More classes of permutation trinomials with Niho exponents

Abstract

This paper presents two classes of permutation trinomials with the form $x^{s(2^{m}-1)+ 1}+x^{t(2^{m}-1)+ 1}+x$ over the finite field $\mathbb {F}_{2^{2m}}$ as a supplement of the recent works of Li and Helleseth, and a class of permutation trinomials like this form over $\mathbb {F}_{3^{2m}}$. Moreover, we give a method to construct permutation polynomials from known ones.

1 Introduction

Let q be a power of a prime, and $\mathbb {F}_{q}$ be a finite field with q elements. A polynomial $f(x)\in \mathbb {F}_{q}[x]$ is called a permutation polynomial of $\mathbb {F}_{q}$ if the associated polynomial function f : a↦f(a) from $\mathbb {F}_{q}$ into $\mathbb {F}_{q}$ is a permutation of $\mathbb {F}_{q}$ [15]. Permutation polynomials over finite fields have important applications in cryptography [17, 20, 21], coding theory [8, 22] and combinatorial design theory [4]. Finding new permutation polynomials is of great interest in both theoretical and applied aspects.

The construction of permutation polynomials with few terms attracts many authors to work on this topic due to their simple form and wide applications. The recent progress on construction of permutation binomials and trinomials can be seen in [2, 3, 5,6,7, 9, 10, 12,13,14, 23, 26] and the references therein. In particular, Li and Helleseth in [12, 13] investigated permutation polynomials with the following form

$$ f(x)=x+x^{s(2^{m}-1)+ 1}+x^{t(2^{m}-1)+ 1}, $$

(1)

where s, t are integers satisfying 1 ≤ s, t ≤ 2^m. For simplicity, when s, t are written as fractions or negative integers, then they are interpreted as modulo 2^m + 1. By solving equations with low degree over finite fields and using the property of linear fractional polynomials over the finite fields $\mathbb {F}_{2^{m}}$, they obtained some new permutation polynomials over $\mathbb {F}_{2^{2m}}$ of the form (1).

In this paper, we will further study the permutation polynomials of the form (1) following the works in [12, 13]. By some delicate operation of solving equations with low degrees over finite fields, we find two new pairs (s, t) such that (1) are permutation polynomials over $\mathbb {F}_{2^{2m}}$. By using the technique provided in [6] we obtain a new class of permutation trinomials like this form over $\mathbb {F}_{3^{2m}}$. Moreover, we present a method to construct permutation polynomials from known ones.

2 Preliminaries

Let $\mathbb {F}_{p^{n}}$ be a finite field with pⁿ elements for a prime p. Let k be a divisor of n. The trace function from $\mathbb { F}_{p^{n}}$ to $\mathbb {F}_{p^{k}}$ is defined by

$$\text{Tr}^{n}_{k}(x)= \sum\limits_{i = 0}^{n/k-1} x^{p^{ik}}, $$

where $x\in \mathbb {F}_{p^{n}}$. Let q be a power of a prime p. A positive integer s is called a Niho exponent with respect to the finite field $\mathbb {F}_{q^{2}}$ if s ≡ pⁱ mod (q − 1) for some nonnegative integer i. s is called a normalized Niho exponent if i = 0. The Niho exponents were first introduced by Niho in [18] for studying the cross-correlation between an m-sequence and its d-decimation. Let d be a positive integer with d | (q − 1), and μ_d denote the set of d-th roots of unity in $\mathbb {F}_{q}^{*}$, i.e.,

$$\begin{array}{@{}rcl@{}} \mu_{d}=\left\{ x\in \mathbb{F}_{q}^{*} \,:\, x^{d}= 1 \right\}. \end{array} $$

A main technique to investigate permutation behavior of polynomials of the form (1) is the following lemma.

Lemma 1 (19, 24, 25)

Let q be a power of a prime. Let$\mathbb {F}_{q}$be a finite field with qelements and$h(x)\in \mathbb {F}_{q}[x]$.Letd,r be positive integers withd | (q − 1).Thenf(x) = x^rh(x^(q− 1)/d) permutes$\mathbb {F}_{q}$if and only if the following two conditions hold:

• (1) $\gcd (r,(q-1)/d)= 1$ ,

• (2) x^rh(x)^(q− 1)/dpermutesμ_d.

For later usage we need the following lemmas.

