Several classes of complete permutation polynomials (CPPs) in the form of (x^{pm} − x + δ)^{s} + ax^{pm} + bx over F_{p2m} 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(p^{m} − 1) + p^{j}, 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).
Let F_{pn} be the finite field with p^{n} elements where p is a prime number and n is a positive integer. A polynomial f (x) ∈ F_{pn} [x] is called a permutation polynomial (PP) if the associated mapping f : c ↦ f (c) from F_{pn} to F_{pn} is a one-to-one mapping [9]. A polynomial f (x) over F_{q} is called a complete permutation polynomial (CPP) if f (x) and f (x) + x are two PPs of F_{q}. 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 p^{n} 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
$a{\left[{\mathit{Tr}}_{m}^{n}(x)\right]}^{k}+u(c+x)({\mathit{Tr}}_{m}^{n}(x)+x)+\mathit{bx}$a{\left[ {Tr_m^n(x)} \right]^k} + u(c + x)(Tr_m^n(x) + x) + bx
and ax^{pm} + bx + h(x^{pm} ± x) over F_{p2m} were investigated where
$a,b,c,u\in {F}_{{p}^{2m}}^{*}$a,b,c,u \in F_{{p^{2m}}}^*
and h(x) ∈ F_{pn} [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 F_{32m} in the form of
${({x}^{{3}^{m}}-x+\delta )}^{\frac{{3}^{2m}-1}{3}+1}+x${({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 F_{p2m} having the form
${({x}^{{p}^{m}}+\mathit{bx}+\delta )}^{\frac{{p}^{2m-1}}{d}+1}-\mathit{bx}${({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 F_{p2m} in the form of (x^{2m} + 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(p^{m} − 1) + p^{j} or fractional exponents.
This article aims to seek out more CPPs in the form of
$${({x}^{{p}^{m}}-x+\delta )}^{s}+{\mathit{ax}}^{{p}^{m}}+\mathit{bx}$${({x^{{p^m}}} - x + \delta )^s} + a{x^{{p^m}}} + bx
over F_{pn}, where a, b,δ ∈ F_{pn}, 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 F_{p2m} with
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
. More precisely, for the fractional exponent s satisfying
$s=\frac{k({p}^{2m}-1)}{d}+1$s = {{k({p^{2m}} - 1)} \over d} + 1
or
$s=\frac{{p}^{m}+1}{2}$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 F_{p2m} into proving that the related piecewise function is a bijection on some subsets of F_{p2m}. For the exponent s = i(p^{m} − 1) + p^{j}, 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, p^{m} − 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 F_{pn} to its subfield F_{pm} is defined as
$${\mathit{Tr}}_{m}^{n}(x)=x+{x}^{{p}^{m}}+{x}^{{p}^{2m}}+\cdots +{x}^{{p}^{(n/m-1)m}},\hspace{1em}x\in {F}_{{p}^{n}}.$$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 F_{pn} defined as
$$\mathbf{J}=\{{r}^{{p}^{m}}-r|r\in {F}_{{p}^{n}}\}$${\bf J} = \{ {r^{{p^m}}} - r\left| {r \in {F_{{p^n}}}} \right.\}
Note that
${\mathit{Tr}}_{m}^{n}(\beta )=0$Tr_m^n(\beta ) = 0
for any β ∈ J.
In the following, we restate the AGW criterion for later use.
Lemma 1
( [1]) Let A, S and$\overline{S}$\overline S denote three finite sets satisfying$\left|S\right|=\left|\overline{S}\right|$\left| S \right| = \left| {\overline S } \right|
, let f : A → A,
$\overline{f}:S\to \overline{S}$\overline f :S \to \overline S
, λ : A → S and$\overline{\lambda}:A\to \overline{S}$\overline \lambda :A \to \overline S be four mappings satisfying the following diagram, that is,
$\overline{\lambda}\circ f=\overline{f}\circ \lambda $\overline \lambda \circ f = \overline f \circ \lambda
.
If λ and
$\overline{\lambda}$\overline \lambda
are two surjections, the following two statements are equivalent:
f is a permutation over A;
$\overline{f}$\overline f
is a one-to-one mapping from S to
$\overline{S}$\overline S
and f is an injection on λ^{−1}(s) for each s ∈ S. Using the AGW criterion, we can give Lemma 2, which can be applied to determine the PP behaviour of g(x) = (x^{pm} − x + δ)^{s} + ax^{pm} + bx with the conditions a + b ≠ 0 and
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
.
Lemma 2
For two positive integers m, s and a prime number p, let a, b, δ ∈ F_{p2m}with a + b ≠ 0 and${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
. Then g(x) = (x^{pm} − x + δ )^{s} + ax^{pm} + bx is a PP over F_{p2m}if and only if$$f(x)=(a+b)(x+\delta {)}^{x.{p}^{m}}-({a}^{{p}^{m}}+{b}^{{p}^{m}})(x+\delta {)}^{s}+({b}^{{p}^{m}+1}{a}^{{p}^{m}+1})(x+\delta )$$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 onJwhich is defined in Eq. (2).
Proof
Let θ(x) = (a + b)x^{pm} − (a^{pm} + b^{pm})x + (b^{pm+1} − a^{pm+1}) and
$\overline{S}=\{\theta (x)|x\in {F}_{{p}^{2m}}\}$\overline S = \{ \theta (x)\left| {x \in {F_{{p^{2m}}}}\} } \right.
. Then the diagram
For p = 2, we have J = F_{pm}. For any
$y\in \overline{S}$y \in \overline S
, we have y^{pm} = y due to
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
, thus
$\overline{S}\subseteq \mathbf{J}$\bar S \subseteq {\bf J}
. Since the degree of θ(x) is p^{m}, each image has not more than p^{m} pre-images under θ (x). Therefore,
$\left|\overline{S}\right|\ge \frac{{p}^{2m}}{{p}^{m}}={p}^{m}$\left| {\overline S } \right| \ge {{{p^{2m}}} \over {{p^m}}} = {p^m}
, and so
$\overline{S}=\mathbf{J}$\overline S = {\bf J}
.
When p is an odd prime, let
$y\in \overline{S}$y \in \overline S
; one gets that there exists some x ∈ F_{p2m} satisfying
$y=\theta (x)=(({a}^{{p}^{m}}+{b}^{{p}^{m}})x-\frac{({b}^{{p}^{m}+1}-{a}^{{p}^{m}+1})\delta}{2}{)}^{{p}^{m}}-\left(({a}^{{p}^{m}}+{b}^{{p}^{m}})x-\frac{({b}^{{p}^{m}+1}-{a}^{{p}^{m}+1})\delta}{2}\right)$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
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
, and so
$\overline{S}\subseteq \mathbf{J}$\bar S \subseteq {\bf J}
. By contrast, for any z ∈ J, due to
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
, there exist some x ∈ F_{p2m} such that z = x^{pm} − x = (a + b)ν^{pm} − (a^{pm} + b^{pm})ν + (b^{pm+1} − a^{pm+1}) where
$\nu ={x}^{{p}^{m}}+\frac{1+a+b}{{a}^{{p}^{m}}+{b}^{{p}^{m}}}x+\frac{{b}^{{p}^{m}+1}-{a}^{{p}^{m}+1}}{2({a}^{{p}^{m}}+{b}^{{p}^{m}})}$\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
$\mathbf{J}\subseteq \overline{S}${\bf J} \subseteq \bar S
. Thus
$\overline{S}=\mathbf{J}$\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 (x^{pm} − x)^{−1} (s) for any s ∈ J. Assume that g(x) = g(y) and x^{pm} − x = y^{pm} − y, then (x^{pm} − x + δ)^{s} + ax^{pm} + bx = (y^{pm} − y + δ)^{s} + ay^{pm} + by, that is, (b + a)(x − y) = 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 (x^{pm} − x)^{−1}(s) for each s ∈ J. Therefore, g(x) permutes F_{p2m} and the result can be obtained.
