Finite Fields and Their Applications 31 (2015) 162–177

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Finite Fields and Their Applications

www.elsevier.com/locate/ffa

Constructing permutations and complete

permutations over finite fields via subfield-valued

polynomials

Zhengbang Zha a,b,∗, Lei Hu b,d, Xiwang Cao c

a School of Mathematical Sciences, Luoyang Normal University, Luoyang 471022, Chinab State Key Laboratory of Information Security, Institute of Information Engineering, Chinese Academy of Sciences, Beijing 100093, Chinac School of Mathematical Sciences, Nanjing University of Aeronautics and Astronautics, Nanjing 210016, Chinad Beijing Center for Mathematics and Information Interdisciplinary Sciences, Beijing 100048, China

a r t i c l e i n f o a b s t r a c t

Article history:Received 5 November 2013Received in revised form 12 June 2014Accepted 1 October 2014Available online xxxxCommunicated by Xiang-dong Hou

MSC:05A0511T06

Keywords:Finite fieldTrace mappingPermutation polynomial

We describe a recursive construction of permutation and com-plete permutation polynomials over a finite field Fpn by using Fpk -valued polynomials for several same or different factors kof n. As a result, we obtain some specific permutation poly-nomials which unify and generalize several previous construc-tions.

© 2014 Elsevier Inc. All rights reserved.

* Corresponding author.E-mail addresses: zhazhengbang@163.com (Z. Zha), hu@is.ac.cn (L. Hu), xwcao@nuaa.edu.cn

(X. Cao).

http://dx.doi.org/10.1016/j.ffa.2014.10.0021071-5797/© 2014 Elsevier Inc. All rights reserved.

Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177 163

1. Introduction

Let p be a prime, n be a positive integer, and Fpn be the finite field of order pn. A poly-nomial f(x) in Fpn [x] is called a permutation polynomial (PP) over Fpn if it induces a one-to-one map from Fpn onto itself. Furthermore, it is called a complete permutation polynomial (CPP) if f(x) +x is also a permutation polynomial over Fpn [16]. A linearized polynomial or p-polynomial over Fpn [13] has the form L(X) =

∑n−1i=0 aiX

pi ∈ Fpn [X]. We note that L(X) permutes Fpn if and only if the zero element is the only root of L(X)in Fpn . PPs and CPPs have been studied extensively and have important applications in coding theory, cryptography, combinatorics, design theory and so on [4,5,8,12,13,15]. In recent years, there has been significant progress in finding new PPs and CPPs, see [1–3,6,7,9,10,17,18,20,21,26] for example.

In [22,23], J. Yuan et al. first introduced PPs of the form (xp − x + δ)s + L(x) with linearized polynomials L(x) and several numerical results. An extension of the above works and two new classes of PPs defined over fields of characteristic 2 were found in [27]. Zha and Hu [28] explained two of the numerical results and showed the permutation properties of two classes of polynomials by using their implicit or explicit piecewise function characteristic. Later, Fernando and Hou [6] presented a piecewise construction of PPs which unify and generalize several known PPs. Wang [19] used cyclotomy to construct new PPs of large indices. His work generates PPs in an algorithmic way and also unifies several previous constructions.

In the other direction, by using functions with additive structure, Charpin and Kyureghyan [3] introduced an effective method to construct permutation polynomials of the shape G(x) + γ Tr(H(x)), where G(x), H(x) ∈ Fpn [x]. In recent papers [11,14,29], the above method is generalized to construct more PPs by using the additive structure of finite fields. Akbary, Ghioca and Wang [1] proposed a recipe for constructing PPs by the following lemma:

Lemma 1.1. (See [1, Lemma 1.2], AGW Lemma.) Let A, S and S be finite sets with �S = �S, and let f : A → A, h : S → S, λ : A → S and λ : A → S be maps such that λ◦f = h ◦λ. If both λ and λ are surjective, then the following statements are equivalent:

(1) f is a bijection (a permutation over A); and(2) h is a bijection from S to S and f is injective on λ−1(s) for each s ∈ S.

P. Yuan and C. Ding [24] gave a unified treatment of some earlier constructions of PPs and got many new specific PPs by using the AGW Lemma. A consequent work [25]further showed a new class of polynomials f(x) = u(x) +v(x) which permutes Fpn if and only if u(x) permutes Fpn .

