New classes of permutation binomials and permutation trinomials over finite fields

Finite Fields and Their Applications 43 (2017) 69–85

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Finite Fields and Their Applications www.elsevier.com/locate/ffa

New classes of permutation binomials and permutation trinomials over finite fields ✩ Kangquan Li, Longjiang Qu ∗ , Xi Chen College of Science, National University of Defense Technology, Changsha, 410073, China

a r t i c l e

i n f o

Article history: Received 13 October 2015 Received in revised form 3 August 2016 Accepted 1 September 2016 Available online xxxx Communicated by Rudolf Lidl MSC: 06E30 11T06 94A60

a b s t r a c t Permutation polynomials over finite fields play important roles in finite fields theory. They also have wide applications in many areas of science and engineering such as coding theory, cryptography, combinatorial design, communication theory and so on. Permutation binomials and permutation trinomials attract people’s interest due to their simple algebraic forms and additional extraordinary properties. In this paper, we find a new result about permutation binomials and construct several new classes of permutation trinomials. Some of them are generalizations of known ones. © 2016 Published by Elsevier Inc.

Keywords: Finite field Permutation polynomial Permutation binomial Permutation trinomial

✩ The research of this paper is supported by the NSFC of China under Grants 61272484, 61572026, the National Basic Research Program of China (Grant No. 2013CB338002), the Basic Research Fund of National University of Defense Technology (No. CJ 13-02-01) and the Program for New Century Excellent Talents in University (NCET). * Corresponding author. E-mail addresses: [email protected] (K. Li), [email protected] (L. Qu), [email protected] (X. Chen).

http://dx.doi.org/10.1016/j.ffa.2016.09.002 1071-5797/© 2016 Published by Elsevier Inc.

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1. Introduction A polynomial f ∈ Fq [x] is called a permutation polynomial of Fq if the associated polynomial function f : c → f (c) from Fq into Fq is a permutation. Permutation polynomials over finite fields play important roles in finite fields theory. They also have wide applications in coding theory, cryptography, combinational design, communication theory and so on. The study of permutation polynomial can date back to Hermite [16] and Dickson [11]. There are numerous books and survey papers on the subject covering different periods in the development of this active area [10, Ch. 18], [25, Ch. 7], [29, Ch. 8], [21,22], while the survey by Hou [21] in 2015 is the most recent one. Permutation binomials and permutation trinomials attract people’s interest due to their simple algebraic forms and additional extraordinary properties. Early works on permutation binomials can be traced to Carlitz [7] in 1962. Since then many existence and nonexistence results and new classes of permutation binomials have been discovered. For examples, and Wells [8] claimed the existence of permutation polynomials   q−1 Carlitz f = x x d + a over Fq for each fixed d when q is sufficiently large. On the basis of Carlitz and Wells’s works, Niederreiter and Robinson [30] gave a criterion for the polynomials with the aforementioned form to permute Fq and Wan and Lidl generalized these results  in [35]. And Masuda and Zieve [27] gave a range of the number of a ∈ Fq such that q−1

xr x d + a is a permutation binomial over Fq . There are also numerous results about nonexistence of permutation binomials proved by several authors, such as Niederreiter and Robinson [30], Masuda and Zieve [27] and so on. Niederreiter and Robinson [30] proved that f = xm + ax ∈ Fq [x], where a = 0 and m > 0 is not a power of charFq , is not  2 a permutation binomial over Fq if q ≥ m2 − 4m + 6 . Later, Turnwald [33], Wan [34], Masuda and Zieve [27] improved and generalized this negative result. In recent years, the study about constructing new explicit permutation binomials is increasingly popular [5,17,23,24,32,36].Wang [36] gave a necessary and sufficient condition for a polynomial q−1 with the form xn x d + 1 to be a permutation over Fq through exploring a connec q−1  tion between permutation polynomials of the form xn f x d and cyclotomic mapping

permutations over finite fields. Recently, Hou and Lappano [17,23,24] determined several explicit classes of permutation binomials with the form ax +x2q−1 , ax +x3q−2 , ax +x5q−4 and ax+x7q−6 over Fq2 by Hermite’s Criterion [16] (cf. Lemma 2.3 in Section 2). The hard  point to use Hermite’s Criterion is to compute x∈F 2 f (x)s for 1 ≤ s ≤ 2n −1 and s ≡ 0 q mod CharFq2 . Another way to obtain binomials is deducing from monomial complete permutation polynomials. A polynomial f (x) ∈ Fq [x] is called a complete permutation polynomial over Fq if both f (x) and f (x) +x are permutation polynomials over Fq . Therefore, if f (x) is a monomial complete permutation polynomial, then f (x)+x is a permutation binomial. Recently, there are many results about complete permutation monomials. Therefore, many permutation binomials are obtained in the meanwhile, such as [5,31,37–40,42]. Most proofs of these permutations utilized the method through computing the exponen Tr (γf (x)) tial sum x∈Fq (−1) q , γ ∈ F∗q , where Trq is the absolute trace function over Fq .

