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# Some classes of complete permutation polynomials in the form of (xpm − x + δ)s + axpm + bx over Fp2m

###### Accettato: 27 Nov 2021
Dettagli della rivista
Formato
Rivista
eISSN
2444-8656
Prima pubblicazione
01 Jan 2016
Frequenza di pubblicazione
2 volte all'anno
Lingue
Inglese

Several classes of complete permutation polynomials (CPPs) in the form of (xpmx + δ)s + axpm + bx over Fp2m are described in this paper. For fractional exponent s, the permutation properties of the given polynomials are proved by the AGW criterion and for determining the bijectivity of the corresponding piecewise functions. For exponent s satisfying s = i(pm − 1) + pj, the main proofs here depend on the AGW criterion and the quantity of solutions for certain linearised equations. Several results generalise some earlier results on permutation polynomials (PPs).

#### Keywords

Introduction

Let Fpn be the finite field with pn elements where p is a prime number and n is a positive integer. A polynomial f (x) ∈ Fpn [x] is called a permutation polynomial (PP) if the associated mapping f : cf (c) from Fpn to Fpn is a one-to-one mapping [9]. A polynomial f (x) over Fq is called a complete permutation polynomial (CPP) if f (x) and f (x) + x are two PPs of Fq. CPPs have been widely used in combinatorial designs, coding theory and cryptography. For example, CPPs can be applied to design S-boxes which are the only nonlinear component of many block ciphers [3, 4, 11]; to construct Bent functions which have the highest nonlinearity and important applications in sequences, cryptography and signal designing theory [27]; to design Hash functions which play a significant role in the field of information security [16]; and to define orthogonal Latin squares which are important building blocks in coding theory and symmetric cryptography [10, 15].

The existence and construction of CPPs are the preconditions of their applications. Therefore, the problems of finding and constructing CPPs are extremely important. Till date, various valuable research results on CPPs have been given, but the number of known CPPs is very limited. A detailed investigation of CPPs over the finite field pn was first conducted by Niederreiter and Robinson [13]. In the recent work [2, 14, 17, 19,20,21], several new classes of CPPs with fewer terms are constructed. Some of these classes are monomial CPPs and proved by means of the polar coordinate representation, which was first used by Tu et al. [17] to construct PPs and CPPs. Some of these classes can be characterised by the AGW criterion [1, 17]. Recently, using the AGW criterion, CPPs of the form $a[Trmn(x)]k+u(c+x)(Trmn(x)+x)+bx$ a{\left[ {Tr_m^n(x)} \right]^k} + u(c + x)(Tr_m^n(x) + x) + bx and axpm + bx + h(xpm ± x) over Fp2m were investigated where $a,b,c,u∈Fp2m*$ a,b,c,u \in F_{{p^{2m}}}^* and h(x) ∈ Fpn [x] are polynomials [7]. Different from previous constructions, multivariate CPPs were constructed by recursive methods from the known PPs or CPPs [12, 23].

Using the piecewise method, a class of PPs over F32m in the form of $(x3m−x+δ)32m−13+1+x$ {({x^{{3^m}}} - x + \delta )^{{{{3^{2m}} - 1} \over 3} + 1}} + x was presented by Li et al. [6]. A consequent work [24] presented several classes of PPs over Fp2m having the form $(xpm+bx+δ)p2m−1d+1−bx$ {({x^{{p^m}}} + bx + \delta )^{{{{p^{2m - 1}}} \over d} + 1}} - bx by using the AGW criterion and piecewise function. Later, PPs of a more general form were proposed by employing the AGW criterion twice [26]. Different from previous ones, PPs over Fp2m in the form of (x2m + x + δ)i(pm−1)+1 + x were proposed by transforming the PP problem into determining the quantity of solutions to some equations with a low degree in Tu et al. [18]. Motivated by these works, we focus our attention on CPPs with exponents s = i(pm − 1) + pj or fractional exponents.

This article aims to seek out more CPPs in the form of $(xpm−x+δ)s+axpm+bx$ {({x^{{p^m}}} - x + \delta )^s} + a{x^{{p^m}}} + bx over Fpn, where a, b,δFpn, s is an integer and n is equal to 2m with a positive integer m. Using the AGW criterion (Lemma 2), we present several classes of CPPs in the same form as Eq. (1) over Fp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 . More precisely, for the fractional exponent s satisfying $s=k(p2m−1)d+1$ s = {{k({p^{2m}} - 1)} \over d} + 1 or $s=pm+12$ s = {{{p^m} + 1} \over 2} , the main proofs here depend on Lemmas 2 and 3, which simplifies the problem of proving some polynomials to be PPs over Fp2m into proving that the related piecewise function is a bijection on some subsets of Fp2m. For the exponent s = i(pm − 1) + pj, we obtain three classes of CPPs by using Lemma 2 and calculating the quantity of solutions to some kind of linearised equations. From the work of Zheng et al. [26], we also give two trivial classes of CPPs for any positive integer s or s satisfying gcd(s, pm − 1) = 1. The approach in this paper is also useful for constructing PPs with this form.

The rest of this article is organised as follows. Section 2 is basic for this paper as it contains several concepts and results which can be used throughout the article. Sections 3 and 4 investigate some classes of CPPs with the form as in Eq. (1) for some restrictions on δ, a, b.

Preliminaries

For a positive divisor m of the positive integer n, the trace function from Fpn to its subfield Fpm is defined as $Trmn(x)=x+xpm+xp2m+⋯+xp(n/m−1)m, x∈Fpn.$ Tr_m^n(x) = x + {x^{{p^m}}} + {x^{{p^{2m}}}} + \cdots + {x^{{p^{(n/m - 1)m}}}},\quad x \in {F_{{p^n}}}. Let J be a subset of Fpn defined as $J={rpm−r|r∈Fpn}$ {\bf J} = \{ {r^{{p^m}}} - r\left| {r \in {F_{{p^n}}}} \right.\} Note that $Trmn(β)=0$ Tr_m^n(\beta ) = 0 for any βJ.

In the following, we restate the AGW criterion for later use.

Lemma 1

( [1]) Let A, S and $S¯$ \overline S denote three finite sets satisfying $|S|=|S¯|$ \left| S \right| = \left| {\overline S } \right| , let f : AA, $f¯:S→S¯$ \overline f :S \to \overline S , λ : AS and $λ¯:A→S¯$ \overline \lambda :A \to \overline S be four mappings satisfying the following diagram, that is, $λ¯∘f=f¯∘λ$ \overline \lambda \circ f = \overline f \circ \lambda .

If λ and $λ¯$ \overline \lambda are two surjections, the following two statements are equivalent:

f is a permutation over A;

$f¯$ \overline f is a one-to-one mapping from S to $S¯$ \overline S and f is an injection on λ−1(s) for each sS. Using the AGW criterion, we can give Lemma 2, which can be applied to determine the PP behaviour of g(x) = (xpmx + δ)s + axpm + bx with the conditions a + b ≠ 0 and $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 .

Lemma 2

For two positive integers m, s and a prime number p, let a, b, δFp2m with a + b ≠ 0 and $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 . Then g(x) = (xpmx + δ )s + axpm + bx is a PP over Fp2m if and only if $f(x)=(a+b)(x+δ)x.pm−(apm+bpm)(x+δ)s+(bpm+1apm+1)(x+δ)$ f(x) = (a + b)(x + \delta {)^{x.{p^m}}} - ({a^{{p^m}}} + {b^{{p^m}}})(x + \delta {)^s} + ({b^{{p^m} + 1}}{a^{{p^m} + 1}})(x + \delta ) is bijective on J which is defined in Eq. (2).

Proof

Let θ(x) = (a + b)xpm − (apm + bpm)x + (bpm+1apm+1) and $S¯={θ(x)|x∈Fp2m}$ \overline S = \{ \theta (x)\left| {x \in {F_{{p^{2m}}}}\} } \right. . Then the diagram

commutes since $(θ∘g)(x)=(a+b)gpm−(apm+bpm)g+(bpm+1−apm+1)=(a+b)(xpm−x+δ)s⋅pm−(apm+bpm)(xpm−x+δ)s+(bpm+1−apm+1)(xpm−x+δ)=f(xpm−x).$ \matrix{ {(\theta \circ g)(x)} \hfill & { = (a + b){g^{{p^m}}} - ({a^{{p^m}}} + {b^{{p^m}}})g + ({b^{{p^m} + 1}} - {a^{{p^m} + 1}})} \hfill \cr {} \hfill & { = (a + b)({x^{{p^m}}} - x + \delta {)^{s \cdot {p^m}}} - ({a^{{p^m}}} + {b^{{p^m}}})({x^{{p^m}}} - x + \delta {)^s} + ({b^{{p^m} + 1}} - {a^{{p^m} + 1}})({x^{{p^m}}} - x + \delta )} \hfill \cr {} \hfill & { = f({x^{{p^m}}} - x).} \hfill \cr } Now, ’let us prove $J=S¯$ {\bf J} = \overline S .

