Some classes of monomial complete permutation polynomials over finite fields of characteristic two

Finite Fields and Their Applications 28 (2014) 148–165

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Finite Fields and Their Applications www.elsevier.com/locate/ffa

Some classes of monomial complete permutation polynomials over finite fields of characteristic two Gaofei Wu a , Nian Li b , Tor Helleseth c , Yuqing Zhang d,∗ a

State Key Laboratory of Integrated Service Networks, Xidian University, Xi’an 710071, China Information Security and National Computing Grid Laboratory, Southwest Jiaotong University, Chengdu 610031, China c Department of Informatics, University of Bergen, N-5020 Bergen, Norway d National Computer Network Intrusion Protection Center, UCAS, Beijing 100043, China b

a r t i c l e

i n f o

Article history: Received 26 October 2013 Received in revised form 14 January 2014 Accepted 19 January 2014 Available online 6 March 2014 Communicated by Rudolf Lidl

a b s t r a c t In this paper, four classes of complete permutation polynomials over finite fields of characteristic two are presented. To consider the permutation property of the first three classes, Dickson polynomials play a key role. The fourth class is a generalization of a known result. In addition, we also calculate the inverses of these bijective monomials. © 2014 Elsevier Inc. All rights reserved.

MSC: 05A05 11T06 Keywords: Finite field Complete permutation polynomials Walsh transform Dickson polynomials

* Corresponding author. E-mail addresses: [email protected] (G. Wu), [email protected] (N. Li), [email protected] (T. Helleseth), [email protected] (Y. Zhang). 1071-5797/$ – see front matter © 2014 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.ffa.2014.01.011

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1. Introduction Let Fq be a finite field of q = pn elements, where p is a prime number, and n is a positive integer. We denote by F∗q the multiplication group of Fq . A polynomial f ∈ Fq [x] is called a permutation polynomial (PP) if the associated polynomial mapping f : c → f (c) from Fq to itself is a permutation over Fq [14]. Permutation polynomials over finite fields have important applications in cryptography, coding theory, and combinatorial design theory. In [12,13], Lidl and Mullen listed many open problems and one of them is to find more permutation polynomials. Recently, permutation polynomials have been studied extensively in the literature, see [1,3–5,7,8,10,11,15,19,24,25,27,29] for example. A polynomial f ∈ Fq [x] is called a complete permutation polynomial (CPP) if both f (x) and f (x) + x are permutations over Fq . CPPs are useful in the study of orthogonal Latin squares. In [17], Niederreiter and Robinson gave a detailed study of CPPs over Fq . Dickson polynomials (see Section 2) are familiar examples of PPs. In [16], Mullen and Niederreiter studied under what conditions a Dickson polynomial can be a CPP. Although there are some results on CPPs over Fq [10,20,23,26], still very few classes of CPPs are known, even for monomial functions. For a positive integer d and a ∈ F∗q , a monomial function axd is a CPP over Fq if and only if gcd(d, q − 1) = 1 and axd + x is a PP over Fq . We call such d a CPP exponent k over Fq . In [2], Charpin and Kyureghyan proved that a−1 x2 +2 is a CPP over F22k for odd k and a ∈ βF2k , where β ∈ F22 \F2 . Very recently, Tu, Zeng, and Hu [21] gave three classes of CPP exponents over F2n : (1) d = 22k + 2k + 2, n = 3k, gcd(k, 3) = 1; (2) d = 2k+1 + 3, n = 2k, k odd; and (3) d = 2k−2 (2k + 3), n = 2k, k odd. Some further results about CPPs can be found in [22,28]. In this paper, four more classes of monomial CPPs are studied: 24k −1 + 1, where gcd(k, 4) = 1. 2k −1 26k −1 6k and d = 2k −1 + 1, where gcd(k, 6) = 1. 10k −1 10k and d = 22k −1 + 1, where gcd(k, 10) = 3k 2 −1 3k and d = 2k −1 + 1, where gcd(k, 9) = 3.

(1) n = 4k and d = (2) n = (3) n = (4) n =

1.

In investigating the permutation behavior of the first three classes of CPPs, Dickson polynomials play an important role. The first three classes of CPP exponents are gen2k 3k −1 −1 + 1 [2], and d = 22k −1 +1 eralizations of two known results, namely, d = 22k −1

−1 [21]. Note that for the exponent d = 22k −1 + 1, the case of gcd(k, 3) = 1 has been discussed in [21], and in this paper we proved that d is also a CPP exponent over F23k for gcd(k, 9) = 3. In addition, the inverses of these CPP exponents are also given. 3k

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2. Preliminaries Let n and k be two integers such that k|n. The trace function from F2n onto F2k is defined by 

n/k−1

Trnk (x) =

ik

x2 ,

x ∈ F2n .

i=0

The Walsh transform of a function f from F2n to F2 is defined by 

Wf (a) =

n

(−1)f (x)+Tr1 (ax) .

x∈F2n

In [14], a criterion for PPs is given by using the additive characters of the underlying finite field. Lemma 1. (See [14, Theorem 7.7].) A mapping g : F2n → F2n is a PP if and only if for every α ∈ F∗2n , 

n

(−1)Tr1 (αg(x)) = 0.

x∈F2n

In the sequel, the Dickson polynomials will play a key role in the proofs of our results, we introduce the definition of Dickson polynomials. For a positive integer n and an element η ∈ F∗q , the Dickson polynomial Dn (x, η) over Fq is defined by n/2

Dn (x, η) =

 j=0

  n−j n (−η)j xn−2j . n−j j

Dickson polynomials of degree n can be obtained via the following recursion: Dn (x, η) = xDn−1 (x, η) − ηDn−2 (x, η), where D0 (x, η) = 2, D1 (x, η) = x. Proposition 1. (See [14, Theorem 7.16].) The Dickson polynomial Dn (x, η), η ∈ F∗q is a PP over Fq if and only if gcd(n, q 2 − 1) = 1. The following lemmas are also needed in the sequel. Lemma 2. (See [14, Corollary 3.47].) An irreducible polynomial over Fq of degree n remains irreducible over Fqk if and only if gcd(k, n) = 1.