Lemma 1.2. (See [20, Theorem 3.2].) Assume that A is a finite field and S, S are finite subsets of A with �S = �S such that the maps ψ : A → S and ψ : A → S are surjective and ψ is additive, i.e.,

164 Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177

ψ(x + y) = ψ(x) + ψ(y), x, y ∈ A.

Let u : A → A and v : A → A be maps such that the following diagram commutes:

Au+v

ψ

A

ψ

Sh

S.

Assume also that ψ(v(x)) = 0 for every x ∈ A and v(x) is a constant on each ψ−1(s) for all s ∈ S. Then the map f(x) = u(x) + v(x) permutes A if and only if u permutes A.

Applying Lemma 1.2, P. Yuan and C. Ding presented new PPs of explicit forms, which generalize several old classes of PPs of the form (xpk − x + δ)s + L(x). Among the new PPs Yuan et al. presented, some involve in their expressions a single Fpk-valued polynomial g(x) ∈ Fpn [x], where Fpk is a subfield of Fpn and an Fpk -valued polynomial is a one such that g(x) ∈ Fpk for any x ∈ Fpn .

Inspired by the work of [1] and [25], in this paper we present some specific PPs and CPPs f by using one or several different Fpk-valued polynomials. We prove their per-mutation properties by using recursive methods. The idea is eliminating the Fpk-valued polynomials appearing in the expression of f and getting the permutation properties of f .

The paper is organized as follows. In Section 2, we introduce PPs which involve Fpk -valued polynomials for different values of k in their expressions. In Section 3, we present permutation polynomials involving Fpk -valued polynomials for a single value of k. Our results generalize several previous discovered families of PPs.

Throughout this paper, we consider permutation and complete permutation polyno-mials over Fpn , and we always let k, n1, n2, · · · , nh be divisors of n and n1 |n2 | · · · |nh |n. Let l = n/k and q = pk. Then pn = ql.

2. PPs constructed from FFFpk-valued polynomials for different values of k

The trace function from Fpn onto its subfield Fpk is defined as

Trnk (x) = x + xpk

+ xp2k+ · · · + xpn−k

.

The absolute trace function (i.e., for the case of k = 1) is simply denoted by Tr (for fixed n) and is then defined by Tr(x) = x + xp + xp2 + · · · + xpn−1 . Obviously, Trnk (x)is an Fpk -valued polynomial over Fpn . An Fpk -valued polynomial g : Fpn → Fpk can be expressed as

g(x) =∑

Trklik

(Aix

i)

+ Aql−1xql−1, Ai ∈ Fqli , Aql−1 ∈ Fq,

i∈Γq(l)

Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177 165

where Γq(l) is the set consisting of all coset leaders of integers modulo ql − 1 and li is the size of the coset Ci = {i, iq, · · · , iql−1 (mod ql − 1)} and kli divides n. In [25], five specific examples of Fpk -valued polynomials are listed as follows:

(1) g(x) = Trnk (x);(2) g(x) = h(x)s(pn−1)/(pk−1), h(x) ∈ Fpn [x] and (pk − 1) � s(pi − 1) for any 1 ≤ i < k;(3) g(x) =

∑t−1i=0 h(x)Mpik , where h(x) ∈ Fpn [x], t is an integer with 0 < t ≤ n

k and Mptk ≡ M (mod pn − 1);

(4) If g1(x) and g2(x) are Fpk -valued polynomials, then both g1(x)g2(x) and g1(x) +g2(x)are Fpk -valued polynomials;

(5) If gi(x), i = 1, 2, · · · , r, are Fpk -valued polynomials over Fpn [x] and

g(x1, . . . , xr) ∈ Fpk [x1, . . . , xr],

then g(g1(x), . . . , gr(x)) is an Fpk -valued polynomial over Fpn .

Below we present PPs constructed from Fpk-valued polynomials for different values of k. Some of them are shown to be CPPs.