For p = 2, we have J = Fpm. For any $y∈S¯$ y \in \overline S , we have ypm = y due to $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 , thus $S¯⊆J$ \bar S \subseteq {\bf J} . Since the degree of θ(x) is pm, each image has not more than pm pre-images under θ (x). Therefore, $|S¯|≥p2mpm=pm$ \left| {\overline S } \right| \ge {{{p^{2m}}} \over {{p^m}}} = {p^m} , and so $S¯=J$ \overline S = {\bf J} .

When p is an odd prime, let $y∈S¯$ y \in \overline S ; one gets that there exists some xFp2m satisfying $y=θ(x)=((apm+bpm)x−(bpm+1−apm+1)δ2)pm−((apm+bpm)x−(bpm+1−apm+1)δ2)$ y = \theta (x) = (({a^{{p^m}}} + {b^{{p^m}}})x - {{({b^{{p^m} + 1}} - {a^{{p^m} + 1}})\delta } \over 2}{)^{{p^m}}} - \left( {({a^{{p^m}}} + {b^{{p^m}}})x - {{({b^{{p^m} + 1}} - {a^{{p^m} + 1}})\delta } \over 2}} \right) due to $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 , and so $S¯⊆J$ \bar S \subseteq {\bf J} . By contrast, for any zJ, due to $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 , there exist some xFp2m such that z = xpmx = (a + bpm − (apm + bpm)ν + (bpm+1apm+1) where $ν=xpm+1+a+bapm+bpmx+bpm+1−apm+12(apm+bpm)$ \nu = {x^{{p^m}}} + {{1 + a + b} \over {{a^{{p^m}}} + {b^{{p^m}}}}}x + {{{b^{{p^m} + 1}} - {a^{{p^m} + 1}}} \over {2({a^{{p^m}}} + {b^{{p^m}}})}} , and so $J⊆S¯$ {\bf J} \subseteq \bar S . Thus $S¯=J$ \bar S = {\bf J} .

By Lemma 1, if f (x) is a one-to-one mapping on J, it remains to prove that g(x) is an injection of (xpmx)−1 (s) for any sJ. Assume that g(x) = g(y) and xpmx = ypmy, then (xpmx + δ)s + axpm + bx = (ypmy + δ)s + aypm + by, that is, (b + a)(xy) = 0.

Due to a + b ≠ 0, the above equation is true if and only if x = y, which proves that g(x) is an injection of (xpmx)−1(s) for each sJ. Therefore, g(x) permutes Fp2m and the result can be obtained.

Lemma 3

For a primitive element ρ of Fp2m and a fixed element δFp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 where p is an odd prime and m is a positive integer, the set defined by Eq. (2) can be expressed as $J={xpm−x+δ|x∈Fp2m}={ρ−pm+12b|b∈Fpm}.$ {\bf J} = \{ {x^{{p^m}}} - x + \delta \left| {x \in {F_{{p^{2m}}}}} \right.\} = \{ {\rho ^{ - {{{p^m} + 1} \over 2}}}b\left| {b \in {F_{{p^m}}}} \right.\} .

Proof

Let $J'={ω−pm+12b|b∈Fpm}$ {{\bf J}^\prime} = \{ {\omega ^{ - {{{p^m} + 1} \over 2}b}}\left| {b \in {F_{{p^m}}}} \right.\} . For any yJ, if y = 0, we have yJ′. If y ≠ 0, $ypm−1=−1=ρ−p2m−12=(ρ−pm+12)pm−1$ {y^{{p^m} - 1}} = - 1 = {\rho ^{ - {{{p^{2m}} - 1} \over 2}}} = {\left( {{\rho ^{ - {{{p^m} + 1} \over 2}}}} \right)^{{p^m} - 1}} and then $y=ρ−pm+12b$ y = {\rho ^{ - {{{p^m} + 1} \over 2}b}} for some bFpm. Thus JJ′. Since $|J|≥p2mpm=pm$ \left| {\bf J} \right| \ge {{{p^{2m}}} \over {{p^m}}} = {p^m} , we have J = J′.

The proof for J = {xpmx + δ |xFp2m} is similar.

For δFp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 , due to Lemma 3, we use the notation J to denote the set {xpmx | xFp2m}, {xpmx + δ |xFp2m} or ${ρ−pm+12b|b∈Fpm}$ \{ {\rho ^{ - {{{p^m} + 1} \over 2}b}}\left| {b \in {F_{{p^m}}}} \right.\} where ρ is a primitive element of Fp2m.

In the sequel, the following lemma will be applied to determine the number of solutions of some linearised equations of J.

Lemma 4

( [8]) For a positive divisor of the positive integer n and a prime with mp ∤ n, let i, j be two nonnegative integers such that i − j is coprime to n and p − 1, respectively. For two given elements α, βFpn, if α is a (p − 1)-th power in Fpm, the equation xpi − axpj + β = 0 has not more than one solution in which is defined in Eq. (2).

CPPS over Fp2m with fractional exponents

This section investigates the CPPs over Fp2m in the same form as Eq. (1) with the fractional exponents and satisfying $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 , where p will be restricted to odd prime. The main method used here is Lemma 2 and the main proof idea stems from Yuan and Zheng [24]. For simplicity of notation, let $A=−(a+b+apm+bpm), B=bpm+1−apm+1A'=−(a+b+apm+bpm+2), B'=(b+1)pm+1−apm+1$ \matrix{ {\;A = - (a + b + {a^{{p^m}}} + {b^{{p^m}}}),\quad B = {b^{{p^m} + 1}} - {a^{{p^m} + 1}}} \hfill \cr {{A^\prime} = - (a + b + {a^{{p^m}}} + {b^{{p^m}}} + 2),\quad {B^\prime} = (b + {{1)}^p}^{^m + 1} - {a^{{p^m} + 1}}} \hfill \cr } In the following proposition, let p be an odd prime and n be a positive integer satisfying pn ≡ 1(mod3). Let ρ denote a primitive element of Fpn, then define three subsets of Fpn by D0 =< ρ3 >, Di = ρiD0 for i = 1, 2, where D0 is the multiplicative subgroup of Fpn generated by ρ3. Then Fpn = {0} ∪ D0D1D2. Note that $xpn−13=ρpn−13i$ {x^{{{{p^n} - 1} \over 3}}} = {\rho ^{{{{p^n} - 1} \over 3}i}} when xDi for i = 0, 1, 2. For the sake of simplicity, let $ω=ρpn−13$ \omega = {\rho ^{{{{p^n} - 1} \over 3}}} . Therefore, ω2 + ω + 1 = 0 and ω ≠ 1.

Proposition 1

For a positive integer m and any odd prime p with pm 1(mod3), let δFp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 and aFp2m, bFp2m \ {−a, −a − 1}, then the polynomial $g(x)=(xpm−x+δ)k(p2m−1)3+1+axpm+bx$ g(x) = ({x^{{p^m}}} - x + \delta {)^{{{k({p^{2m}} - 1)} \over 3} + 1}} + a{x^{{p^m}}} + bx is a CPP of for k = 1, 2 if and only if $(A+B,ωkA+B,ω2kA+B)∈∪i=02Di×Di×Di∪Di×Di+1(mod 3)×Di+2(mod 3)$ (A + B,{\omega ^k}A + B,{\omega ^{2k}}A + B) \in \bigcup\limits_{i = 0}^2 {D_i} \times {D_i} \times {D_i} \cup {D_i} \times {D_{i + 1({\rm{mod}}{\kern 1pt} 3)}} \times {D_{i + 2({\rm{mod}}{\kern 1pt} 3)}} and $(A'+B',ωkA'+B',ω2kA'+B')∈∪i=02Di×Di×Di∪Di×Di+1(mod 3)×Di+2(mod 3)$ ({A^\prime} + {B^\prime},{\omega ^k}{A^\prime} + {B^\prime},{\omega ^{2k}}{A^\prime} + {B^\prime}) \in \bigcup\limits_{i = 0}^2 {D_i} \times {D_i} \times {D_i} \cup {D_i} \times {D_{i + 1({\rm{mod}}{\kern 1pt} 3)}} \times {D_{i + 2({\rm{mod}}{\kern 1pt} 3)}} , where A,B,A′,Bare as defined in Eq. (3).