Lemma 3
For a primitive element ρ of F_{p2m}and a fixed element δ ∈ F_{p2m}with${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0where p is an odd prime and m is a positive integer, the set defined by Eq. (2) can be expressed as$$\mathbf{J}=\{{x}^{{p}^{m}}-x+\delta |x\in {F}_{{p}^{2m}}\}=\{{\rho}^{-\frac{{p}^{m}+1}{2}}b|b\in {F}_{{p}^{m}}\}.$${\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
${\mathbf{J}}^{\text{'}}=\{{\omega}^{-\frac{{p}^{m}+1}{2}b}|b\in {F}_{{p}^{m}}\}${{\bf J}^\prime} = \{ {\omega ^{ - {{{p^m} + 1} \over 2}b}}\left| {b \in {F_{{p^m}}}} \right.\}
. For any y ∈ J, if y = 0, we have y ∈ J′. If y ≠ 0,
${y}^{{p}^{m}-1}=-1={\rho}^{-\frac{{p}^{2m}-1}{2}}={\left({\rho}^{-\frac{{p}^{m}+1}{2}}\right)}^{{p}^{m}-1}${y^{{p^m} - 1}} = - 1 = {\rho ^{ - {{{p^{2m}} - 1} \over 2}}} = {\left( {{\rho ^{ - {{{p^m} + 1} \over 2}}}} \right)^{{p^m} - 1}}
and then
$y={\rho}^{-\frac{{p}^{m}+1}{2}b}$y = {\rho ^{ - {{{p^m} + 1} \over 2}b}}
for some b ∈ F_{pm}. Thus J ⊆ J′. Since
$\left|\mathbf{J}\right|\ge \frac{{p}^{2m}}{{p}^{m}}={p}^{m}$\left| {\bf J} \right| \ge {{{p^{2m}}} \over {{p^m}}} = {p^m}
, we have J = J′.
The proof for J = {x^{pm} − x + δ |x ∈ F_{p2m}} is similar.
For δ ∈ F_{p2m} with
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
, due to Lemma 3, we use the notation J to denote the set {x^{pm} − x | x ∈ F_{p2m}}, {x^{pm} − x + δ |x ∈ F_{p2m}} or
$\{{\rho}^{-\frac{{p}^{m}+1}{2}b}|b\in {F}_{{p}^{m}}\}$\{ {\rho ^{ - {{{p^m} + 1} \over 2}b}}\left| {b \in {F_{{p^m}}}} \right.\}
where ρ is a primitive element of F_{p2m}.
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, β ∈ F_{pn}, if α is a (p − 1)-th power in F_{pm}, the equation x^{pi}− ax^{pj} + β = 0 has not more than one solution in which is defined in Eq. (2).
CPPS over F_{p2m} with fractional exponents
This section investigates the CPPs over F_{p2m} in the same form as Eq. (1) with the fractional exponents and satisfying
${\mathit{Tr}}_{m}^{2m}(\delta )=0$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
$$\begin{array}{l}\hspace{0.17em}A=-(a+b+{a}^{{p}^{m}}+{b}^{{p}^{m}}),\hspace{1em}B={b}^{{p}^{m}+1}-{a}^{{p}^{m}+1}\hfill \\ {A}^{\text{'}}=-(a+b+{a}^{{p}^{m}}+{b}^{{p}^{m}}+2),\hspace{1em}{B}^{\text{'}}=(b+{1)}^{p}{}^{{}^{m}+1}-{a}^{{p}^{m}+1}\hfill \end{array}$$\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 p^{n} ≡ 1(mod3). Let ρ denote a primitive element of F_{pn}, then define three subsets of F_{pn} by D_{0} =< ρ^{3}>, D_{i} = ρ^{i}D_{0} for i = 1, 2, where D_{0} is the multiplicative subgroup of F_{pn} generated by ρ^{3}. Then F_{pn} = {0} ∪ D_{0} ∪ D_{1} ∪ D_{2}. Note that
${x}^{\frac{{p}^{n}-1}{3}}={\rho}^{\frac{{p}^{n}-1}{3}i}${x^{{{{p^n} - 1} \over 3}}} = {\rho ^{{{{p^n} - 1} \over 3}i}}
when x ∈ D_{i} for i = 0, 1, 2. For the sake of simplicity, let
$\omega ={\rho}^{\frac{{p}^{n}-1}{3}}$\omega = {\rho ^{{{{p^n} - 1} \over 3}}}
. Therefore, ω^{2} + ω + 1 = 0 and ω ≠ 1.
Proposition 1
For a positive integer m and any odd prime p with p^{m}≡ 1(mod3), let δ ∈ F_{p2m}with${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0and a ∈ F_{p2m}, b ∈ F_{p2m} \ {−a, −a − 1}, then the polynomial$g(x)=({x}^{{p}^{m}}-x+\delta {)}^{\frac{k({p}^{2m}-1)}{3}+1}+{\mathit{ax}}^{{p}^{m}}+\mathit{bx}$g(x) = ({x^{{p^m}}} - x + \delta {)^{{{k({p^{2m}} - 1)} \over 3} + 1}} + a{x^{{p^m}}} + bxis a CPP of for k = 1, 2 if and only if$(A+B,{\omega}^{k}A+B,{\omega}^{2k}A+B)\in {\displaystyle \underset{i=0}{\overset{2}{\cup}}}{D}_{i}\times {D}_{i}\times {D}_{i}\cup {D}_{i}\times {D}_{i+1(\text{mod}\u200a3)}\times {D}_{i+2(\text{mod}\u200a3)}$(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$({A}^{\text{'}}+{B}^{\text{'}},{\omega}^{k}{A}^{\text{'}}+{B}^{\text{'}},{\omega}^{2k}{A}^{\text{'}}+{B}^{\text{'}})\in {\displaystyle \underset{i=0}{\overset{2}{\cup}}}{D}_{i}\times {D}_{i}\times {D}_{i}\cup {D}_{i}\times {D}_{i+1(\text{mod}\u200a3)}\times {D}_{i+2(\text{mod}\u200a3)}$({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′,B′ are 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
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
and
$\frac{{p}^{2m}-1}{3}+1${{{p^{2m}} - 1} \over 3} + 1
is odd, by Lemma 2, g(x) is a PP if and only if
$f(x)=A{(x+\delta )}^{\frac{{p}^{2m}-1}{3}+1}+B(x+\delta )$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
${f}^{\text{'}}(u)=u\left({\mathit{Au}}^{\frac{{p}^{2m}-1}{3}}+B\right)${f^\prime}(u) = u\left( {A{u^{{{{p^{2m}} - 1} \over 3}}} + B} \right)
is a bijection on J where u = x + δ.
When p^{m}≡ 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 D_{0}, D_{1} and D_{2} such that each of these subsets have
$\frac{{p}^{m}-1}{3}${{{p^m} - 1} \over 3}
elements.