Theorem 2.1. Let n1 |n2 | · · · |nh |n, gi(x) ∈ Fpn [x] be an Fpni -valued polynomial for i = 1, · · · , h, and let L(x) ∈ Fpn1 [x] be a linearized polynomial. Let θi ∈ F∗

pn such that θp

ni−1i ∈ Fpn1 for i = 1, · · · , h and θi+1

θi∈ Fpni+1 for 1 ≤ i ≤ h − 1. Then

f(x) =h∑

i=1θigi

(xpni − θp

ni−1i x

)+ L(x)

is a PP (resp. CPP) over Fpn if and only if L(x) is a PP (resp. CPP) over Fpn .

Proof. We make induction on h.

Case h = 1. Let

u(x) = L(x), v(x) = θ1g1(xpn1 − θp

n1−11 x

),

ψ(x) = ψ(x) = xpn1 − θpn1−1

1 x, h(x) = L(x),

and S = S = {xpn1 − θpn1−1

1 x | x ∈ Fpn}. Since g1(x) is an Fpn1 -valued polynomial, we can easily check that ψ(v(x)) = 0 for every x ∈ Fpn and v(x) is a constant on each ψ−1(s)for all s ∈ S. By a direct computation, we know that the assumption of θp

n1−11 ∈ Fpn1

assures that

ψ(u(x) + v(x)

)= ψ

(u(x)

)= L(x)p

n1 − θpn1−1

1 L(x) = L(xpn1 − θp

n1−11 x

)= h

(ψ(x)

)

166 Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177

holds for any x ∈ Fpn , namely, the corresponding diagram of Lemma 1.2 commutes. Thus, by Lemma 1.2, we have that f(x) = θ1g1(xpn1 − θp

n1−11 x) +L(x) is a PP over Fpn

if and only if L(x) is a PP over Fpn .

Case h > 1. We deduce θhθi ∈ Fpnh and θpnh−1

i = θpnh−1

h from θi+1θi

∈ Fpni+1 . Let

u(x) =h−1∑i=1

θigi(xpni − θp

ni−1i x

)+ L(x), v(x) = θhgh

(xpnh − θp

nh−1h x

),

ψ(x) = ψ(x) = xpnh − θpnh−1

h x, h(x) = L(x),

and S = S = {xpnh − θpnh−1

h x | x ∈ Fpn}. Similarly as in the case of h = 1, it is easy to check that ψ(v(x)) = 0 for every x ∈ Fpn and v(x) is a constant on each ψ−1(s) for all s ∈ S. From θp

nh−1i = θp

nh−1h and θp

nh−11 ∈ Fpn1 we have that

ψ(u(x) + v(x)

)= ψ

(u(x)

)

=h−1∑i=1

(θp

nh

i − θpnh−1

h θi)gi(xpni − θp

ni−1i x

)+ L(x)p

nh − θpnh−1

h L(x)

= L(xpnh − θp

nh−1h x

)= h

(ψ(x)

)

holds for any x ∈ Fpn and the corresponding diagram of Lemma 1.2 commutes. By Lemma 1.2 again, we get that f(x) is a PP over Fpn if and only if u(x) is a PP over Fpn .

Now by induction we derive that f(x) is a PP over Fpn if and only if L is a PP over Fpn . Similarly, we can show that f(x) + x is a PP over Fpn if and only if L(x) + x

is a PP over Fpn . �Theorem 2.1 is a generalization of Theorems 3.4 and 3.6 in [25]. As an application of

Theorem 2.1, we have the following corollary.

Corollary 2.1. Let l = n/k be even and s be an integer with s(p2k − 1) ≡ 0 (mod pn− 1). Let θ, β, γ, δ ∈ Fpn with θp

2k = θ, δpk = −δ and γpk = −γ. Let L(x) ∈ Fpk [x] be a linearized polynomial. Then

f(x) = θ(xp2k − x + β

)s + Trnk(δx1+pk)

+ Tr(γx) + L(x)

is a PP (resp. CPP) over Fpn if and only if L(x) is a PP (resp. CPP) over Fpn .