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Lemma 3. Let f (x) = x3 + x + 1 ∈ F2 [x] be an irreducible polynomial over F2 . Let θ be a root of f (x) and g(x) = x3 + x + θ ∈ F23 [x]. Then g(x) is irreducible over F2k , where gcd(k, 9) = 3. Proof. Let k = 3k  . From gcd(3k , 9) = 3, we have gcd(k , 3) = 1. It is easy to verify that g(x) has no root in F23 , so that g(x) is irreducible over F23 . From Lemma 2, g(x) is irreducible over F23k if and only if gcd(k , 3) = 1, which completes the proof. 2 Lemma 4. If gcd(r, k) = 1, then {2ik (mod 2r −1) | 1  i  r − 1} = {2i | 1  i  r − 1}. Proof. Since gcd(r, k) = 1, we have {k, 2k, 3k, . . . , (r−1)k} (mod r) = {1, 2, 3, . . . , r−1}. Then the desired conclusion follows. 2 3. Three classes of CPPs Throughout this section, let n, r and k be positive integers with n = rk. Let d=

    where gcd d − 1, 2k − 1 = gcd r, 2k − 1 = 1.

2rk − 1 + 1, 2k − 1

(1)

In this section, a sufficient and necessary condition for the exponent d to be a CPP exponent over F2n is given. As an application, we give three classes of CPPs. Dickson polynomials play an important role in the proofs of these classes of CPPs. Lemma 5. Let d be defined as in (1). If xd + ax is a PP over F2n , then a ∈ / F∗2k . Proof. Assume that a ∈ F∗2k , then a = (θd−1 )l for some integer l, where θ is a primitive element in F2n . In this case, xd + ax = x(xd−1 + a) = 0 has two different solutions x = 0 and x = θl , which shows that xd + ax is not a PP over F2n . 2 ik

For any a ∈ F2n , let ai = a2 , where 0  i  r − 1. Define ha (y) = y

r−1  i=0

where λ0 = 1, and λi = k a2i



(y + ai ) = y

r 

λi y r−i ,

(2)

i=0

0j1
 i  r. Note that

= ai+1 (mod r) , we have · · · aji +1 = λi . Hence λi ∈ F2k for 0  i  r, and ha (y) ∈ F2k [y]. The following lemma is a special case of [27, Lemma 2.1], for self-contained, we give a different proof here. Lemma 6. Let d be defined as in (1). Then for any a ∈ F2n \F2k , ha (y) ∈ F2k [y] is a PP over F2k if and only if xd + ax ∈ F2n [x] is a PP over F2n .

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Proof. First we show that if ha (y) is a PP over F2k , then xd + ax ∈ F2n [x] is a PP over F2n . Using the fact y ∈ F2k , we have 

y r+1 + ay

d−1

r−1   d−1  r−1 2ik    = y d−1 y r + a = y r y r + a i=0 = yr y r + ai = ha y r , i=0 (r−1)k

(r−2)k

k

+2 +···+2 +1 where the second equal sign holds due to y d−1 = y 2 = y r . Since k r r+1 gcd(r, 2 − 1) = 1, then if ha (y) is a PP over F2k , so do ha (y ) and (y + ay)d−1 . From k gcd(d − 1, 2 − 1) = 1, every nonzero x ∈ F2n can be represented uniquely as x = yz, where z ∈ Λ, y ∈ F2k , and Λ = {z ∈ F2n : z d−1 = 1}. Then we have

xd + ax = (yz)d + ayz = zy d + ayz   = z y d + ay   = z y r+1 + ay .

(3)

Let x1 = z1 y1 and x2 = z2 y2 . We need to show that if xd1 + ax1 = xd2 + ax2 , then x1 = x2 . From (3), we have     z1 y1r+1 + ay1 = z2 y2r+1 + ay2 .

(4)

Raising both sides of (4) to the (d − 1)-th power, we have z1d−1 (y1r+1 + ay1 )d−1 = z2d−1 (y2r+1 + ay2 )d−1 . Note that since z1d−1 = z2d−1 = 1, we have (y1r+1 + ay1 )d−1 = (y2r+1 + ay2 )d−1 . Using the fact (y r+1 + ay)d−1 is a PP over F2k , we conclude that y1 = y2 . Since a ∈ / F2k and y ∈ F2k , which implies y r+1 + ay = 0, then z1 = z2 follows from (4). Hence, we have x1 = x2 . Next we show that if xd + ax ∈ F2n [x] is a PP over F2n , then ha (y) is a PP over F2k . From {xd + ax | x ∈ F2n } = F2n , we have {(xd + ax)d−1 | x ∈ F2n } = F2k , which is equivalent to {xd−1 (xd−1 + a)d−1 | x ∈ F2n } = F2k . When x runs through F∗2n , xd−1 runs through F∗2k , and each element in F∗2k occurs (d − 1) times. Then {ha (y) = y(y + a)d−1 | y ∈ F2k } = F2k . This completes the proof. 2 The following corollary can be easily obtained from Lemma 6. Corollary 1. Let d be defined as in (1). Then d is a CPP exponent over F2n if and only if gcd(d, 2n − 1) = 1, and there exist a’s ∈ F2n \F2k such that ha (y) ∈ F2k [y] are PPs over F2k . 3.1. r = 4 and d =

24k −1 2k −1

+1

Let r = 4 and n = 4k, where k is odd. By Lemma 2, for an odd integer k, the polynomial x4 + x + 1 is irreducible over F2k as it is irreducible over F2 . Let β be a root

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of x4 + x + 1, then we have β 15 = 1, and each x ∈ F2n can be represented uniquely as x = x0 + x1 β + x2 β 2 + x3 β 3 , xi ∈ F2k . 4k 4k 4 2 Because Tr4k k (β) is the sum of the roots of x +x+1, we have Trk (β) = Trk (β ) = 0. 4k 3 3 6 12 9 3 6 6 4 3 Moreover, Trk (β ) = β + β + β + β = β + β + (β + β ) + (β + β ) = 1. Since 4k Tr4k k (1) = 0, then Trk (x) = x3 . Note that k is odd, then gcd(k, 4) = 1. By Lemma 4, for both k ≡ 1 (mod 4) and k ≡ 3 (mod 4), we have

ik

β2



 

0  i  3 = β 2i 0  i  3 .