Note that
${f}^{\text{'}}(u)=\{\begin{array}{l}0,\hspace{1em}\text{if}\hspace{1em}u=0,\hfill \\ (A+B)u,\hspace{1em}\text{if}\hspace{1em}u\in {D}_{0},\hfill \\ (\omega A+B)u,\hspace{1em}\text{if}\hspace{1em}u\in {D}_{1},\hfill \\ ({\omega}^{2}A+B)u,\hspace{1em}\text{if}\hspace{1em}u\in {D}_{2}.\hfill \end{array}${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 + B ∈ D_{i}, then (A + B)D_{0} = D_{i}. If ωA + B ∈ D_{i+2(mod3)}, we have (ωA + B)D_{1} = D_{i} which contradicts the assumption that f′(u) is a bijection. It follows that ωA + B ∉ D_{i+2(mod3)}. When ωA + B ∈ D_{i}, we have (ωA + B)D_{1} = D_{i+2(mod3)} and then (ω^{2}A + B)D_{2} = D_{i+1(mod3)} due to the assumption that f′(u) is a bijection. Hence ω^{2}A + B ∈ D_{i}. When ωA + B ∈ D_{i+1(mod3)}, we have (ωA + B)D_{1} = D_{i+2(mod3)} and then (ω^{2}A + B)D_{2} = D_{i+1(mod3)} due to the assumption that f′(u) is a bijection. Thus ω^{2}A + B ∈ D_{i+2(mod3)}. Therefore if f′(u) is a bijection, then
$(A+B,{\omega}^{k}A+B,{\omega}^{2k}A+B)\in {\displaystyle \underset{i=0}{\overset{2}{\cup}}}{D}_{i}\times {D}_{i}\times {D}_{i}\cup {D}_{i}\times {D}_{i+1(\text{mod}\u200a3)}\times {D}_{i+2(\text{mod}\u200a3)}$(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
$(A+B,{\omega}^{k}A+B,{\omega}^{2k}A+B)\in {\displaystyle \underset{i=0}{\overset{2}{\cup}}}{D}_{i}\times {D}_{i}\times {D}_{i}\cup {D}_{i}\times {D}_{i+1(\text{mod}\u200a3)}\times {D}_{i+2(\text{mod}\u200a3)}$(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 F_{p2m} iff
$(A+B,{\omega}^{k}A+B,{\omega}^{2k}A+B)\in {\displaystyle \underset{i=0}{\overset{2}{\cup}}}{D}_{i}\times {D}_{i}\times {D}_{i}\cup {D}_{i}\times {D}_{i+1(\text{mod}\u200a3)}\times {D}_{i+2(\text{mod}\u200a3)}$(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 F_{p2m} iff (A′ + B′, ωA′ + B′, ω^{2}A′ + B′) ∈ D_{j} × D_{j} × D_{j} or D_{j} × D_{j+1(mod3)} × D_{j+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$(\delta ,a,b)\in {F}_{{3}^{4}}^{3}$(\delta ,a,b) \in F_{{3^4}}^3satisfying that${\mathit{Tr}}_{1}^{2}(\delta )=0$Tr_1^2(\delta ) = 0
, (A + B, ωA + B, ω^{2}A + B) ∈ D_{i} × D_{i} × D_{i}or D_{i} × D_{i+1(mod3)} × D_{i+2(mod3)}for i = 0, 1, 2 and$({A}^{\text{'}}+{B}^{\text{'}},{\omega}^{k}{A}^{\text{'}}+{B}^{\text{'}},{\omega}^{2k}{A}^{\text{'}}+{B}^{\text{'}})\in {\displaystyle \underset{i=0}{\overset{2}{\cup}}}{D}_{i}\times {D}_{i}\times {D}_{i}\cup {D}_{i}\times {D}_{i+1(\text{mod}\u200a3)}\times {D}_{i+2(\text{mod}\u200a3)}$({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${\mathit{Tr}}_{1}^{2}(\delta )=0$Tr_1^2(\delta ) = 0
, these are exactly all triples which can make be a CPP of F_{72}.
Next, when p^{n}≡ 1(mod4), the letter ρ denotes a primitive element of F_{pn}. In the following proposition, the multiplicative subgroup of F_{pn} generated by ρ^{4} is defined as D_{0} =< ρ^{4}>, and then let for i = 1, 2, 3. We can obtain that F_{pn} ∪ {0} ∪ D_{0} ∪ D_{1} ∪ D_{2} ∪ D_{3}. For i = 0, 1, 2, 3, if x ∈ D_{i}, we have
${x}^{\frac{{p}^{n}-1}{4}}={\rho}^{\frac{{p}^{n}-1}{4}i}={\omega}^{i}${x^{{{{p^n} - 1} \over 4}}} = {\rho ^{{{{p^n} - 1} \over 4}i}} = {\omega ^i}
where
$\omega ={\rho}^{\frac{{p}^{n}-1}{4}}$\omega = {\rho ^{{{{p^n} - 1} \over 4}}}
satisfies ω^{2} + 1 = 0 and ω ≠ 1.
Proposition 2
For a positive integer m and any odd prime p with p^{m}≡ 1(mod 4), let δ ∈ F_{p2m}with${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0and a ∈ F_{p2m}, b ∈ F_{p2m} \ {−a, −a − 1}. Let$s\in \left\{\frac{{p}^{m}+1}{2},\frac{{p}^{2m}-1}{4}+1,\frac{3\left({p}^{2m}-1\right)}{4}+1\right\}$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) = (x^{pm} − x + δ )^{s} + ax^{pm} + bx is a CPP of F_{p2m}when and only when (ωA + B, −ωA + B) ∈ D_{0} × D_{0} ∪ D_{2} × D_{2}and (ωA′ + B′, −ωA′ + B′) ∈ D_{0} × D_{0} ∪ D_{2} × D_{2}where A,B,A′,B′ are as defined in Eq. (3).
Proof
We just consider the case for
$s=\frac{{p}^{m}+1}{2}$s = {{{p^m} + 1} \over 2}
since the proofs for other cases are similar. Since p^{m}≡ 1(mod4), we have that
$\frac{{p}^{m}+1}{2}${{{p^m} + 1} \over 2}
is odd. Together with
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
, by Lemma 2, we prove that g(x) is a PP by showing that for any d ∈ J, the equation
$A{(x+\delta )}^{\frac{{p}^{m}+1}{2}}+B(x+\delta )=d$A{(x + \delta )^{{{{p^m} + 1} \over 2}}} + B(x + \delta ) = d
has only one solution in J, which is equivalent to show that
$$(x+\delta )(A{(x+\delta )}^{\frac{{p}^{m}-1}{2}}+B)=d$$(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
$f(u)=u({\mathit{Au}}^{\frac{{p}^{m}-1}{2}}+B)$f(u) = u(A{u^{{{{p^m} - 1} \over 2}}} + B)
is a bijection on J.
By Lemma 3, due to the fact that
$\frac{{p}^{m}+1}{2}${{{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 D_{1} and D_{3} such that each of these subsets have
$\frac{{p}^{m}-1}{2}${{{p^m} - 1} \over 2}
elements.