Lemma 2 (5, 26)

Let m be a positive integer. Each of the polynomials 1 + x + x³, 1 + x³ + x⁴,1 + x + x⁴and1 + x + x⁵(mis even) has no roots in$\mu _{2^{m}+ 1}$.

Lemma 3

Letm be an odd positive integer. Each of the polynomialsx² + x + 1 ,x⁴ + x³ + 1 andx⁴ + x + 1 has no roots in$\mathbb {F}_{2^{m}}$.

Lemma 4 (Theorem 2 in 1)

Letq = 2^mand$\mathbb {F}_{q}$be a finite field. Let$a,b\in \mathbb {F}_{q}$withb ≠ 0.Then the cubic equationx³ + ax + b = 0 has a unique solution in$\mathbb {F}_{q}$if and only if$\text {Tr}_{1}^{m}(\frac {a^{3}}{b^{2}}+ 1)\neq 0$.

Lemma 5 (16)

Let$\mathbb {F}_{q}$be afinite field withq = 2^m.Leth (x) = x⁴ + a₂x² + a₁x + a₀andg (y) = y³ + a₂y + a₁be polynomialsover$\mathbb {F}_{q}$witha₀a₁ ≠ 0. Letγ_ibe the rootsofg (y) = 0 whenthey exist in$\mathbb {F}_{q}$and$\omega _{i}=a_{0}{\gamma _{i}^{2}}/{a_{1}^{2}}$fori = 1,2,3 . Thenh (x) has nosolution in$\mathbb {F}_{q}$if one of the following conditions is satisfied:

• (1) g (y) = 0 has exactly one solution in$\mathbb {F}_{q}$and$\text {Tr}_{1}^{m}(\omega _{1})= 1$;

• (2) g (y) = 0 has exactly three solution in$\mathbb {F}_{q}$and$\text {Tr}_{1}^{m}(\omega _{1})= 0, \text {Tr}_{1}^{m}(\omega _{2})= \text {Tr}_{1}^{m}(\omega _{3})= 1$.

Lemma 6 (Theorem 2.10 in 10)

Letq = 2^mand$\mathbb {F}_{q}$be a finite field. The polynomialg (x) = x⁷(1 + x⁴ + x⁶)^q− 1permutes the unite circleμ_q+ 1in$\mathbb {F}_{q^{2}}$if$\gcd (m,3)= 1$.

Lemma 7 (Lemma 4.1 in 26)

The mapping $g(x)=x^{3}(1+x^{2}+x^{3})^{2^{m}-1}$ permutes $\mu _{2^{m}+ 1}$ in the finite field $\mathbb {F}_{2^{2m}}$ .

Lemma 8 (Lemma 4 in 13)

Letq = 2^mand$\mathbb {F}_{q}$be a finitefield. Let (s, t) = (i, j) be apair such thatf (x) defined by (1) is a permutation polynomial over$\mathbb {F}_{q^{2}}$, thenf(x) defined by (1) is also a permutation polynomial over$\mathbb {F}_{q^{2}}$forthe following pairs:

• (1) $(s,t)=(\frac {i}{2i-1},\frac {i-j}{2i-1})$ if $\gcd (2i-1,2^{m}+ 1)= 1$ , or

• (2) $(s,t)=(\frac {j}{2j-1},\frac {j-i}{2j-1})$ if $\gcd (2j-1,2^{m}+ 1)= 1$ .

These pairs are called equivalent pairs if they exist.

If (s, t) is a pair such that (1) is a permutation polynomial, then from Lemma 8 one can easily get equivalent pairs of (s, t) such that (1) is a permutation polynomial. So, from this property of Niho exponents one easily checks multiplicative equivalence of two permutation trinomials of the form (1) [12].

3 Two classes of permutation trinomials over $\mathbb {F}_{2^{2m}}$

In this section, we investigate the permutation behavior of two classes of trinomials of the form (1) over $\mathbb {F}_{2^{2m}}$, and present a method to construct new permutation polynomials from known ones.

Theorem 1

Let$\mathbb {F}_{q}$be a finite field withq = 2^mand$\gcd (m,2)= 1$.The trinomialf (x) defined by (1) is a permutation over$\mathbb {F}_{q^{2}}$if$(s,t)=(\frac {2}{7},\frac {8}{7})$.