(5)

In the following we show that λi is the same for k ≡ 1 (mod 4) and k ≡ 3 (mod 4), where 1  i  4. So that ha (y) is the same for k ≡ 1 (mod 4) and k ≡ 3 (mod 4). ik Let a = u0 + u1 β + u2 β 2 + u3 β 3 , ui ∈ F2k . Recall that ai = a2 . Define g(x) = ik ik u0 + u1 x + u2 x2 + u3 x3 . Then ai = a2 = g(β 2 ). From (5), we have

ik

a2

  

  



0  i  3 = g β 2ik 0  i  3 = g β 2i 0  i  3 .

(6)

By the definition of λi in (2), the element λi can be seen as an elementary symmetric polynomial in r variables a0 , a1 , . . . , ar−1 . From (6) and the symmetry of λi , we have that λi is the same for k ≡ 1 (mod 4) and k ≡ 3 (mod 4), where 1  i  4. Theorem 1. Let n = 4k and d = F2n for the following a’s:

24k −1 2k −1

+ 1, where k is odd. Then a−1 xd is a CPP over

1) a = u(1 + β + β 2 ) + vβ 3 ; 2) a = u(1 + β + β 3 ) + v(β + β 2 ), or a = u(1 + β 3 ) + v(1 + β + β 2 ); 3) a = u(β + β 3 ) + v(β 2 + β 3 ). Where u, v ∈ F2k , u and v cannot be zero at the same time. Proof. Note that gcd(d, 23k +22k +2k +1) = gcd(23k +22k +2k +2, 23k +22k +2k +1) = 1. Since k is odd, we have gcd(d, 2k −1) = gcd(23k +22k +2k +2, 2k −1) = gcd(5, 2k −1) = 1. Hence, gcd(d, 24k − 1) = 1. Thus, we only need to prove that xd + ax is a PP for all the a’s listed in the theorem. Now we prove that for all these a’s, ha (y) ∈ F2k [y] is a PP over F2k . Then by Lemma 6, xd + ax is a PP over F2n . By (2), we have ha (y) = y

4 

λi y 4−i = y 5 + λ1 y 4 + λ2 y 3 + λ3 y 2 + λ4 y.

i=0

Replacing y by y + λ1 in (7), we get that     ha (y) = y 5 + λ2 y 3 + (λ1 λ2 + λ3 )y 2 + λ41 + λ2 λ21 + λ4 y + λ2 λ31 + λ3 λ21 + λ4 λ1 .

(7)

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Note that if ha (y) is a PP, so does ha (y) + λ2 λ31 + λ3 λ21 + λ4 λ1 . Thus, to complete the proof, it is sufficient to show that ha (y) = y 5 + λ2 y 3 + (λ1 λ2 + λ3 )y 2 + (λ41 + λ2 λ21 + λ4 )y is a PP over F2k . Observe that if a satisfies the following two conditions: a) λ1 λ2 = λ3 , and b) λ22 = λ41 + λ2 λ21 + λ4 , then ha (y) = y 5 + λ2 y 3 + λ22 y = D5 (y, λ2 ) is a PP over F2k due to Proposition 1. Now our task is to verify that all the a’s listed in Theorem 1 satisfy the conditions a) and b). 1) a = u(1 + β + β 2 ) + vβ 3 . Using the fact β 4 = β + 1, one can compute λi ’s directly. As an example, we show the details of the computation of λ2 for this case. Since a = u(1 + β + β 2 ) + vβ 3 = uβ 10 + vβ 3 and Tr4k k (a) = a0 + a1 + a2 + a3 = v, we have λ2 = a0 (a1 + a2 + a3 ) + a1 (a2 + a3 ) + a2 a3 = a0 (a0 + v) + a1 (a2 + a3 ) + a2 a3       = uβ 10 + vβ 3 uβ 10 + vβ 3 + v + uβ 5 + vβ 6 u + vβ 8    + uβ 10 + vβ 12 uβ 5 + vβ 9         = u2 β 5 + v 2 β 6 + v uβ 10 + vβ 3 + u2 β 5 + v 2 β 14 + uv β 6 + β 13   + u2 + v 2 β 6 + uv β 4 + β 2     = u2 + uv β 10 + β 6 + β 13 + β 4 + β 2 + v 2 β 3 + β 14 = u2 + uv + v 2 . 2 2 3 4 3 4 Similarly, we can compute λ1 = Tr4k k (a) = v, λ3 = u v +uv +v , λ4 = u +uv +v . And it can be verified that they satisfy the two conditions a) and b). 2) a = u(1+β+β 3 )+v(β+β 2 ) = uβ 7 +vβ 5 or a = u(1+β 3 )+v(1+β+β 2 ) = uβ 14 +vβ 10 . k In these two cases, note that if k ≡ 1 (mod 4), then uβ 14 + vβ 10 = (uβ 7 + vβ 5 )2 ; 3k if k ≡ 3 (mod 4), then uβ 14 + vβ 10 = (uβ 7 + vβ 5 )2 . This implies that for both of them, we have λ1 = u, λ2 = v 2 + uv, λ3 = u2 v + v 2 u, λ4 = u4 + u3 v + v 4 , and it can be verified that they satisfy the two conditions a) and b). 3) a = u(β + β 3 ) + v(β 2 + β 3 ). Similarly, in this case, we have λ1 = u + v, λ2 = u2 + uv + v 2 , λ3 = u3 + v 3 , λ4 = u4 + u3 v + u2 v 2 + uv 3 + v 4 , and it can be verified that they satisfy the two conditions a) and b).

