On permutation trinomials of type $x^{2p^s+r}+x^{p^{s}+r} +\lambda x^r$

TL;DR

This paper characterizes all permutation trinomials of a specific form over finite fields under certain size constraints, extending previous results for the case when p=2 and r=1.

Contribution

It provides a complete classification of permutation trinomials of the given form over finite fields under the condition $(2p^s+r)^4 < p^t$, extending earlier work for special cases.

Findings

All such permutation trinomials are explicitly determined.

The classification applies when $(2p^s+r)^4 < p^t$, covering a broad class of finite fields.

Partial extension of previous results for the case $p=2$, $r=1$.

Abstract

We determine all permutation trinomials of type $x^{2p^s+r}+x^{p^s+r}+\lambda x^r$ over the finite field $\mathbb{F}_{p^t}$ when $(2p^s+r)^4<p^t$. This partially extends a previous result by Bhattacharya and Sarkar in the case $p=2$, $r=1$.

Equations

$$f_{\lambda }(x)=x^{2p^s+r}+x^{p^s+r}+\lambda x^r\in {\mathbb{F}}_{p^t}[x].$$

$$d=(2p^s+r)/p^m,m:=\max \{n\ge 0:p^n\mid (2p^s+r),p^n\mid (p^s+r),p^n\mid r\}.$$

$$\frac{f_{\lambda }(x)-f_{\lambda }(y)}{x-y}=x^2+xy+y^2+x+y+\lambda$$

$$\frac{f_{\lambda }(x)-f_{\lambda }(y)}{x-y}=\left(x+\frac{1+\sqrt{-3}}{2}y+\frac{1}{2}+\frac{\sqrt{-3}}{6}\right)\cdot \left(x+\frac{1-\sqrt{-3}}{2}y+\frac{1}{2}-\frac{\sqrt{-3}}{6}\right)$$

$$f_{\lambda }(x)=x{\left(x+b\right)}^{p^s}(x+b^{p^t}{)}^{p^s}=x{\left(x+b\right)}^{(p^t{)}^{2p^s-1}+(p^t{)}^{2p^s-2}+\cdots +p^t+1}.$$

$$f(x)={\left(x^{p^{w-s}}+x^{p^{w-s}-1}+\lambda x^{p^{w-s}-2}\right)}^{p^s};$$

$$\frac{f(x)-f(y)}{x-y}=(x-y{)}^{p^w-1}+x^{p^w-2}+x^{p^w-3}y+\cdots +y^{p^w-2}+\lambda (x^{p^w-3}+x^{p^w-4}y+\cdots +y^{p^w-3}).$$

$$\frac{f_{\lambda }(-x)-f_{\lambda }(0)}{(-x{)}^r}=\frac{h(k(-x))-h(k(0))}{(-x{)}^r}=\frac{(k(-x)-k(0)){\prod }_{i=1}^{\mathrm{deg}(h)-1}(k(-x)-{\beta }_i)}{(-x{)}^r}$$

Full text

Keywords: Permutation trinomials, exceptional polynomials.

1 Introduction

Let $p$ be a prime number, $t$ be a positive integer, $\mathbb{F}_{p^{t}}$ be the finite field with $p^{t}$ elements, and $f(x)$ be a polynomial over $\mathbb{F}_{p^{t}}$. If $f:x\mapsto f(x)$ is a permutation of $\mathbb{F}_{p^{t}}$, then $f(x)$ is a permutation polynomial (PP) of $\mathbb{F}_{p^{t}}$. If $f(x)$ is a PP of $\mathbb{F}_{(p^{t})^{m}}$ for infinitely many $m$, then $f(x)$ is an exceptional polynomial over $\mathbb{F}_{p^{t}}$. If both $f(x)$ and $f(x)+x$ are PPs of $\mathbb{F}_{p^{t}}$, then $f(x)$ is a complete permutation polynomial (CPP) of $\mathbb{F}_{p^{t}}$.

The study of permutation polynomials over finite fields is motivated not only by their theoretical importance, but also by their remarkable applications to cryptography, combinatorial designs, and coding theory; see for instance [8, 10, 13]. For a detailed introduction to old and new developments on permutation polynomials, see the survey [7] and the references therein. Permutation polynomials of monomial and binomial type have been intensively investigated, while much less is known on permutation trinomials; see [4, 6].

