Planar Polynomials arising from Linearized polynomials

TL;DR

This paper constructs and classifies a family of planar polynomials over finite fields, revealing their structure through algebraic curves and providing a comprehensive understanding of their properties.

Contribution

It introduces a new class of planar polynomials derived from linearized polynomials and fully classifies the parameter pairs that produce planar functions.

Findings

Complete classification of pairs (A,B) for planarity

Connection between planar polynomials and algebraic curves

New family of planar polynomials over finite fields

Abstract

In this paper we construct planar polynomials of the type $f_{A,B}(x)=x(x^{q^2}+Ax^{q}+Bx)\in \mathbb{F}_{q^3}[x]$, with $A,B \in \mathbb{F}_{q}$. In particular we completely classify the pairs $(A,B)\in \mathbb{F}_{q}^2$ such that $f_{A,B}(x)$ is planar using connections with algebraic curves over finite fields.

Equations

$$D_{f,a}:{\mathbb{F}}_{q^n}\to {\mathbb{F}}_{q^n},D_{f,a}(x)=f(x+a)-f(x)$$

$$||\mathcal{X}({\mathbb{F}}_q)|-q-1)|\le (d-1)(d-2)\sqrt{q},$$

$$\sum _{i,j}{a}_{ij}{X}^{p^i+p^j},$$

$$f_{A,B}(x)=x(x^{q^2}+A{x}^q+Bx),$$

$$f_{A,B}(x+C)-f_{A,B}(x)-f_{A,B}(C)=C{x}^{q^2}+AC{x}^q+(C^{q^2}+A{C}^q+2BC)x$$

$$\left(\begin{array}[]{ccc}AC^{q}+2BC+C^{q^{2}}&AC&C\\ C^{q}&AC^{q^{2}}+2BC^{q}+C&AC^{q}\\ AC^{q^{2}}&C^{q^{2}}&AC+2BC^{q^{2}}+C^{q}\\ \end{array}\right)$$

$$\left(\begin{array}[]{ccc}AY+2BX+T&AX&X\\ Y&AT+2BY+X&AY\\ AT&T&AX+2BT+Y\\ \end{array}\right),$$

$$\det (M_{A,B,C})=G_{A,B}(C,C^q,C^{q^2}).$$

$$F_{A,B}(X,Y)$$

$$B(A+2Bb+b^2)(Ab+1)$$

$$B(A+a)(A{a}^2+2Ba+1)$$

$$A^{3}b+A^{2}Ba+A^{2}B{b}^2+4A{B}^{2}ab+2A{B}^2+3ABa{b}^2$$

$$+4B^{3}b+2B^{2}a+2B^2{b}^2+2Bab+B+b$$

$$A^{3}a+2A^{2}Bab+A^{2}B+2A{B}^2{a}^2+2A{B}^{2}b+3AB{a}^{2}b$$

$$+4B^{3}a+4B^{2}ab+2B^2+B{a}^2+Bb+a$$

$$(Y+X+1)\mid F_{A,B}(X,Y).$$

$$F_{A,B}(X,Y)=(A^2+AX+Y)(A^{2}Y+A+X)(A^{2}X+AY+1).$$

$$F_{A,B}(X,Y)=(B^2+BY+X)(B^{2}Y+BX+1)(B^{2}X+B+Y).$$

$$(2X-Y-1+\alpha (Y-1))\mid F_{A,B}(X,Y),$$

$$2B(2BY+X)XY+2B{Y}^2+(8B^3+2)XY+4B^2{X}^2+2B(2BY+X).$$

$$\left(\begin{array}[]{ccc}\gamma&\beta&\alpha\\ \alpha&\gamma&\beta\\ \beta&\alpha&\gamma\\ \end{array}\right)$$

$$f_{A,B}(x)=x(x^{q^2}+A{x}^q+Bx),$$

$$\det (M_{A,B,C})=(A^2{C}^{q^2}+AC+C^q)(A^2{C}^q+A{C}^{q^2}+C)(A^{2}C+A{C}^q+C^{q^2}).$$

$$\det (M_{A,B,C})=(B^2{C}^{q^2}+B{C}^q+C)(B^2{C}^q+BC+C^{q^2})(B^{2}C+B{C}^{q^2}+C^q).$$