This completes the proof. 2

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Remark 1. The number of a’s in Theorem 1 is Nk = (2k+2 + 2)(2k − 1). For k = 3, Nk = 238, and it is confirmed by Magma that these are all the a’s such that a−1 xd is a CPP over F2n . Remark 2. In [17, Lemma 1], Niederreiter and Robinson give the conditions under which q−1 the polynomial x m +1 + ax is a PP over Fq , where m|(q − 1). However, they do not give q−1 the explicit a’s such that x m +1 + ax is a PP. Note that for some m|(q − 1), there do not q−1 exist a’s such that x m +1 + ax is a PP. For example, let n = 11, 211 − 1 = 23 · 89, then 11 11 d = 2 23−1 + 1 = 90 and d = 2 89−1 + 1 = 24 are not CPP exponents over F211 . As another example, let n = 12, 212 − 1 = 32 · 5 · 7 · 13, then d = 21 + 1 = 22 and d = 91 + 1 = 92 are not CPP exponents over F212 . 3.2. r = 6 and d =

26k −1 2k −1

+1

In this subsection, we study the second class of CPPs, namely, n = 6k and d = + 1, where gcd(k, 6) = 1. From Lemma 2, for an integer k such that gcd(k, 6) = 1, the polynomial x6 + x + 1 is irreducible over F2k as it is irreducible over F2 . Let β be a root of x6 + x + 1, then we have β 63 = 1, and each x ∈ F2n can be represented uniquely as x = x0 + x1 β + x2 β 2 + x3 β 3 + x4 β 4 + x5 β 5 , xi ∈ F2k . Note that gcd(k, 6) = 1, thus by Lemma 4, we have 26k −1 2k −1



ik

β2



 

1  i  5 = β 2i 1  i  5 .

Similar as the case of r = 4, since gcd(k, 6) = 1, λi and ha (y) are the same for k ≡ 1, 5 (mod 6). Theorem 2. Let n = 6k and d = over F2n for the following a’s:

26k −1 2k −1

+ 1, where gcd(k, 6) = 1. Then a−1 xd is a CPP

1) a = u(1 + β + β 2 + β 3 ) + vβ 5 ; 2) a = u(β + β 2 + β 5 ) + v(β 4 + β 5 ); 3) a = u(β 2 + β 5 ) + v(β 3 + β 4 ). Where u, v ∈ F2k , u and v cannot be zero at the same time. Proof. Since gcd(k, 6) = 1, we have gcd(d − 1, 2k − 1) = gcd(6, 2k − 1) = 1. Then gcd(d, 26k − 1) = gcd(d, d − 1) · gcd(d, 2k − 1) = 1 · gcd(7, 2k − 1) = 1. Thus, we only need to prove that xd + ax is a PP for all the a’s listed in the theorem. Similarly, according to Lemma 6, it is sufficient to prove that for all these a’s, ha (y) ∈ F2k [y] is a PP over F2k .

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By (2), we have

ha (y) = y

6 

λi y 6−i = y 7 + λ1 y 6 + λ2 y 5 + λ3 y 4 + λ4 y 3 + λ5 y 2 + λ6 y.

(8)

i=0

Replacing y by y + λ1 in (8), we get that     ha (y) = y 7 + λ21 + λ2 y 5 + (λ1 λ2 + λ3 )y 4 + λ41 + λ4 y 3 + (λ5 + λ4 λ1 )y 2     + λ4 λ21 + λ6 + λ2 λ41 + λ61 y + λ2 λ51 + λ3 λ41 + λ4 λ31 + λ5 λ21 + λ6 λ1 . Similar as the proof in Theorem 1, observe that if a satisfies the following four conditions: a) b) c) d)

λ1 λ2 = λ3 , λ41 = λ4 , λ5 = λ4 λ1 , and λ4 λ21 + λ6 + λ2 λ41 + λ61 = (λ21 + λ2 )3 ,

then ha (y) = y 7 + (λ21 + λ2 )y 5 + (λ21 + λ2 )3 y + λ1 λ6 = D7 (y, λ21 + λ2 ) + λ1 λ6 is a PP over F2k due to Proposition 1. In the following, we give the values of λi ’s for each case, where 1  i  r. 1) a = u(1 + β + β 2 + β 3 ) + vβ 5 . In this case, we have λ1 = v, λ2 = u(u + v), λ3 = uv(u + v), λ4 = v 4 , λ5 = v 5 , λ6 = u6 + u5 v + u3 v 3 + u2 v 4 + v 6 . 2) a = u(β + β 2 + β 5 ) + v(β 4 + β 5 ). In this case, we have λ1 = u + v, λ2 = uv, λ3 = uv(u + v), λ4 = u4 + v 4 , λ5 = u5 + uv(u3 + v 3 ) + v 5 , λ6 = u6 + u3 v 3 + v 6 . 3) a = u(β 2 + β 5 ) + v(β 3 + β 4 ). Similarly, in this case, we have λ1 = u, λ2 = v(u + v), λ3 = uv(u + v), λ4 = u4 , λ5 = u5 , λ6 = u6 + u4 v 2 + u3 v 3 + uv 5 + v 6 . It can be verified that they all satisfy the four conditions. This completes the proof. 2 Remark 3. The number of a’s in Theorem 2 is Nk = 3(22k − 1). For k = 5, these are all the a’s such that ha (y) is a Dickson polynomial over F2k , but it is not clear that whether these are all the a’s such that ha (y) is a PP over F2k . 3.3. r = 10 and d =

210k −1 2k −1

+1

In this subsection, we study the third class of CPPs, namely, n = 10k and d = + 1, where gcd(k, 10) = 1.

210k −1 2k −1

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Let r = 10 and n = 10k. By Lemma 2, for an integer k such that gcd(k, 10) = 1, the polynomial x10 + x3 + 1 is irreducible over F2k as it is irreducible over F2 . Let β be a root of x10 + x3 + 1, then we have β 1023 = 1, and each x ∈ F2n can be represented uniquely as x = x0 + x1 β + x2 β 2 + x3 β 3 + x4 β 4 + x5 β 5 + x6 β 6 + x7 β 7 + x8 β 8 + x9 β 9 , xi ∈ F2k . Similar as the case of r = 4, since gcd(k, 10) = 1, λi and ha (y) are the same for k ≡ 1, 3, 7, 9 (mod 10). Theorem 3. Let n = 10k and d = over F2n for the following a’s: 1) 2) 3) 4) 5)

210k −1 , 2k −1

where gcd(k, 10) = 1. Then a−1 xd is a CPP

a = u(1 + β 2 + β 4 + β 6 + β 7 ) + v(β + β 2 + β 3 + β 5 + β 6 + β 7 + β 8 ); a = u(1 + β 3 + β 5 + β 6 + β 7 + β 9 ) + v(β + β 5 + β 6 + β 9 ); a = u(1 + β 3 + β 5 + β 7 + β 9 ) + v(β 2 + β 3 + β 8 + β 9 ); a = u(β + β 2 + β 5 + β 6 + β 7 + β 8 ) + v(β 4 + β 5 + β 7 + β 9 ); a = u(β + β 8 + β 9 ) + v(β 3 + β 6 + β 7 + β 9 ).