In this note we characterize a certain class of permutation trinomials. Let $s$ and $r$ be non-negative integers. For $\lambda\in\mathbb{F}_{p^{t}}$, denote by $f_{\lambda}(x)$ the polynomial

[TABLE]

If $r=0$, define $d=0$. If $r\neq 0$, write $r=p^{u}v$ with $u\geq 0$ and $p\nmid v$, and define $d=2p^{s-u}+v$ if $u\leq s$, $d=2+p^{u-s}v$ if $u>s$; that is,

[TABLE]

We prove the following result.

Theorem 1.1**.**

Assume that $d^{4}<p^{t}$. Then $f_{\lambda}(x)$ is a PP of $\mathbb{F}_{p^{t}}$ if and only if one of the following cases holds:

• •

$p=2$, $t$ is odd, and $f_{\lambda}(x)=x^{3}+x^{2}+x$ or $f_{\lambda}(x)=x^{5}+x^{3}+x$;

• •

$p\equiv 2\pmod{3}$, $t$ is odd, and $f_{\lambda}(x)=x^{3}+x^{2}+\frac{1}{3}x$.

The case $p=2$ and $r=1$ was already considered by Bhattacharya and Sarkar [3], where the result proved for $f_{\lambda}(x)$ was then used to characterize permutation binomials of $\mathbb{F}_{p^{2tp^{s}}}$ of type $g_{b}(x)=x^{\frac{p^{2tp^{s}}-1}{p^{t}-1}+1}+bx$. Here for $p>2$ and $r=1$ we go the opposite direction, using the characterization in [1] for permutation binomials of type $g_{b}(x)$ to deduce the result for $f_{\lambda}(x)$.

Every permutation polynomial of $\mathbb{F}_{p^{t}}$ with degree less than $(p^{t})^{1/4}$ is exceptional over $\mathbb{F}_{p^{t}}$; thus, the condition $d^{4}<p^{t}$ allows us to consider only exceptional polynomials. For $r>1$, this leads to the non-existence of permutation trinomials of type $f_{\lambda}(x)$.

2 Proof of Theorem 1.1

Since the maps $x\mapsto x^{p^{u}}$ and $x\mapsto x^{p^{s}}$ are permutations of $\mathbb{F}_{p^{t}}$, we can assume that $u=0$ if $u\leq s$, and $s=0$ if $u>s$.

• •

Case $r=0$. Since $f_{\lambda}(x)=f_{\lambda}(-x-1)$, $f_{\lambda}(x)$ is not a PP of $\mathbb{F}_{p^{t}}$.

• •

Case $r=1$ and $p=2$. The claim is proved in [3, Theorem 1.3].

• •

Case $r=1$ and $p>2$.

Assume first that $s=0$, so that $f_{\lambda}(x)=x^{3}+x^{2}+\lambda x$. By direct computation,

[TABLE]

splits into two linear components if and only if $p\neq 3$ and $\lambda=\frac{1}{3}$. In this case

[TABLE]

and the two components are not defined over $\mathbb{F}_{p^{t}}$ if and only if $-3$ is a non-square in $\mathbb{F}_{p^{t}}$. From [5, Lemma 4.5], this is equivalent to require $t$ odd and $p\equiv 2\pmod{3}$.

Now assume that $s>0$. Let $\mu\in\mathbb{F}_{p^{t}}$ with $\mu^{p^{s}}=\lambda$, so that $f(x)=x\left(x^{2}+x+\mu\right)^{p^{s}}$. Let $b,b^{p^{t}}\in\mathbb{F}_{p^{2t}}$ be the zeros of $x^{2}+x+\mu$; then, for any $x\in\mathbb{F}_{p^{t}}$,

[TABLE]