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As for permutation trinomials, there are many interesting constructions, while the existence and nonexistence results seem to be quite rare. For instances, Dobbertin used m m m a trick called as the multivariate method to show that f (x) = x2 + x2 +2 + x3·2 +4 m−1 +1 [15] is an o-polynomial of F22m−1 and he also proved that f (x) = x2 + x3 + x [22] is a permutation polynomial over Fq , where q = 22m−1 , through considering the number of the solutions of the equation f (x) = a in Fq for any a ∈ Fq . Recently, through AGW Criterion [1] raised by Akbary, Ghioca and Wang in 2011, Tu, Zeng and Hu [31] proved 2m m that f (x) = x2 +1 + x2 +1 + vx, where v ∈ F2m /{0, 1}, is a complete permutation polynomial over F23m . In [18–20], Hou determined several types of permutation trinomials over finite fields also by Hermite’s criterion. In addition, Ball and Zieve [2], Ding, Qu, Wang, et al. [13] among others obtained some explicit constructions. For the convenience of the reader, we list all the known permutation trinomials over finite fields with even characteristic in Section 4. For more relevant progresses about permutation binomials and permutation trinomials, readers can consult a recent survey [22] presented by Hou. Although quite a few permutation binomials and permutation trinomials have been found, however, an explicit and unified characterization of them is still missing and seems to be elusive today. Therefore, it is both interesting and important to find more explicit classes of them and try to find the hidden reason for their permutation property. In this paper, we find a new result about permutation binomials over finite fields with even extensions and present several permutation trinomials over finite fields with even characteristic. To prove these results, we use three classical approaches. The first one is the Hermite’s criterion. We use it to determine one class of permutation binomials in Section 3. Secondly, to prove that a trinomial f permutes Fq , we try to show that f (x) = a has at most one solution in Fq for any a ∈ Fq . Lastly, we try to prove that f (x) = f ((1 + c)x) can not hold for any x ∈ F∗q and any c ∈ F∗q after showing that x = 0 is the only solution of f (x) = 0 in Fq . The last class of permutation trinomials will be proved by the third approach while other classes will be proved by the second one. Clearly, some skills or tricks will be needed in these proofs. In Section 2, we introduce the definition of the multiplicative equivalence of permutation polynomials and some lemmas. In Section 3, we provide one class of permutation binomial. In Section 4, we give four new classes of permutation trinomials over finite fields with even characteristic after listing all the known such polynomials. Section 5 is the conclusion. In the following, Trk (x) is treated as the absolute trace function on F2k . The algebraic closure of Fq is denoted by Fq . For integer d > 0, μd = {x ∈ Fq : xd = 1}. Other symbols follow the standard notation of finite fields. 2. Preparation As we all know, f (x) = g(h(x)) is a permutation polynomial over Fq if and only if both g(x) and h(x) permute Fq . Notice that what we consider in this paper are binomials and trinomials, for assuring that the number of the terms of f is equal to that of g, we let h(x) be a monomial. It is trivial that a monomial xd is a permutation polynomial