Where u, v ∈ F2k , u and v cannot be zero at the same time. Proof. Since gcd(k, 10) = 1, we have gcd(d − 1, 2k − 1) = gcd(10, 2k − 1) = 1. Then gcd(d, 210k − 1) = gcd(d, d − 1) · gcd(d, 2k − 1) = 1 · gcd(11, 2k − 1) = 1. Thus, we only need to prove that xd +ax is a PP for all the a’s listed in the theorem. Similarly, as in the proofs of Theorems 1 and 2, we prove that for all these a’s, ha (y) ∈ F2k [y] is a PP over F2k . By (2), we have ha (y) = y =y

10 

λi y 10−i

i=0 11

+ λ1 y 10 + λ2 y 9 + λ3 y 8 + λ4 y 7 + λ5 y 6 + λ6 y 5 + λ7 y 4 + λ8 y 3

+ λ9 y 2 + λ10 y.

(9)

Replacing y by y + λ1 in (9), we get that     ha (y) = y 11 + λ21 + λ2 y 9 + (λ3 + λ2 λ1 )y 8 + λ4 y 7 + (λ5 + λ4 λ1 )y 6 + λ4 λ21 + λ6 y 5     + λ4 λ31 + λ5 λ21 + λ7 + λ6 λ1 y 4 + λ4 λ41 + λ81 + λ8 y 3     8 6 2 + λ9 + λ5 λ41 + λ4 λ51 + λ8 λ1 y 2 + λ6 λ41 + λ10 1 + λ2 λ1 + λ4 λ1 + λ10 + λ8 λ1 y + λ2 λ91 + λ3 λ81 + λ4 λ71 + λ5 λ61 + λ6 λ51 + λ7 λ41 + λ8 λ31 + λ9 λ21 + λ10 λ1 . Note that if ha (y) is a PP, so does ha (y) = ha (y) + λ2 λ91 + λ3 λ81 + λ4 λ71 + λ5 λ61 + λ6 λ51 + λ7 λ41 + λ8 λ31 + λ9 λ21 + λ10 λ1 . Through observing, if a satisfies the following seven conditions: a) λ1 λ2 = λ3 , b) λ4 = λ5 = 0,

158

c) d) e) f) g)

G. Wu et al. / Finite Fields and Their Applications 28 (2014) 148–165

λ4 λ21 + λ6 = (λ21 + λ2 )3 , λ7 = λ6 λ1 , λ81 + λ8 = (λ21 + λ2 )4 , λ9 = λ8 λ1 , and 8 6 2 2 5 λ6 λ41 + λ10 1 + λ2 λ1 + λ4 λ1 + λ10 + λ8 λ1 = (λ1 + λ2 ) ,

then ha (y) = y 11 +(λ21 +λ2 )y 9 +(λ21 +λ2 )3 y 5 +(λ21 +λ2 )4 y 3 +(λ21 +λ2 )5 y = D11 (y, λ21 +λ2 ) is a PP over F2k due to Proposition 1. It remains to verify that all the a’s satisfy the seven conditions. We give the values of λi ’s for each case as follows, where 1  i  r. 1) a = u(1+β 2 +β 4 +β 6 +β 7 )+v(β+β 2 +β 3 +β 5 +β 6 +β 7 +β 8 ) = uβ 420 +vβ 141 (item 1) in the theorem) or a = u(β+β 2 +β 5 +β 6 +β 7 +β 8 )+v(β 4 +β 5 +β 7 +β 9 ) = uβ 291 +vβ 105 (item 4) in the theorem). In these two cases, note that if k ≡ 1 (mod 10), then uβ 291 + vβ 105 = (uβ 420 + 3k k vβ 141 )2 ; if k ≡ 3 (mod 10), then uβ 291 + vβ 105 = (uβ 420 + vβ 141 )2 ; if k ≡ 9k 7 (mod 10), then uβ 291 + vβ 105 = (uβ 420 + vβ 141 )2 ; and if k ≡ 9 (mod 10), 7k then uβ 291 + vβ 105 = (uβ 420 + vβ 141 )2 . This implies that for both of them, we have λ1 = u + v, λ2 = u2 + uv + v 2 , λ3 = u3 + v 3 , λ4 = λ5 = 0, λ6 = u3 v 3 , λ7 = u3 v 4 + u4 v 3 , λ8 = u8 + u4 v 4 + v 8 , λ9 = u9 + u8 v + u5 v 4 + u4 v 5 + uv 8 + v 9 , λ10 = u10 + u9 v + u8 v 2 + u7 v 3 + u6 v 4 + u5 v 5 + u4 v 6 + u3 v 7 + u2 v 8 + uv 9 + v 10 . 2) a = u(1 + β 3 + β 5 + β 6 + β 7 + β 9 ) + v(β + β 5 + β 6 + β 9 ) = uβ 582 + vβ 792 (item 2) in the theorem), or a = u(1 + β 3 + β 5 + β 7 + β 9 ) + v(β 2 + β 3 + β 8 + β 9 ) = uβ 282 + vβ 99 (item 3) in the theorem). In these two cases, note that if k ≡ 1 (mod 10), then uβ 282 + vβ 99 = (uβ 582 + 2k 4k vβ 792 )2 ; if k ≡ 3 (mod 10), then uβ 282 + vβ 99 = (uβ 582 + vβ 792 )2 ; if k ≡ 6k 7 (mod 10), then uβ 282 + vβ 99 = (uβ 582 + vβ 792 )2 ; and if k ≡ 9 (mod 10), then 8k uβ 282 +vβ 99 = (uβ 582 +vβ 792 )2 . This implies that for both of them, we have λ1 = u, λ2 = u2 + uv + v 2 , λ3 = u3 + u2 v + uv 2 , λ4 = λ5 = 0, λ6 = u3 v 3 + u2 v 4 + uv 5 + v 6 , λ7 = u4 v 3 + u3 v 4 + u2 v 5 + uv 6 , λ8 = u8 + u4 v 4 + v 8 , λ9 = u9 + u5 v 4 + uv 8 , λ10 = u10 + u9 v + u8 v 2 + u7 v 3 + u2 v 8 + uv 9 + v 10 . 3) a = u(β + β 8 + β 9 ) + v(β 3 + β 6 + β 7 + β 9 ). In this case, we have λ1 = v, λ2 = u2 + uv + v 2 , λ3 = u2 v + uv 2 + v 3 , λ4 = λ5 = 0, λ6 = u 6 + u 5 v + u4 v 2 + u 3 v 3 , λ7 = u 6 v + u 5 v 2 + u4 v 3 + u 3 v 4 , λ8 = u 8 + u 4 v 4 + v 8 , λ9 = u8 v + u4 v 5 + v 9 , λ10 = u10 + u9 v + u8 v 2 + u3 v 7 + u2 v 8 + uv 9 + v 10 . Then it can be verified that they satisfy the seven conditions. This completes the proof. 2 Remark 4. The number of a’s in Theorem 3 is Nk = 5(22k − 1).