Proof. We first show that Trnk (δx1+pk) +Tr(γx) can be expressed as a function of xpk−x.Let β1 = xpk − x. Since δp

k = −δ, then we get that xpik = x + β1 + · · · + βp(i−1)k

1 for 1 ≤ i < l and

Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177 167

Trnk(δx1+pk)

= δ(x1+pk − xpk+p2k

+ · · · − xpn−k+1)

= δ(x(x + β1) − (x + β1)

(x + β1 + βpk

1)

+ · · · −(x + β1 + · · · + βp(l−2)k

1)x)

= δg(β1),

where g(β1) is a function of β1. That is to say, Trnk (δx1+pk) = δg(xpk − x). Similarly, we have

Tr(γx) = Trk1(Trnk (γx)

)= Trk1

(γ(−β1 − βp2k

1 − · · · − βp(l−2)k

1))

and it is also a function of xpk − x.Since s(p2k − 1) ≡ 0 (mod pn − 1) and θp

2k = θ, we get that θ(xp2k − x + β)s is an Fp2k -valued polynomial on xp2k − x. Obviously, Trnk (δx1+pk) + Tr(γx) is Fpk -valued. By Theorem 2.1, we get the desired conclusion. �

Similarly as for Theorem 2.1, we can prove the following theorem and we omit its proof.

Theorem 2.2. Let δi ∈ F∗pn with Trnnh

(δi) = 0. Let L(x) ∈ Fpn1 [x] be a linearized polyno-mial and gi(x) ∈ Fpn [x] be an Fpni -valued polynomial for i = 1, · · · , h. Then

f(x) =h∑

i=1δigi

(Trnni

(x))

+ L(x)

is a PP (resp. CPP) over Fpn if and only if L(x) is a PP (resp. CPP) over Fpn .

The following theorem is a slight generalization of Theorems 3.9 and 3.10 in [25]. We repeat some of their proofs for completeness.

Theorem 2.3. Let 4k |n, n1 = k, n2 = 2k, n2 |n3 | · · · |nh |n, t = n4k , and 0 ≤ s < n. Let

g(x) =∑t

i=1 x(q2(i−1)+q2(i−1)+2t)ps , and let gi(x) ∈ Fpn [x] be an Fpni -valued polynomial

for any i = 1, · · · , h. Assume that δ ∈ Fpn and θ ∈ F∗q . Then

f(x) = g(xq − x + δ

)+

h∑i=1

gi(xpni − x

)+ θx

is a PP over Fpn if and only if

(i) s = 0 and θ �= Trnn1(δ); or

(ii) s = n1 and θ �= − Trnn1(δ); or

(iii) s �= 0, n1 and either θ−1 Trnn1(δ)ps is not a (ps − 1)-th power or −θ−1 Trnn1

(δ)ps is not a (ps−n1 − 1)-th power.

168 Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177

Proof. It is easy to check that g(x)q2 = g(x), i.e., g(x) is an Fp2n1 -valued polynomial over Fpn . By using the induction on h and the proof for the case of h > 1 in the proof of Theorem 2.1 in the same way, we can prove that f(x) is a PP over Fpn if and only if

f1(x) := g(xq − x + δ

)+ g1

(xq − x

)+ θx

is a PP over Fpn .To prove the permutation property of f1(x), we form the following commutative dia-

gram

Fpn

f1

xq−x+δ

Fpn

xq−x

Sh

S,

where S = {bq − b + δ | b ∈ Fpn}, S = {bq − b | b ∈ Fpn} and

h(x) = g(x)q − g(x) + θx− θδ.

Here h is indeed a map from S to S. By Lemma 1.1, f1 permutes Fpn if and only if his a bijection from S to S, or equivalently, if and only if h is an injection from S to Ssince S and S have the same cardinality.

Since θ ∈ F∗q we can write an element of S as θb, where b ∈ S. We need to prove that

equation

g(x)q − g(x) + θx− θδ = θb (1)

has at most one solution over S. Let c = b + δ. Then Trnn1(c) = Trnn1

(δ). By adding (1)and its q-th power, we have

xq + x = c + cq, (2)

which leads to xqj = (−1)j(x − c) + cqj for 0 ≤ j < 4t − 1. Substituting it into (1), we

obtain

θ(x− c) = g(x) − g(x)q

=t∑

i=1

(x− c + cq

2i−2)ps(x− c + cq

2t+2i−2)ps

−t∑

i=1

(x− c− cq

2i−1)ps(x− c− cq

2t+2i−1)ps

= Trnn1(c)p

s

(x− c)ps −

t∑(c(q

2i−1+q2t+2i−1)ps − c(q2i−2+q2t+2i−2)ps)

.

i=1

Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177 169

Thus, Eq. (1) is equivalent to the system of equations (x − c)q + (x − c) = 0 and θ(x − c) − Trnn1

(δ)ps(x − c)ps = 0. By some trivial computation, the latter has at most one solution if and only if one of the three conditions in Theorem 2.3 holds. �Remark 2.1. For any s, the function f in Theorem 2.3 is indeed a PP when Trnn1

(δ) = 0. Moreover, if q �= 2 and θ �= −1, then we can easily check that the function f in Theo-rem 2.3 is a CPP over Fpn .