Full text

Keywords: Planar polynomials; Perfect nonlinear functions; Finite fields

MSC 2010 Codes: 05B10, 05B25, 51A35, 94A60

1. Introduction

Let $q$ be a power of a prime, $n\geq 1$, $\mathbb{F}_{q^{n}}$ be the finite field with $q^{n}$ elements and $\mathbb{F}_{q^{n}}^{*}=\mathbb{F}_{q^{n}}\setminus\{0\}$. A function $f:\mathbb{F}_{q^{n}}\to\mathbb{F}_{q^{n}}$ is called planar if for each $a\in\mathbb{F}_{q^{n}}^{*}$, the difference function

[TABLE]

is a permutation of $\mathbb{F}_{q^{n}}$. Functions satisfying the previous condition exist only for $q$ odd, since for $q$ even $D_{f,a}(x)=D_{f,a}(x+a)$ holds. In the even characteristic case, functions $f$ such that $D_{f,a}$ is a $2$-to-$1$ map are called almost perfect nonlinear (APN) and they are connected with the construction of S-boxes in block ciphers [15, 16]. In the odd characteristic case, planar functions, also called perfect nonlinear functions (or 1-differentiable), are never bijections, since the corresponding derivative $D_{f,a}$ is bijective and there exists a unique $x\in\mathbb{F}_{q^{n}}$ such that $f(x+a)-f(x)=0$. Planar functions in odd characteristic achieve the best possible differential properties, which makes them useful in the construction of cryptographic protocols, in particular for the design of S-boxes in block chypers; see [15, 16]. Such functions have close connections with projective planes and have been investigated since the seminal paper [6] where the authors showed that, considering two groups $G$ and $H$ of order $n$, every planar function $f$ from $G$ into $H$ gives rise to an affine plane $S(G,H,f)$ of order $n$. Finally, these functions are strictly related to the construction of relative difference sets [10], optimal constant-composition codes [8], secret sharing schemes arising from certain linear codes [2], signal sets with good correlation properties [7], finite semifields [3]. More recently, Zhou [19] defined a natural analogue of planar functions for even characteristic: If $q$ is even, a function $f:\mathbb{F}_{q}\to\mathbb{F}_{q}$ is planar if, for each nonzero $a\in\mathbb{F}_{q}$, the function $x\to f(x+a)+f(x)+ax$ is a permutation of $\mathbb{F}_{q}$. As shown by Zhou [19] and Schmidt and Zhou [18], such planar functions have similar properties and applications as their odd characteristic counterparts.

In the past years, many papers have been devoted to existence and non-existence results for planar mappings, using a variety of methods; see [4, 11, 13, 14, 5, 1, 17].

A fundamental tool in our investigations is the Hasse-Weil bound.

Theorem 1.1**.**

Let $\mathcal{X}$ be an absolutely irreducible projective curve defined over $\mathbb{F}_{q}$. Then the number $\mathcal{X}(\mathbb{F}_{q})$ of $\mathbb{F}_{q}$-rational points of $\mathcal{X}$ satisfies

[TABLE]

where $d$ is the degree of $\mathcal{X}$.

In this paper we investigate the planarity of $xL(x)$, where $L$ is a linearized polynomial. In this direction, the following results provides a necessary condition.

Proposition 1.2**.**

[14, Proposition 1]** Let $L_{1},L_{2}:\mathbb{F}_{p^{n}}\to\mathbb{F}_{p^{n}}$ be $\mathbb{F}_{q}$-linear mappings. If the mapping $L_{1}\cdot L_{2}:\mathbb{F}_{p^{n}}\to\mathbb{F}_{p^{n}}$ is planar then necessarily the mappings $L_{1}$ and $L_{2}$ are bijective on $\mathbb{F}_{p^{n}}$.