Assume that
$\frac{{p}^{m}+1}{2}\equiv 1(\text{mod}\u200a4)${{{p^m} + 1} \over 2} \equiv 1({\rm{mod}}{\kern 1pt} 4)
. By Lemma 3, we have that for any u ∈ J, there exists a unique element b ∈ F_{pm} satisfying
$u={\rho}^{\frac{-{p}^{m}+1}{2}b}$u = {\rho ^{{{ - {p^m} + 1} \over 2}b}}
. Thus, one gets
$u\in {D}_{1}\iff {b}^{\frac{{p}^{m}-1}{2}}=-1$u \in {D_1} \Leftrightarrow {b^{{{{p^m} - 1} \over 2}}} = - 1
and
$u\in {D}_{3}\iff {b}^{\frac{{p}^{m}-1}{2}}=1$u \in {D_3} \Leftrightarrow {b^{{{{p^m} - 1} \over 2}}} = 1
. Then we obtain
$$f(u)=\{\begin{array}{l}0,\hspace{1em}\text{if}\hspace{1em}u=0,\hfill \\ (\omega A+B)u,\hspace{1em}\text{if}\hspace{1em}u\in {D}_{1},\hfill \\ (-\omega A+B)u,\hspace{1em}\text{if}\hspace{1em}u\in {D}_{3}.\hfill \end{array}$$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 + b ∈ D_{0}, we have (ωA + B)D_{1} = D_{1} and then (−ωA + B)D_{3} = D_{3} due to f (u) is a bijection. Hence −ωA + B ∈ D_{0}. Similarly, when ωA + B ∈ D_{2}, we have (ωA + B)D_{1} = D_{3} and then (−ωA + B)D_{3} = D_{1} due to the assumption that f (u) is a bijection. Hence −ωA + B ∈ D_{2}. For i = 1, 3, if ωA + B ∈ D_{i}, we can obtain that (ωA + B)D_{1} equals D_{0} or D_{2}. It follows that ωA + B ∉ D_{i} for i = 1, 3 due to the assumption that f (u) is a bijection. Using the same method, we have −ωA + B ∉ D_{i} for i = 1, 3. Therefore if f (u) is a bijection, then (ωA + B, −ωA + B) ∈ D_{0} × D_{0} or D_{2} × D_{2}. Conversely, when (ωA + B, −ωA + B) ∈ D_{0} × D_{0} or D_{2} × D_{2}, we have f (u) is a bijection on J.
The case for
$\frac{{p}^{m}+1}{2}\equiv 3(\text{mod}\u200a4)${{{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) ∈ D_{0} × D_{0} or D_{2} × D_{2}. Similarly, we can prove that g(x) + x is a PP iff (ωA′ + B′, −ωA′ + B′) ∈ D_{0} × D_{0} or D_{2} × D_{2}.
Example 2
Take p = 3, we have s ∈ {5, 21, 61}. By Magma software, one can verify that there exist 9,693 different triples$(\delta ,a,b)\in {F}_{{3}^{4}}^{3}$(\delta ,a,b) \in F_{{3^4}}^3satisfying that${\mathit{Tr}}_{2}^{4}(\delta )=0$Tr_2^4(\delta ) = 0
, (ωA + B, −ωA + B) ∈ D_{0} × D_{0} ∪ D_{2} × D_{2}and (ωA′ + B′, −ωA′ + B′) ∈ D_{0} × D_{0} ∪ D_{2} × D_{2}. These (δ, a, b) are exactly all elements of${F}_{{3}^{4}}^{3}$F_{{3^4}}^3such that g(x) = (x^{9} − x + δ )^{s} + ax^{9} + bx is a CPP of${F}_{{3}^{4}}^{3}$F_{{3^4}}^3which implies that the complete permutation behaviour for these s does not hold for${\mathit{Tr}}_{2}^{4}(\delta )\ne 0$Tr_2^4(\delta ) \ne 0
.
Similarly, let n be a positive integer with p^{n}≡ 1(mod6), where p is an odd prime and let ρ denote a primitive element in F_{pn}. In the following propositions, the notation D_{0} =< ρ^{6}> represents the multiplicative subgroup of F_{pn} which is generated by ρ^{6}, and we define five disjoint subsets D_{i} = ρ^{i}D_{0} of F_{pn} for i = 1, 2, 3, 4, 5. Therefore, F_{pn} = {0} ∪ D_{0} ∪ D_{1} ∪ D_{2} ∪ D_{3} ∪ D_{4} ∪ D_{5}. Note that
${x}^{\frac{{p}^{n}-1}{6}}={\rho}^{\frac{{p}^{n}-1}{6}i}${x^{{{{p^n} - 1} \over 6}}} = {\rho ^{{{{p^n} - 1} \over 6}i}}
if x ∈ D_{i} for i = 0, 1, 2, 3, 4, 5. For the sake of simplicity, denote
$\omega =\rho \frac{{p}^{n}-1}{6}$\omega = \rho {{{p^n} - 1} \over 6}
. Therefore, ω^{2} − ω + 1 = 0 and ω ∉ 1. Define the set
$$S={\displaystyle \underset{i\in \{0,2,4\}}{\cup}}\left({D}_{i}^{3}\cup {D}_{i}\times {D}_{i+2(\text{mod}\u200a6)}\times {D}_{i+4(\text{mod}\u200a6)}\right)$$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 p^{m}≡ 1(mod12) where p is an odd prime and let δ ∈ F_{p2m}with${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
. For a ∈ F_{p2m}, b ∈ F_{p2m} \{−a, −a − 1} and k = 1, 5, the polynomial$$g(x)=({x}^{{p}^{m}}-x+\delta {)}^{\frac{k({p}^{2m}-1)}{6}+1}+{\mathit{ax}}^{{p}^{m}}+\mathit{bx}$$g(x) = ({x^{{p^m}}} - x + \delta {)^{{{k({p^{2m}} - 1)} \over 6} + 1}} + a{x^{{p^m}}} + bx
is a CPP of F_{p2m} if and only if (ω^{k}A + B, −A + B, ω^{5k}A + B) ∈ S and (w^{k}A′ + B′, −A′ + B′, ω^{5k}A′ + 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
$A{(x+\delta )}^{\frac{{p}^{2m}-1}{6}+1}+B(x+\delta )$A{(x + \delta )^{{{{p^{2m}} - 1} \over 6} + 1}} + B(x + \delta )
is a bijection on J, that is,
$f(u)=u({\mathit{Au}}^{\frac{{p}^{2m}-1}{6}}+B)$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 D_{1}, D_{3} and D_{5} such that each of these subsets have
$\frac{{p}^{m}-1}{3}${{{p^m} - 1} \over 3}
elements since
$\frac{{p}^{m}+1}{2}${{{p^m} + 1} \over 2}
odd for p^{m}≡ 1(mod12).
Note that
$$f(u)=\{\begin{array}{l}0,\hspace{1em}\text{if}\hspace{1em}u=0,\hfill \\ (\omega A+B)u,\hspace{1em}\text{if}\hspace{1em}u\in {D}_{1},\hfill \\ (-\omega A+B)u,\hspace{1em}\text{if}\hspace{1em}u\in {D}_{3},\hfill \\ \left({\omega}^{5}A+B\right)u,\hspace{1em}\text{if}\hspace{1em}u\in {D}_{5}.\hfill \end{array}$$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 + B ∈ D_{i} for any i = 1, 3, 5, then (ωA + B)D_{1} = D_{j} 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}}D_{i}. Similarly, we can prove that −A + B, ω^{5}A + B ∉ ∪_{i∈(1, 3, 5}}D_{i}. Assume that ωA + B ∈ D_{i} for any i = 0, 2, 4, then (ωA + B)D_{1} =D_{i+1(mod6)}.
If −A + B ∈ D_{i+4(mod6)}, we have (−A + B)D_{3} = D_{i+1(mod6)} which is contradictory to the assumption that f (u) is a bijection. It follows that −A + B ∉ D_{i+4(mod6)} when f (u) is a bijection. When −A + B ∈ D_{i} for any i = 0, 2, 4. Similarly, we can obtain that ω^{5}A + B ∈ D_{i} for any i = 0, 2, 4. When −A + B ∈ D_{i+2(mod6)}, we have (−A + B)D_{3} = D_{i+3(mod6)} and then ω^{5}A + B ∈ D_{i+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, ω^{5}A + B) ∈ S.
