa2k Trk 1 λa2k r+ar 2k +r 2k +1 1 [ 2 k + 1 Trk 1 λa2k +1 a F 2 k + 1 Trk 1 λr2k +1 + a F 2 k a F 2 k 1 Trk 1 λa+r2k +1 ]. 1 Trk 1 a2k λr+λ 2k r 22k Note that λ + λ 23k 1 r F 2. It then follows from k Lemma 2 that λr+λ 2k r 22k 0, which implies that 1 Trk 1 a2k λr+λ 2k r 22k 0. Clearly, a F 2 k a F 2 k 1 Trk 1 λa2k +1 a F 2 k χ g 0 2 2k. 1 Trk 1 λa+r2k +1 Thus, N 2 2k 2 k k 1. Similarly, it follows from 21 that N 2 2k c c3 1 a F 2 k a F 2 k 1 1 Trk 1 λa2k Trk 1 λa2k r+ar 2k +r 2k +1 2 k 2 2 2k 1. Let N 2 2k+1 resp., N 2 2k+1 denote the number of a F 2 k such that χ f a 2 2k+1 resp., 2 2k+1. From 22, we know that if c 3 1, c 1 c 2 0 c 0 0, then χ f a 2 2k+1. Hence we have N 2 2k c0 1 1 c c2 a F 2 k 1+ 1 c1 1 [ 1+ 1 c c2 + 1 c1+c2 1 c3 1 c3+c1 1 c3+c2 1 c3+c2+c1 + 1 c0 + 1 c0+c1 + 1 c0+c2 + 1 c0+c2+c1 1 c0+c3 1 c0+c3+c1 1 c0+c3+c2 1 c0+c3+c1+c2]. Similar to above, we have 1 c1 1 c2 1 c3 0 a F 2 k a F 2 k a F 2 k a F 2 k 1 c0 1 c0+c1 1 c0+c2 1 c0+c3 χ g 0 2 2k. Recalling that u,v,r are pairwise distinct u+v +r 0 by lemma 2 again, we have a F 2 k 1 c1+c2 1 Trk 1 λu2k +1 +v 2k +1 a F 2 k 0 a F 2 k 1 c1+c2+c3 1 Trk 1 a2k λu+v+λ 2k u+v 22k 1 Trk 1 λu2k +1 +v 2k +1 +r 2k +1 1 Trk 1 a2k λu+v+r+λ 2k u+v+r 22k 0 a F 2 k Similarly, a F 2 k 1c3+c1 a F 0. 2 k 1c3+c2 Since t 1 t 2 0, we have 1 c0+c3+c1 1 c0+c3+c1+t1 a F 2 k a F 2 k a F 2 k a F 2 k 1 c0+c3+c2 χ g 0 2 2k a F 2 k a F 2 k χ g 0 2 2k. Since t 3 1, then 1 c0+c2+c1 a F 2 k a F 2 k a F 2 k 1 Trk 1 λa+r+v2k +1 1 c0+c3+c2+t2 χ g 0 2 2k 1 Trk 1 λa+r+u2k +1 1 c0+c2+c1+t3 1 Trk 1 λa+u+v2k +1

9 9 a F 2 k 1 c0+c3+c1+c2 Then we have a F 2 k a F 2 k χ g 0 2 2k. N 2 m+1 1 2k 2 k. a F 2 k 1 c0+c3+c1+c2+t1+t2+t3 1 Trm 1 λa+r+u+v2k +1 Similarly, we have N 2 m c0 1 1 c c c1 2 k. Finally, the number of a F 2 k such that χ f a 0 is equal to 2 k 2 k 1 2 k 3. By analyses similar to those in Theorems 2, we get the following result. Theorem : Let k be a positive integer such that k > 1 let u,v F 2 n. Assume that λ F 2 k such that λ+λ 23k 1. Define a Boolean function as fx Tr k 1 λx2k +1 +Tr k 1 uxtrk 1 vx over F 2 k. Then the following hold: 1 If Tr k 1 v +uv λu2k 2k 0, then f is bent. 2 If Tr k 1 λu 2k v + uv 2k 1, then f is semi-bent. Moreover, if Tr k 1 λu 2k +1 1 or Tr k 1 λv 2k +1 1, then f is a balanced semi-bent function. Example 3: Let k 2, F 2 8 be generated by the primitive polynomial x 8 +x +x 3 +x 2 +1 ξ be a primitive element of F Let P be the set of all permutations on 0,1,2. If one takes λ ξ 3, u ξ 212, v ξ 10 r ξ, then by a Magma program, one can get λ + λ 26 1, Tr 8 1 λu v + uv Tr 8 1λr u + ru Tr 8 1λr v + rv 0 i,j,k P u2i v 2j v 2k ξ 8 0. Computer experiment shows that fx Tr 8 1ξ 3 x 5 + Tr 8 1ξ 212 xtr k 1 ξ 10 xtr k 1 ξ x given by in Theorem 3 is a cubic bent function, which is consistent with the results given in Theorem 3. 