Polynomial of the type $xL(x)$ belong to the family of so called Dembowski-Ostrom polynomials, that is polynomials of the type

[TABLE]

see [6]. It has been conjectured for many years that all planar polynomials belong to such family. Counterexamples have been provided in [4], where the authors prove that the monomial $X^{(3^{\alpha}+1)/2}$ is planar over $\mathbb{F}_{3^{e}}$ if and only if $GCD(\alpha,2e)=1$. In the monomial case, the Dembowski-Ostrom conjecture remains still open for $p\geq 5$.

There are different possibile definitions of equivalence between planar functions. Two functions $F$ and $F^{\prime}$ from $\mathbb{F}_{q}$ to itself are called extended affine equivalent (EA-equivalent) if $F=A_{1}\circ F\circ A_{2}+A$, where the mappings $A,A_{1},A_{2}$ are affine, and $A_{1}$, $A_{2}$ are permutations of $\mathbb{F}_{q}$. The following theorem lists the currently known families of EA-inequivalent planar functions.

Theorem 1.3**.**

Let $p$ be an odd prime number. The following are the currently known EA-inequivalent planar functions.

1. (1)

$x^{2}$* in $\mathbb{F}_{p^{n}}$;* 2. (2)

$x^{p^{k}+1}$* in $\mathbb{F}_{p^{n}}$, $k\leq n/2$ and $n/GCD(k,n)$ odd [4, 6];* 3. (3)

$x^{10}+x^{6}-x^{2}$* in $\mathbb{F}_{3^{n}}$, $n\geq 5$ odd [4];* 4. (4)

$x^{10}-x^{6}-x^{2}$* in $\mathbb{F}_{3^{n}}$, $n\geq 5$ odd [9];* 5. (5)

$x^{p^{s}+1}-u^{p^{k}-1}x^{p^{k}+p^{2k+s}}$* in $\mathbb{F}_{p^{3k}}$, where $GCD(k,3)=1$, $k\equiv s\pmod{3}$, $s\neq k$ and $k/GCD(k,s)$ odd, and $u$ a primitive element of $\mathbb{F}_{p^{3k}}$ [20, Theorem 1];* 6. (6)

$x^{(3^{k}+1)/2}$* in $\mathbb{F}_{3^{n}}$, $k\geq 3$ odd and $GCD(k,n)=1$ [4, 12].*

In this paper we investigate the planarity of $xL_{3}(x)$ on $F_{q^{3}}$, a particular example of Dembowski-Ostrom polynomials. We make use of the study of curves associated with polynomials $xL_{3}(x)$. The main result of this paper is the following.

Theorem**.**

Let $q$ be a power of an odd prime and consider the function $f_{A,B}(x)\in\mathbb{F}_{q^{3}}[x]$ of the type

[TABLE]

where $A,B\in\mathbb{F}_{q}$. Then, $f_{A,B}$ is planar if and only if one of the following condition holds

• •

$A^{3}-2AB+1=0$* and $A^{3}\neq\pm 1$;*

• •

$A=B^{2}$, $B^{3}\neq 1$;

• •

$B=0$, $A^{3}+1\neq 0$.

2. Construction of planar polynomials of type $x(x^{q^{2}}+Ax^{q}+Bx)\in\mathbb{F}_{q^{3}}[x],A,B\in\mathbb{F}_{q}$

In this section we provide the complete classification of planar polynomial of the type $f_{A,B}(x)=x(x^{q^{2}}+Ax^{q}+Bx)\in\mathbb{F}_{q^{3}}[x]$, where $A,B\in\mathbb{F}_{q}$. Let us denote by $\Box_{q}$ and $\overline{\mathbb{F}_{q}}$ the set of squares in $\mathbb{F}_{q}$ and the algebraic closure of $\mathbb{F}_{q}$, respectively.