Suppose that $f_{\lambda}(x)$ is a PP of $\mathbb{F}_{p^{t}}$; in particular, $b\notin\mathbb{F}_{p^{t}}$. Since $(\deg f_{\lambda}(x))^{4}=d^{4}<p^{t}$, $f_{\lambda}(x)$ is an exceptional polynomial over $\mathbb{F}_{p^{t}}$ from [11, Theorem 8.4.19]. Also, from [1, Proposition 2.4], $f_{\lambda}(x)$ is indecomposable as exceptional polynomial over $\mathbb{F}_{p^{t}}$. Hence, from [11, Theorem 8.4.11], $\deg f_{\lambda}(x)=2p^{s}+1$ is a prime not dividing $p^{t}-1$. From the Niederreiter-Robinson criterion [12, Lemma 1], the polynomial $x^{\frac{(p^{t})^{2p^{s}}-1}{p^{t}-1}+1}+bx$ is a PP of $\mathbb{F}_{(p^{t})^{2p^{s}}}$; equivalently, the monomial $b^{-1}x^{\frac{p^{2tp^{s}}-1}{p^{t}-1}+1}$ is a CPP of $\mathbb{F}_{p^{2tp^{s}}}$. Thus, from [1, Theorem 3.1], one of the following cases hold, where $\zeta$ is a primitive $(2p^{s}+1)$-th root of unity and $\alpha:=\zeta+\zeta^{-1}$, $\beta:=\zeta-\zeta^{-1}$:

• –

$p^{t}$ has order $2p^{s}$ modulo $2p^{s}+1$ and $b=\zeta-1$ up to multiplication by a non-zero element in $\mathbb{F}_{p^{t}}$. Since $b\in\mathbb{F}_{p^{2t}}\setminus\mathbb{F}_{p^{t}}$ and $2p^{s}+1$ is prime, we have $(2p^{s}+1)\mid(p^{t}+1)$. Hence $p^{t}\equiv-1\pmod{2p^{s}+1}$ has order $2\neq 2p^{s}$ modulo $2p^{s}+1$, a contradiction.

• –

$p^{t}$ has order $2p^{s}$ modulo $2p^{s}+1$ and $b=e(\alpha-1)\sqrt{\beta^{2}(e^{2}-4a)}$ up to multiplication by a non-zero element in $\mathbb{F}_{p^{t}}$, for some $e,a\in\mathbb{F}_{p^{t}}$, $a\neq 0$, such that $e^{2}-4a$ is a square in $\mathbb{F}_{p^{t}}$. Since $b\in\mathbb{F}_{p^{2t}}\setminus\mathbb{F}_{p^{t}}$ we have $\alpha\in\mathbb{F}_{p^{t}}$, which implies $\zeta^{p^{t}-1}=1$ or $\zeta^{p^{t}+1}=1$. As $2p^{s}+1$ is prime and $b\notin\mathbb{F}_{p^{t}}$, this yields $(2p^{s}+1)\mid(p^{t}+1)$; hence, $p^{t}$ has order $2$ modulo $2p^{s}+1$, a contradiction to $s>0$.

• –

$p^{t}$ has order $p^{s}$ modulo $2p^{s}+1$ and $b=e(\alpha-1)\sqrt{\beta^{2}(e^{2}-4a)}$ up to multiplication by a non-zero element in $\mathbb{F}_{p^{t}}$, for some $e,a\in\mathbb{F}_{p^{t}}$, $a\neq 0$, such that $e^{2}-4a$ is [math] or a non-square in $\mathbb{F}_{p^{t}}$. From $b\in\mathbb{F}_{p^{2t}}\setminus\mathbb{F}_{p^{t}}$ we have $\alpha\in\mathbb{F}_{p^{t}}$ which implies $\zeta^{p^{t}-1}=1$ or $\zeta^{p^{t}+1}=1$; hence, $p^{t}$ has order $1$ or $2$ modulo $2p^{s}+1$, a contradiction to $s>0$.

Therefore, $f_{\lambda}(x)$ is not a PP of $\mathbb{F}_{p^{t}}$.

• •

Case $r>1$.

Assume first $u\leq s$, so that we can take $u=0$ and $d=2p^{s}+r$. Suppose by contradiction that $f_{\lambda}(x)$ is a PP of $\mathbb{F}_{p^{t}}$. As $(\deg f_{\lambda}(x))^{4}<p^{t}$, $f_{\lambda}(x)$ is exceptional over $\mathbb{F}_{p^{t}}$, see [11, Theorem 8.4.19]. Note that $f_{\lambda}(x)$ has exactly three distinct zeros, one in $\mathbb{F}_{p^{t}}$ with multiplicity $r$ and two in $\mathbb{F}_{p^{2t}}\setminus\mathbb{F}_{p^{t}}$ with multiplicity $p^{s}$.

• –

Suppose that $f(x)$ is indecomposable as exceptional polynomial over $\mathbb{F}_{p^{t}}$. From [11, Theorem 8.4.10], one of the following cases holds.