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over Fq if and only if gcd(d, q − 1) = 1. Let f and g be two polynomials in Fq [x] satisfying that f (x) = ag(xd ), where a ∈ F∗q and 1 ≤ d ≤ q − 1 is an integer such that gcd(d, q − 1) = 1. Then f is a permutation polynomial if and only if so is g. Particularly, f is a permutation binomial resp. permutation trinomial if and only if so is g. Hence we introduce the following definition of multiplicative equivalence. Definition 2.1. Two permutation polynomials f (x) and g(x) in Fq [x] are called multiplicative equivalent if there exists an integer 1 ≤ d ≤ q − 1 such that gcd(d, q − 1) = 1 and f (x) = ag(xd ) for some nonzero element a in Fq . It should be noted that in [21] and many references before, two permutation polynomials f and g of Fq are called equivalent if f (x) = cg(ax + b) + d, where a, c ∈ F∗q , b, d ∈ Fq . In our opinion, this type of equivalence can be called linear equivalence. The following lemmas will be useful in our future discussion. The first one is a corollary following directly from [41, Lemma 2.1]. Lemma 2.2. Let f (x) = xr0 (x(q−1)/d + a) ∈ Fq [x], where d | q − 1, a ∈ F∗q . If f (x) is a permutation polynomial over Fq and gcd(r0 + d, (q − 1)/d) = 1, then g(x) = xr0 +d (x(q−1)/d + a) is also a permutation polynomial over Fq . Lemma 2.3. [16,25] (Hermite’s Criterion) Let Fq be of characteristic p. Then f ∈ Fq [x] is a permutation polynomial of Fq if and only if the following two conditions hold: (i) f has exactly one root in Fq ; (ii) for each integer t with 1 ≤ t ≤ q − 2 and t ≡ 0 f (x)t mod xq − x has degree ≤ q − 2.

mod p, the reduction of

Lemma 2.4. (Lucas formula) Let n, i be positive integers and p be a prime.  Asn sume n = am pm + · · · + a1 p + a0 and i = bm pm + · · · + b1 p + b0 . Then ≡ i      am−1 a0 am ··· (mod p). bm bm−1 b0 Lemma 2.5. Let q = 2k and k > 0 be an integer. Then f (x) = x +x2 +x4 is a permutation polynomial over Fq if and only if k ≡ 0 (mod 3). Proof. This lemma can be easily proved. We omit it here.

2

Lemma 2.6. [3] Let a ∈ F∗2k . Then the cubic equation x3 + x = a has a unique solution   in F2k if and only if Trk a−1 = Trk (1).

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3. Permutation binomials Given f (x) = xm + axn ∈ Fq [x], where 0 < n < m < q and a ∈ F∗q , there exist integers r, t, d > 0 with gcd(t, q − 1) = 1 and d | q − 1 such that f (xt ) ≡ xr (x(q−1)/d + a) (mod xq − x) [21]. Therefore, w.l.o.g., we only consider permutation binomials with the form xr (x(q−1)/d + a). Due to Lemma 2.2, we can assume 1 ≤ r ≤ d. Inspired by Hou and Lappano [23], in this section, we consider permutation binomials over finite fields with even extensions, that is, over Fq2 . Theorem 3.1. Let f (x) = xr (xq−1 +a) ∈ Fq2 [x], 1 ≤ r ≤ q+1. Then f (x) is a permutation binomial over Fq2 if and only if r = 1 and aq+1 = 1. Proof. First, it is clear that f (x) = xr (xq−1 + a) has only one root in Fq2 if and only if aq+1 = 1. Next, let 0 ≤ α, β ≤ q − 1 with (α, β) = (0, 0), (q − 1, q − 1). We have

α+βq

f (x)



=

x∈Fq2

xr(α+βq) (xq−1 + a)

(α+βq)

x∈Fq2



=

α

β

xr(α+βq) (xq−1 + a) (x1−q + aq )

x∈F∗ q2



=

x

r(α+βq)

x∈F∗ q2

= aα+βq

β   α   α β β−j i(q−1) α−i x xj(1−q) (aq ) a i j i=0

0≤i≤α,0≤j≤β

j=0

   α β a−i−qj xr(α+βq)+(i−j)(q−1) . i j ∗ x∈Fq2

The inner sum is 0 unless r(α + βq) + (i − j)(q − 1) ≡ 0 (mod q 2 − 1), which can happen only if α + βq ≡ 0 (mod q − 1), or equivalently, α + β = q − 1. Let β = q − 1 − α. Then we have

α+(q−1−α)q

f (x)

x∈Fq2



(α+1)(1−q)

=a

0≤i≤α,0≤j≤q−1−α

= −a(α+1)(1−q)

   α q−1−α a−i−qj x(q−1)(−r(α+1)+i−j) i j ∗ x∈Fq2

   α q−1−α a−i−qj . i j

−r(α+1)+i−j≡0

(mod q+1)

As i runs over the interval [0, α] and j over the interval [0, q−1−α], −r(α+1)+i−j ∈ Sα,r , where Sα,r = [−r(α + 1) − (q − 1 − α), −r(α + 1) + α].