Note that $f_{A,B}(x)$ is planar if and only if for each $C\in\mathbb{F}_{q^{3}}^{*}$ the linearized polynomial

[TABLE]

is a permutation polynomial. This happens if and only if the following matrix $M_{A,B,C}$

[TABLE]

has rank $3$ for each choice of $C\in\mathbb{F}_{q^{3}}^{*}$. Let $G_{A,B}(X,Y,T)$ be the determinant of

[TABLE]

and define $F_{A,B}(X,Y):=G_{A,B}(X,Y,1)$. Clearly,

[TABLE]

It is readily seen that

[TABLE]

Our main approach involves specific algebraic curves of degree three, whose reducibility is investigated in the following propositions.

Proposition 2.1**.**

Let $\mathcal{C}_{A,B}$ be a curve defined by $F_{A,B}(X,Y)=0$, $A,B\in\mathbb{F}_{q}^{*}$, with

[TABLE]

If $\mathcal{C}$ has a line of the form $Y-aX-b$ as a component, then one of the following holds

1. (1)

$A-2B+1=0$* or $(A,B)=(1,-1/2)$; or* 2. (2)

$A^{3}-2AB+1=0$* and $A^{3}+1\neq 0$; or* 3. (3)

$A^{2}+A+1=0$, $B\in\{A^{2},-A^{2}/2\}$ and $-3\in\square_{q}$; or 4. (4)

$A=B^{2}$; or 5. (5)

$A^{2}+2AB-A+4B^{2}+2B+1=0$* and $-3\in\square_{q}$.*

Proof.

First we deal with the case $Y-aX-b=Y+X+1$. If such a line is a component, by direct computations either $A-2B+1=0$ or $(A,B)=(1,-1/2)$. On the other hand, it easily seen that if $A-2B+1=0$ or $(A,B)=(1,-1/2)$ then $Y+X+1$ is a factor of $F_{A,B}(X,Y)$.

By direct computations, $F(X,aX+b)\equiv 0$ if and only if

[TABLE]

We distinguish two cases.

• •

$b=-1/A$.

• –

If $a=-A$ then $A^{3}-2AB+1=0$ or $A^{2}B+A-2B^{2}=A^{2}-2AB^{2}+B=0$. In this last case, either $(A,B)\in\{(1,1),(1,-1/2)\}$ (already done) or $A^{2}+A+1=0$ (and then $-3\in\square_{q}$) and $B=A^{2},-A^{2}/2$.

• –

If $Aa^{2}+2Ba+1=0$ then $A=B^{2}$ or $A^{3}-2AB+1=0$. In this last case $A^{3}+1\neq 0$ otherwise $B=0$.

• •

$A+2Bb+b^{2}=0$.

• –

If $a=-A$ then $A^{3}-2AB+1=0$ or $A=B^{2}$.

• –

If $Aa^{2}+2Ba+1=0$ then $A-2B+1=0$ (already done), or $A=B^{2}$, or $A^{2}+2AB-A+4B^{2}+2B+1=0$ or $A^{3}-2AB+1=0$. Note that $A^{2}+2AB-A+4B^{2}+2B+1=0$ implies $B=(-A-1+\alpha(A-1))/4$, with $\alpha^{2}=-3$. Since $A=1$ would imply $(A,B)\in\{(1,1),(1,-1/2)\}$ (already done) then $A\neq 1$ and $-3\in\square_{q}$.

∎

Proposition 2.2**.**

Suppose that $\mathcal{C}_{A,B}$, $AB\neq 0$, contains a line of the form $Y-aX-b$ as a component with $(a,b)\neq(-1,-1)$. Then one of the following cases occurs.

1. (1)

$A-2B+1=0$* or $(A,B)=(1,-1/2)$ and*

[TABLE] 2. (2)

$A^{3}-2AB+1=0$, $A^{3}+1\neq 0$ and

[TABLE] 3. (3)

$A=B^{2}$* and*

[TABLE] 4. (4)

$-3\in\square_{q}$, $A^{2}+2AB-A+4B^{2}+2B+1=0$ and

[TABLE]

where $\alpha^{2}=-3$. 5. (5)

$-3\in\square_{q}$, $A^{2}+A+1=0$, $B\in\{A^{2},-A^{2}/2\}$, and

[TABLE]

where $\alpha^{2}=-3$.