$2p^{s}+r=p^{w}$ for some $w\geq 1$. In this case

[TABLE]

since $x\mapsto x^{p^{s}}$ is a permutation of $\mathbb{F}_{p^{t}}$, we can assume $s=0$. Then

[TABLE]

Let $\mathcal{C}$ be the plane curve of degree $p^{w}-1$ defined over $\mathbb{F}_{p^{t}}$ by affine equation $\frac{f(x)-f(y)}{x-y}=0$. From Equation (1), $\mathcal{C}$ has a unique point at infinity $P_{\infty}$. Moreover, $\mathcal{C}$ intersects the line $x=y$ at the affine points $(0,0)$ and $(-2\lambda,-2\lambda)$ with multiplicity $p^{w}-3$ and $1$, respectively; hence, $P_{\infty}$ is a simple point for $\mathcal{C}$. This implies that $\mathcal{C}$ is absolutely irreducible, a contradiction to the exceptionality of $f(x)$ (see [11, Theorem 8.4.4]).

• *

$2p^{s}+r=\frac{p^{a}(p^{a}-1)}{2}$, with $p\in\{2,3\}$ and $a>1$ odd; this is not possible, since $p\nmid r$.

• *

$2p^{s}+r$ is coprime with $p$. From [11, Theorem 8.4.11], one of the following holds:

• ·

$f_{\lambda}(x)$ is linear. This is not possible by the assumptions.

• ·

$f_{\lambda}(x)=x^{2p^{s}+r}$ where $2p^{s}+r$ is a prime not dividing $p^{t}-1$, up to composition with linear functions. Then $f_{\lambda}(x)$ has either one or $n$ distinct roots in $\overline{\mathbb{F}}_{p^{t}}$, a contradiction.

• ·

$f_{\lambda}(x)=\ell_{1}\circ D_{2p^{s}+r}(\ell_{2}(x),a)$, where $2p^{s}+r$ is a prime not dividing $p^{2t}-1$, $D_{2p^{s}+r}(x,a)$ is a Dickson polynomial with $a\neq 0$ of degree $2p^{s}+r$, and $\ell_{1},\ell_{2}\in\mathbb{F}_{p^{t}}[x]$ are linear permutations. If $(2p^{s}+r)\nmid(p^{2t}+1)$, then $D_{2p^{s}+r}(x,a)$ is a PP of $\mathbb{F}_{p^{2t}}$; see [11, Theorem 8.4.11]. This is not possible, as $f_{\lambda}(x)$ has three distinct zeros in $\mathbb{F}_{p^{2t}}$. Thus, $(2p^{s}+r)\mid(p^{2t}+1)$. Denote $\ell_{1}(x)=bx+c$. As $\ell_{2}$ permutes $\mathbb{F}_{p^{2t}}$, the number of zeros of $f_{\lambda}(x)$ in $\mathbb{F}_{p^{2t}}$ is equal to the number $Z$ of preimages of $-c/b$ under $D_{2p^{s}+r}(x,a)$; hence, $Z=3$. On the other hand, from [9, Theorems $3.26$ and $3.26^{\prime}$] we have $Z\in\{1,2p^{s}+r,\frac{2p^{s}+r}{2},\frac{2p^{s}+r+1}{2}\}$. Then $s=0$ and $r=3$, so that $f_{\lambda}(x)=x^{5}+x^{4}+\lambda x^{3}$ with $p\neq 5$. We have $D_{5}(x,a)=x^{5}-5ax^{3}+5a^{2}x$; by direct inspection, the polynomial $\ell_{1}\circ D_{5}(\ell_{2}(x),a)$ cannot have the form $x^{5}+x^{4}+\lambda x^{3}$ for any $\ell_{1},\ell_{2}$.

• –

Now suppose that $f_{\lambda}(x)$ is a decomposable exceptional polynomial over $\mathbb{F}_{p^{t}}$, say $f_{\lambda}(x)=h(k(x))$ for some exceptional polynomials $h,k\in\mathbb{F}_{p^{t}}[x]$ with $\deg(h),\deg(k)>1$. The roots of $f_{\lambda}(x)/x^{r}$ are conjugated under the Frobenius map $x\mapsto x^{p^{t}}$; hence, the polynomial

[TABLE]

is a power of a unique irreducible factor over $\mathbb{F}_{p^{t}}$.