Proof

By Lemma 1 we only need to show ϕ(x) = x(1 + x^s + x^t)^q− 1 permutes the unite circle μ_q+ 1. This is equivalent to showing that g(x) = ϕ(x⁷) = x⁷(1 + x^7s + x^7t)^q− 1 = x⁷(1 + x² + x⁸)^q− 1 permutes the unite circle μ_q+ 1 since $\gcd (7,q + 1)= 1$. By Lemma 2 we know that 1 + x² + x⁸≠ 0 for all x ∈ μ_q+ 1. So g(μ_q+ 1) ⊆ μ_q+ 1. For x ∈ μ_q+ 1, g(x) is reduced to the following fraction,

$$\begin{array}{@{}rcl@{}} g(x)=x^{7}(1+x^{2}+x^{8})^{q-1}=\frac{1+x^{6}+x^{8}}{x+x^{3}+x^{9}}. \end{array} $$

Next we show that there is no pair $(x,y)\in \mu _{q + 1}^{2}$ with x≠y satisfying g(x) = g(y). From

$$\frac{1+x^{6}+x^{8}}{x+x^{3}+x^{9}}=\frac{1+y^{6}+y^{8}}{y+y^{3}+y^{9}}, $$

we get

$$ \begin{array}{l} (x+y)+xy(x^{5}+y^{5})+xy(x^{7}+y^{7})+(x^{3}+y^{3})+x^{3}y^{3}(x^{3}+y^{3})\\ +x^{3}y^{3}(x^{5}+y^{5})+(x^{9}+y^{9})+x^{6}y^{6}(x^{3}+y^{3})+x^{8}y^{8}(x+y)= 0. \end{array} $$

(2)

Set x + y = u and xy = v. We know that u≠ 0 and v≠ 0 since x≠y. It follows that $u^{1-q}=\frac {x+y}{x^{-1}+y^{-1}}=xy=v$ since x, y ∈ μ_q+ 1. Plugging them into (2) and simplifying it, we get

$$ \begin{array}{l} 1+vu^{4}+v^{2}u^{2}+v^{3}+u^{2}+v+v^{3}u^{2}+v^{4} \\ +v^{3}u^{4}+v^{4}u^{2}+v^{5}+u^{8}+v^{6}u^{2}+v^{7}+v^{8}= 0. \end{array} $$

(3)

Let u = z^− 1 for $z\in \mathbb {F}_{q^{2}}^{*}$, and then v = z^q− 1. Multiplying both sides of (3) by z⁸ we get

$$\begin{array}{@{}rcl@{}} \begin{array}{l} (z^{8}+z^{8q})+z^{q + 1}(z^{2}+z^{2q})+z^{2(q + 1)}(z^{2}+z^{2q})+z^{3(q + 1)}(z^{2}+z^{2q})\\ +(z^{6}+z^{6q})+z^{q + 1}(z^{6}+z^{6q})+z^{3(q + 1)}+z^{4(q + 1)}+ 1 = 0. \end{array} \end{array} $$

(4)

Let α, β denote z + z^q and z^q+ 1 respectively. Then $\alpha ,\beta \in \mathbb {F}_{q}$ and from (4) we have

$$ \alpha^{8}+\alpha^{6}+\beta\alpha^{6}+\beta\alpha^{2}+ 1+\beta^{3}+\beta^{4}= 0. $$

(5)

Next we show that (5) has no solutions $(\alpha ,\beta ) \in { \mathbb {F}_{q}^{2}}$.

If $z\in \mathbb {F}_{q}$ then α = z + z^q = 0. From (5) we obtain

$$\begin{array}{@{}rcl@{}} \beta^{4}+\beta^{3}+ 1 = 0. \end{array} $$

By Lemma 3 we know this is a contradiction since $\beta \in \mathbb {F}_{q}$ and $\gcd (2,m)= 1$.

If $z\in \mathbb {F}_{q^{2}}\backslash \mathbb {F}_{q}$ then α = z + z^q≠ 0. Let ν = α². From (5) we have

$$ \nu^{4}+\nu^{3}+\beta\nu^{3}+\beta\nu+ 1+\beta^{3}+\beta^{4}= 0. $$

(6)

Multiplying both sides of (6) by ν^− 4 and let $\gamma =\frac {1}{\nu }, \delta =\frac {\beta }{\nu }$, we get

$$ \gamma^{4}+\delta\gamma^{2}+(1+\delta^{3})\gamma+ 1+\delta+\delta^{4}= 0. $$

(7)

From Lemma 3 we know that 1 + δ + δ⁴≠ 0 for $\delta \in \mathbb {F}_{q}$ since $\gcd (2,m)= 1$. Next we show that there is no pair $(\gamma , \delta )\in { \mathbb {F}_{q}^{2}}$ satisfying (7).