Conversely, when (A + B, ωA + B, ω^{2}A + 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, ω^{5}A + B) ∈ S. A similar argument shows that g(x) + x a PP iff (ω^{k}A′ + B′, −A′ + B′, ω^{5k}A′ + 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$(\delta ,a,b)\in {F}_{{13}^{2}}^{3}$(\delta ,a,b) \in F_{{{13}^2}}^3satisfying that${\mathit{Tr}}_{1}^{2}(\delta )=0$Tr_1^2(\delta ) = 0
, (A + B, ωA + B, ω^{2}A + B) ∈ S and (ωA + B, −A + B, ω^{5}A + B) ∈ S. When${\mathit{Tr}}_{1}^{2}(\delta )=0$Tr_1^2(\delta ) = 0
, these (δ, a, b) are exactly all elements such that the polynomial g(x) = (x^{13} − x + δ )^{29} + ax^{13} + bx is a CPP of F_{134}.
The following proposition can be shown in a similar way.
Proposition 4
For any odd prime p and a positive integer m with p^{m}≡ 7(mod12), let δ ∈ F_{p2m}with${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0and a ∈ F_{p2m}, b ∈ F_{p2m} \ {−a, −a−1}, then the polynomial$g(x)=({x}^{{p}^{m}}-x+\delta {)}^{\frac{k({p}^{2m}-1)}{6}+1}+{\mathit{ax}}^{{p}^{m}}+\mathit{bx}$g(x) = ({x^{{p^m}}} - x + \delta {)^{{{k({p^{2m}} - 1)} \over 6} + 1}} + a{x^{{p^m}}} + bxis a CPP of F_{p2m}for k = 1, 5 iff (A + B, ω^{2k}A + B, −ω^{k}A + B) ∈ S and (A′ + B′, ω^{2k}A′ + B′, −ω^{k}A′ + B′) ∈ S, where A,B,A′,B′ are 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) = (x^{pm} − x + δ)s + ax^{pm} satisfying
${\mathit{Tr}}_{m}^{2m}(\delta )=0$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 p^{2m}≡ 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 p^{m}≡ 1(modd) where m is a positive integer, let ρ denote the primitive element of${F}_{p}^{2m}$F_p^{2m}and$\omega ={\rho}^{\frac{{p}^{n}-1}{d}}$\omega = {\rho ^{{{{p^n} - 1} \over d}}}
. For i = 1, 2, ···, d − 1, let D_{0} =< ρ^{d}>, D_{i} = ρ^{i}< D_{0}>, where D_{0}is the multiplicative subgroup of${F}_{p}^{2m}$F_p^{2m}which is generated by ρ^{d}. The set T consists of all elements of the form (i_{0},i_{1},···, i_{d−1}) where the components ij are elements in {0,1,···, d −1} such that (i_{0},i_{1} +1(modd),···, i_{d−1} + d − 1(modd)) is a permutation of the integer set {0,1,···, d − 1}.
Let δ ∈ F_{p2m} with
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
and a ∈ F_{p2m}, b ∈ F_{p2m}\{−a, −a − 1}. Then the polynomial
$g(x)=({x}^{{p}^{m}}-x+\delta {)}^{\frac{k({p}^{2m}-1)}{d}+1}+{\mathit{ax}}^{{p}^{m}}+\mathit{bx}$g(x) = ({x^{{p^m}}} - x + \delta {)^{{{k({p^{2m}} - 1)} \over d} + 1}} + a{x^{{p^m}}} + bx
is a CPP of F_{p2m} for k = 1, 2, ···, d − 1 if and only if the following two conditions hold:
For an odd prime number p and positive integers m, d with d ≡ 0(mod4) and p^{m}≡ 1(modd), let ρ denote the primitive element of F_{p2m}and$\omega ={\rho}^{\frac{{p}^{n}-1}{d}}$\omega = {\rho ^{{{{p^n} - 1} \over d}}}
, then define d subsets of F_{p2m}by D_{0} =< ρ^{d}>,D_{i} = ρ^{i}D_{0}for i = 1, 2, ···, d − 1, where D_{0}is the multiplicative subgroup of F_{p2m}generated by ρ^{d}. The set T consists of all elements of the form (i_{1},i_{3},···, i_{d−1}) where the components ij are elements of {0,2,···, d − 2} such that (i_{1} + 1(modd),i_{3} + 3(modd),···, i_{d−1} + d − 1(modd)) is a permutation of the integer set {1, 3, ···, d − 1}. Let δ ∈ F_{p2m}with${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0and a ∈ F_{p2m},b ∈ F_{p2m} \{−a, −a − 1}. Then, for k = 1, 2, ···, d − 1, the polynomial$g(x)=({x}^{{p}^{m}}-x+\delta {)}^{\frac{k({p}^{2m}-1)}{d}}+{\mathit{ax}}^{{p}^{m}}+\mathit{bx}$g(x) = ({x^{{p^m}}} - x + \delta {)^{{{k({p^{2m}} - 1)} \over d}}} + a{x^{{p^m}}} + bxis a CPP of F_{p2m}if and only if the following two conditions hold:
This section proposes five classes of CPPs in the same form as in Eq. (1) over F_{p2m} with
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
. More precisely, for any positive integer s or some integers s satisfying gcd(s, p^{m} − 1) = 1, we can obtain two trivial classes of CPPs, and for the exponent s = i(p^{m} − 1) + p^{j}, 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, b ∈ F_{p2m}satisfy${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
. Then, the polynomial g(x) = (x^{pm} − x + δ )^{s} + ax^{pm} + bx is a CPP over F_{p2m}if (a + b)^{pm} + (−1)^{1−s}(a + b) = 0, (a + b + 1)^{pm} + (−1)^{1−s}(a + b + 1) = 0 and a^{pm+1} ∉ {b^{pm+1}, (b + 1)^{ pm+1}}^{.}
When gcd(s, p^{m} −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, p^{m} − 1) = 1 and a fixed element δ with${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
, let b ∈ F_{p2m}, b ∈ F_{p2m} \{−a, −a − 1} and g(x) = (x^{pm} − x + δ )^{s} + ax^{pm} + bx.
For p = 2, g(x) is a CPP over F_{22m} if either a + b + a^{2m} + b^{2m} = 0, a^{2m+1} ∉ {b^{2m+1}, (b + 1)^{2m+1}} or a + a^{2m} + b + b^{2m} ≠ 0, a^{2m+1} = b^{2m+1}, b + b^{2m} + 1 = 0.
For odd p, s, g(x) is a CPP over F_{p2m} if one of the following conditions is true:
a + b + a^{pm} + b^{pm} = 0,a^{pm+1} ≠ b^{pm+1},b + b^{pm} + 1 ≠ 0;
a + b + a^{pm} + b^{pm} = −2,a^{pm+1} = b^{pm+1},b + b^{pm} + 1 ≠ 0;
a + b + a^{pm} + b^{pm} ≠ 0,a^{pm+1} = b^{pm+1},b + b^{pm} + 1 = 0.
When s = i(p^{m} − 1) + p^{j}, one gets gcd(s, p^{m} − 1) = 1, but Proposition 8 cannot contain all possible
$(\delta ,a,b)\in {F}_{{p}^{2m}}^{3}$(\delta ,a,b) \in F_{{p^{2m}}}^3
, which can make g(x) a CPP for
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
. Thus, under some restrictions in the process of selecting the elements a, b in F_{p2m}, three classes of CPPs over F_{p2m} for s = i(p^{m} − 1) + p^{j} and
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
are discussed in the following.
The following proposition is on the assumption that s = i(2^{m} − 1) + 2^{j} for j > 0.