Proof

We only consider the case for k = 1 since the case for k = 2 follows similarly. Due to the fact that $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 and $p2m−13+1$ {{{p^{2m}} - 1} \over 3} + 1 is odd, by Lemma 2, g(x) is a PP if and only if $f(x)=A(x+δ)p2m−13+1+B(x+δ)$ f(x) = A{(x + \delta )^{{{{p^{2m}} - 1} \over 3} + 1}} + B(x + \delta ) is a bijection on J defined in Eq. (2). Together with Lemma 3, f (x) is a bijection on J iff $f'(u)=u(Aup2m−13+B)$ {f^\prime}(u) = u\left( {A{u^{{{{p^{2m}} - 1} \over 3}}} + B} \right) is a bijection on J where u = x + δ.

When pm 1(mod 3), by Lemma 3, we can obtain that the non-zero elements of J can be decomposed into a pairwise disjoint union of the subsets of D0, D1 and D2 such that each of these subsets have $pm−13$ {{{p^m} - 1} \over 3} elements.

Note that $f'(u)={0, if u=0,(A+B)u, if u∈D0,(ωA+B)u, if u∈D1,(ω2A+B)u, if u∈D2.$ {f^\prime}(u) = \left\{ {\matrix{ {0,\quad {\rm{if}}\quad u = 0,} \hfill \cr {(A + B)u,\quad {\rm{if}}\quad u \in {D_0},} \hfill \cr {(\omega A + B)u,\quad {\rm{if}}\quad u \in {D_1},} \hfill \cr {({\omega ^2}A + B)u,\quad {\rm{if}}\quad u \in {D_2}.} \hfill \cr } } \right.

Assume that f′(u) is a bijection and A + BDi, then (A + B)D0 = Di. If ωA + BDi+2(mod3), we have (ωA + B)D1 = Di which contradicts the assumption that f′(u) is a bijection. It follows that ωA + BDi+2(mod3). When ωA + BDi, we have (ωA + B)D1 = Di+2(mod3) and then (ω2A + B)D2 = Di+1(mod3) due to the assumption that f′(u) is a bijection. Hence ω2A + BDi. When ωA + BDi+1(mod3), we have (ωA + B)D1 = Di+2(mod3) and then (ω2A + B)D2 = Di+1(mod3) due to the assumption that f′(u) is a bijection. Thus ω2A + BDi+2(mod3). Therefore if f′(u) is a bijection, then $(A+B,ωkA+B,ω2kA+B)∈∪i=02Di×Di×Di∪Di×Di+1(mod 3)×Di+2(mod 3)$ (A + B,{\omega ^k}A + B,{\omega ^{2k}}A + B) \in \bigcup\limits_{i = 0}^2 {D_i} \times {D_i} \times {D_i} \cup {D_i} \times {D_{i + 1({\rm{mod}}{\kern 1pt} 3)}} \times {D_{i + 2({\rm{mod}}{\kern 1pt} 3)}} . Conversely, when $(A+B,ωkA+B,ω2kA+B)∈∪i=02Di×Di×Di∪Di×Di+1(mod 3)×Di+2(mod 3)$ (A + B,{\omega ^k}A + B,{\omega ^{2k}}A + B) \in \bigcup\limits_{i = 0}^2 {D_i} \times {D_i} \times {D_i} \cup {D_i} \times {D_{i + 1({\rm{mod}}{\kern 1pt} 3)}} \times {D_{i + 2({\rm{mod}}{\kern 1pt} 3)}} we can easily validate that f ′(u) is a bijection.

Therefore, g(x) is a PP of Fp2m iff $(A+B,ωkA+B,ω2kA+B)∈∪i=02Di×Di×Di∪Di×Di+1(mod 3)×Di+2(mod 3)$ (A + B,{\omega ^k}A + B,{\omega ^{2k}}A + B) \in \bigcup\limits_{i = 0}^2 {D_i} \times {D_i} \times {D_i} \cup {D_i} \times {D_{i + 1({\rm{mod}}{\kern 1pt} 3)}} \times {D_{i + 2({\rm{mod}}{\kern 1pt} 3)}} . Similarly, we can prove that g(x) + x is a PP of Fp2m iff (A′ + B′, ωA′ + B′, ω2A′ + B′) ∈ Dj × Dj × Dj or Dj × Dj+1(mod3) × Dj+2(mod3) for j = 0, 1, 2. Hence the result follows.

Example 1

Take p = 7, m = 1 and k = 1. By Magma software, we can verify that there exist 714 different triples $(δ,a,b)∈F343$ (\delta ,a,b) \in F_{{3^4}}^3 satisfying that $Tr12(δ)=0$ Tr_1^2(\delta ) = 0 , (A + B, ωA + B, ω2A + B) ∈ Di × Di × Di or Di × Di+1(mod3) × Di+2(mod3) for i = 0, 1, 2 and $(A'+B',ωkA'+B',ω2kA'+B')∈∪i=02Di×Di×Di∪Di×Di+1(mod 3)×Di+2(mod 3)$ ({A^\prime} + {B^\prime},{\omega ^k}{A^\prime} + {B^\prime},{\omega ^{2k}}{A^\prime} + {B^\prime}) \in \bigcup\limits_{i = 0}^2 {D_i} \times {D_i} \times {D_i} \cup {D_i} \times {D_{i + 1({\rm{mod}}{\kern 1pt} 3)}} \times {D_{i + 2({\rm{mod}}{\kern 1pt} 3)}} , When $Tr12(δ)=0$ Tr_1^2(\delta ) = 0 , these are exactly all triples which can make be a CPP of F72.

Next, when pn 1(mod4), the letter ρ denotes a primitive element of Fpn. In the following proposition, the multiplicative subgroup of Fpn generated by ρ4 is defined as D0 =< ρ4 >, and then let for i = 1, 2, 3. We can obtain that Fpn ∪ {0} ∪ D0D1D2D3. For i = 0, 1, 2, 3, if xDi, we have $xpn−14=ρpn−14i=ωi$ {x^{{{{p^n} - 1} \over 4}}} = {\rho ^{{{{p^n} - 1} \over 4}i}} = {\omega ^i} where $ω=ρpn−14$ \omega = {\rho ^{{{{p^n} - 1} \over 4}}} satisfies ω2 + 1 = 0 and ω ≠ 1.

Proposition 2

For a positive integer m and any odd prime p with pm 1(mod 4), let δFp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 and aFp2m, bFp2m \ {−a, −a − 1}. Let $s∈{pm+12,p2m−14+1,3(p2m−1)4+1}$ s \in \left\{ {{{{p^m} + 1} \over 2},{{{p^{2m}} - 1} \over 4} + 1,{{3\left( {{p^{2m}} - 1} \right)} \over 4} + 1} \right\} . Then the polynomial g(x) = (xpmx + δ )s + axpm + bx is a CPP of Fp2m when and only when (ωA + B, −ωA + B) ∈ D0 × D0D2 × D2 and (ωA′ + B′, −ωA′ + B′) ∈ D0 × D0D2 × D2 where A,B,A′,Bare as defined in Eq. (3).

Proof

We just consider the case for $s=pm+12$ s = {{{p^m} + 1} \over 2} since the proofs for other cases are similar. Since pm 1(mod4), we have that $pm+12$ {{{p^m} + 1} \over 2} is odd. Together with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 , by Lemma 2, we prove that g(x) is a PP by showing that for any dJ, the equation $A(x+δ)pm+12+B(x+δ)=d$ A{(x + \delta )^{{{{p^m} + 1} \over 2}}} + B(x + \delta ) = d has only one solution in J, which is equivalent to show that $(x+δ)(A(x+δ)pm−12+B)=d$ (x + \delta )(A{(x + \delta )^{{{{p^m} - 1} \over 2}}} + B) = d has one and only one solution in J, where J is defined in Eq. (2).

Therefore, g(x) is a PP iff $f(u)=u(Aupm−12+B)$ f(u) = u(A{u^{{{{p^m} - 1} \over 2}}} + B) is a bijection on J.

By Lemma 3, due to the fact that $pm+12$ {{{p^m} + 1} \over 2} is odd, we can obtain that the non-zero elements of J can be decomposed into a pairwise disjoint union of the subsets of D1 and D3 such that each of these subsets have $pm−12$ {{{p^m} - 1} \over 2} elements.