2If one takes λ ξ 3, u ξ 212, v ξ 10 r ξ 12, then by a Magma program, one can get Tr 8 1λr v+rv Tr 8 1λr u + ru 1 Tr 8 1λu v + uv 0 Computer experiment shows that fx Tr 8 1 ξ3 x 5 + Tr 8 1ξ 212 xtr k 1 ξ 10 xtr k 1 ξ 12 x given by in Theorem 3 is a five-valued function its distribution of the Walsh spectrum is 0, occurs 96 times, occurs 6 times χ f a, occurs 6 times 32, occurs 20 times 32, occurs 12 times. This is compatible with the results given in Theorem 3. IV. NEW INFINITE FAMILY OF BENT FUNCTIONS FROM THE The bent function NIHO EXPONENTS g Tr m 1 x 2m +1 +Tr n 1 2 k 1 1 i1 x 2m 1 i 2 k +1 via 2 k Niho exponents was found by Leer Kholosha [3], wheregcdk,m 1. Take anyα F 2 n with α+α 2m 1. It was shown in [35] that g is gx Tr m 1 α1+a+a 2 m +α 2n k +a 2m 1+a+a 2m 1/2k Now using Lemma 1 23, we can present the following class of bent functions via 2 k Niho exponents. Theorem 5: Let n 2m, k be a positive with gcdk,m 1 u,v,r F 2m such that u+v+r 0. Then the Boolean function fx Tr m 1 x 2m +1 +Tr n 1 2 k 1 1 i1 +Tr n 1 uxtrn 1 vxtrn 1 rx is a bent function. Proof: Let gx Tr m Tr n x2m 1 2 k 1 1 i1 For each a F 2 n, by Lemma 1, we have x 2m 1 i 2 k +1 x 2m 1 i 2 k +1. χ f a 1 [3 χ ga+ χ g a+v+ χ g a+u χ g a+u+v where + χ g a+r χ g a+r+v χ g a+r+u + χ g a+r+u+v] 1 + 2, 1 1 [3 χ ga+ χ g a+v+ χ g a+u χ g a+u+v] 2 1 [ χ ga+r χ g a+r+v χ g a+r+u + χ g a+r+u+v]. Set A 1+a+a 2m, where α F 2 n such that α+α 2m 1. It follows from 23 that χ g a 2 m 1 Trm 1 αa+α 2n k +a 2m A 1/2k 1. Now we compute 1 2 respectively. Note that u,v,r F 2m. Then we have

10 [3+ χ ga+v+ χ g a+u χ g a+u+v] 1 χ ga [ 3+ 1 Trm 1 va 1/2k Trm 1 va 1/2k 1 +Tr m 1 ua 1 ] 1/2k 1 Trm 1 1 2m 1 Trm 1 [ 3+ 1 Trm 1 1 Trm 1 Similarly, we have 2 1 2m 1 Trm 1 [ 1 1 Trm Trm 1 αa+α 2n k +a 2m A 1/2k 1 va 1/2k 1 va 1/2k 1 +Tr m 1 ua 1/2k Trm 1 ua 1/2k 1 ]. 2 ua 1/2k 1 αa+α 2n k +a 2m +ra 1/2k 1 va 1/2k 1 va 1/2k 1 +Tr m 1 1 Trm 1 ua 1/2k 1 ]. 25 ua 1/2k 1 polynomial in two variables over F 2 m. For a a 1,a 2,b b 1,b 2 F 2 n, the scalar product in F 2 n can be defined as a 1,a 2,b 1,b 2 Tr m 1 a 1 b 1 +a 2 b 2. The well-known Maiorana-McFarl class of bent functions can be defined as follows. gx,y Tr m 1 xπy+hy,x,y F 2 m F 2 m where π : F 2 m F 2 m is a permutation h is a Boolean function over F 2 m, its dual is given by gx,y Tr m 1 yπ 