Proposition 9
For three positive integers m, i, j with 0 < j < m and a fixed element δ ∈ F_{2m}, let a ∈ F_{22m}and b ∈ F_{22m}\ {a,a + 1}. The polynomial g(x) = (x^{2m} + x + δ )^{i(2m−1)+2j} + ax^{2m} + bx is a CPP of F_{2m}if one of the following conditions is satisfied:
a + b + a^{2m} + b^{2m} = 0, a^{2m+1} ≠ b^{2m+1}, a^{2m+1} ≠ (b + 1)^{2m+1};
a + b + a^{2m} + b^{2m} ≠ 0,
$\frac{{a}^{{2}^{m}+1}+{b}^{{2}^{m}+1}}{a+b+{a}^{{2}^{m}}+{b}^{{2}^{m}}}${{{a^{{2^m} + 1}} + {b^{{2^m} + 1}}} \over {a + b + {a^{{2^m}}} + {b^{{2^m}}}}}
and
$\frac{{a}^{{2}^{m}+1}+{(b+1)}^{{2}^{m}+1}}{a+b+{a}^{{2}^{m}}+{b}^{{2}^{m}}}${{{a^{{2^m} + 1}} + {{(b + 1)}^{{2^m} + 1}}} \over {a + b + {a^{{2^m}}} + {b^{{2^m}}}}}
are not (2^{j} − 1)-th elements in F_{2m}.
Proof
In order to show that g(x) is a permutation of F_{22m}, we only need to show that
$$(a+b+{a}^{{2}^{m}}+{b}^{{2}^{m}})(x+\delta {)}^{i{(2}^{m}-1)+{2}^{j}}+({a}^{{2}^{m}+1}+{b}^{{2}^{m}+1})(x+\delta )=d$$(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 F_{2m} for any d ∈ F_{2m} by Lemma 2.
We can reduce Eq. (6) to
$$(a+b+{a}^{{2}^{m}}+{b}^{{2}^{m}})(x+\delta {)}^{{2}^{j}}+({a}^{{2}^{m}+1}+{b}^{{2}^{m}+1})(x+\delta )=d$$(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 + a^{2m} + b^{2m} = 0 and a^{2m+1} + b^{2m} + 1 ≠ 0, Eq. (7) has only one solution.
When a + b + a^{2m} + b^{2m} ≠ 0 and
$\frac{{a}^{{2}^{m}+1}+{b}^{{2}^{m}+1}}{a+b+{a}^{{2}^{m}}+{b}^{{2}^{m}}}${{{a^{{2^m} + 1}} + {b^{{2^m} + 1}}} \over {a + b + {a^{{2^m}}} + {b^{{2^m}}}}}
is not a (2^{j} − 1)-th element in F_{2m}, we have that (a + b + a^{2m} + b^{2m})x^{2j−1} + a^{2m+1} + b^{2m+1} = 0 has no solution in F_{2m}, and so Eq. (7) has a unique solution in F_{2m}. In this way, g(x) is a permutation of F_{22m}.
In the same way, we can get that g(x) + x is a PP if either a + b + a^{2m} + b^{2m} = 0, a^{2m+1} + (b + 1)^{2m+1} ≠ 0 or a + b + a^{2m} + b^{2m} ≠ 0 and
$\frac{{a}^{{2}^{m}+1}+{(b+1)}^{{2}^{m}+1}+1}{a+b+{a}^{{2}^{m}}+{b}^{{2}^{m}}}${{{a^{{2^m} + 1}} + {{(b + 1)}^{{2^m} + 1}} + 1} \over {a + b + {a^{{2^m}}} + {b^{{2^m}}}}}
is not a (2^{j} − 1)-th element in F_{2m}.
Therefore, g(x) is a CPP of F_{22m} if either a + b + a^{2m} + b^{2m} = 0, a^{2m+1} ≠ b^{2m+1}, a^{2m+1} ≠ (b + 1)^{2m+1} or a + b + a^{2m} + b^{2m} ≠ 0 such that
$\frac{{a}^{{2}^{m}+1}+{b}^{{2}^{m}+1}}{a+b+{a}^{{2}^{m}}+{b}^{{2}^{m}}}${{{a^{{2^m} + 1}} + {b^{{2^m} + 1}}} \over {a + b + {a^{{2^m}}} + {b^{{2^m}}}}}
and
$\frac{{a}^{{2}^{m}+1}+{(b+1)}^{{2}^{m}+1}+1}{a+b+{a}^{{2}^{m}}+{b}^{{2}^{m}}}${{{a^{{2^m} + 1}} + {{(b + 1)}^{{2^m} + 1}} + 1} \over {a + b + {a^{{2^m}}} + {b^{{2^m}}}}}
are not (2^{j} − 1)-th elements in F_{2m}.
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) ∈ F_{24} × F_{28} × F_{28}satisfying that a + a^{16} + b + b^{16} = 0, b^{17} ≠ a^{17}, (b + 1)^{17} ≠ a^{17}and 384,000 different triples (δ, a, b) ∈ F_{24} × F_{28} × F_{28}satisfying that a + a^{16} + b + b^{16} ≠ 0,
$\frac{{b}^{17}+{a}^{17}}{a+{a}^{16}+b+{b}^{16}}${{{b^{17}} + {a^{17}}} \over {a + {a^{16}} + b + {b^{16}}}}and$\frac{{(b+1)}^{17}+{a}^{17}}{a+{a}^{16}+b+{b}^{16}}${{{{(b + 1)}^{17}} + {a^{17}}} \over {a + {a^{16}} + b + {b^{16}}}}are not cube elements in F_{24}. These (δ, a, b) are exactly all elements in F_{24} × F_{28} × F_{28}such that g(x) = (x^{16} + x + δ )^{49} + ax^{16} + bx is a CPP of F_{28}. Furthermore, it can be verified that g(x) cannot be a CPP for δ ∉ F_{24}.
For any old prime p and s = i(p^{m} − 1) + p^{j} with j > 0, CPPs over F_{p2m} are discussed in the next proposition.
Proposition 10
For three positive integers m, i, j with 0 < j < m and a fixed element δ ∈ F_{p2m}with${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0let b ∈ F_{p2m}\{−a, −a − 1} and g(x) = (x^{pm} − x + δ)^{i(pm−1)+pj} + ax^{pm} + bx where gcd( j,2m) = gcd( j, p − 1) = 1 and the prime p is odd. Hence g(x) is a CPP of F_{p2m}if one of the following conditions is satisfied:
a + b + a^{pm} + b^{pm} ≠ 0, −2 and
$\frac{(-1)({b}^{{p}^{m}+1}-{a}^{{p}^{m}+1}}{a+b+{a}^{{p}^{m}}+{b}^{{p}^{m}}}${{( - 1)({b^{{p^m} + 1}} - {a^{{p^m} + 1}}} \over {a + b + {a^{{p^m}}} + {b^{{p^m}}}}}
,
$\frac{{(-1)}^{i}((b+{1)}^{{p}^{m}+1}-{a}^{{p}^{m}+1})}{a+b+{a}^{{p}^{m}}+{b}^{{p}^{m}}+2}${{{{( - 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 F_{pm} ;
a + b + a^{pm} + b^{pm} = 0, b^{pm+1} ≠ a^{pm+1},
${(-1)}^{i}\frac{{(b+1)}^{{p}^{m}+1}-{a}^{{p}^{m}+1}}{2}${( - 1)^i}{{{{(b + 1)}^{{p^m} + 1}} - {a^{{p^m} + 1}}} \over 2}
is a (p − 1)-th element in F_{pm} ;
a + b + a^{pm} + b^{pm} = −2, (b + 1)^{ pm+1} ≠ a^{pm+1} and
${(-1)}^{i+1}\frac{({b}^{{p}^{m}+1}-{a}^{{p}^{m}+1})}{2}${( - 1)^{i + 1}}{{({b^{{p^m} + 1}} - {a^{{p^m} + 1}})} \over 2}
is a (p − 1)-th element in F_{pm}.