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If r = 1, then Sα,r = [−q, −1], and the equation −(α+1) +i −j ≡ 0 (mod q +1) has no  α+(q−1−α)q solution. Therefore, f (x) = 0. Hence, f (x) is a permutation polynomial x∈Fq2

over Fq2 . If r ∈ [2, q + 1], let α = 0. Then S0,r = [−r − q + 1, −r]. In this case, the only multiple of q + 1 in S0,r is −(q + 1). Therefore, we have

 (q−1)q

f (x)

= −a(1−q)

x∈Fq2

q−1 q+1−r



a−1−q+rq = −



q−1 q+1−r

 a(r−2)q .

  q−1 ≡ 0 (mod p) if r ∈ [2, q + 1]. Therefore, It follows from Lucas formula that q+1−r  (q−1)q f (x) = 0. Hence, when r ∈ [2, q + 1], f (x) is not a permutation polynomial x∈Fq2

over Fq2 . The proof is finished. 2 In Theorem 3.1, let r = 1, then we get f (x) = xr (xq−1 + a) = xq + ax, which is a linearized polynomial. Hence Theorem 3.1 does not provide a new permutation binomial. However, to the authors’ best knowledge, the characterization of the sufficient and necessary condition for the polynomial with this form to be a permutation is new. 4. Permutation trinomials 4.1. The known permutation trinomials over F2m First, we review the known permutation trinomials over F2m . The results before 2014 can be summarized in [13]. We copy their list into the following theorem for the readers’ conveniences, Theorem 4.1. (1) (2) (3) (4) (5) (6) (7) (8)

Some linearized permutation trinomials described in [25]. x + x3 + x5 over F2m , where m is odd (the Dickson polynomial of degree 5). x + x5 + x7 over F2m , where m ≡ 0 (mod 3) (the Dickson polynomial of degree 7). (m+1)/2 +1 over F2m , where m is odd. [14] x + x3 + x2 2k m k 2 +1 2k +1 [4] x + (ax) + ax2 over F2m , where m = 3k and a(2 −1)/(2 −1) = 1. (m+1)/2 (m+1)/2 (m+1)/2 +4 +2 + x2 + x2 over F2m , where m is odd. [9,15] x3·2 2k k 2 +1 2 +1 +x + vx over F2m , where m = 3k and v ∈ F∗2k . [31] x [26] Let q ≡ 1 (mod 3) be a prime power. Let α be a generator of F∗q , s = (q − 1)/3, and let ω = αs . Define f (x) = ax2 + bx + c ∈ Fq [x]. Then h(x) := xr f (xs )

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is a permutation polynomial over Fq if and only if the following conditions are satisfied: a) gcd(r, s) = 1; b) f (ω i ) = 0 for 0 ≤ i ≤ 2; c) logα (f (1)/f (ω)) ≡ logα (f (ω)/f (ω 2 )) ≡ r (mod 3). The following results are new classes of permutation trinomials constructed in [13]. −1 +x2 Theorem 4.2. [13] Let m > 1 be an odd integer. Then both x+x2 3 2m −2(m+3)/2 +2 and x + x + x are permutation polynomials over F2m . (m+1)/2

(m+2)/2

Theorem 4.3. [13] Let m be a positive even integer. Then x + x2 is a permutation polynomial over F2m .

m

−1

−2(m+1)/2 +1

m

+ x2

−2m/2 +1

Theorem 4.4. [13] Let k be a positive integer and q be a prime power with q ≡ 0 (mod 3). m/2 m/2 Let m be a positive even integer. Then x + xkq −(k−1) + xk+1−kq is a permutation polynomial over Fqm if and only if one of the following three conditions holds: (i) m ≡ 0 (mod 4); (ii) q ≡ 1 (mod 4); (iii) m ≡ 2 (mod 4), q ≡ 2 (mod 3), and exp3 (q m/2 + 1), where exp3 (i) denotes the exponent of 3 in the canonical factorization of i. Recently, some particular types of permutation trinomials have been determined by Hou [18–20]. We list these permutation trinomials in even characteristic. In fact, the results in [18,19] are special cases of those in [20]. Therefore, we only list the result in [20]. Theorem 4.5. [20] Let f = ax + bxq + x2q−1 ∈ Fq2 [x], where q is even. Then f is a permutation polynomial over Fq2 if and only if one of the following is satisfied: (i) a = b = 0, q = 22k ;  0, a = b1−q , Trq/2 (b−1−q ) = 0; (ii) ab = (iii) ab(a − b1−q ) = 0, ba2 ∈ Fq , Trq/2 ( ba2 ) = 0, b2 + a2 bq−1 + a = 0. Very recently, Ma, Zhang, Feng, et al. introduced two classes of permutation trinomials. Theorem 4.6. [28] Let m > 1 be an odd integer such that m = 2k − 1. Then both x + k k m k+1 k−1 k k−1 k ux2 −1 +u2 x2 −2 +2 and x +u2 −1 x2 −1 +u2 x2 +1 are permutation polynomials over F2m for any u ∈ F2m .