Suppose that $k(-x)-k(0)$ has a monic absolutely irreducible factor different from $x$ and defined over $\mathbb{F}_{p^{t}}$. Since the roots of $f_{\lambda}(x)/x^{r}$ are conjugated under $x\mapsto x^{p^{t}}$, we have $\beta_{i}=k(0)$ for all $i$. Hence, $\frac{f_{\lambda}(-x)-f_{\lambda}(0)}{(-x)^{r}}=\frac{(k(-x)-k(0))^{\deg(h)}}{(-x)^{r}}$. Also, $m\cdot\deg(h)=r$, where $x^{m}$ is the maximum power of $x$ which divides $k(-x)-k(0)$; in particular, $p\nmid\deg(h)$. On the other hand, $f_{\lambda}(x)$ has just two distinct non-zero roots (the ones of $x^{2}+x+\mu$ where $\mu^{p^{s}}=\lambda$) with multiplicity $p^{s}$; hence, $\deg(h)\mid p^{s}$. This is a contradiction, either to $p\nmid\deg(h)$ or to $\deg(h)>1$.

Suppose that $k(-x)-k(0)=ax^{m}$, for some $a\in\mathbb{F}_{p^{t}}$ and $m>1$ with $\gcd(m,p^{t}-1)=1$. If $p\mid m$, then $f_{\lambda}(x)$ is invariant under $x\mapsto\gamma x$ when $\gamma\in\mathbb{F}_{p}$; this is a contradiction to $\gcd(2p^{s}+r,p^{s}+r)=1$. Then $p\nmid m$. Let $\bar{x}$ be a non-zero root of $f_{\lambda}(x)$; for any $\delta$ with $\delta^{m}=1$, $k(\delta\bar{x})=k(\bar{x})$ and $f_{\lambda}(\delta\bar{x})=0$. Thus, the number of distinct non-zero roots of $f_{\lambda}(x)$ is a multiple of $m$; hence, $m=2$. This implies $p=2$. Therefore $f_{\lambda}(x)=h(k(0)+ax^{2})=h((\ell_{0}+\ell_{1}x)^{2})$ with $\ell_{0},\ell_{1}\in\mathbb{F}_{p^{t}}$, so that the polynomial $h(x^{2})$ is also exceptional of degree $\deg(f_{\lambda})$. Since $\deg(h)$ is odd, this is not possible.

We have shown that $f_{\lambda}(x)$ is not a PP of $\mathbb{F}_{p^{t}}$ under the assumption $u\leq s$. If $u>s$, then we can take $s=0$ so that $d=r+2$ and $f_{\lambda}(x)=x^{r}(x^{2}+x+\lambda)$. The same arguments as in the case $u\leq s$ still apply and show that $f_{\lambda}(x)$ is not a PP of $\mathbb{F}_{p^{t}}$.

Remark 2.1**.**

Theorem 1.1 yields the characterization also of permutation trinomials of $\mathbb{F}_{p^{t}}$ of type $g_{\alpha,\beta}(x)=x^{2p^{s}+r}+\alpha x^{p^{s}+r}+\beta x^{r}$, under the assumptions $\alpha\neq 0$ and $d^{4}<p^{t}$ (with $d$ defined as in Theorem 1.1).

In fact, let $\gamma\in\mathbb{F}_{p^{t}}$ satisfy $\gamma^{p^{s}}=\alpha$. Then $g_{\alpha,\beta}(x)$ is a PP of $\mathbb{F}_{p^{t}}$ if and only if $\frac{1}{\gamma^{2p^{s}+r}}g(\gamma x)=f_{\beta/\alpha^{2}}(x)$ is a PP of $\mathbb{F}_{p^{t}}$. Thus, $g_{\alpha,\beta}(x)$ is a PP of $\mathbb{F}_{p^{t}}$ exactly in the following cases:

• •

$p=2$, $t$ is odd, and $g_{\alpha,\beta}(x)=x^{3}+\alpha x^{2}+\alpha^{2}x$ or $g_{\alpha,\beta}(x)=x^{5}+\alpha x^{3}+\alpha^{2}x$ for some $\alpha\in\mathbb{F}_{p^{t}}$;

• •

$p\equiv 2\pmod{3}$, $t$ is odd, and $g_{\alpha,\beta}(x)=x^{3}+\alpha x^{2}+\frac{\alpha^{2}}{3}x$ for some $\alpha\in\mathbb{F}_{p^{t}}$.