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159

4. The fourth class of CPPs In [21], Tu, Zeng and Hu gave a class of CPP exponents over F23k with the form of d = 22k + 2k + 2, where gcd(k, 3) = 1. Using a similar method as in [21], we prove that d is also a CPP exponent over F23k , where gcd(k, 9) = 3. Throughout this section, let n = 3k. Let x3 + x + 1 ∈ F2 [x] be a primitive polynomial over F2 . Let θ be a root of x3 + x + 1 ∈ F2 [x]. We have θ7 = 1. From Lemma 3, f (x) = x3 + x + θ ∈ F23 [x] is irreducible over F2k , where gcd(k, 9) = 3. Note that the roots of f (x) are in F29 . Let ω ∈ F29 be a root of f (x). We have ω 511 = 1. Since F2n = F2k [ω], each x ∈ F2n can be represented uniquely as x 0 + x1 ω + x 2 ω 2 ,

(10)

where xi ∈ F2k . Since gcd(k, 9) = 3, then k ≡ 3, 6 (mod 9), and k

ω2 =



ω8 ,

if k ≡ 3 (mod 9),

ω , if k ≡ 6 (mod 9), 64

2k

and ω 2

=

ω 64 ,

if k ≡ 3 (mod 9),

ω8 ,

if k ≡ 6 (mod 9).

(11)

By (10) and (11), we have 2k

x2

+2k +2

 22k +2k +2 = x0 + x1 ω + x2 ω 2   2k 2k+1  k k+1  = x0 + x1 ω 2 + x2 ω 2 x 0 + x 1 ω 2 + x2 ω 2 x20 + x21 ω 2 + x22 ω 4     (12) = x0 + x1 ω 64 + x2 ω 128 x0 + x1 ω 8 + x2 ω 16 x20 + x21 ω 2 + x22 ω 4 .

3 Because Tr3k k (ω) is the sum of the roots of x + x + θ, we have

 2 3k Tr3k = 0. k (ω) = Trk ω

(13)

By (12) and (13), a straight computation gives  22k +2k +2      Tr3k = x40 + x2 x1 θ + x22 + x21 x20 + x32 θ2 + x1 x22 θ + x31 θ x0 . k x

(14)

Theorem 4. Let n = 3k and gcd(k, 9) = 3. Then d = 22k + 2k + 2 is a CPP exponent over F2n . Moreover, a−1 xd is a CPP over F2n , where a ∈ V , and  V = v1 ω + v2 ω 2 : v1 , v2 ∈ F2k \ {0}. Proof. Note that gcd(22k + 2k + 2, 22k + 2k + 1) = 1 and gcd(22k + 2k + 2, 2k − 1) = gcd(4, 2k − 1) = 1, then gcd(22k + 2k + 2, 23k − 1) = 1, which implies that xd is a PP over F2n . To complete the proof, we only need to prove that xd + ax is a PP over F2n for a ∈ V .

G. Wu et al. / Finite Fields and Their Applications 28 (2014) 148–165

160

From Lemma 1, we need to show that 

3k

(−1)Tr1

2k +2k +2

(γ(x2

+ax))

=0

x∈F23k

for any nonzero γ ∈ F23k and a ∈ V . Since gcd(d, 2n − 1) = 1, for any γ ∈ F2n , we can represent γ as γ = δ d for a unique nonzero element δ ∈ F2n . Then 

3k

(−1)Tr1

2k +2k +2

(γ(x2

+ax))



=

x∈F23k

2k +2k +2

3k

((δx)2

3k

(x2

(−1)Tr1

2k +2k +1

+δ 2

a(δx))

x∈F23k



=

(−1)Tr1

2k +2k +2

2k +2k +1

+δ 2

ax)

.

(15)

x∈F23k

Since (δ 2 +2 +1 )2 −1 = 1, then δ 2 +2 +1 ∈ F2k . 2k k 2k k For a ∈ V , we have δ 2 +2 +1 a ∈ V . Let δ 2 +2 +1 a = u = u1 ω + u2 ω 2 with ui ∈ F2k . i 3k 3k 2 3 Recall that ω 3 + ω + θ = 0. Since Tr3k k (ω ) = 0 for 0  i  2 and Trk (ω ) = Trk (ω + 3k 3k 2 2 θ) = θ, we have Trk (x · u) = Trk ((x0 + x1 ω + x2 ω ) · (u1 ω + u2 ω )) = (u1 x2 + u2 x1 )θ. Thus, by (14), we have 2k



k

k

3k

(−1)Tr1

x∈F23k

2k +2k +2

(x2



=

2k

k

k

+ux)

4

2

2 2

(−1)Tr1 (x0 (1+x1 x2 θ

8 4 8 4 12 4 +x41 +x42 +x12 2 θ +x1 x2 θ +x1 θ )+(u1 x2 +u2 x1 )θ)

x0 ,x1 ,x2 ∈F2k



=

k

(−1)Tr1 ((u1 x2 +u2 x1 )θ)

x1 ,x2 ∈F2k

×



k

4

2

2 2

(−1)Tr1 (x0 (1+x1 x2 θ

8 4 8 4 12 4 +x41 +x42 +x12 2 θ +x1 x2 θ +x1 θ ))

x0 ∈F2k

= 2k



k

(−1)Tr1 ((u1 x2 +u2 x1 )θ) ,

(16)

(x1 ,x2 )∈Ω 8 4 8 4 12 4 where Ω = {(x1 , x2 ): x1 , x2 ∈ F2k , 1 + x21 x22 θ2 + x41 + x42 + x12 2 θ + x1 x2 θ + x1 θ = 0}. Note that



3k

(−1)Tr1

x∈F23k

=



x0 ,x1 ,x2 ∈F2k

= 2k |Ω|.