1 x+hπ 1 x where π 1 denotes the inverse mapping of the permutation π []. This together with the definition of the dual function implies that for each a a 1,a 2 F 2 n χ g a 1,a 2 2 m 1 Trm 1 a2π 1 a 1+hπ 1 a Let c 1 Tr m 1 va 1/2 k 1 c 2 Tr m 1 ua 1/2 k 1. When Tr m 1 ra 1/2 k 1 0, by Eqs we have χ f a m 1 Trm 1 αa+α 2n k +a 2m A 1/2k 1. When Tr m 1 ra 1/2 k 1 1, by 2 25 again, we have χ f a m 1 Trm c2 1 c1+c2] 2 m 1 Trm 1 2 m 1 Trm 1 αa+α 2n k +a 2m A 1/2k 1 [1+ 1 c 1 αa+α 2n k +a 2m A 1/2k 1, if c 1 1, c 2 1 αa+α 2n k +a 2m A 1/2k 1, otherwise. Therefore, fx is a bent function. Remark : This result generalizes the case in [31, Theorem 11] for r v. It may be noted that we can not construct more bent functions for the case u,v,r / F 2m according to our numerical results. Example : Let m, k 3 F 2 8 be generated by the primitive polynomial x 8 + x + x 3 + x ξ be a primitive element of F 2 8. If we take u ξ 3, v ξ 17, r ξ 51, then by a Magma program, we can see that fx Tr 1 λx17 + Tr 8 1 x226 + Tr 8 1 x196 + Tr 8 1 x6 + Tr 8 1ξ 3 xtr 8 1ξ 17 xtr 8 1ξ 51 x given by in Theorem 5 is a bent function, which is consistent with the results given in Theorem 5. V. NEW INFINITE FAMILIES OF BENT, SEMI-BENT AND FIVE-VALUED FUNCTIONS FROM THE CLASS OF MAIORANA-MCFARLAND In this section, we identify F 2 n where n 2m with F 2 m F 2 m consider Boolean functions with bivariate representation fx,y Tr m 1 Px,y, where Px,y is a In what follows, by choosing suitable permutations π, we will construct some new bent, semi-bent five-valued functions from the class of Maiorana-McFarl. It is well known that the compositional inverse of a linearized permutation polynomial is also a linearized polynomial. The following two theorems will employ the linearized permutation polynomial over F 2 m to give new Boolean functions with few Walsh transform values. Theorem 6: Let n 2m u u 1,u 2,v v 1,v 2,r r 1,r 2 are three pairwise distinct nonzero elements in F 2 m F 2 m such that u + v + r 0. Assume that π is a linearized permutation polynomial over F 2 m. Let fx,y be the Boolean function given by fx,y Tr m 1 xπy+tr m 1 y +Tr m 1 u 1x+u 2 ytr m 1 v 1x+v 2 ytr m 1 r 1x+r 2 y. If Tr m 1 r 2π 1 v 1 + v 2 π 1 r 1 0, Tr m 1 r 2π 1 u 1 + u 2 π 1 r 1 0 Tr m 1 u 2 π 1 v 1 + v 2 π 1 u 1 0, then fx, y is bent. Otherwise, fx, y is five-valued the Walsh spectrum of fx,y is 0,±2 m,±2 m+1 }. Moreover, if Tr m 1 r 2 π 1 v 1 + v 2 π 1 r 1, Tr m 1 r 2π 1 u 1 + u 2 π 1 r 1, Tr m 1 u 2π 1 v 1 + v 2 π 1 u 1 0,0,1,1,0,0,0,1,0,1,1,1}, when a 1,a 2 runs through all elements in F 2 m F 