Assume that $pm+12≡1(mod 4)$ {{{p^m} + 1} \over 2} \equiv 1({\rm{mod}}{\kern 1pt} 4) . By Lemma 3, we have that for any uJ, there exists a unique element bFpm satisfying $u=ρ−pm+12b$ u = {\rho ^{{{ - {p^m} + 1} \over 2}b}} . Thus, one gets $u∈D1⇔bpm−12=−1$ u \in {D_1} \Leftrightarrow {b^{{{{p^m} - 1} \over 2}}} = - 1 and $u∈D3⇔bpm−12=1$ u \in {D_3} \Leftrightarrow {b^{{{{p^m} - 1} \over 2}}} = 1 . Then we obtain $f(u)={0, if u=0,(ωA+B)u, if u∈D1,(−ωA+B)u, if u∈D3.$ f(u) = \left\{ {\matrix{ {0,\quad {\rm{if}}\quad u = 0,} \hfill \cr {(\omega A + B)u,\quad {\rm{if}}\quad u \in {D_1},} \hfill \cr {( - \omega A + B)u,\quad {\rm{if}}\quad u \in {D_3}.} \hfill \cr } } \right.

Assume that f (u) is a bijection. If ωA + bD0, we have (ωA + B)D1 = D1 and then (−ωA + B)D3 = D3 due to f (u) is a bijection. Hence −ωA + BD0. Similarly, when ωA + BD2, we have (ωA + B)D1 = D3 and then (−ωA + B)D3 = D1 due to the assumption that f (u) is a bijection. Hence −ωA + BD2. For i = 1, 3, if ωA + BDi, we can obtain that (ωA + B)D1 equals D0 or D2. It follows that ωA + BDi for i = 1, 3 due to the assumption that f (u) is a bijection. Using the same method, we have −ωA + BDi for i = 1, 3. Therefore if f (u) is a bijection, then (ωA + B, −ωA + B) ∈ D0 × D0 or D2 × D2. Conversely, when (ωA + B, −ωA + B) ∈ D0 × D0 or D2 × D2, we have f (u) is a bijection on J.

The case for $pm+12≡3(mod 4)$ {{{p^m} + 1} \over 2} \equiv 3({\rm{mod}}{\kern 1pt} 4) is similar, so we omit the proof for this case.

Therefore, g(x) is a PP iff (ωA + B, −ωA + B) ∈ D0 × D0 or D2 × D2. Similarly, we can prove that g(x) + x is a PP iff (ωA′ + B′, −ωA′ + B′) ∈ D0 × D0 or D2 × D2.

Example 2

Take p = 3, we have s ∈ {5, 21, 61}. By Magma software, one can verify that there exist 9,693 different triples $(δ,a,b)∈F343$ (\delta ,a,b) \in F_{{3^4}}^3 satisfying that $Tr24(δ)=0$ Tr_2^4(\delta ) = 0 , (ωA + B, −ωA + B) ∈ D0 × D0D2 × D2 and (ωA′ + B′, −ωA′ + B′) ∈ D0 × D0D2 × D2. These (δ, a, b) are exactly all elements of $F343$ F_{{3^4}}^3 such that g(x) = (x9x + δ )s + ax9 + bx is a CPP of $F343$ F_{{3^4}}^3 which implies that the complete permutation behaviour for these s does not hold for $Tr24(δ)≠0$ Tr_2^4(\delta ) \ne 0 .

Similarly, let n be a positive integer with pn 1(mod6), where p is an odd prime and let ρ denote a primitive element in Fpn. In the following propositions, the notation D0 =< ρ6 > represents the multiplicative subgroup of Fpn which is generated by ρ6, and we define five disjoint subsets Di = ρiD0 of Fpn for i = 1, 2, 3, 4, 5. Therefore, Fpn = {0} ∪ D0D1D2D3D4D5. Note that $xpn−16=ρpn−16i$ {x^{{{{p^n} - 1} \over 6}}} = {\rho ^{{{{p^n} - 1} \over 6}i}} if xDi for i = 0, 1, 2, 3, 4, 5. For the sake of simplicity, denote $ω=ρpn−16$ \omega = \rho {{{p^n} - 1} \over 6} . Therefore, ω2ω + 1 = 0 and ω ∉ 1. Define the set $S=∪i∈{0,2,4}(Di3∪Di×Di+2(mod 6)×Di+4(mod 6))$ S = \bigcup\limits_{i \in \{ 0,2,4\} } \left( {D_i^3 \cup {D_i} \times {D_{i + 2({\rm{mod}}{\kern 1pt} 6)}} \times {D_{i + 4({\rm{mod}}{\kern 1pt} 6)}}} \right)

Proposition 3

Let m be a positive integer with pm 1(mod12) where p is an odd prime and let δFp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 . For aFp2m, bFp2m \{−a, −a − 1} and k = 1, 5, the polynomial $g(x)=(xpm−x+δ)k(p2m−1)6+1+axpm+bx$ g(x) = ({x^{{p^m}}} - x + \delta {)^{{{k({p^{2m}} - 1)} \over 6} + 1}} + a{x^{{p^m}}} + bx is a CPP of Fp2m if and only if (ωkA + B, −A + B, ω5kA + B) ∈ S and (wkA′ + B′, −A′ + B′, ω5kA′ + B′) ∈ S, where A,B,A′,B′ are defined in Eq. (3) and S is defined in Eq. (5).

Proof

We only show the case for k = 1, since the case for k = 5 can be checked in the same way. By Lemmas 2 and 3, to prove that g(x) is a PP, it is enough to show that $A(x+δ)p2m−16+1+B(x+δ)$ A{(x + \delta )^{{{{p^{2m}} - 1} \over 6} + 1}} + B(x + \delta ) is a bijection on J, that is, $f(u)=u(Aup2m−16+B)$ f(u) = u(A{u^{{{{p^{2m}} - 1} \over 6}}} + B) is a bijection on J.

By Lemma 3, we can obtain that the non-zero elements of J can be decomposed into pairwise disjoint union of the subsets of D1, D3 and D5 such that each of these subsets have $pm−13$ {{{p^m} - 1} \over 3} elements since $pm+12$ {{{p^m} + 1} \over 2} odd for pm 1(mod12).

Note that $f(u)={0, if u=0,(ωA+B)u, if u∈D1,(−ωA+B)u, if u∈D3,(ω5A+B)u, if u∈D5.$ f(u) = \left\{ {\matrix{ {0,\quad {\rm{if}}\quad u = 0,} \hfill \cr {(\omega A + B)u,\quad {\rm{if}}\quad u \in {D_1},} \hfill \cr {( - \omega A + B)u,\quad {\rm{if}}\quad u \in {D_3},} \hfill \cr {\left( {{\omega ^5}A + B} \right)u,\quad {\rm{if}}\quad u \in {D_5}.} \hfill \cr } } \right.

Assume that f (u) is a bijection. If ωA + BDi for any i = 1, 3, 5, then (ωA + B)D1 = Dj for any j = 0, 2, 4 which contradicts the assumption that f (u) is a bijection. It follows that ωA + B ∉ ∪i∈{1, 3, 5}Di. Similarly, we can prove that −A + B, ω5A + B ∉ ∪i∈(1, 3, 5}Di. Assume that ωA + BDi for any i = 0, 2, 4, then (ωA + B)D1 =Di+1(mod6).

If −A + BDi+4(mod6), we have (−A + B)D3 = Di+1(mod6) which is contradictory to the assumption that f (u) is a bijection. It follows that −A + BDi+4(mod6) when f (u) is a bijection. When −A + BDi for any i = 0, 2, 4. Similarly, we can obtain that ω5A + BDi for any i = 0, 2, 4. When −A + BDi+2(mod6), we have (−A + B)D3 = Di+3(mod6) and then ω5A + BDi+4(mod6) due to the assumption that f (u) is a one-to-one mapping. Thus, when f (u) is a one-to-one mapping, we have (ωA + B, −A + B, ω5A + B) ∈ S.

Conversely, when (A + B, ωA + B, ω2A + B) ∈ S, we can easily check that f (u) is a one-to-one mapping. Therefore, g(x) is a PP iff (ωA + B, −A + B, ω5A + B) ∈ S. A similar argument shows that g(x) + x a PP iff (ωkA′ + B′, −A′ + B′, ω5kA′ + B′) ∈ S.

Example 3

Take p = 13, m = 1, k = 1 and s = 29. By Magma software, one can check that there exist 5,460 different triples $(δ,a,b)∈F1323$ (\delta ,a,b) \in F_{{{13}^2}}^3 satisfying that $Tr12(δ)=0$ Tr_1^2(\delta ) = 0 , (A + B, ωA + B, ω2A + B) ∈ S and (ωA + B, −A + B, ω5A + B) ∈ S. When $Tr12(δ)=0$ Tr_1^2(\delta ) = 0 , these (δ, a, b) are exactly all elements such that the polynomial g(x) = (x13x + δ )29 + ax13 + bx is a CPP of F134.