Proof
To obtain that is a PP, we only need to show that for any d ∈ J, the equation
$$(a+b)(x+\delta {)}^{{p}^{m}(i({p}^{m}-1)+{p}^{j})}-({a}^{{p}^{m}}+{b}^{{p}^{m}})(x+\delta {)}^{i({p}^{m}-1)+{p}^{j}}+({b}^{{p}^{m}+1}-{a}^{{p}^{m}+1})(x+\delta )=d$$(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
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
and x ∈ J, 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
$${(-1)}^{i+1}(a+b+{a}^{{p}^{m}}+{b}^{{p}^{m}})(x+\delta {)}^{{p}^{j}}+({b}^{{p}^{m}+1}-{a}^{{p}^{m}+1})(x+\delta )=d$${( - 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 + a^{pm} + b^{pm} = 0, for b^{pm+1} − a^{pm+1} ≠ 0, Eq. (9) has a unique solution
$x=\frac{d}{{b}^{{p}^{m}+1}-{a}^{{p}^{m}+1}}-\delta $x = {d \over {{b^{{p^m} + 1}} - {a^{{p^m} + 1}}}} - \delta
. When a + b + a^{pm} + b^{pm} ≠ 0, Eq. (9) can be simplified to
$${x}^{{p}^{j}}-\frac{{(-1)}^{i}\left({b}^{{p}^{m}+1}-{a}^{{p}^{m}+1}\right)}{a+b+{a}^{{p}^{m}}+{b}^{{p}^{m}}}x+{\delta}^{{p}^{j}}+\frac{{(-1)}^{i+1}\left({b}^{{p}^{m}+1}-{a}^{{p}^{m}+1}\right)}{a+b+{a}^{{p}^{m}}+{b}^{{p}^{m}}}+\frac{{(-1)}^{i}d}{a+b+{a}^{{p}^{m}}+{b}^{{p}^{m}}}=0$${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
$\frac{{(-1)}^{i}({b}^{{p}^{m}+1}-{a}^{{p}^{m}+1})}{a+b+{a}^{{p}^{m}}+{b}^{{p}^{m}}}${{{{( - 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 F_{pm}. Thus g(x) is a PP of F_{p2m} under the conditions that either a + b + a^{pm} + b^{pm} = 0, b^{pm+1} − a^{pm+1} ≠ 0 or a + b + a^{pm} + b^{pm} ≠ 0,
$\frac{{(-1)}^{i}({b}^{{p}^{m}+1}-{a}^{{p}^{m}+1})}{a+b+{a}^{{p}^{m}}+{b}^{{p}^{m}}}${{{{( - 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 F_{pm}.
In the same way, one can show that g(x) + x is a PP of F_{p2m} under the conditions that either a + b + a^{pm} + b^{pm} + 2 = 0, (b + 1)^{ pm+1} − a^{pm+1} ≠ 0 or a + b + a^{pm} + b^{pm} ≠ −2,
$\frac{{(-1)}^{i}((b+{1)}^{{p}^{m}+1}-{a}^{{p}^{m}+1})}{a+b+{a}^{{p}^{m}}+{b}^{{p}^{m}}+2}${{{{( - 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 F_{pm}. Thus, we can complete the proof.
Example 5
Take m = 2, p = 3, j = 1, i = 2. For${\mathit{Tr}}_{2}^{4}(\delta )=0$Tr_2^4(\delta ) = 0
, it can be verified by computer that there exist 14,301 different triples$(\delta ,a,b)\in {F}_{{3}^{4}}^{3}$(\delta ,a,b) \in F_{{3^4}}^3satisfying that a + b + a^{9} + b^{9} ≠ 0, −2 and$\frac{{b}^{10}-{a}^{10}}{a+b+{a}^{9}+{b}^{9}}${{{b^{10}} - {a^{10}}} \over {a + b + {a^9} + {b^9}}}
,
$\frac{{(b+1)}^{10}-{a}^{10}}{a+b+{a}^{9}+{b}^{9}+2}${{{{(b + 1)}^{10}} - {a^{10}}} \over {a + b + {a^9} + {b^9} + 2}}are all square elements in F_{32}, 2,880 different triples$(\delta ,a,b)\in {F}_{{3}^{4}}^{3}$(\delta ,a,b) \in F_{{3^4}}^3satisfying that a + b + a^{9} + b^{9} = 0, b^{10} ≠ a^{10}and −(b + 1)^{10} + a^{10}is a square element in F_{32}, 1,440 different triples$(\delta ,a,b)\in {F}_{{3}^{4}}^{3}$(\delta ,a,b) \in F_{{3^4}}^3satisfying that a + b + a^{9} + b^{9} = 1, (b + 1)^{10} ≠ a^{10}and b^{10} − a^{10}is a square element in F_{32}. These triples can make g(x) = (x^{9} − x + δ)^{19} + ax^{9} + bx a CPP of F_{34}.
The following result is a class of CPPs for the exponent s = i(p^{m} − 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 δ ∈ F_{p2m}with${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0and a ∈ F_{p2m}, b ∈ F_{p2m} \{−a, −a −1} where p is any prime. Then, the polynomial g(x) = (x^{pm} − x + δ )^{i(pm−1)+1} + ax^{pm} + bx is a CPP over F_{p2m}if and only if (−1)^{i+1}(a + b + a^{pm} + b^{pm}) + b^{pm+1} − a^{pm+1} ≠ 0 and (−1)^{i+1}(a + b + a^{pm} + b^{pm} + 2) + (b + 1)^{ pm+1} − a^{pm+1} ≠ 0.
Proof
Using the similar proof method of Proposition 10, to show that g(x) is a PP over F_{p2m}, it is enough to prove that for any d ∈ J, the equation
$$((-{1)}^{i+1}(a+b+{a}^{{p}^{m}}+{b}^{{p}^{m}})+{b}^{{p}^{m}+1}-{a}^{{p}^{m}+1})(x+\delta )=d$$(( - {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 + a^{pm} + b^{pm}) + b^{pm+1} − a^{pm+1} ≠ 0, Eq. (11) has only one solution in J, and so g(x) permutes F_{p2m}.
If (−1)^{i+1}(a + b + a^{pm} + b^{pm}) + b^{pm+1} − a^{pm+1} = 0, let d ≠ 0, then Eq. (11) cannot hold for any x in J. This is a contradiction.
Thus, when
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
, g(x) is a PP over F_{p2m} if and only if (−1)^{i+1}(a + b + a^{pm} + b^{pm}) + b^{pm+1} − a^{pm+1} ≠ 0.
Similarly, we can prove that g(x) + x permutes F_{p2m} if and only if (−1)^{i+1}(a + b + a^{pm} + b^{pm} + 2) + (b + 1)^{ pm+1} − a^{pm+1} ≠ 0. Thus, we can complete the proof.
It is worth noting that the complete permutation behaviour for some s = i(p^{m} − 1) + 1 does not hold for
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
. This fact is demonstrated by taking m = 2, i = 2 and p ∈ {2,3}, that is, g(x) = (x^{p2} − x + δ )^{2(p2−1)+1} + ax^{p2} + bx is not a CPP of F_{p4} 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) = (x^{pm} − x + δ)^{i(pm−1)+pj} + ax^{pm} + bx with
${\mathit{Tr}}_{m}^{2m}(\delta )=0$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 (x^{pm} − x + δ )^{s} + ax^{pm} + bx over F_{p2m} with
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
, as shown in Table 1.