Lemma 2.1. Let x ∈ Fpn and t ∈ F∗pk with xp−1 = t. If gcd(l, p(p − 1)) = 1, then

Trnk (x) �= 0.

Proof. Since xp−1 = t, then we get x �= 0 and xpk = tpk−1p−1 x. Let β := t

pk−1p−1 . From

t ∈ F∗pk , we deduce that βp−1 = 1, β ∈ F∗

pk , and

Trnk (x) = x + xpk

+ · · · + xpn−k

= x + βx + · · · + βl−1x.

If β = 1, then we have Trnk (x) = lx �= 0 for p � l. If β �= 1, then we obtain Trnk (x) = βl−1β−1 x.

Assume on the contrary that Trnk (x) = 0. Then βl = 1 and β = βgcd(l,p−1) = 1 since βp−1 = 1, which is a contradiction. �

Utilizing Lemma 2.1, we can prove the following theorem.

Theorem 2.4. Let t be an integer with gcd(t + 1, pn1 − 1) = 1. Let p � nn1

and gcd( n

n1, p − 1) = 1. Let δ0 ∈ F∗

pn1 and g(x) ∈ Fpn [x] be an Fpnh -valued polynomial. Let γ ∈ Fpnh with Trnnh

(γ) = 0. Let δi ∈ F∗pni , 1 ≤ i ≤ h, such that

∑ij=1 δj �= 0 for all i.

Then

f(x) =h∑

i=1δi(xp − xTrnni

(x)p−1) + δ0xTrnn1(x)t + γg

(Trnnh

(x))

is a PP over Fpn .

Proof. To show f(x) is a PP, we assume on the contrary that for some x �= y ∈ Fpn , f(x) = f(y), namely,

h∑i=1

δi(xp − xTrnni

(x)p−1) + δ0xTrnn1(x)t + γg

(Trnnh

(x))

=h∑

i=1δi(yp − yTrnni

(y)p−1) + δ0yTrnn1(y)t + γg

(Trnnh

(y)). (3)

Case h = 1. Applying Trnn1to the both sides of Eq. (3) and using the fact that

g(Trnn (x)) ∈ Fpn1 and Trnn (γ) = 0, we have

1 1

170 Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177

δ0 Trnn1(x)t+1 = δ0 Trnn1

(y)t+1.

Since δ0 ∈ F∗pn1 and gcd(t + 1, pn1 − 1) = 1, then we have Trnn1

(x) = Trnn1(y). From (3),

we have

δ1xp − δ1 Trnn1

(x)p−1x + δ0 Trnn1(x)tx = δ1y

p − δ1 Trnn1(y)p−1y + δ0 Trnn1

(y)ty.

Let θ := δ0 Trnn1(x)t − δ1 Trnn1

(x)p−1 ∈ Fpn1 . Then we have

δ1(x− y)p = −θ(x− y),

which leads to (x − y)p−1 = − θδ1

∈ Fpn1 . By Lemma 2.1, we have that Trnn1(x − y) �= 0,

i.e., Trnn1(x) �= Trnn1

(y), which is a contradiction.

Case h > 1. Applying Trnnhto the both sides of Eq. (3), using Trnnh

(γ) = 0 and Trnni

(x) = Trnhni

(Trnnh(x)), and writing x1 = Trnnh

(x) and y1 = Trnnh(y), we obtain

h−1∑i=1

δi(xp

1 − x1 Trnhni

(x1)p−1) + δ0x1 Trnhn1

(x1)t

=h−1∑i=1

δi(yp1 − y1 Trnh

ni(y1)p−1) + δ0y1 Trnh

n1(y1)t, (4)

which holds as an equation over Fpnh . By induction on h, we have Trnnh(x) = x1 = y1 =