The following proposition can be shown in a similar way.

Proposition 4

For any odd prime p and a positive integer m with pm 7(mod12), let δFp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 and aFp2m, bFp2m \ {−a, −a−1}, then the polynomial $g(x)=(xpm−x+δ)k(p2m−1)6+1+axpm+bx$ g(x) = ({x^{{p^m}}} - x + \delta {)^{{{k({p^{2m}} - 1)} \over 6} + 1}} + a{x^{{p^m}}} + bx is a CPP of Fp2m for k = 1, 5 iff (A + B, ω2kA + B, −ωkA + B) ∈ S and (A′ + B′, ω2kA′ + B′, −ωkA′ + B′) ∈ S, where A,B,A′,Bare defined in Eq. (3) and S is defined in Eq. (5).

Remark 1

From the proof of Propositions 1–4, a lot of PPs in the same form as g(x) = (xpmx + δ)s + axpm satisfying $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 can be obtained. Some results here partly generalise some discovered PPs in the same form for the case a = b = 1 in Yuan and Zheng [24] and Zheng and Chen [25].

Using the same method, we can obtain the following results for p2m 1(modd), where d is an odd prime or a positive integer satisfying d ≡ 0 (mod4).

Proposition 5

Let p,d be two odd primes with pm 1(modd) where m is a positive integer, let ρ denote the primitive element of $Fp2m$ F_p^{2m} and $ω=ρpn−1d$ \omega = {\rho ^{{{{p^n} - 1} \over d}}} . For i = 1, 2, ···, d − 1, let D0 =< ρd >, Di = ρi < D0 >, where D0 is the multiplicative subgroup of $Fp2m$ F_p^{2m} which is generated by ρd. The set T consists of all elements of the form (i0,i1,···, id−1) where the components ij are elements in {0,1,···, d −1} such that (i0,i1 +1(modd),···, id−1 + d − 1(modd)) is a permutation of the integer set {0,1,···, d − 1}.

Let δFp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 and aFp2m, bFp2m \{−a, −a − 1}. Then the polynomial $g(x)=(xpm−x+δ)k(p2m−1)d+1+axpm+bx$ g(x) = ({x^{{p^m}}} - x + \delta {)^{{{k({p^{2m}} - 1)} \over d} + 1}} + a{x^{{p^m}}} + bx is a CPP of Fp2m for k = 1, 2, ···, d − 1 if and only if the following two conditions hold:

A + B, ωkA + B, ω2kA + B,···, $ω(d−1)kA+B∈∪(i0,i1, ⋯,id−1)∈TDi0×Di1×⋯×Did−1$ {\omega ^{(d - 1)k}}A + B \in \bigcup\limits_{({i_0},{i_1},\; \cdots ,{i_{d - 1}}) \in T} {D_{{i_0}}} \times {D_{{i_1}}} \times \cdots \times {D_{{i_{d - 1}}}} ;

A′ + B′, ωkA′ + B′, ω2kA′ + B′,···, $ω(d−1)kA'+B'∈∪(i0,i1, ⋯,id−1)∈TDi0×Di1×⋯×Did−1$ {\omega ^{(d - 1)k}}{A^\prime} + {B^\prime} \in \bigcup\limits_{({i_0},{i_1},\; \cdots ,{i_{d - 1}}) \in T} {D_{{i_0}}} \times {D_{{i_1}}} \times \cdots \times {D_{{i_{d - 1}}}} ,

where A,B,A′,B′ are defined in Eq. (3).

Proposition 6

For an odd prime number p and positive integers m, d with d ≡ 0(mod4) and pm 1(modd), let ρ denote the primitive element of Fp2m and $ω=ρpn−1d$ \omega = {\rho ^{{{{p^n} - 1} \over d}}} , then define d subsets of Fp2m by D0 =< ρd >,Di = ρiD0 for i = 1, 2, ···, d − 1, where D0 is the multiplicative subgroup of Fp2m generated by ρd. The set T consists of all elements of the form (i1,i3,···, id−1) where the components ij are elements of {0,2,···, d − 2} such that (i1 + 1(modd),i3 + 3(modd),···, id−1 + d − 1(modd)) is a permutation of the integer set {1, 3, ···, d − 1}. Let δFp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 and aFp2m,bFp2m \{−a, −a − 1}. Then, for k = 1, 2, ···, d − 1, the polynomial $g(x)=(xpm−x+δ)k(p2m−1)d+axpm+bx$ g(x) = ({x^{{p^m}}} - x + \delta {)^{{{k({p^{2m}} - 1)} \over d}}} + a{x^{{p^m}}} + bx is a CPP of Fp2m if and only if the following two conditions hold:

ωkA + B, ω3kA + B, ω5kA + B···, $ω(d−1)kA+B∈∪(i1,i3, ⋯,id−1)∈TDi1×Di3×⋯×Did−1$ {\omega ^{(d - 1)k}}A + B \in \bigcup\limits_{({i_1},{i_3},\; \cdots ,{i_{d - 1}}) \in T} {D_{{i_1}}} \times {D_{{i_3}}} \times \cdots \times {D_{{i_{d - 1}}}} ;

ωkA′ + B′, ω3kA′ + B′, ω5kA′ + B′ ···, $ω(d−1)kA'+B'∈∪(i1,i3, ⋯,id−1)∈TDi1×Di3×⋯×Did−1$ {\omega ^{(d - 1)k}}{A^\prime} + {B^\prime} \in \bigcup\limits_{({i_1},{i_3},\; \cdots ,{i_{d - 1}}) \in T} {D_{{i_1}}} \times {D_{{i_3}}} \times \cdots \times {D_{{i_{d - 1}}}} ,

where A, B, A′, B′ are defined in Eq. (3).

Other CPPS g(x) over Fp2m

This section proposes five classes of CPPs in the same form as in Eq. (1) over Fp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 . More precisely, for any positive integer s or some integers s satisfying gcd(s, pm − 1) = 1, we can obtain two trivial classes of CPPs, and for the exponent s = i(pm − 1) + pj, we give three classes of CPPs.

For any positive integer s, the following result can be obtained easily from Corollary 6 of Zheng et al. [26].

Proposition 7

For two positive integers m, s, let δ, a, bFp2m satisfy $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 . Then, the polynomial g(x) = (xpmx + δ )s + axpm + bx is a CPP over Fp2m if (a + b)pm + (−1)1−s(a + b) = 0, (a + b + 1)pm + (−1)1−s(a + b + 1) = 0 and apm+1 ∉ {bpm+1, (b + 1) pm+1}.

When gcd(s, pm −1) = 1, the following proposition can be obtained easily from Corollaries 6 and 7 of Zheng et al. [26].

Proposition 8

For two positive integers m, s which satisfy gcd(s, pm − 1) = 1 and a fixed element δ with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 , let bFp2m, bFp2m \{−a, −a − 1} and g(x) = (xpmx + δ )s + axpm + bx.

For p = 2, g(x) is a CPP over F22m if either a + b + a2m + b2m = 0, a2m+1 ∉ {b2m+1, (b + 1)2m+1} or a + a2m + b + b2m ≠ 0, a2m+1 = b2m+1, b + b2m + 1 = 0.

For odd p, s, g(x) is a CPP over Fp2m if one of the following conditions is true:

a + b + apm + bpm = 0,apm+1bpm+1,b + bpm + 1 ≠ 0;

a + b + apm + bpm = −2,apm+1 = bpm+1,b + bpm + 1 ≠ 0;

a + b + apm + bpm ≠ 0,apm+1 = bpm+1,b + bpm + 1 = 0.

When s = i(pm − 1) + pj, one gets gcd(s, pm − 1) = 1, but Proposition 8 cannot contain all possible $(δ,a,b)∈Fp2m3$ (\delta ,a,b) \in F_{{p^{2m}}}^3 , which can make g(x) a CPP for $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 . Thus, under some restrictions in the process of selecting the elements a, b in Fp2m, three classes of CPPs over Fp2m for s = i(pm − 1) + pj and $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 are discussed in the following.

The following proposition is on the assumption that s = i(2m − 1) + 2j for j > 0.