Proof.

Proposition 2.1 applies here. The first three cases are just easy computations.

Now suppose that $A^{2}+2AB-A+4B^{2}+2B+1=0$ and $-3\in\square_{q}$. Then $B=\frac{-A-1+\alpha(A-1)}{4}$, where $\alpha^{2}+3=0$ and $F_{A,B}\left(\frac{Y+1+\alpha(Y-1)}{2},Y\right)$ vanishes.

If $-3\in\square_{q}$, $A^{2}+A+1=0$, and $B\in\{A^{2},-A^{2}/2\}$ then $A=(-1+\alpha)/2$, where $\alpha^{2}+3=0$. It is easily seen that $F_{A,B}\left(\frac{Y+1+\alpha(Y-1)}{2},Y\right)$ vanishes. ∎

Proposition 2.3**.**

The curve $\mathcal{C}_{A,0}$, decomposes in $(A^{3}+1)XY=0$. The curve $\mathcal{C}_{0,B}$ is reducible only if $B^{3}+1/8=0$. In this case $(2BX+4B^{2}Y+1)\mid F_{0,B}(X,Y)$.

Proof.

If $B=0$, then there is nothing to prove. Suppose now $A=0$ and then $B\neq 0$. Then $F_{0,B}(X,Y)$ reads

[TABLE]

If $F_{0,B}(X,Y)$ is not absolutely irreducible then it contains a line of type $X=a$, $Y=a$, or $X+2BY+a=0$ for some $a\in\overline{\mathbb{F}_{q}}$. By direct computations, the first two cases imply $B=0$, a contradiction. The case $X+2BY+a=0$ gives $B^{3}+1/8=0$ and $a=\frac{1}{2B}$. ∎

Finally, we need this easy results on linearized polynomials.

Proposition 2.4**.**

A linearized polynomial $\alpha X^{q^{2}}+\beta X^{q}+\gamma X$, with $\alpha,\beta,\gamma\in\mathbb{F}_{q}$ has nonzero roots in $\mathbb{F}_{q^{3}}$ if and only if $\alpha^{3}+\beta^{3}+\gamma^{3}-3\alpha\beta\gamma=0$.

Proof.

The polynomial $\alpha X^{q^{2}}+\beta X^{q}+\gamma X$ has a nonzero root if and only if the rank of the matrix

[TABLE]

is smaller than three, that is, $\alpha^{3}+\beta^{3}+\gamma^{3}-3\alpha\beta\gamma=0$. ∎

We are now in a position to prove our main results concerning planar polynomials of type $f_{A,B}(x)$ of $\mathbb{F}_{q^{3}}$ as in (2).

Theorem 2.5**.**

Let $q$ be a power of an odd prime and consider the function $f_{A,B}(x)\in\mathbb{F}_{q^{3}}[x]$ of the type

[TABLE]

where $A,B\in\mathbb{F}_{q}$. Then, $f_{A,B}$ is planar if and only if one of the following condition holds

• •

$A^{3}-2AB+1=0$, $A^{3}\neq\pm 1$;

• •

$A=B^{2}$, $B^{3}\neq 1$;

• •

$B=0$, $A^{3}+1\neq 0$.

Proof.

Recall that $f_{A,B}(x)$ is planar if and only if $\det(M_{A,B,C})$ has no roots $C\in\mathbb{F}_{q^{3}}^{*}$. We consider first the case $AB\neq 0$.

1. (1)

If $A^{3}-2AB+1=0$ and $A^{3}+1\neq 0$ then $F_{A,B}(X,Y)=(A^{2}+AX+Y)(A^{2}Y+A+X)(A^{2}X+AY+1)$ by Proposition 2.2. Therefore

[TABLE]

By Proposition 2.4, all factors have no roots $C\in\mathbb{F}_{q^{3}}^{*}$ if and only if $A^{3}-1\neq 0$. 2. (2)

If $A=B^{2}$ then $F_{A,B}(X,Y)=(B^{2}+BY+X)(B^{2}Y+BX+1)(B^{2}X+B+Y)$ by Proposition 2.2. Therefore

[TABLE]

By Proposition 2.4, all factors has no roots $C\in\mathbb{F}_{q^{3}}^{*}$ if and only if $B^{3}-1\neq 0$.