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4.2. New classes of permutation trinomials In this subsection, we introduce four new classes of permutation trinomials over F2m . Motivated by [13], we first consider permutation trinomials with trivial coefficients, that is, whose nonzero coefficients are all 1. k

2k−1

Theorem 4.7. Let q = 22k and k be a positive integer. Then f (x) = x+x2 +x2 is a permutation trinomial over Fq if and only if k ≡ 0 (mod 3).

−2k−1 +1

Proof. First of all, we show that x = 0 is the only solution of f (x) = 0 in Fq when k 2k−1 −2k−1 k ≡ 0 (mod 3). If f (x) = 0, then either x = 0 or 1 + x2 −1 + x2 = 0. Therefore, we only need to prove that the equation k

1 + x2

−1

2k−1

+ x2

−2k−1

=0

(1)

has no solution in Fq . k k Let y = x2 −1 . Then computing (y ∗ (1))2 and simplifying it by y 2 +1 = 1, we get y + y 2 + y 4 = 0.

(2)

It follows from Lemma 2.5 that (2) has no nonzero solution in Fq when k ≡ 0 (mod 3). Since x = 0 is not the solution of (1), f (x) = 0 has only one solution in Fq when k ≡ 0 (mod 3). Then we prove that f (x) = a has at most one solution in Fq for any a ∈ F∗q if and only if k ≡ 0 (mod 3). Let d = 22k−1 − 2k−1 + 1. Let s be an integer satisfying 1 ≤ s ≤ 22k − 2 and 4s ≡ 2k + 3 (mod 22k − 1). Then it is easy to verify that ds ≡ 1 (mod 22k − 1). And we have the following equation from f (x) = a: k

x + x2 + xd = a.

(3)

k

Let u = xd + a = x + x2 . Then u ∈ F2k and x = (a + u)s . Plugging it into (3), we have k

(a + u)s + (a + u)2

·s

= u.

Raising the above equation to the 4-th power and simplifying it, we get k+1

u4 + (a2 k

k

Let b = a2 + a, c = a2

k

k

+ a2 )u2 + (a2 + a)3 u + a2

+1

+1

k+1

(a2 + a2

) = 0.

(4)

. Then b, c ∈ F2k , and the above equation reduces to u4 + b2 u2 + b3 u + b2 c = 0.

If b = 0, then there is only one solution u = 0 of the above equation. Hence, f (x) = a has at most one solution x = as in Fq for any a ∈ F∗q .

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If b = 0, let u = bv. Rewriting the above equation, we get k

v4 + v2 + v =

c a2 +1 = 2k . 2 b (a + a)2

(5)

Since x = (a + u)s = (a + bv)s , it suffices to prove that (5) has at most one solution in F2k for any a ∈ F∗q if and only if k ≡ 0 (mod 3). Then the result follows from Lemma 2.5. We finish the proof. 2 k

Theorem 4.8. Let q = 22k and k > 0 be an odd integer. Then f (x) = x + x2 2k−1 +2k−1 +1 x2 is a permutation trinomial over Fq .

+2

+

Proof. First of all, we show that x = 0 is the only solution of f (x) = 0 in Fq . If f (x) = 0, k 2k−1 +2k−1 then either x = 0 or 1 + x2 +1 + x2 = 0. Therefore, we only need to prove the equation k

1 + x2

+1

2k−1

+ x2

+2k−1

=0

(6)

has no solution in Fq . Raising (6) to its 2-th power, we have k+1

1 + x2 k

Let y = x2

+1

+2

k

+ x2

+1

= 0.

. Then y ∈ F2k , and we get 1 + y + y 2 = 0.