Assume that 1 + δ³ = 0. We have δ = 1 since $\delta \in \mathbb {F}_{q}$ and $\gcd (2,m)= 1$. Substituting δ = 1 into (7) we get

$$\begin{array}{@{}rcl@{}} \gamma^{4}+\gamma^{2}+ 1 = 0. \end{array} $$

This is a contradiction from Lemma 3 since $\gamma \in \mathbb {F}_{q}$ and $\gcd (2,m)= 1$.

Assume that 1 + δ³≠ 0. Let h(y) = y³ + δy + (1 + δ³). Since $\gcd (2,m)= 1$ we have

$$\begin{array}{@{}rcl@{}} \text{Tr}_{1}^{m}\left( \frac{\delta^{3}}{(1+\delta^{3})^{2}}+ 1\right)=\text{Tr}_{1}^{m}\left( \frac{1}{1+\delta^{3}}+\frac{1}{(1+\delta^{3})^{2}}+ 1\right)=\text{Tr}_{1}^{m}(1)\neq 0. \end{array} $$

From Lemma 4 we know that h(y) has exactly one solution, which is δ + 1. Let

$$\begin{array}{@{}rcl@{}} \omega=\frac{(1+\delta+\delta^{4})(1+\delta)^{2}}{(1+\delta^{3})^{2}}= 1+\frac{\delta}{1+\delta^{3}}+ \frac{\delta^{2}}{(1+\delta^{3})^{2}}+\frac{1}{1+\delta^{3}}+\frac{1}{(1+\delta^{3})^{2}}. \end{array} $$

It is clear that $\text {Tr}_{1}^{m}(\omega )= 1$ since $\gcd (m,2)= 1$. By Lemma 5 we know that there is no pair $(\gamma , \delta )\in { \mathbb {F}_{q}^{2}}$ satisfying (7). Therefore, there is no pair (x, y) in $\mu _{q + 1}^{2}$ with x≠y satisfying g(x) = g(y), that is, g(x) permutes the unite circle μ_q+ 1. □

Theorem 2

Letm be a positive integer withm ≡ 2,4 mod 6.Let$\mathbb {F}_{q}$be a finite field withq = 2^m.The trinomialf (x) defined by (1) is a permutation over$\mathbb {F}_{q^{2}}$if$(s,t)=(-\frac {2}{7},\frac {8}{7})$.

Proof

By Lemma 1 we need to show ϕ(x) = x(1 + x^s + x^t)^q− 1 permutes the unite circle μ_q+ 1. This is equivalent to showing that g(x) = ϕ(x⁷) = x⁷(1 + x^− 2 + x⁸)^q− 1 permutes the unite circle μ_q+ 1 since $\gcd (7,q + 1)= 1$. By Lemma 2 we know that 1 + x² + x¹⁰≠ 0 for all x ∈ μ_q+ 1 when m is even. So g(μ_q+ 1) ⊆ μ_q+ 1. For x ∈ μ_q+ 1, g(x) is reduced to the following form,

$$\begin{array}{@{}rcl@{}} g(x)=x^{7}(1+x^{-2}+x^{8})^{q-1}=\frac{x(1+x^{8}+x^{10})}{1+x^{2}+x^{10}}. \end{array} $$

Dividing the common divisor 1 + x² + x⁴ of numerator and denominator of the above fraction we get

$$g(x) =\frac{x(1+x^{8}+x^{10})}{1+x^{2}+x^{10}}=\frac{x(1+x^{2}+x^{6})}{1+x^{4}+x^{6}}= x^{7}(1+x^{4}+x^{6})^{q-1}.$$

Since m ≡ 2,4 mod 6, we have that $\gcd (m, 3)= 1$. By Lemma 6 we know that g(x) permutes the unite circle μ_q+ 1. So, f(x) is a permutation over $\mathbb {F}_{q^{2}}$. □

Inspired by the proof of Theorem 2 we present a method to construct new permutation polynomials from known ones. To this end, we introduce some symbols as follows. Let n and k_i be integers satisfying n ≥ 2, k₀ = 0 and k_i + k_n−i = k_n for 0 ≤ i ≤ n. Let q = 2^m and $R(x) = {\sum }_{i = 0}^{n} x^{k_{i}}$ be a polynomial over $\mathbb {F}_{q^{2}}$.