Trnnh(y) and hence, Trnni

(x) = Trnni(y) for all i. Therefore, f(x) = f(y) implies δ(x −y)p =

−θ(x − y), where δ =∑h

i=1 δi and θ = δ0 Trnn1(x)t −

∑hi=1 δi Trnni

(x)p−1. Similarly, we have (x − y)p−1 = − θ

δ ∈ Fpnh . By Lemma 2.1 again, we have that Trnnh(x − y) �= 0, i.e.,

Trnnh(x) �= Trnnh

(y), which is a contradiction. �Remark 2.2. Theorem 2.4 can also be proved by applying the AGW Lemma recursively. We show iteratively the permutation property in the above proof. Actually, we can get more PPs by the same method. For example, replacing the condition

∑ij=1 δj �= 0 for

all i’s in Theorem 2.4 by δ0 +∑i

j=1 δj �= 0 for all i’s, we get that

f1(x) =h∑

i=1δix

p −h∑

i=1δixTrnni

(x)p−1 + δ0xp + γg

(Trnnh

(x))

is also a PP over Fpn .

Remark 2.3. In Theorem 2.4, let p = 2, γ = 0 and t = 2l. Then

f(x) =h∑

δix2 −

h∑δixTrnni

(x) + δ0xTrnn1(x)2

l

i=1 i=1

Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177 171

is exactly the PP in [5, Theorem 3.3]. Furthermore, if t = 0, pn1 �= 2 and δ0 �= −1, the function f in Theorem 2.4 is a CPP over Fpn .

3. PPs constructed from FFFpk-valued polynomials for a same value of k

In this section, we present some sporadic PPs constructed by using Fpk -valued poly-nomials for a same value of k. Some CPPs are constructed simultaneously.

Theorem 3.1. Let g(x) ∈ Fpn [x] be an Fpk-valued polynomial, θ ∈ F∗pk , and β ∈ F∗

pn . Then the polynomial

f(x) = θx + βg(xpk − βpk−1x

)Trnk (x)

permutes Fpn if and only if θ + g(bpk − βpk−1b) Trnk (β) �= 0 for any b ∈ Fpn .

Proof. We can easily get the following commutative diagram:

Fpn

f

xpk−βpk−1x

Fpn

xpk−βpk−1x

Sθx

S,

where S = S = {xpk − βpk−1x : x ∈ Fpn}. By the AGW Lemma, we have that f(x) =θx + βg(xpk − βpk−1x) Trnk (x) permutes Fpn if and only if for any b ∈ Fpn , f1(x) :=θx + βg(bpk − βpk−1b) Trnk (x) is injective on

S1 :={x∣∣ xpk − βpk−1x = bp

k − βpk−1b}.

Let S′1 = {f1(x) | x ∈ S1}. It can be verified that the following diagram commutes

S1θx+βg(bp

k−βpk−1b) Trnk (x)

Trnk (x)

S′1

Trnk (x)

S(θ+g(bp

k−βpk−1b) Trnk (β))xS,

where S = {Trnk (x) : x ∈ S1} and S = {(θ + g(bpk − βpk−1b) Trnk (β)) Trnk (x) : x ∈ S1}. The desired conclusion of this theorem follows from the AGW Lemma. �

Similarly, we have the following theorem and omit its proof.

172 Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177

Theorem 3.2. Let g(x1, x2) ∈ Fpn [x1, x2] be a bivariate Fpk-valued polynomial, θ ∈ F∗pk ,

and β ∈ F∗pn with Trnk (β) = 0. Then the polynomial

f(x) = θx + βg(xpk − βpk−1x,Trnk (x)

)

is a PP over Fpn .

We note that the functions f in Theorems 3.1 and 3.2 are also CPPs if in addition pk �= 2, θ �= −1 and Trnk (β) = 0.

Theorem 3.3. Let n/k ≡ 0 (mod p), θ, β ∈ F∗pk and g(x) ∈ Fpn [x] be an Fpk-valued

polynomial. Then the polynomial

f(x) =(g(xpk − x

)+ x

)(β + θTrnk (x)p

k−1)

is a PP over Fpn if and only if θ + β �= 0.

Proof. Since n/k ≡ 0 (mod p), θ, β ∈ F∗pk and g(x) ∈ Fpn [x] is an Fpk -valued polynomial,

we can easily check that

Trnk(g(xpk − x

)(β + θTrnk (x)p

k−1)) = 0.