2k +2k +2

(x2

k

)

4

2

2 2

(−1)Tr1 (x0 (1+x1 x2 θ

8 4 8 4 12 4 +x41 +x42 +x12 2 θ +x1 x2 θ +x1 θ ))

G. Wu et al. / Finite Fields and Their Applications 28 (2014) 148–165

161

 2k k 3k 22k +2k +2 ) By Lemma 1, and from the fact x2 +2 +2 permutes F23k , x∈F 3k (−1)Tr1 (x = 2 0, which implies the set Ω is empty. It then follows from (15) and (16) that 

3k

(−1)Tr1

2k +2k +2

(γ(x2

+ax))

=0

x∈F23k

for any nonzero γ ∈ F23k and a ∈ V . This completes the proof. 2 5. The inverses of some known CPP exponents In [17], Niederreiter and Robinson proved the following theorem. Theorem 5. (See [17].) If f (x) is a CPP over Fq , then so are the following polynomials: (1) f (x + a) + b, for all a, b ∈ Fq ; (2) af (a−1 x), for all a ∈ Fq ; (3) any polynomial representing the inverse mapping of c → f (c). Remark 5. From Theorem 5(2), if v −1 xd is a CPP over F2k , so is v −1 a1−d xd , for all a ∈ F∗2k . From Theorem 5(3), if d is a CPP exponent over Fq , so is d−1 (mod q − 1). We list some known CPP exponents and their inverses over F2n as follows. We refer the reader to [9] for some recent work on the computation of inverses of bijective monomials. • d = 2k + 2, n = 2k, k odd [2], d−1 = 43 · (2k−1 − 1)2 + 2k − 1. • d = 2k−2 (2k + 3), n = 2k, k odd [21], d−1 = (2k−1 + 1)(2k + 1) − 2. • d = 2k+1 + 3, n = 2k, k odd [21], d−1 = −[( 85 · (4k−1 − 1) − 2k ) · (2k−1 − 1) + 25 · (4k−1 − 1) − 1]. • d = 22k + 2k + 2, n = 3k, gcd(k, 3) = 1 [21], or gcd(k, 9) = 3 (this paper), d−1 = 2k−2 − (2k−2 · (2k + 2) + 1) · (2k − 1). • d = 23k + 22k + 2k + 2, n = 4k, gcd(k, 4) = 1, d−1 = −[( 25 · (4k+1 − 1) − 1) · (23k + 22k + 2k − 4) + 8(22k − 1)]. • d = 25k + 24k + 23k + 22k + 2k + 2, n = 6k, gcd(k, 6) = 1, d−1 =



(BC + 1)(2k − 1) − C,

if k ≡ 1 (mod 6),

2C + 1 − (2BC + B + 2)(2 − 1), k

if k ≡ 5 (mod 6),

where B = 24k + 23k+1 + 3 · 22k + 4 · 2k + 5, and

C=

2k −2 7 , 2k −4 7 ,

if k ≡ 1 (mod 6), if k ≡ 5 (mod 6).

G. Wu et al. / Finite Fields and Their Applications 28 (2014) 148–165

162

• d=

210k −1 , 2k −1

d−1

n = 10k, gcd(k, 10) = 1, ⎧ (BC + 1)(2k − 1) − C, ⎪ ⎪ ⎪ ⎨ −(8C + 5) + (8BC + 5B + 8)(2k − 1), = ⎪ 9C + 5 − (9BC + 5B + 9)(2k − 1), ⎪ ⎪ ⎩ −(9C + 4) + (9BC + 4B + 9)(2k − 1),

if k ≡ 1 (mod 10), if k ≡ 3 (mod 10), if k ≡ 7 (mod 10), if k ≡ 9 (mod 10),

where B = 28k + 27k+1 + 3 · 26k + 4 · 25k + 5 · 24k + 6 · 23k + 7 · 22k + 8 · 2k + 9, and

C=

⎧ 2k −2 , ⎪ ⎪ ⎪ k11 ⎪ ⎨ 2 −8 ,

if k ≡ 1 (mod 10),

⎪ ⎪ ⎪ ⎪ ⎩

if k ≡ 7 (mod 10),

11 2k −7 11 , k 2 −6 11 ,

if k ≡ 3 (mod 10), if k ≡ 9 (mod 10).

Remark 6. Let n = 9 and k = 3. By Theorem 4, 74 is a CPP exponent over F29 . By computer search, there are only two nonlinear CPP exponents over F29 , namely, 74 and 366. It can be verified that 74−1 = 366 (mod 29 − 1). So both of them can be explained. Let n = 2k, and define the unit circle of F2n as

k  U = λ ∈ F2n λ2 +1 = 1 .

(17)

In [22], the authors gave a class of CPP exponents over F2n of Niho type, and Zieve [28] generalized their results to Fpn , where p can be any prime. A positive integer d (always understood modulo pn − 1) is a Niho exponent if d ≡ pj (mod pn − 1) for some j < n [18]. Theorem 6. (See [22, Proposition 2].) Let n = 2k, d = (2k − 1) · r + 1. If the following conditions are satisfied, then d is a CPP exponent over F2n . 1) gcd(r, 2k + 1) = l > 1, 2) gcd(2r − 1, 2k + 1) = 1, and 3) gcd(r − 1, 2k + 1) = 1. Moreover, v −1 xd is a CPP over F2n , where v ∈ V , and

 V = U \ v v ∈ U, v = θl , for all θ ∈ U . Where U is defined by (17). We show in the following theorem that the inverse of a Niho type CPP exponent in Theorem 6 also satisfies the three conditions in the theorem.