Proposition 9

For three positive integers m, i, j with 0 < j < m and a fixed element δF2m, let aF22m and bF22m \ {a,a + 1}. The polynomial g(x) = (x2m + x + δ )i(2m−1)+2j + ax2m + bx is a CPP of F2m if one of the following conditions is satisfied:

a + b + a2m + b2m = 0, a2m+1b2m+1, a2m+1 ≠ (b + 1)2m+1;

a + b + a2m + b2m ≠ 0, $a2m+1+b2m+1a+b+a2m+b2m$ {{{a^{{2^m} + 1}} + {b^{{2^m} + 1}}} \over {a + b + {a^{{2^m}}} + {b^{{2^m}}}}} and $a2m+1+(b+1)2m+1a+b+a2m+b2m$ {{{a^{{2^m} + 1}} + {{(b + 1)}^{{2^m} + 1}}} \over {a + b + {a^{{2^m}}} + {b^{{2^m}}}}} are not (2j − 1)-th elements in F2m.

Proof

In order to show that g(x) is a permutation of F22m, we only need to show that $(a+b+a2m+b2m)(x+δ)i(2m−1)+2j+(a2m+1+b2m+1)(x+δ)=d$ (a + b + {a^{{2^m}}} + {b^{{2^m}}})(x + \delta {)^{i{{(2}^m} - 1) + {2^j}}} + ({a^{{2^m} + 1}} + {b^{{2^m} + 1}})(x + \delta ) = d has a unique solution in F2m for any dF2m by Lemma 2.

We can reduce Eq. (6) to $(a+b+a2m+b2m)(x+δ)2j+(a2m+1+b2m+1)(x+δ)=d$ (a + b + {a^{{2^m}}} + {b^{{2^m}}})(x + \delta {)^{{2^j}}} + ({a^{{2^m} + 1}} + {b^{{2^m} + 1}})(x + \delta ) = d When a + b + a2m + b2m = 0 and a2m+1 + b2m + 1 ≠ 0, Eq. (7) has only one solution.

When a + b + a2m + b2m ≠ 0 and $a2m+1+b2m+1a+b+a2m+b2m$ {{{a^{{2^m} + 1}} + {b^{{2^m} + 1}}} \over {a + b + {a^{{2^m}}} + {b^{{2^m}}}}} is not a (2j − 1)-th element in F2m, we have that (a + b + a2m + b2m)x2j−1 + a2m+1 + b2m+1 = 0 has no solution in F2m, and so Eq. (7) has a unique solution in F2m. In this way, g(x) is a permutation of F22m.

In the same way, we can get that g(x) + x is a PP if either a + b + a2m + b2m = 0, a2m+1 + (b + 1)2m+1 ≠ 0 or a + b + a2m + b2m ≠ 0 and $a2m+1+(b+1)2m+1+1a+b+a2m+b2m$ {{{a^{{2^m} + 1}} + {{(b + 1)}^{{2^m} + 1}} + 1} \over {a + b + {a^{{2^m}}} + {b^{{2^m}}}}} is not a (2j − 1)-th element in F2m.

Therefore, g(x) is a CPP of F22m if either a + b + a2m + b2m = 0, a2m+1b2m+1, a2m+1 ≠ (b + 1)2m+1 or a + b + a2m + b2m ≠ 0 such that $a2m+1+b2m+1a+b+a2m+b2m$ {{{a^{{2^m} + 1}} + {b^{{2^m} + 1}}} \over {a + b + {a^{{2^m}}} + {b^{{2^m}}}}} and $a2m+1+(b+1)2m+1+1a+b+a2m+b2m$ {{{a^{{2^m} + 1}} + {{(b + 1)}^{{2^m} + 1}} + 1} \over {a + b + {a^{{2^m}}} + {b^{{2^m}}}}} are not (2j − 1)-th elements in F2m.

Example 4

Take m = 4, i = 3, j = 2, then s = 49. Through Magma software, we can confirm that there exist 50,176 different triples (δ, a, b) ∈ F24 × F28 × F28 satisfying that a + a16 + b + b16 = 0, b17a17, (b + 1)17a17 and 384,000 different triples (δ, a, b) ∈ F24 × F28 × F28 satisfying that a + a16 + b + b16 ≠ 0, $b17+a17a+a16+b+b16$ {{{b^{17}} + {a^{17}}} \over {a + {a^{16}} + b + {b^{16}}}} and $(b+1)17+a17a+a16+b+b16$ {{{{(b + 1)}^{17}} + {a^{17}}} \over {a + {a^{16}} + b + {b^{16}}}} are not cube elements in F24. These (δ, a, b) are exactly all elements in F24 × F28 × F28 such that g(x) = (x16 + x + δ )49 + ax16 + bx is a CPP of F28. Furthermore, it can be verified that g(x) cannot be a CPP for δF24.

For any old prime p and s = i(pm − 1) + pj with j > 0, CPPs over Fp2m are discussed in the next proposition.

Proposition 10

For three positive integers m, i, j with 0 < j < m and a fixed element δFp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 let bFp2m \{−a, −a − 1} and g(x) = (xpmx + δ)i(pm−1)+pj + axpm + bx where gcd( j,2m) = gcd( j, p − 1) = 1 and the prime p is odd. Hence g(x) is a CPP of Fp2m if one of the following conditions is satisfied:

a + b + apm + bpm ≠ 0, −2 and $(−1)(bpm+1−apm+1a+b+apm+bpm$ {{( - 1)({b^{{p^m} + 1}} - {a^{{p^m} + 1}}} \over {a + b + {a^{{p^m}}} + {b^{{p^m}}}}} , $(−1)i((b+1)pm+1−apm+1)a+b+apm+bpm+2$ {{{{( - 1)}^i}((b + {{1)}^{{p^m} + 1}} - {a^{{p^m} + 1}})} \over {a + b + {a^{{p^m}}} + {b^{{p^m}}} + 2}} are all (p − 1)-th elements in Fpm ;

a + b + apm + bpm = 0, bpm+1apm+1, $(−1)i(b+1)pm+1−apm+12$ {( - 1)^i}{{{{(b + 1)}^{{p^m} + 1}} - {a^{{p^m} + 1}}} \over 2} is a (p − 1)-th element in Fpm ;

a + b + apm + bpm = −2, (b + 1) pm+1apm+1 and $(−1)i+1(bpm+1−apm+1)2$ {( - 1)^{i + 1}}{{({b^{{p^m} + 1}} - {a^{{p^m} + 1}})} \over 2} is a (p − 1)-th element in Fpm.

Proof

To obtain that is a PP, we only need to show that for any dJ, the equation $(a+b)(x+δ)pm(i(pm−1)+pj)−(apm+bpm)(x+δ)i(pm−1)+pj+(bpm+1−apm+1)(x+δ)=d$ (a + b)(x + \delta {)^{{p^m}(i({p^m} - 1) + {p^j})}} - ({a^{{p^m}}} + {b^{{p^m}}})(x + \delta {)^{i({p^m} - 1) + {p^j}}} + ({b^{{p^m} + 1}} - {a^{{p^m} + 1}})(x + \delta ) = d has at most one solution in J by Lemma 2.

Since $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 and xJ, we have (x + δ )pm(i(pm−1)+pj) = (−1)i+1(x + δ)pj, (x + δ )i(pm−1)+pj = (−1)i(x + δ )pj.

Eq. (8) is equivalent to $(−1)i+1(a+b+apm+bpm)(x+δ)pj+(bpm+1−apm+1)(x+δ)=d$ {( - 1)^{i + 1}}(a + b + {a^{{p^m}}} + {b^{{p^m}}})(x + \delta {)^{{p^j}}} + ({b^{{p^m} + 1}} - {a^{{p^m} + 1}})(x + \delta ) = d When a + b + apm + bpm = 0, for bpm+1apm+1 ≠ 0, Eq. (9) has a unique solution $x=dbpm+1−apm+1−δ$ x = {d \over {{b^{{p^m} + 1}} - {a^{{p^m} + 1}}}} - \delta . When a + b + apm + bpm ≠ 0, Eq. (9) can be simplified to $xpj−(−1)i(bpm+1−apm+1)a+b+apm+bpmx+δpj+(−1)i+1(bpm+1−apm+1)a+b+apm+bpm+(−1)ida+b+apm+bpm=0$ {x^{{p^j}}} - {{{{( - 1)}^i}\left( {{b^{{p^m} + 1}} - {a^{{p^m} + 1}}} \right)} \over {a + b + {a^{{p^m}}} + {b^{{p^m}}}}}x + {\delta ^{{p^j}}} + {{{{( - 1)}^{i + 1}}\left( {{b^{{p^m} + 1}} - {a^{{p^m} + 1}}} \right)} \over {a + b + {a^{{p^m}}} + {b^{{p^m}}}}} + {{{{( - 1)}^i}d} \over {a + b + {a^{{p^m}}} + {b^{{p^m}}}}} = 0 By Lemma 4, we can draw the conclusion that Eq. (10) has at most one solution in J, if $(−1)i(bpm+1−apm+1)a+b+apm+bpm$ {{{{( - 1)}^i}({b^{{p^m} + 1}} - {a^{{p^m} + 1}})} \over {a + b + {a^{{p^m}}} + {b^{{p^m}}}}} is a (p − 1)-th element in Fpm. Thus g(x) is a PP of Fp2m under the conditions that either a + b + apm + bpm = 0, bpm+1apm+1 ≠ 0 or a + b + apm + bpm ≠ 0, $(−1)i(bpm+1−apm+1)a+b+apm+bpm$ {{{{( - 1)}^i}({b^{{p^m} + 1}} - {a^{{p^m} + 1}})} \over {a + b + {a^{{p^m}}} + {b^{{p^m}}}}} is a (p − 1)-th element in Fpm.