Then we have the following commutative diagram

Fpn

f

Trnk (x)

Fpn

Trnk (x)

Fpk

(θ+β)xFpk .

Note that (θ+β)x is a bijection of Fpk if and only if θ+β �= 0. From Theorem 2.1, we get that g(xpk − x) + x is a PP over Fpn , which implies that f(x) is injective on (Trnk )−1(s)for each s ∈ Fpk while β �= 0 and θ + β �= 0. Then we can obtain the desired conclusion by the AGW Lemma. �

Similarly to the proof of Theorem 3.3, we can prove the following theorem.

Theorem 3.4. Let n/k ≡ 0 (mod p), θ, β ∈ F∗pk and g(x) ∈ Fpn [x] be an Fpk-valued

polynomial. Then the polynomial

f(x) = g(x)(1 − Trnk (x)p

k−1) + x(β + θTrnk (x)p

k−1)

is a PP over Fpn if and only if θ + β �= 0.

Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177 173

We remark that the functions f in Theorems 3.3 and 3.4 are also CPPs if in addition pk �= 2, β �= −1 and θ + β �= −1.

Theorem 3.5. Let p be odd, g1(x), g2(x) ∈ Fpn [x] be Fq-valued polynomials, θ, β ∈ Fpn , θq = θ, and βq = −β. If both 2θg1(x) + x and 2βg2(x) − x are permutation polynomials over Fpn , then the polynomial

f(x) = θg1(xq + x

)+ βg2

(xq − x

)+ x

is a PP over Fpn .

Proof. We consider the solutions of the equation

θg1(xq + x

)+ βg2

(xq − x

)+ x = b (5)

for b ∈ Fpn . Raising the both sides of Eq. (5) to q-th powers, we get

θg1(xq + x

)− βg2

(xq − x

)+ xq = bq. (6)

Then we have 2θg1(xq +x) +(xq +x) = bq +b and 2βg2(xq−x) +x −xq = b −bq from (5)and (6). Since 2θg1(x) + x and 2βg2(x) − x are both permutations over Fpn , the values of xq − x and xq + x can be determined uniquely, which leads to exactly one solution of (5). �Remark 3.1. If g1(x) = g̃1(xq − x), where g̃1(x) is an Fq-valued polynomial, then 2θg1(x) + x permutes Fpn . Similarly, if we choose g2(x) = g̃2(xq + x) with g̃2(x) anFq-valued polynomial, we can also get that 2βg2(x) − x is a permutation over Fpn . Therefore, the conditions of Theorem 3.5 are easily satisfied.

In the following, we present two specific PPs by using the trace mapping Trnk (x), which is a familiar Fpk -valued polynomial.

Theorem 3.6. Let q = pk, p � l, gcd(l, q − 1) = 1, θ ∈ F∗q and δ ∈ F∗

pn with Trnk (δ) = 0. Assume g2(x) ∈ Fq[x] and g1(x) ∈ Fpn [x] with g1(0) = 0. If g1(x) is a permutation over Fq, then the polynomial

f(x) = θ(xq − x

)+ δg2

(Trnk (x)

)+ x

(Trnk (x)

)pn−2g1(Trnk (x)

)

is a PP over Fpn .

Proof. If g1(x) is a permutation over Fq, then we have that g1(Trnk (x)) ∈ Fq for each x ∈ Fpn , which implies the following commutative diagram:

174 Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177

Fpn

f

Trnk (x)

Fpn

Trnk (x)

Fqxpn−1g1(x)

Fq.

Notice that xpn−1 = 1 for any x ∈ F∗pn . Therefore, xpn−1g1(x) is a bijection of Fq if and

only if g1(x) is a permutation over Fq.Suppose x, y ∈ Fpn with Trnk (x) = Trnk (y)(:= t). If f(x) = f(y), then we obtain

θ(xq − x

)+ xtp

n−2g1(t) = θ(yq − y

)+ ytp

n−2g1(t). (7)

Let β = 1 − g1(t)tpn−2θ−1. We obviously have β ∈ Fq and (x − y)q = β(x − y), which

implies that

Trnk (x− y) = (x− y)(1 + β + · · · + βl−1) = 0. (8)