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163

Theorem 7. Let n = 2k and d = r · (2k − 1) + 1, where r satisfy the three conditions listed in Theorem 6. Then, the inverse of d (mod 2n − 1) is d = r · (2k − 1) + 1, where r and r satisfy the following equation:   2r r − r − r = 0 mod 2k + 1 . Furthermore, r also meets all the three conditions listed in Theorem 6. Proof. Let N = 2n − 1. Since gcd(d, 2n − 1) = 1, the inverse of d exists. It is known that the inverse of a Niho exponent is again a Niho exponent [6]. Let d−1 = d = r ·(2k −1)+1. From dd = 1 (mod N ), we have         d · d (mod N ) = r · 2k − 1 + 1 · r · 2k − 1 + 1  2    = rr 2k − 1 + r + r 2k − 1 + 1 = 1 (mod N ).

(18)

From (18), we have (rr (2k − 1) + r + r ) = 0 (mod 2k + 1), which is equivalent to   2rr − r − r = 0 mod 2k + 1 .

(19)

Recall that for r we have: gcd(r, 2k + 1) = l > 1, gcd(2r − 1, 2k + 1) = 1, and gcd(r − 1, 2k + 1) = 1. First, from 2rr − r − r = 0 (mod 2k + 1), it follows that gcd(r , 2k + 1)| gcd(r, 2k + 1), and by symmetry in r and r we obtain gcd(r, 2k + 1)| gcd(r , 2k + 1) and therefore gcd(r , 2k + 1) = gcd(r, 2k + 1) = l. So r meets condition 1) in Theorem 6. Second, by (19), we have     r r − 1 + r (r − 1) = 0 mod 2k + 1 , which implies gcd(r −1, 2k +1)| gcd(r −1, 2k +1), and by symmetry in r and r we obtain gcd(r−1, 2k +1)| gcd(r −1, 2k +1) and therefore gcd(r −1, 2k +1) = gcd(r−1, 2k +1) = 1. This means that the r meets condition 3) in Theorem 6. Finally, by (19), we have     r 2r − 1 − r = 0 mod 2k + 1 , which implies gcd((2r − 1)r, 2k + 1)| gcd(r , 2k + 1). Then we have gcd(2r − 1, 2k + 1)| gcd(r , 2k + 1), together with gcd(2r − 1, r ) = 1, we get gcd(2r − 1, 2k + 1) = 1. This shows that r meets condition 2) in Theorem 6. We complete the proof. 2 6. Concluding remarks −1 In this paper, we give four classes of CPP exponents of the form d = 22k −1 +1 over F2rk . When r = 3 and gcd(k, 9) = 3, or when r ∈ {4, 6, 10} and gcd(k, r) = 1, rk

164

G. Wu et al. / Finite Fields and Their Applications 28 (2014) 148–165

the exponents d are proved to be CPP exponents over F2rk . Naturally, one may ask that when r take other values, whether the exponent d is a CPP exponent or not. By computer experiments, we find that for some other values of r and k, the exponents d can be CPP exponents over F2rk . However, these cases cannot be treated by the method presented in this paper. Thus, we need to explore new methods to deal with the following open question: −1 Open Question. Under what conditions for r and k is the exponent d = 22k −1 + 1 a CPP −1 d exponent over F2rk ? Give the parameter v such that v x is a CPP over F2rk . rk

Acknowledgments The authors would like to thank the anonymous referees for their valuable comments. The work of G. Wu was sponsored in part by the China Scholarship Council. The work of G. Wu and Y. Zhang was supported in part by the National Science Foundation of China under Grant No. 61272481. The work of T. Helleseth was supported in part by the Norwegian Research Council. References [1] A. Akbary, D. Ghioca, Q. Wang, On constructing permutations of finite fields, Finite Fields Appl. 17 (1) (2011) 51–67. [2] P. Charpin, G.M. Kyureghyan, Cubic monomial bent functions: a subclass of M, SIAM J. Discrete Math. 22 (2) (2008) 650–665. [3] R. Coulter, M. Henderson, R. Matthews, A note on constructing permutation polynomials, Finite Fields Appl. 15 (5) (2009) 553–557. [4] C. Ding, Q. Xiang, J. Yuan, P. Yuan, Explicit classes of permutation polynomials of F33m , Sci. China Ser. A 53 (4) (2009) 630–647. [5] H. Dobbertin, One-to-one highly nonlinear power functions on GF (2n ), Appl. Algebra Eng. Commun. Comput. 9 (2) (1998) 139–152. [6] H. Dobbertin, P. Felke, T. Helleseth, P. Rosendahl, Niho type cross-correlation functions via Dickson polynomials and Kloosterman sums, IEEE Trans. Inf. Theory 52 (2) (2006) 613–627. [7] X. Hou, A new approach to permutation polynomials over finite fields, Finite Fields Appl. 18 (3) (2012) 492–521. [8] X. Hou, T. Ly, Necessary conditions for reversed Dickson polynomials to be permutational, Finite Fields Appl. 16 (6) (2010) 436–448. [9] G. Kyureghyan, V. Suder, On inversion in Z2n −1 , Finite Fields Appl. 25 (2014) 234–254. [10] Y. Laigle-Chapuy, Permutation polynomials and applications to coding theory, Finite Fields Appl. 13 (1) (2007) 58–70. [11] N. Li, T. Helleseth, X. Tang, Further results on a class of permutation polynomials over finite fields, Finite Fields Appl. 22 (2013) 16–23. [12] R. Lidl, G.L. Mullen, When does a polynomial over a finite field permute the elements of the field?, Am. Math. Mon. 95 (3) (1988) 243–246. [13] R. Lidl, G.L. Mullen, When does a polynomial over a finite field permute the elements of the field? II, Am. Math. Mon. 100 (1) (1993) 71–74. [14] R. Lidl, H. Niederreiter, Finite Fields, Encycl. Math. Appl., Cambridge University Press, 1997. [15] J.E. Marcos, Specific permutation polynomials over finite fields, Finite Fields Appl. 17 (2) (2011) 105–112. [16] G.L. Mullen, H. Niederreiter, Dickson polynomials over finite fields and complete mappings, Can. Math. Bull. 30 (1) (1987) 19–27.

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