(7)

It is clear that y = 1 is not the solution of (7). Then multiplying (1 + y) to both sides, we have y 3 = 1. We know gcd(2k − 1, 3) = 1 since k is odd. Therefore the above equation has no solution in F2k . Hence, x = 0 is the only solution of f (x) = 0 in Fq . Next, we prove that f (x) = a has at most one solution in Fq for any a ∈ F∗q . Considering the following equation: k

k

Let u = xa . Then u = 1 + x2 (8), we have

+1

+1

+ x2

2k−1

+ x2

k+1

1+ 2k +1

2k−1

1 + x2

a2

u4

+2k−1

+2k−1

=

a . x

∈ F∗2k . Plugging x =

(8) a u

into the square of

k

+2

+

a2 +1 = u2 . u2

(9) k

Let y = a u2 . Then y ∈ F2k . Plugging it into (9), we have y 3 + y 2 + y = a2 +1 . Since g(y) = y 3 + y 2 + y = (y + 1)3 + 1 is a permutation polynomial over F2k when k > 0 is odd, we know that f (x) = a has at most one solution in Fq for any a ∈ F∗q . Hence the proof is complete. 2

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Next we introduce two new classes of permutation trinomials with the form f (x) = x + axα + bxβ ∈ F2m [x], where a, b ∈ F∗2m . Theorem 4.9. Let q = 22k and k > 0 be an integer, f (x) = x +ax2 −1 +a2 x2 −2 +1 , where a ∈ Fq and the order of a is 2k + 1. Then f (x) is a permutation trinomial over Fq . k+1

k−1

2k

k

Proof. First of all, we show that x = 0 is the only solution to f (x) = 0. If f (x) = 0, k+1 k−1 2k k then either x = 0 or 1 + ax2 −2 + a2 x2 −2 = 0. Therefore, we only need to prove the equation k+1

1 + ax2

−2

k−1

+ a2

2k+1

has no solution in Fq . Adding (10) to (10) x3·2

k

−3

2k

x2

−2k

= 0,

(10)

, we have k−1

= a3·2

.

Therefore, x2 −1 = a2 or x2 −1 = a2 ω, where ω 2 + ω + 1 = 0. k 2k−1 2k −1 2k−1 +2k−1 If x =a , then by plugging it into (10), one get 1 + a2 +1 + a2 = 0. Recalling the order of a is 2k + 1, we have 1 = 0. It is a contradiction. k k−1 k If x2 −1 = a2 ω, then by plugging it into (10), one have ω 2 + ω 2 + 1 = 0. Hence k k−1 k k ω 2 = ω, which means that k is even. However, in this case, (a2 ω)2 +1 = ω 2 +1 = k k−1 ω 2 = 1. It then follows that the equation x2 −1 = a2 ω has no solution in Fq . Contradicts! Hence, x = 0 is the only solution to f (x) = 0 in Fq . Next we prove that f (x) = c has at most one solution in Fq for any c ∈ F∗q . Let us consider the equation k

k−1

k

k−1

k+1

x + ax2

−1

k−1

+ a2

k

x2−2 = c.

Raising it to its square, we have k+2

x 2 + a2 x 2 2k

Computing (11) + (11)

−2

k

k+1

+ a2 x4−2

= c2 . k

∗ a, and simplifying it by using a2 k+1

x2 + ax2

k+1

= c2 + ac2

+1

(11) = 1, we have

.

k

Let u = x2 + c2 . Then u = au2 . Plugging x2 = u + c2 into (11), we get k+1

u + a2 (u + c2 )2

−1

k

k

+ a2 (u + c2 )2−2 = 0.

(12)

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k

Multiplying (u + c2 )2 +1 across both sides of the above equation and then substituting k k u2 = u/a and a2 +1 = 1 into it, one can get the following equation after simplification. u3 + αu + β = 0, k+1

k+2

(13)

k+1

where α = ac2 +2 + c4 + a2 c2 , β = a3 c3·2 + c6 . k+1 + c2 . Then β = γα. We distinguish two cases. Now let γ = ac2 Case 1: γ = 0. Then α = c4 and β = 0. Therefore, the solutions of (13) are u = 0 or u = c2 . So x = 0 and x = c may be the solutions of the equation f (x) = c. However, f (0) = c. Hence, in this case, f (x) = c has one unique nonzero solution x = c for each nonzero c ∈ Fq . Case 2: γ = 0. If α = 0, then u = 0 is the only solution of (13). k

If α = 0, then let u = α1/2 ε. Further, α2 =

α a2 ,

k

ε2 =

k

u2 (α1/2 )2k

=

u a a α1/2

= ε. Therefore

3 2

ε ∈ F2k . Dividing (13) by α results in γ

ε3 + ε +

α1/2

= 0.