Known CPPs (x^{pm} − x + δ )^{s} + ax^{pm} + bx over F_{p2m} with
${\mathit{Tr}}_{m}^{2m}(\delta )=0$Tr_m^{2m}(\delta ) = 0
Akbary, A., Ghioca, D., & Wang, Q. (2011). On constructing permutations of finite fields. Finite Fields Appl. 17, 51–67.AkbaryA.GhiocaD.WangQ.2011On constructing permutations of finite fieldsSearch in Google Scholar
Bartoli, D., Giulietti, M., Quoos, L., & Zini, G. (2017). Complete permutation polynomials from exceptional polynomials. J. Number Theory. 176, 46–66.BartoliD.GiuliettiM.QuoosL.ZiniG.2017Complete permutation polynomials from exceptional polynomialsSearch in Google Scholar
Diffie, W., & Ledin, G. (translators). (2008). SMS4 encryption algorithm for wireless networks. From https://eprint.iacr.org/2008/329.pdf.DiffieW.LedinG.(translators).2008Search in Google Scholar
Feng, D., Feng, X., & Zhang, W., et al. (2011). Loiss: a byte-oriented stream cipher. IWCC’11 Proceedings of the Third international conference on Coding and Cryptology, Springer, 109–125.FengD.FengX.ZhangW.2011Search in Google Scholar
Gupta, R., & Sharma, R. K. (2018). Further results on permutation polynomials of the form (x^{pm} − x + δ)^{s} + x over F_{p2m}. Finite Fields Appl. 50, 196–208.GuptaR.SharmaR. K.2018Further results on permutation polynomials of the form (x^{pm} − x + δ)^{s} + x over F_{p2m}Search in Google Scholar
Li, N., Helleseth, T., & Tang, X. (2013). Further results on a class of permutation polynomials over finite fields. Finite Fields Appl. 22, 16–23.LiN.HellesethT.TangX.2013Further results on a class of permutation polynomials over finite fieldsSearch in Google Scholar
Li, L., Li, C., Li, C., & Zeng, X. (2019). New classes of complete permutation polynomials. Finite Fields Appl. 55, 177–201.LiL.LiC.LiC.ZengX.2019New classes of complete permutation polynomialsSearch in Google Scholar
Li, L., Wang, S., Li, C., & Zeng, X. (2018). Permutation polynomials (x^{pm} − x + δ )^{s1} + (x^{pm} − x + δ )^{s2} + x over F_{pn}, Finite Fields Appl. 51, 31–61.LiL.WangS.LiC.ZengX.2018Permutation polynomials (x^{pm} − x + δ )^{s1} + (x^{pm} − x + δ )^{s2} + x over F_{pn}Search in Google Scholar
Lidl, R., & Niederreiter, H. (1997). Finite Fields, second ed. Encyclopedia of Mathematics and its Applications, vol. 20, Cambridge University Press, Cambridge.LidlR.NiederreiterH.1997Search in Google Scholar
Mann, H. B. (1942). The construction of orthogonal Latin squares. Annals of Mathematical Statistics 13 (4), 418–423.MannH. B.1942The construction of orthogonal Latin squaresSearch in Google Scholar
Mittenthal, L. (1995). Block substitutions using orthomorphic mappings. Adv. in Appl. Math. 16 (10), 59–71.MittenthalL.1995Block substitutions using orthomorphic mappingsSearch in Google Scholar
Mileva, A., & Markovski, S. (2012). Shapeless quasigroups derived by Feistel orthomorphisms. Glasnik Matematicki 47 (67), 333–349.MilevaA.MarkovskiS.2012Shapeless quasigroups derived by Feistel orthomorphismsSearch in Google Scholar
Niederreiter, H., & Robinson, K. H. (1982). Complete mappings of finite fields. J. Aust. Math. Soc. A 33(2), 197–212.NiederreiterH.RobinsonK. H.1982Complete mappings of finite fieldsSearch in Google Scholar
Ongan, P., & Temür, B. G. (2019). Some permutations and complete permutation polynomials over finite fields. Turk. J. Math., 43, 2154–2160.OnganP.TemürB. G.2019Some permutations and complete permutation polynomials over finite fieldsSearch in Google Scholar
Samardjiska, S., & Gligoroski, D. (2017). Quadratic permutation polynomials, complete mappings and mutually orthogonal Latin squares. Math. Slovaca, 67(5), 1129–1146.SamardjiskaS.GligoroskiD.2017Quadratic permutation polynomials, complete mappings and mutually orthogonal Latin squaresSearch in Google Scholar
Schnorr, C. P., & Vaudenay, S. (1995). Black box cryptanalysis of hash networks based on multipermutations. Advances in Cryptology-Eurocrypt’94, Springer, 47–57.SchnorrC. P.VaudenayS.1995Black box cryptanalysis of hash networks based on multipermutationsSearch in Google Scholar
Tu, Z., Zeng, X., & Hu, L. (2014). Several classes of complete permutation polynomials. Finite Fields Appl. 25, 182–193.TuZ.ZengX.HuL.2014Several classes of complete permutation polynomialsSearch in Google Scholar
Tu, Z., Zeng, X., & Jiang, Y. (2015). Two classes of permutation polynomials having the form (x^{2m} + x + δ )^{s} + x. Finite Fields Appl. 31, 12–24.TuZ.ZengX.JiangY.2015Two classes of permutation polynomials having the form (x^{2m} + x + δ )^{s} + xSearch in Google Scholar
Tuxanidy, A., & Wang, Q. (2017). Compositional inverses and complete mappings over finite fields. Discrete Appl. Math. 217, 318–329.TuxanidyA.WangQ.2017Compositional inverses and complete mappings over finite fieldsSearch in Google Scholar
Wu, G., Li, N., Helleseth, T., & Zhang, Y. (2015). Some classes of complete permutation polynomials over F_{q}. Sci. China math. 58, 2081–2094.WuG.LiN.HellesethT.ZhangY.2015Some classes of complete permutation polynomials over F_{q}Search in Google Scholar
Wu, B., & Lin, D. (2015). On constructing complete permutation polynomials over finite fields of even characteristic. Discrete Applied Mathematics 184, 213–222.WuB.LinD.2015On constructing complete permutation polynomials over finite fields of even characteristicSearch in Google Scholar
Xu, X., Feng, X., & Zeng, X. (2019). Complete permutation polynomials with the form (x^{pm} − x + δ )^{s} + ax^{pm} + bx over F_{pn}. Finite Fields Appl., 57, 309–343.XuX.FengX.ZengX.2019Complete permutation polynomials with the form (x^{pm} − x + δ )^{s} + ax^{pm} + bx over F_{pn}Search in Google Scholar
Xu, X., Li, C., Zeng, X., & Helleseth, T. (2018). Constructions of complete permutation polynomials. Designs, Codes and Cryptography, 86(12), 2869–2892.XuX.LiC.ZengX.HellesethT.2018Constructions of complete permutation polynomialsSearch in Google Scholar
Yuan, P., & Zheng, Y. (2015). Permutation polynomials from piecewise functions. Finite Fields Appl. 35, 215–230.YuanP.ZhengY.2015Permutation polynomials from piecewise functionsSearch in Google Scholar
Zheng, D., & Chen, Z. (2017). More classes of permutation polynomials of the form (x^{pm} − x + δ )^{s} + L(x). App Algebra Engrg Comm Comp. 28(3), 215–223.ZhengD.ChenZ.2017More classes of permutation polynomials of the form (x^{pm} − x + δ )^{s} + L(x)Search in Google Scholar
Zheng, Y., Yuan, P., & Pei, D. (2016). Large classes of permutation polynomials over F_{q2}. Designs Codes and Cryptography, 81(3), 505–521.ZhengY.YuanP.PeiD.2016Large classes of permutation polynomials over F_{q2}Search in Google Scholar
Zhou, Y., & Qu, L. (2017). Constructions of negabent functions over finite fields. Cryptography and Communications, 9(2), 165–180.ZhouY.QuL.2017Constructions of negabent functions over finite fieldsSearch in Google Scholar