2 m, the distribution of the Walsh spectrum of five-valued function fx,y is given by 0, occurs 2 n 2 n 1 2 n 3 times 2 m, occurs 2 n 2 +2 m 1 times χ f a 2 m, occurs 2 n 2 2 m 1 times 2 m+1, occurs 2 n times 2 m+1, occurs 2 n times. If Tr m 1 r 2 π 1 v 1 + v 2 π 1 r 1, Tr m 1 r 2π 1 u 1 + u 2 π 1 r 1, Tr m 1 u 2 π 1 v 1 + v 2 π 1 u 1 1,1,0,1,0,1,0,1,1}, when a 1,a 2 runs through all elements in F 2 m F 2 m, the distribution of the Walsh

11 11 spectrum of five-valued function fx,y is given by 0, occurs 2 n 2 n 1 2 n 3 times 2 m, occurs 2 n 2 times χ f a 2 m, occurs 2 n 2 times 2 m+1, occurs 2 n +2 m 2 times 2 m+1, occurs 2 n 2 m 2 times. Proof: Let gx,y Tr m 1 xπy +Tr m 1 y. From 26, for each a 1,a 2 F 2 m F 2 m, we get χ g a 1,a 2 2 m 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a Applying Lemma 1 again, for each a 1,a 2 F 2 m F 2 m, we have where χ f a 1,a , χ g a 1 +r 1 +u 1,a 2 +r 2 +u 2 + χ g a 1 +r 1 +v 1 +u 1,a 2 +r 2 +v 2 +u 2 ]. Note that π 1 is a linearized polynomial. It then follows from 27 that χ g a 1 +v 1,a 2 +v 2 2 m 1 Trm 1 a2+v2π 1 a 1+v 1+Tr m 1 π 1 a 1+v 1 2 m 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1 1 Trm 1 a2π 1 v 1+v 2π 1 a 1+v 2+1π 1 v 1 χ g a 1,a 2 1 Trm 1 a2π 1 v 1+v 2π 1 a 1+v 2+1π 1 v 1. Similarly, we can compute χ g a 1 +u 1,a 2 +u 2 χ g a 1,a 2 1 Trm 1 a2π 1 u 1+u 2π 1 a 1+u 2+1π 1 u 1, χ g a 1 +v 1 +u 1,a 2 +v 2 +u 2 χ g a 1,a 2 1 Trm 1 a2π 1 v 1+v 2π 1 a 1+v 2+1π 1 v 1 1 Trm 1 a2π 1 u 1+u 2π 1 a 1+u 2+1π 1 u 1 1 Trm 1 u2π 1 v 1+v 2π 1 u 1. On the other h, we can compute χ g a 1 +r 1 +v 1,a 2 +r 2 +v 2 χ g a 1 +r 1,a 2 +r 2 1 Trm 1 a2π 1 v 1+v 2π 1 a 1+v 2+1π 1 v 1 1 Trm 1 r2π 1 v 1+v 2π 1 r 1 Similarly, we can compute the values of χ g a 1 +r 1 +u 1,a 2 + r 2 +u 2 χ g a 1 +r 1 +v 1 +u 1,a 2 +r 2 +v 2 +u 2 as follows respectively. χ g a 1 +r 1 +u 1,a 2 +r 2 +u 2 χ g a 1 +r 1,a 2 +r 2 1 Trm 1 a2π 1 u 1+u 2π 1 a 1+u 2+1π 1 u 1 1 Trm 1 r2π 1 u 1+u 2π 1 r 1 χ g a 1 +r 1 +v 1 +u 1,a 2 +r 2 +v 2 +u 2 χ g a 1 +r 1,a 2 +r 2 1 Trm 1 a2π 1 v 1+v 2π 1 a 1+v 2+1π 1 v 1 1 Trm 1 a2π 1 u 1+u 2π 1 a 1+u 2+1π 1 u 1 1 Trm 1 r2π 1 v 1+v 2π 1 r 1+Tr m 1 r2π 1 u 1+u 2π 1 r [3 χ ga 1,a 2 + χ g a 1 +v 1,a 2 +v 2 1 Trm 1 u2π 1 v 1+v 2π 1 u 1. + χ g a 1 +u 1,a 2 +u 2 Let c 1 Tr m 1 a2 π 1 v 1 +v 2 π 1 a 1 +v 2 +1π 1 v 1, c 2 Tr m 1 χ g a 1 +v 1 +u 1,a 2 +v 2 +u 2 ] a 2π 1 u 1 +u 2 π 1 a 1 +u 2 +1π 1 u 1 c 3 Tr m 1 a 2π 1 r 1 +r 2 π 1 a 1 +r 2 +1π 1 r 1. Denote t 1 Tr m 1 r 2 π 1 v 1 +v 2 π 1 r 1, t 2 Tr m 1 r 2 π 1 u u 2 π 1 r 1 t 3 Tr m [ χ 1 u 2π 1 v 1 +v 2 π 1 u 1. Summarizing the discussion above, we can ga 1 +r 1,a 2 +r 2 χ g a 1 +r 1 +v 1,a 2 +r 2 +v 2 get 1 1 2m 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1 [ 3+ 1 c1 + 1 c2 1 c1+c2+t3] m 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1+c 3 [ 1 1 c1+t1 1 c2+t2 + 1 c1+c2+t1+t2+t3]. 