Theorem 3

Letr andℓbe positive integers. Leth (x) be a polynomial over$\mathbb {F}_{q^{2}}$.Assume thatR (x) ≠ 0 forx ∈ μ_q+ 1,then the polynomial$F(x)=x^{r+k_{n}\ell } R(x^{q-1})^{\ell } h(x^{q-1})$permutes$\mathbb {F}_{q^{2}}$if and only if$\gcd (r+k_{n}\ell , q-1)= 1$and the polynomialg (x) = x^rh (x) ^q− 1permutesμ_q+ 1.

Proof

By Lemma 1 we know that F(x) permutes $\mathbb {F}_{q^{2}}$ if and only if $\gcd (r+k_{n}l,q-1)= 1$ and the polynomial $G(x)=x^{r+k_{n}\ell } R(x)^{(q-1)\ell } h(x)^{q-1}$ permutes μ_q+ 1. Since R(x)≠ 0 for x ∈ μ_q+ 1 and k_i + k_n−i = k_n for 0 ≤ i ≤ n, we have

$$\begin{array}{@{}rcl@{}} G(x)&=& x^{r+k_{n} \ell} \left( R(x)^{\ell} h(x) \right)^{q-1} =x^{r+k_{n}\ell}\left( \sum\limits_{i = 0}^{n} x^{k_{i}}\right)^{(q-1)\ell} h(x)^{q-1} \\ &=& x^{r}\left( \frac{ \sum\limits_{i = 0}^{n} x^{k_{n}-k_{i}}}{ \sum\limits_{i = 0}^{n} x^{k_{i}} } \right)^{\ell} h(x)^{q-1} \\ &=&g(x). \end{array} $$

So G(x) permutes μ_q+ 1 if and only if g(x) = x^rh(x)^q− 1 permutes μ_q+ 1. This completes the proof. □

By this theorem we can obtain a new permutation polynomial F(x) over $\mathbb {F}_{q^{2}}$ from a permutation polynomial over the unite circle μ_q+ 1 if the desired polynomial R(x) is provided. The following corollary presents a method to construct R(x), and then obtains new permutation polynomials over $\mathbb {F}_{q^{2}}$ from known permutation polynomials over the unit circle μ_q+ 1.

Corollary 1

Letk, r, ℓbe positive integers andn be an even number. Let$R(x)=\sum \limits _{i = 0}^{n} x^{ki}$andh (x) be a polynomialover$\mathbb {F}_{q^{2}}$. Then thepolynomialF (x) = x^{r + knℓ}R (x^q− 1)^ℓh (x^q− 1) permutes$\mathbb {F}_{q^{2}}$if and only if the following three conditions hold:

• (1) $\gcd \left ((n + 1)k, q + 1\right )= 1$ ,

• (2) $\gcd \left (r+kn\ell , q-1\right )= 1$ ,

• (3) g (x) = x^rh (x) ^q− 1permutesμ_q+ 1.

Proof

If the condition (1) holds then R(x)≠ 0 for x ∈ μ_q+ 1. In fact, assume that α ∈ μ_q+ 1 is a root of R(x), i.e., $\sum \limits _{i = 0}^{n} \alpha ^{ki}= 0$. We have

$$\alpha^{(n + 1)k}+ 1=(\alpha^{k}+ 1)\left( \sum\limits_{i = 0}^{n} \alpha^{ki}\right) = 0. $$

From $\gcd ((n + 1)k, q + 1)= 1$, we get α = 1. But α = 1 is not a root of R(x) since n is even. This proves that R(x)≠ 0 for x ∈ μ_q+ 1. By Theorem 3 and conditions (2) and (3) we know that F(x) permutes $\mathbb {F}_{q^{2}}$.