γ ∈ F2k . Hence h(ε) = ε3 + ε + α1/2 ∈ F2k [ε]. According to   Lemma 2.6, h(ε) = 0 has only one solution in F2k if and only if Trk γα2 = Trk (1) + 1.   Next, we show that Trk γα2 = Trk (1) + 1. We know

It is routine to verify that

γ α1/2

 Trk

α γ2



 = Trk

k+1

ac2



k+2

+2

+ c4 + a2 c2 c4 + a2 c2k+2  k+1



ac2 +2 +1 c4 + a2 c2k+2

 k+1 ac2 +2 = Trk (1) + Trk c4 + a2 c2k+2 = Trk

and

Trk

k+1

ac2 +2 4 c + a2 c2k+2



= Trk

= Trk

Let v =

1 . 1+ac2k+1 −2

k

Then v 2 =

ac2 −2 1 + a2 c2k+2 −4 k+1



1 1 + 2 2 1 + ac2k+1 −2 (1 + ac k+1 −2 )

1 k+1 1+a2k c2−22



= v + 1. Therefore we have

.

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80

Trk  Hence, Trk

α γ2

k+1

ac2 +2 4 c + a2 c2k+2



  k = Trk v + v 2 = v + v 2 = 1.



= Trk (1) + 1.    12 Then, Trk ( α γ ) = Trk γα2 = Trk (1). Hence, h(ε) = 0 has only one solution in F2k according to Lemma 2.6. Therefore f (x) = c has at most one solution in Fq for any c ∈ F∗q . The proof is complete. 2 1/2

The last class of permutation trinomial is a generalization of the second permutation trinomial in Theorem 4.2 [13, Theorem 2.2]. 2k+1

Theorem 4.10. Let q = 22k+1 , f (x) = x + ax3 + a2 f (x) is a permutation trinomial over Fq .

−2k+1 22k+1 −2k+2 +2

x

, a ∈ Fq . Then

Proof. The case a = 0 is trivial. If a = 1, then it reduces to [13, Theorem 2.2]. Therefore, we assume a = 0, 1 in the following. k+1 k+1 k+1 k+1 Let y = x2 and b = a2 . Then y 2 = x2 and b2 = a2 . For all x ∈ F∗q , we have f (x) = x + ax3 +

a x3 x(abx2 y 2 + by 2 + ax2 ) = . b y2 by 2

(14)

First, we show that x = 0 is the only solution of f (x) = 0 in Fq . From (14), we only need to prove the equation abx2 y 2 + by 2 + ax2 = 0

(15)

has no solution in F∗q . Raising (15) to its 2k+1 -th power, we have ba2 x4 y 2 + a2 x4 + by 2 = 0.

(16)

2

Computing (15) + (16), we have k+1

(1 + ax2 )2

+2

= 0.

Then x2 = a1 and y 2 = 1b . Plugging them into (15), we get 1 = 0. It is a contradiction. If f (x) is not a permutation polynomial of F∗q , then there exists x ∈ F∗q and c ∈ F∗q k+1 such that f (x) = f ((1 + c)x). Let d = c2 . It is clear that c, d = 0, 1 and c = d. Then 2

2

2

2

(1 + c)x(b(1 + d) y 2 + ab(1 + c) (1 + d) x2 y 2 + a(1 + c) x2 ) x(by 2 + abx2 y 2 + ax2 ) = . 2 by 2 b(1 + d) y 2

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81

After simplifying, we get A1 x2 y 2 + A2 y 2 + A3 x2 = 0,

(17)

where A1 = abc(1 + d)2 (c2 + c + 1), A2 = bc(1 + d)2 , A3 = a[(1 + d)2 + (1 + c)3 ]. Raising (17) to its 2k+1 -th power, we have k+1

A21 k+1

k+1

x4 y 2 + A23

k+1

Computing (17)*(A21

x4 +A23

k+1

y 2 + A22

x4 = 0.

(18)

) + (18)*(A1 x2 + A2 ), we get

B1 x4 + B2 x2 + B3 = 0,

(19)

where k+1

k+1

B1 = A3 A21

k+1

+1

k+1

+1

B2 = A22 B3 = A32

+ A1 A22

= a3 bd2 (1 + c)4 [(1 + c)3 + (1 + d)3 ],

= a2 bcd(1 + c)4 (1 + d)2 , .