29 Similar as in Theorem 1, we can show that when t 1 t 2 t 3 0 c 3 0 χ f a 1,a 2 2 m 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1 when t 1 t 2 t 3 0 c 3 1 χ f a 1,a 2 2 m 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1, if c 1 c m 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1, otherwise. Hence, fx,y is bent if t 1 t 2 t 3 0. Next we will prove that fx, y is five-valued determine the distribution of its Walsh transform in the case of t 1 t 2 1 t 3 0 others can be proved by a similar manner. In this case, become 1 1 2m 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1 [ 3+ 1 c1 + 1 c2 1 c1+c2] 2 1 2m 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1+c 3 [ 1+ 1 c c2 + 1 c1+c2].

12 12 When c 3 0, we have χ f a 1,a m 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1 [2+ 1 c1 + 1 c2 ] 2 m+1 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1, if c 1 c 2 0 0, if c 1 c m 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1, otherwise. 30 When c 3 1, we have χ f a 1,a c1+c2 1 Trm 1 a2π 1 v 1+v 2π 1 a 1+v 2+1π 1 v 1 1 Trm 1 a2π 1 u 1+u 2π 1 a 1+u 2+1π 1 u 1 1 Trm 1 v2+1π 1 v 1+u 2+1π 1 u 1 1 Trm 1 π 1 a 1v 2+u 2 a 1 F 2 m a 2 F 2 m 1 Trm 1 a2π 1 u 1+v m 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1 [1 1 c1+c2 ] 0, if c 1 c 2 1 or c 1 c m 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1, otherwise. 31 Combing 30 31, we conclude that fx, y is five-valued the Walsh spectrum of fx,y is 0,±2 m,±2 m+1 }. Let c 0 Tr m 1 a 2π 1 a 1 +Tr m 1 π 1 a 1 denote by N i the number ofa 1,a 2 F 2 m F 2 m such that χ f a 1,a 2 2 m i, where i 0,1, 1,2,2. Firstly, we compute N 2 m N 2 m. From 30 31, we have N 2 m c c1 1 1 c c0 1 1 c c c0 1 1 c1+c2 1 [ 2 n 1 c1+c2 + 1 c0+c1+c2]. 1 c0 For a 1,a 2 F 2, we have m 1c0 1 c0 1 Trm 1 a2π 1 a 1+Tr m 1 π 1 a 1 χ g 0 2 m. 32 Since π is linearized permutation polynomial over F 2 m t 3 0, we have 1 c0+c1+c2 1 c0+c1+c2+t3 1 Trm 1 a2+v2+u2π 1 a 1+v 1+u 1 1 Trm 1 π 1 a 1+v 1+u 1 χ g 0 2 m. 33 where the last identity holds since u 1,u 2 v 1,v 2 implies that either u 1 +v 1 0 or u 2 +v 2 0. Based on the analysis above, we have N 2 m 2 n 2. Similarly, we can get N 2 m 2 n 2. Secondly, we compute N 2 m+1 N 2 m+1. It follows from 30 that N 2 m c c c c3 1 [ 1+ 1 c c2 + 1 c1+c2 + 1 c3 + 1 c3+c1 + 1 c3+c2 + 1 