Conversely, if F(x) permutes $\mathbb {F}_{q^{2}}$ then (2) holds and the polynomial G(x) = x^{r + knℓ}(R(x)^ℓh(x))^q− 1 permutes μ_q+ 1 by Lemma 1. From G(x) permutes μ_q+ 1 we know that conditions (1) and (3) hold. □

From Corollary 1 we list some examples of permutation polynomials over $\mathbb {F}_{q^{2}}$ from permutation polynomials over the unite circle μ_q+ 1.

Example 1

Let m be a positive integer and q = 2^m. The following polynomials permute the finite field $\mathbb {F}_{q^{2}}$.

• (1) x⁷ + x^3q+ 4 + x^4q+ 3 + x^6q+ 1 + x^7q for an m satisfying $\gcd (m,3)= 1$ and m ≢ 2 mod 4.

• (2) x¹¹ + x^q+ 10 + x^2q+ 9 + x^4q+ 7 + x^6q+ 5 + x^8q+ 3 + x^11q for an m satisfying m ≢ 0 mod 10 and m ≢ 2 mod 4.

• (3) x¹¹ + x^q+ 10 + x^3q+ 8 + x^4q+ 7 + x^5q+ 6 + x^6q+ 5 + x^7q+ 4 + x^8q+ 3 + x^11q for an even m satisfying m ≢ 0 mod 10.

• (4) $x^{2^{k}+ 3}+x^{3q + 2^{k}}+x^{2^{k}\cdot q + 3}+x^{(2^{k}+ 1)q + 2}+x^{(2^{k}+ 2)q + 1}$ for a positive integer k and an even m satisfying $\gcd (2^{k}+ 3,q-1)= 1$ and $\gcd (2^{k}+ 1,q + 1)= 1$.

• (5) $x^{2^{k}+ 3}+x^{3q + 2^{k}}+x^{(2^{k}+ 1)q + 2}+x^{(2^{k}+ 2)q + 1}+x^{(2^{k}+ 3)q}$ for a positive integer k and an even m satisfying $\gcd (2^{k}+ 3,q-1)= 1$ and $\gcd (2^{k}-1,q + 1)= 1$.

Proof

We only prove the case (1) and the proofs of the other cases are similar, and omit the details here.

Let R(x) = 1 + x + x² + x³ + x⁴ and h(x) = 1 + x + x³. It is verified that F(x) = x⁷ + x^3q+ 4 + x^4q+ 3 + x^6q+ 1 + x^7q = x⁷R(x^q− 1)h(x^q− 1). This is F(x) in Corollary 1 for n = 4, r = 3, k = 1 and ℓ = 1. We easily verify that $\gcd ((n + 1)k, q + 1)=\gcd (5, 2^{m}+ 1)= 1$ since m ≢ 2 mod 4, and $\gcd (r+kn\ell , q-1)= \gcd (7, 2^{m}-1) = 1$ since $\gcd (m,3)= 1$. By Lemma 7, we know that g(x) = x^rh(x)^q− 1 = x³(1 + x + x³)^q− 1 permutes μ_q+ 1. So, the polynomial in (1) is a permutation polynomial over $\mathbb {F}_{q^{2}}$ from Corollary 1. □

In the end of this section, we list all the known pairs (s, t) such that the polynomials of the form (1) are permutations in Table 1.

Table 1 Known pairs (s, t) such that f(x) defined by (1) are permutation polynomials

4 A class of permutation trinomials over $\mathbb {F}_{3^{2m}}$

It is clear that polynomials of the form (1) are a special case of polynomials of the form $x^{r}h(x^{\frac {q-1}{d}})$, where r, d are positive integers satisfying $d \,|\, (q-1), 1 \leq r<\frac {q-1}{d}$ and $h(x)\in \mathbb {F}_{q}[x]$. Recently, some new permutation trinomials of this form were obtained in [9,10,11] over finite fields with odd characteristic. Permutation property of trinomials in those papers is derived by using Lemma 1. At the same time, Hou [6, 7] proposed several classes of permutation trinomials with this form by highly technical calculations. Following some techniques in [6] we obtain a new class of permutation trinomials having the form,

$$ f(x) = x^{4(q-1)+ 1} + x^{(q-1)^{2}+ 1} - x, $$

(8)

over $\mathbb {F}_{q^{2}}$, where q = 3^m.

To this end, we need some preparations. Denote by $\lambda =\text {Tr}_{m}^{2m}(x)=x+x^{q}$ and μ = x ⋅ x^q for $x\in \mathbb {F}_{q^{2}}$. The following lemma can be verified by routine calculations.