By Propositions 2.1 and 2.2 if $F_{A,B}(X,Y)$ has a linear factor but $A^{3}-2AB+1\neq 0$ and $A\neq B^{2}$ then such a factor corresponds to a linearized factor of $\det(M_{A,B,C})$ which is either $C+C^{q}+C^{q^{2}}$ or $2C-C^{q}-C^{q^{2}}+\alpha(C^{q}-C^{q^{2}})$, with $\alpha^{2}+3=0$ and $-3\in\square_{q}$. It can be easily verified, using Proposition 2.4, that such a factor has a nonzero root in $\mathbb{F}_{q^{3}}$ and therefore $f_{A,B}(x)$ is not planar.

We deal now with the case $AB=0$.

1. (1)

If $B=0$, by Proposition 2.3 $F_{A,B}(X,Y)=(A^{3}+1)XY$, that is $\det(M_{A,B,C})=(A^{3}+1)C^{q+1}$ and its unique root is $C=0$ and thus $f_{A,B}(x)$ is planar. 2. (2)

Consider now the case $A=0$ and $B\neq 0$. By Proposition 2.3, if $F_{A,B}(X,Y)$ is reducible then $B^{3}+1/8=0$ and $(2BX+4B^{2}Y+1)$ is a factor. Such a factor corresponds to a factor $2BC+4B^{2}C^{q}+C^{q^{2}}$ in $\det(M_{A,B,C})$ and by Proposition 2.4 it has nonzero roots $C\in\mathbb{F}_{q^{3}}$ and so $f_{A,B}(x)$ is not planar.

We are left with the cases in which $F_{A,B}(X,Y)$ is absolutely irreducible. Now consider the polynomial $G_{A,B}(X,Y,T)$ as in (4). Let us fix a normal basis $\{\xi,\xi^{q},\xi^{q^{2}}\}$ of $\mathbb{F}_{q^{3}}$ over $\mathbb{F}_{q}$. Then each element $C\in\mathbb{F}_{q^{3}}^{*}$ can be written in a unique way as $C=X\xi+Y\xi^{q}+T\xi^{q^{2}}$, with $(X,Y,T)\neq(0,0,0)$, $X,Y,T\in\mathbb{F}_{q}$. The curves defined by $H_{A,B}(X,Y,T)=G_{A,B}(X\xi+Y\xi^{q}+T\xi^{q^{2}},X\xi^{q}+Y\xi^{q^{2}}+T\xi,X\xi^{q^{2}}+Y\xi+T\xi^{q})\in\mathbb{F}_{q}[X,Y,Z]$ and $G_{A,B}(X,Y,T)$ are $\mathbb{F}_{q^{3}}$-isomorphic. Since $F_{A,B}(X,Y)$ and therefore $G_{A,B}(X,Y,T)$ are absolutely irreducible, so it is $H_{A,B}(X,Y,T)$. Since $q>3$, by Hasse-Weil Theorem the curve defined by $H_{A,B}(X,Y,T)=0$ has $\mathbb{F}_{q}$-rational points $(x_{0},y_{0},t_{0})$ which correspond to nonzero roots $C=x_{0}\xi+y_{0}\xi^{q}+t_{0}\xi^{q^{2}}$ of $\det(M_{A,B,C})$. Thus, in all these cases $f_{A,B}(x)$ is not planar. ∎

Remark 2.6**.**

By Theorem 2.5 the number of pairs $(A,B)\in\mathbb{F}_{q}$ such that $f_{A,B}(x)$ is a planar polynomial in $\mathbb{F}_{q^{3}}$ is $3q-2-4GCD(3,q-1)$.

3. Acknowledgments

The research was supported by the Italian National Group for Algebraic and Geometric Structures and their Applications (GNSAGA - INdAM).