c3+c2+c1 + 1 c0 + 1 c0+c1 + 1 c0+c2 + 1 c0+c2+c1 + 1 c0+c3 + 1 c0+c3+c1 + 1 c0+c3+c2 + 1 c0+c3+c1+c2]. Note that u,v,r are pairwise distinct. Similar to 3, we have 1 c1 1 c2 1 c3 0 1 c1+c2 1 c3+c1 1 c3+c2 0. Since u+v +r 0 implies that either u 1 +v 1 +r 1 0 or u 2 +v 2 +r 2 0, we have 1 c1+c2+c3 1 Trm 1 v2+1π 1 v 1+u 2+1π 1 u 1+r 2+1π 1 r 1 1 Trm 1 π 1 a 1v 2+u 2+r 2 0. a 1 F 2 m a 2 F 2 m 1 Trm 1 a2π 1 u 1+v 1+r 1

13 13 Similar to 32, we can get 1 c0 1 c0+c1 1 c0+c3 χ g 0 2 m. Similar to 33, since t 1 t 2 1, we have 1 c0+c3+c1 1 Trm 1 π 1 a 1+r 1+v 1 χ g 0 2 m 1 c0+c2 1 c0+c3+c1+t1 1 Trm 1 a2+r2+v2π 1 a 1+r 1+v 1 1 c0+c3+c2 1 c0+c3+c2+t2 1 Trm 1 a2+r2+u2π 1 a 1+r 1+u 1 1 Trm 1 π 1 a 1+r 1+u 1 χ g 0 2 m. By t 1 t 2 1 t 3 0, we have 1 c0+c3+c1+c2 1 c0+c3+c1+c2+t1+t2+t3 1 Trm 1 a2+r2+v2+u2π 1 a 1+r 1++v 1+u 1 1 Trm 1 π 1 a 1+r 1+v 1+u 1 χ g 0 2 m. Based on the discussions above, we can get N 2 m+1 1 2n +2 m+2 2 n +2 m 2. Similarly, we have N 2 m+1 1 a F 2 k 1+ 1 c1 2 n 2 m c c c2 It should be noted that two of u,v,r F 2n can be equal. Without loss of generality, we assume that r v, then the following result can be obtained. Theorem 7: Let n 2m u u 1,u 2,v v 1,v 2 are two distinct nonzero elements in F 2 m F 2 m. Assume that π is a linearized permutation polynomial of F 2 m. Let fx,y be the Boolean function given by fx,y Tr m 1 xπy+tr m 1 y +Tr m 1 u 1x+u 2 ytr m 1 v 1x+v 2 y. If Tr m 1 u 2π 1 v 1 +v 2 π 1 u 1 0, then f is bent. Otherwise, f is semi-bent. Proof: The proof is similar to Theorem 2 we omit it here. Remark 5: To obtain our constructions in Theorems 6 7, we need to determine the compositional inverse of a given linearized permutation polynomial over F 2 m. Information on the compositional inverses of certain linearized permutation polynomials could be found in [36], [37], [38]. Clearly, the simplest suitable linearized permutation polynomial π over F 2 m in Theorems 6 7 is x 2k where 0 k n 1. Theorem 8: Let n 2m s be a divisor of m with m s is odd. Assume that u u 1,u 2,v v 1,v 2 are two distinct nonzero elements in F 2 s F 2 s such that u 1 v 2 +v 1 u 2 0. Let fx,y be the Boolean function given by fx,y Tr m 1 xyd +Tr m 1 u 1x+u 2 ytr m 1 v 1x+v 2 y where d2 s +1 1 mod 2 m 1. If Tr m 1 u2 1 v 2+u 2 v 2 1 0, then fx,y is bent. Otherwise, fx,y is semi-bent. Proof: Let πy y d gx,y Tr m 1 xπy. Since d2 s mod2 m 1, then π 1 y y 2s +1. This together with 26 implies that for each a a 1,a 2 F 2 n χ g a 1,a 2 2 m 1 Trm 1 a2a2s According to Lemma 1, for each a 1,a 2 F 2 n, we have χ f a 1,a [ χ ga 1,a 2 + χ g a 1 +v 1,a 2 +v 2 + χ g a 1 +u 1,a 2 +u 2 χ g a 1 +v 1 +u 1,a 2 +v 2 +u 2 ]. Now we compute χ g a 1 +v 1,a 2 +v 2, χ g a 1 +u 1,a 2 +u 2 χ g a 1 +v 1 +u 1,a 2 +v 2 +u 2 respectively. By 35, we have χ g a 1 +v 1,a 2 +v 2 2 m 1 Trm 1 a2+v2a1+v12s +1 2 m 1 Trm 1 a2a2s Tr m 1 a2a2s 1 v1+a2a1v2s 1 +a2v2s Trm 1 a2s +1 1 v 2+a 2s 1 v1v2+a1v2s 1 v2+v2s +1 1 v 2 χ g a 1,a 2 1 Trm 1 a2a2s 1 v1+a2a1v1+a2v2 1 1 Trm 1 a2s +1 1 v 2+a 2s 1 v1v2+a1v1v2+v2 1 v2 36 where the last identity holds since v v 1,v 2 is a nonzero element in F 2 s F 2 s. Similarly, we can show that χ g a 1 +u 1,a 2 +u 2 χ g a 1,a 2 1 Trm 1 a2a2s 1 u1+a2a1u1+a2u2 1 1 Trm 1 a2s +1 1 u 2+a 2s 1 u1u2+a1u1u2+u2 1 u2 37

14 1 χ g a 1 +v 1 +u 1,a 2 +v 2 +u 2 χ g a 1,a 2 1 Trm 1 a2a2s 1 v1+a2a1v1+a2v2 1 1 Trm 1 a2s +1 1 v 2+a 2s 1 v1v2+a1v1v2+v2 1 v2 1 Trm 1 a2a2s 1 u1+a2a1u1+a2u2 1 1 a2s +1 1 u 2+a 2s 1 u1u2+a1u1u2+u2 1 u2 1 Trm 1 a2s 1 +a1u1v2+v1u2+u2 1 v2+v2 1 u2. 38 Let c 1 Tr m 1 a 2a 2s 1 v 1+a 2 a 1 v 1 +a 2 v a2s 1 v 2 +a 2s 1 v 1v 2 + a 1 v 1 v 2 + v1 2v 2 c 2 Tr m 1 a 2 a 2s 1 u 1 + a 2 a 1 u 1 + a 2 u a 2s +1 1 u 2 +a 2s 1 u 1 u 2 +a 1 u 1 u 2 +u 2 1u 2. Note that u 1 v 2 + v 1 u 2 0. If Tr m 1 u2 1 v 2 + u 2 v1 2 0, combing 36, 37 38, we get χ f a 1,a m 1 Trm 1 a2a2s +1 1 [1+ 1 c1 + 1 c2 1 c1+c2 ] 2 m 1 Trm 1 a2a2s +1 1, if c 1 c m 1 Trm 1 a2a2s +1 1, otherwise. If Tr m 1 u 2 1v 2 +u 2 v 2 1 1, then χ f a 1,a m 1 Trm 1 a2a2s +1 1 [1+ 1 c1 + 1 c2 + 1 c1+c2 ] 2 m+1 1 Trm 1 a2a2s +1 1, if c 1 c 2 0 0, otherwise. The desired conclusion follows from the definitions of bent semi-bent function. Example 5: Let m 9, s 3 F 2 9 be generated by the primitive polynomial x 9 +x +1 ξ be a primitive element of F Take u u 1,u 2 ξ 219,ξ 73 v v 1,v 2 ξ,1. Clearly, u 1 v 2 + u 2 v mod 512. By help of a computer, we can get Tr 9 1u 2 1v 2 + u 2 v the function fx Tr9 1 xy28 +Tr 9 1 ξ219 x+ ξ 73 ytr 9 1 ξ x+y is a bent function over F 2 9 F 2 9, which is consistent with the results given in Theorem 8; 2 Take u u 1,u 2 ξ,ξ 73 v v 1,v 2 ξ 73,1. Clearly, u 1 v 2 + u 2 v mod 512. By help of a computer, we can get Tr 9 1u 2 1v 2 + u 2 v the function fx Tr9 1 xy28 +Tr 9 1 ξ x+ ξ 73 ytr 9 1ξ 73 x+y is semi-bent function overf 2 9 F 2 9, which is consistent with the results given in Theorem 8. VI. CONCLUSION Several new classes of Boolean functions with few Walsh transform values, including bent, semi-bent five-valued functions are provided, the distribution of the Walsh spectrum of five-valued functions presented in this paper are also completely determined. 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