Lemma 9

Letq = 3^mand$\mathbb {F}_{q^{2}}$be a finite field. For$x\in \mathbb {F}_{q^{2}}$,

$$\text{Tr}_{m}^{2m}(x^{7})=\lambda^{7}-\mu \lambda^{5} -\mu^{2} \lambda^{3} -\mu^{3}\lambda, \quad \text{Tr}_{m}^{2m}(x^{5})= \lambda^{5}+\mu \lambda^{3}-\mu^{2}\lambda, $$

$$\text{Tr}_{m}^{2m}(x^{4})= \lambda^{4} - \mu \lambda^{2} - \mu^{2}, \quad \text{Tr}_{m}^{2m}(x^{12})= \lambda^{12}-\mu^{3} \lambda^{6}-\mu^{6}, $$

$$\text{Tr}_{m}^{2m}(x^{8})= \lambda^{8} + \mu \lambda^{6} - \mu^{2}\lambda^{4} -\mu^{3}\lambda^{2}-\mu^{4}. $$

Using above lemma we can prove the following theorem.

Theorem 4

Letm be a positive integer withm ≢ 0 (mod 6).Letq = 3^mand$\mathbb {F}_{q^{2}}$be a finite field withq²elements. The polynomial$f(x)=x^{4q-3}+x^{q^{2}-2q + 2}-x$is a permutation trinomial over$\mathbb {F}_{q^{2}}$.

Proof

We first show that f(x) permutes $\mathbb {F}_{q}$. If $f(x)\in \mathbb {F}_{q}$ for $x\in \mathbb {F}_{q^{2}}$ then f(x)^q = f(x), i.e.,

$$ x^{4-3q}+x^{3q-2}-x^{q}= x^{4q-3}+ x^{3-2q}-x. $$

(9)

Denote by y = x^q− 1. Assume x≠ 0 then y≠ 0. From (9) we get

$$ y^{4}+y^{-2}-1=y^{-3}-y^{3}+y. $$

(10)

Equation (10) is reduced to

$$(y-1)^{7}= 0. $$

So, y = 1, i.e., x^q− 1 = 1. Hence, $x\in \mathbb {F}_{q}$. On the other hand, if $x\in \mathbb {F}_{q}$ then f(x) = x^4q− 3 + x^{3 − 2q} − x = x. So, f(x) permutes $\mathbb {F}_{q}$.

From above analysis we know that $f(\mathbb {F}_{q^{2}}\setminus \mathbb {F}_{q})\subset \mathbb {F}_{q^{2}}\setminus \mathbb {F}_{q}$. Next, for any $\alpha \in \mathbb {F}_{q^{2}}\setminus \mathbb {F}_{q}$, we show that the equation

$$ f(x)=x^{4q-3}+x^{q^{2}-2q + 2}-x =\alpha $$

(11)

has one solution in $\mathbb {F}_{q^{2}}\setminus \mathbb {F}_{q}$ only depending on α. Taking trace function from $\mathbb {F}_{q^{2}}$ to $\mathbb {F}_{q}$ on both sides of (11) and using Lemma 9 we have

$$\begin{array}{@{}rcl@{}} \beta =\text{Tr}_{m}^{2m}(\alpha) & =&\text{Tr}_{m}^{2m}(x^{4q-3})+\text{Tr}_{m}^{2m}(x^{3-2q})-\text{Tr}_{m}^{2m}(x) \\ & =&\frac{\text{Tr}_{m}^{2m}(x^{7})}{x^{3(1+q)}}+\frac{\text{Tr}_{m}^{2m}(x^{5})}{x^{2(1+q)}}-\text{Tr}_{m}^{2m}(x)\\ &=& \frac{\lambda^{7}-\mu \lambda^{5} -\mu^{2} \lambda^{3} -\mu^{3}\lambda}{\mu^{3}} +\frac{\lambda^{5}+\mu \lambda^{3}-\mu^{2}\lambda}{\mu^{2}} -\lambda \\ &=& \frac{\lambda^{7}}{\mu^{3}}. \end{array} $$

(12)