In the same way, one can show that g(x) + x is a PP of Fp2m under the conditions that either a + b + apm + bpm + 2 = 0, (b + 1) pm+1apm+1 ≠ 0 or a + b + apm + bpm ≠ −2, $(−1)i((b+1)pm+1−apm+1)a+b+apm+bpm+2$ {{{{( - 1)}^i}((b + {{1)}^{{p^m} + 1}} - {a^{{p^m} + 1}})} \over {a + b + {a^{{p^m}}} + {b^{{p^m}}} + 2}} is a (p − 1)-th element in Fpm. Thus, we can complete the proof.

Example 5

Take m = 2, p = 3, j = 1, i = 2. For $Tr24(δ)=0$ Tr_2^4(\delta ) = 0 , it can be verified by computer that there exist 14,301 different triples $(δ,a,b)∈F343$ (\delta ,a,b) \in F_{{3^4}}^3 satisfying that a + b + a9 + b9 ≠ 0, −2 and $b10−a10a+b+a9+b9$ {{{b^{10}} - {a^{10}}} \over {a + b + {a^9} + {b^9}}} , $(b+1)10−a10a+b+a9+b9+2$ {{{{(b + 1)}^{10}} - {a^{10}}} \over {a + b + {a^9} + {b^9} + 2}} are all square elements in F32, 2,880 different triples $(δ,a,b)∈F343$ (\delta ,a,b) \in F_{{3^4}}^3 satisfying that a + b + a9 + b9 = 0, b10a10 and −(b + 1)10 + a10 is a square element in F32, 1,440 different triples $(δ,a,b)∈F343$ (\delta ,a,b) \in F_{{3^4}}^3 satisfying that a + b + a9 + b9 = 1, (b + 1)10a10 and b10a10 is a square element in F32. These triples can make g(x) = (x9x + δ)19 + ax9 + bx a CPP of F34.

The following result is a class of CPPs for the exponent s = i(pm − 1) + 1 with any prime p, which has been given by Xu et al. [22]. Now, we restate the result for convenience and give a different proof to ensure the completeness of the paper.

Proposition 11

For two positive integers i and m, let δFp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 and aFp2m, bFp2m \{−a, −a −1} where p is any prime. Then, the polynomial g(x) = (xpmx + δ )i(pm−1)+1 + axpm + bx is a CPP over Fp2m if and only if (−1)i+1(a + b + apm + bpm) + bpm+1apm+1 ≠ 0 and (−1)i+1(a + b + apm + bpm + 2) + (b + 1) pm+1apm+1 ≠ 0.

Proof

Using the similar proof method of Proposition 10, to show that g(x) is a PP over Fp2m, it is enough to prove that for any dJ, the equation $((−1)i+1(a+b+apm+bpm)+bpm+1−apm+1)(x+δ)=d$ (( - {1)^{i + 1}}(a + b + {a^{{p^m}}} + {b^{{p^m}}}) + {b^{{p^m} + 1}} - {a^{{p^m} + 1}})(x + \delta ) = d has only one solution in J, where J is defined in Eq. (2).

When (−1)i+1(a + b + apm + bpm) + bpm+1apm+1 ≠ 0, Eq. (11) has only one solution in J, and so g(x) permutes Fp2m.

If (−1)i+1(a + b + apm + bpm) + bpm+1apm+1 = 0, let d ≠ 0, then Eq. (11) cannot hold for any x in J. This is a contradiction.

Thus, when $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 , g(x) is a PP over Fp2m if and only if (−1)i+1(a + b + apm + bpm) + bpm+1apm+1 ≠ 0.

Similarly, we can prove that g(x) + x permutes Fp2m if and only if (−1)i+1(a + b + apm + bpm + 2) + (b + 1) pm+1apm+1 ≠ 0. Thus, we can complete the proof.

It is worth noting that the complete permutation behaviour for some s = i(pm − 1) + 1 does not hold for $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 . This fact is demonstrated by taking m = 2, i = 2 and p ∈ {2,3}, that is, g(x) = (xp2x + δ )2(p2−1)+1 + axp2 + bx is not a CPP of Fp4 for p ∈ {2,3}, which can be verified by computer.

Remark 2

For b = 1 and a ∈ {0,1}, under some special i, j, the permutation behaviour of polynomials in similar form as that in Propositions 9–11 have been described in detail [5, 6, 18]. From the proof of these propositions, we can also get a lot of PPs having the form g(x) = (xpmx + δ)i(pm−1)+pj + axpm + bx with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 , which partly unifies and generalises several discovered PPs with a similar form.

We close this section by summarising the known CPPs of the same form (xpmx + δ )s + axpm + bx over Fp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0 , as shown in Table 1.

Known CPPs (xpmx + δ )s + axpm + bx over Fp2m with $Trm2m(δ)=0$ Tr_m^{2m}(\delta ) = 0

Values of p, m and s Conditions on a, b Reference
p = 2, any s a, bF2m with a + b ≠ 0,1 [26]
Odd p with pm 1(mod3), $s=k(p2m−1)3+1$ s = {{k({p^{2m}} - 1)} \over 3} + 1 , k = 1, 2. bFp2m\{−a, −a−1}, $(A+B,ωkA+B,ω2kA+B)∈Di3$ (A + B,{\omega ^k}A + B,{\omega ^{2k}}A + B) \in D_i^3 or Di × Di+1(mod3) × Di+2(mod3) and $(A'+B',ωkA'+B',ω2kA'+B')∈Dj3$ ({A^\prime} + {B^\prime},{\omega ^k}{A^\prime} + {B^\prime},{\omega ^{2k}}{A^\prime} + {B^\prime}) \in D_j^3 or Dj × Dj+1(mod3) × Dj+2(mod3) where i, j = 0, 1, 2 Proposition 1
Odd p with pm 1(mod4), $s∈{pm+12,p2m−14+1,3(p2m−1)4+1}$ s \in \{ {{{p^m} + 1} \over 2},{{{p^{2m}} - 1} \over 4} + 1,{{3({p^{2m}} - 1)} \over 4} + 1\} bFp2m\{−a, −a − 1}, $(ωA+B,−ωA+B)∈D03∪D22$ (\omega A + B, - \omega A + B) \in D_0^3 \cup D_2^2 and $(ωA'+B',−ωA'+B')∈D03∪D22$ (\omega {A^\prime} + {B^\prime}, - \omega {A^\prime} + {B^\prime}) \in D_0^3 \cup D_2^2 Proposition 2
Odd p with pm 1(mod12), $s=k(p2m−1)6+1$ s = {{k({p^{2m}} - 1)} \over 6} + 1 bFp2m \{−a, −a − 1}, (ωkA + B, −A + B, ω5kA + B), (ωkA′ + B′, −A′ + B′, ω5kA′ + B′) ∈ S Proposition 3
Odd p, pm 7(mod12), $s=k(p2m−1)6+1$ s = {{k({p^{2m}} - 1)} \over 6} + 1 , k = 1, 5 bFp2m \{−a, −a − 1}, (A + B, ω2kA + B, −ωkA + B), (A′ + B′, ω2kA′ + B′, −ωkA′ + B′) ∈ s Proposition 4
Any s (a + b)pm + (−1)1−s(a + b) = (a + b + 1) pm + (−1)1−s(a + b + 1) = 0, apm+1 ∉ {bpm+1, (b + 1) pm+1} Proposition 7
p = 2, gcd(s, 2m − 1) = 1 bF22m \{−a, −a − 1}, either a + a2m + b + b2m = 0, a2m+1 ∉ {b2m+1, (b + 1)2m+1} or a + a2m + b + b2m ≠ 0, a2m+1 = b2m+1, b + b2m + 1 = 0 Proposition 8
Odd p, odd s with gcd(s, 2m − 1) = 1 bF22m \{−a, −a − 1}, with a + apm + b + bpm = 0, apm+1 ∉ {bpm+1, (b + 1)pm+1}Or a + apm + b + bpm = −2, or a + apm + b + bpm ≠ 0, −2, apm+1 = bpm+1, b + bpm + 1 = 0 Proposition 8
p = 2, s = i(2m − 1) + 2j with 0 < j < m bF22m \{a,a + 1} either a + apm + b + b2m = 0, b2m+1a2m+1, (b + 1)2m+1a2m+1 or a + apm + b + b2m ≠ 0, $b2m+1+a2m+1a+a2m+b+b2m$ {{{b^{{2^m} + 1}} + {a^{{2^m} + 1}}} \over {a + {a^{{2^m}}} + b + {b^{2m}}}} and $(b+1)2m+1+a2m+1a+a2m+b+b2m$ {{{{(b + 1)}^{{2^m} + 1}} + {a^{{2^m} + 1}}} \over {a + {a^{{2^m}}} + b + {b^{{2^m}}}}} are not (2j − 1)-th elements in F2m Proposition 9
Odd p, s = i(pm − 1) + pj with 0 < j < m and gcd( j, 2m) = gcd( j, p − 1) = 1 bFp2m \{−a, −a − 1}, b + bpm + a + apm ≠ 0, −2 and $(−1)i(bpm+1−apm+1)b+bpm+a+apm$ {{{{( - 1)}^i}({b^{{p^m} + 1}} - {a^{{p^m} + 1}})} \over {b + {b^{{p^m}}} + a + {a^{{p^m}}}}} , $(−1)i((b+1)pm+1−apm+1)b+bpm+a+apm+2$ {{{{( - 1)}^i}((b + {{1)}^{{p^m} + 1}} - {a^{{p^m} + 1}})} \over {b + {b^{{p^m}}} + a + {a^{{p^m}}} + 2}} are all (p − 1)-th elements in Fpm or b + bpm + a + apm = 0, bpm+1apm+1, $(−1)i(b+1)pm+1−apm+12$ {( - 1)^i}{{{{(b + 1)}^{{p^m} + 1}} - {a^{{p^m} + 1}}} \over 2} is a (p − 1)-th element in Fpm or b + bpm + a + apm = −2, (b + 1) pm+1apm+1 and $(−1)i+1(bpm+1−apm+1)2$ {( - 1)^{i + 1{{({b^{{p^m} + 1}} - {a^{{p^m} + 1}})} \over 2}}} is a (p − 1)-th element in Fpm Proposition 10
Any p, s = i(pm − 1) + 1 bFp2m \{−a, −a − 1}, (−1)i+1(b + bpm + a + apm) + bpm+1apm+1 ≠ 0 and (−1)i+1(2+b + bpm + a + apm)+(b + 1) pm+1apm+1 ≠ 0 Proposition 11