If β = 1, from (8) we get x = y since p � l. If β �= 1, then we get 1 − βl �= 0 since gcd(l, q − 1) = 1. It follows that

1 + β + · · · + βl−1 = 1 − βl

1 − β�= 0

and x = y from (8). Thus f(x) is injective on (Trnk )−1(s) for each s ∈ Fq. The desired conclusion of this theorem follows from the AGW Lemma. �

If g1(x) is a CPP over Fq, we can easily check that the function f in Theorem 3.6 is a CPP over Fpn . Furthermore, if we add a condition g1(Trnk (x)) ∈ Fq for any x ∈ Fpn , Theorem 3.6 can be improved to that: f is a PP (resp. CPP) of Fpn if and only if g1(x)is a PP (resp. CPP) of Fq.

Lemma 3.1. Let x ∈ Fpn and t ∈ F∗pk with xp = tp−1x. Let p � l. Then Trnk (x) = 0 if and

only if x = 0.

Proof. Obviously, we have Trnk (x) = 0 if x = 0. Below we prove the necessary condition. Since xp = tp−1x, we have x = 0 or xp−1 = tp−1. The latter implies xpk−1 = tp

k−1 = 1for t ∈ F∗

pk . Then we have xpk = x and Trnk (x) = lx. Therefore, we have x = 0 from Trnk (x) = 0 and p � l. �Theorem 3.7. Let q = pk and gcd(l, p(p − 1)) = 1. Let g1(x), g2(x) ∈ Fpn [x] satisfying g1(Trnk (x)), g2(Trnk (x)) ∈ Fq for any x ∈ Fpn . Let a, θ ∈ F∗

q and δ ∈ F∗pn with Trnk (δ) = 0.

Then the polynomial

Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177 175

f(x) = g1(Trnk (x)

)+ δg2

(Trnk (x)

)+ ax

(Trnk (x)

)p−1 − axp + θx

is a PP over Fpn if and only if lg1(x) + θx is a permutation over Fq.

Proof. It is easy to get the following commutative diagram

Fpn

f

Trnk (x)

Fpn

Trnk (x)

Fqlg1(x)+θx

Fq.

We need to prove that f(x) is injective on (Trnk )−1(s) for each s ∈ Fq. Assume x, y ∈ Fpn

with f(x) = f(y) and Trnk (x) = Trnk (y) = s. Then we get

(asp−1 + θ

)x− axp =

(asp−1 + θ

)y − ayp. (9)

Since a, θ ∈ F∗q , we can deduce that sp−1 + θa−1 ∈ Fq and

(x− y)p −(sp−1 + θa−1)(x− y) = 0 (10)

from (9). If x �= y, by Lemma 2.1 and the known condition gcd(l, p(p −1)) = 1 we derive Trnk (x − y) �= 0 from (10). It contradicts the assumption Trnk (x) = Trnk (y). Hence we get x = y.

Utilizing the AGW Lemma, we can verify that f is a PP over Fpn if and only if lg1(x) + θx is a permutation over Fq. �Remark 3.2. We note that Theorem 3.7 also holds if we replace the conditions that gcd(l, p(p − 1)) = 1, a ∈ F∗

q by a = 0. Moreover, if we replace the conditions that gcd(l, p(p − 1)) = 1, θ ∈ F∗

q by p � l and θ = 0, Theorem 3.7 also holds from Lemma 3.1. The function f in Theorem 3.7 is a CPP over Fpn if and only if lg1(x) + θx is a CPP over Fq.

Remark 3.3. In Theorem 3.7, if p = 2, l = 3, a = 1, θ �= 1 and g1(x) = g2(x) = 0, then f(x) = x Tr3kk (x) + x2 + θx, which is exactly the PP in [15, Theorem 4].

Acknowledgments

The authors would like to thank the anonymous reviewers for their detailed com-ments and suggestions which improved both the quality and presentation of this paper. The work of this paper was supported by the National Basic Research Programme un-der Grant 2013CB834203, the National Natural Science Foundation of China (Grants 61472417, 11201214 and 11371011), the Strategic Priority Research Program of Chinese

176 Z. Zha et al. / Finite Fields and Their Applications 31 (2015) 162–177

Academy of Sciences under Grant XDA06010702, and the NUAA Fundamental Research Funds, No. 2013202.

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