Taking norm function from $\mathbb {F}_{q^{2}}$ to $\mathbb {F}_{q}$ on both sides of (11) and using Lemma 9 we have

$$\begin{array}{@{}rcl@{}} \gamma=\alpha \cdot \alpha^{q} &= &(x^{4q-3}+ x^{3-2q}-x)(x^{4-3q}+x^{3q-2}-x^{q})\\ &=& \frac{\text{Tr}_{m}^{2m}(x^{12})}{ x^{5(1+q)}}-\frac{\text{Tr}_{m}^{2m}(x^{8})}{ x^{3(1+q)}}-\frac{\text{Tr}_{m}^{2m}(x^{4})}{x^{1+q}}\\ &=&\frac{\lambda^{12}-\mu^{3} \lambda^{6}-\mu^{6}}{\mu^{5}} -\frac{\lambda^{4} - \mu \lambda^{2} - \mu^{2}}{\mu}\\ &&-\frac{\lambda^{8} + \mu \lambda^{6} - \mu^{2}\lambda^{4} -\mu^{3}\lambda^{2}-\mu^{4}}{\mu^{3}} \\ &=&\frac{\lambda^{12}}{\mu^{5}}-\frac{\lambda^{8}}{\mu^{3}} + \frac{\lambda^{6}}{\mu^{2}} -\lambda^{2} +\mu. \end{array} $$

(13)

From (12) and (13) we calculate $\frac {\gamma }{\beta ^{2}}$ as follows,

$$\begin{array}{@{}rcl@{}} \frac{\gamma}{\beta^{2}} &=& \frac{\mu}{\lambda^{2}}-\frac{\mu^{3}}{\lambda^{6}}+ \frac{\mu^{4}}{\lambda^{8}}-\frac{\mu^{6}}{\lambda^{12}}+\frac{\mu^{7}}{\lambda^{14}} \\ &=& \left( \frac{\mu}{\lambda^{2}}-1\right)^{7} + 1. \end{array} $$

(14)

It is clear that $\lambda , \mu \in \mathbb {F}_{q}$. Since m ≢ 0 (mod 6), we have $\gcd (7, q-1)= 1$, and denote the inverse of 7 modulo q − 1 by $\frac {1}{7}$. So, from (14) we have

$$ \frac{\mu}{\lambda^{2}} = \left( \frac{\gamma}{\beta^{2}}-1 \right)^{\frac{1}{7}}+ 1. $$

(15)

Combining (12) and (15) we have

$$ x+ x^{q} = \lambda = \beta\left( (\frac{\gamma}{\beta^{2}}-1)^{\frac{1}{7}}+ 1\right)^{3}. $$

(16)

Substituting the value of λ into (15) we get

$$ x \cdot x^{q} = \mu = \beta^{2} \left( (\frac{\gamma}{\beta^{2}}-1)^{\frac{1}{7}}+ 1\right)^{7} . $$

(17)

It is know that $x\in \mathbb {F}_{q^{2}} \setminus \mathbb {F}_{q}$. So, x^q≠x. From (16) and (17) we know that x and x^q are two distinct solutions of the following quadratic equation

$$z^{2} -\beta\left( (\frac{\gamma}{\beta^{2}}-1)^{\frac{1}{7}}+ 1\right)^{3} z + \beta^{2} \left( (\frac{\gamma}{\beta^{2}}-1)^{\frac{1}{7}}+ 1\right)^{7}= 0. $$

So, (11) has only one solution x in $\mathbb {F}_{q^{2}}\setminus \mathbb {F}_{q}$, i.e., f(x) permutes the set $\mathbb {F}_{q^{2}}\setminus \mathbb {F}_{q}$. Combining above two cases we know that f(x) permutes $\mathbb {F}_{q^{2}}$. □

5 Concluding remark

In this paper, we present two classes of permutation trinomials with the form (1) over the finite field $\mathbb {F}_{2^{2m}}$ and a class of permutation trinomials like this form over $\mathbb {F}_{3^{2m}}$. Moreover, a method to construct permutation polynomials from known ones is provided. During the study of permutation trinomials of the form (1), we have come across the pair $(s,t)=(\frac {4}{11},\frac {10}{11})$ such that the polynomial of the form (1) for an m with $\gcd (m,5)= 1$ can possibly be a permutation trinomial over $\mathbb {F}_{2^{2m}}$. We have verified the conjecture for m from 2 to 12 using computers. It would be nice if this conjecture can be settled.