CPP, complete permutation polynomial.

#### Known CPPs (xpm − x + δ )s + axpm + bx over Fp2m with Trm2m(δ)=0 Tr.m^{2m}(\delta ) = 0

Values of p, m and s Conditions on a, b Reference
p = 2, any s a, bF2m with a + b ≠ 0,1 [26]
Odd p with pm 1(mod3), s=k(p2m1)3+1 s = {{k({p^{2m}} - 1)} \over 3} + 1 , k = 1, 2. bFp2m\{−a, −a−1}, (A+B,ωkA+B,ω2kA+B)Di3 (A + B,{\omega ^k}A + B,{\omega ^{2k}}A + B) \in D_i^3 or Di × Di+1(mod3) × Di+2(mod3) and (A'+B',ωkA'+B',ω2kA'+B')Dj3 ({A^\prime} + {B^\prime},{\omega ^k}{A^\prime} + {B^\prime},{\omega ^{2k}}{A^\prime} + {B^\prime}) \in D_j^3 or Dj × Dj+1(mod3) × Dj+2(mod3) where i, j = 0, 1, 2 Proposition 1
Odd p with pm 1(mod4), s{pm+12,p2m14+1,3(p2m1)4+1} s \in \{ {{{p^m} + 1} \over 2},{{{p^{2m}} - 1} \over 4} + 1,{{3({p^{2m}} - 1)} \over 4} + 1\} bFp2m\{−a, −a − 1}, (ωA+B,ωA+B)D03D22 (\omega A + B, - \omega A + B) \in D_0^3 \cup D_2^2 and (ωA'+B',ωA'+B')D03D22 (\omega {A^\prime} + {B^\prime}, - \omega {A^\prime} + {B^\prime}) \in D_0^3 \cup D_2^2 Proposition 2
Odd p with pm 1(mod12), s=k(p2m1)6+1 s = {{k({p^{2m}} - 1)} \over 6} + 1 bFp2m \{−a, −a − 1}, (ωkA + B, −A + B, ω5kA + B), (ωkA′ + B′, −A′ + B′, ω5kA′ + B′) ∈ S Proposition 3
Odd p, pm 7(mod12), s=k(p2m1)6+1 s = {{k({p^{2m}} - 1)} \over 6} + 1 , k = 1, 5 bFp2m \{−a, −a − 1}, (A + B, ω2kA + B, −ωkA + B), (A′ + B′, ω2kA′ + B′, −ωkA′ + B′) ∈ s Proposition 4
Any s (a + b)pm + (−1)1−s(a + b) = (a + b + 1) pm + (−1)1−s(a + b + 1) = 0, apm+1 ∉ {bpm+1, (b + 1) pm+1} Proposition 7
p = 2, gcd(s, 2m − 1) = 1 bF22m \{−a, −a − 1}, either a + a2m + b + b2m = 0, a2m+1 ∉ {b2m+1, (b + 1)2m+1} or a + a2m + b + b2m ≠ 0, a2m+1 = b2m+1, b + b2m + 1 = 0 Proposition 8
Odd p, odd s with gcd(s, 2m − 1) = 1 bF22m \{−a, −a − 1}, with a + apm + b + bpm = 0, apm+1 ∉ {bpm+1, (b + 1)pm+1}Or a + apm + b + bpm = −2, or a + apm + b + bpm ≠ 0, −2, apm+1 = bpm+1, b + bpm + 1 = 0 Proposition 8
p = 2, s = i(2m − 1) + 2j with 0 < j < m bF22m \{a,a + 1} either a + apm + b + b2m = 0, b2m+1a2m+1, (b + 1)2m+1a2m+1 or a + apm + b + b2m ≠ 0, b2m+1+a2m+1a+a2m+b+b2m {{{b^{{2^m} + 1}} + {a^{{2^m} + 1}}} \over {a + {a^{{2^m}}} + b + {b^{2m}}}} and (b+1)2m+1+a2m+1a+a2m+b+b2m {{{{(b + 1)}^{{2^m} + 1}} + {a^{{2^m} + 1}}} \over {a + {a^{{2^m}}} + b + {b^{{2^m}}}}} are not (2j − 1)-th elements in F2m Proposition 9
Odd p, s = i(pm − 1) + pj with 0 < j < m and gcd( j, 2m) = gcd( j, p − 1) = 1 bFp2m \{−a, −a − 1}, b + bpm + a + apm ≠ 0, −2 and (1)i(bpm+1apm+1)b+bpm+a+apm {{{{( - 1)}^i}({b^{{p^m} + 1}} - {a^{{p^m} + 1}})} \over {b + {b^{{p^m}}} + a + {a^{{p^m}}}}} , (1)i((b+1)pm+1apm+1)b+bpm+a+apm+2 {{{{( - 1)}^i}((b + {{1)}^{{p^m} + 1}} - {a^{{p^m} + 1}})} \over {b + {b^{{p^m}}} + a + {a^{{p^m}}} + 2}} are all (p − 1)-th elements in Fpm or b + bpm + a + apm = 0, bpm+1apm+1, (1)i(b+1)pm+1apm+12 {( - 1)^i}{{{{(b + 1)}^{{p^m} + 1}} - {a^{{p^m} + 1}}} \over 2} is a (p − 1)-th element in Fpm or b + bpm + a + apm = −2, (b + 1) pm+1apm+1 and (1)i+1(bpm+1apm+1)2 {( - 1)^{i + 1{{({b^{{p^m} + 1}} - {a^{{p^m} + 1}})} \over 2}}} is a (p − 1)-th element in Fpm Proposition 10
Any p, s = i(pm − 1) + 1 bFp2m \{−a, −a − 1}, (−1)i+1(b + bpm + a + apm) + bpm+1apm+1 ≠ 0 and (−1)i+1(2+b + bpm + a + apm)+(b + 1) pm+1apm+1 ≠ 0 Proposition 11

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