It is routine to verify that A1 , A2 , B1 , B2 = 0. Let x2 = have

B2 B1 ε.

Plugging it into (19), we

ε2 + ε + D = 0, where D =

B1 B3 . B22

(20)

As for D, we know k+1

B1 B3 A2 D= = 3 2 B2 k+1

= D1 + D12

+1

k+1

(A3 A21

2k+2 +2

A2

k+1

+ A1 A22

)

k+1

=

A1 A32

2k+1 +2

A2

k+1

+1

+

A21

k+1

A23

+2

2k+2 +2

A2

,

where k+1

D1 =

A1 A32

+1

2k+1 +2

A2

=

A1 B3 (c2 + c + 1)[(1 + c)4 + (1 + d)3 ][(1 + d)2 + (1 + c)3 ] = . A2 B2 cd(1 + d)2 (1 + c)4

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Now we claim that Tr2k+1 (D1 ) = 1. It is the same claim appeared in the proof of [13, Theorem 2.2]. The proof of this claim is a bit long and intricate. We omit it here. The interested reader please refer [13] for details. Raising (20) to its 2i -th power, where i = 0, 1, · · · , k, and summing them, we get k+1

ε2

=ε+

k

k+1

(D1 + D12

2i

)

=ε+

i=0

2k+1

i

D12 = ε + D1 + Tr2k+1 (D1 ) = ε + D1 + 1,

i=0

and k+1

ε2 B2 B1 ε,

Substituting x2 = obtain

k+1

A1 B22

+1

k+1 +1

B12

+1

= ε(ε + D1 + 1) = D1 ε + D. k+1

2 2 y2 = ( B B1 )

k+1

ε2

and the above two equations into (17), we

k+1

A2 B22 A3 B2 (D1 ε + D) + (ε + D1 + 1) + ε = 0. k+1 2 B1 B1

k+1

k+1

Multiplying B12 +1 across the two sides of the above equation and using B22 k+1 A22 +2 = A2 B2 , we have

=

C1 ε + C2 = 0, where k+1

C1 = A1 A2 B2 D1 + A22 B1 + A3 B12

k+1

= A1 A22

+2

,

C2 = A1 A2 B2 D + A22 B1 (D1 + 1) = A22 B1 . Therefore, ε =

C2 C1

=

B1 k+1 A1 A22

=

A2 B1 A1 B2 .

Plugging it into (20), we have

B1 A22 + A1 A2 B2 = A21 B3 . k+1

k+1

That is A21 A22 = A21 A23 equation, we get

. Substituting the definitions of A1 , A2 , A3 into the above

a2 bd(1 + c)4 (d2 + d + 1)b2 c2 (1 + d)4 = a2 b2 c2 (1 + d)4 (c2 + c + 1)2 b[(1 + c)4 + (1 + d)3 ]. 3

6

Simplifying it, we have (1 + d) = (1 + c) . Since gcd(3, 22k+1 − 1) = 1, one get d = c2 . After raising it to the 2k+1 -th power, we obtain c2 = d2 = d, which means d = 0, 1. It is a contradiction since d = 0, 1. Hence, the proof is complete. 2

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4.3. Multiplicative inequivalence of the new permutation trinomials with known ones In this subsection, we briefly discuss the multiplicative inequivalence of these new permutation trinomials with known ones. First, the last two classes of permutation trinomials are not with trivial coefficients. As far as the authors know, there exist only seven such classes of permutation trinomials in F2m . They are the functions in Theorem 4.1 (5), (7), (8), Theorem 4.5 and those (two) functions in Theorem 4.6. Any of our newly constructed permutation trinomial is multiplicative inequivalent to those in Theorem 4.1 (5), (7) since the latter permutations are defined over F23k . The three conditions in Theorem 4.1 (8) are further investigated in [26]. In general, no simple conditions can make h(x) into a permutation trinomial. Hence it is multiplicative inequivalent to any of our newly constructed permutation trinomial. Considering the dimension parity of the defining field of the permutation, we know that the permutation in Theorem 4.10 is multiplicative inequivalent to those in Theorem 4.5, and the permutation in Theorem 4.9 is multiplicative inequivalent to those in Theorem 4.6. Further, it can be easily verified that the function in Theorem 4.9 is multiplicative inequivalent to those in Theorem 4.5, and that in Theorem 4.10 is multiplicative inequivalent to those in Theorem 4.6. Hence the permutation trinomials in Theorems 4.9 and 4.10 are indeed multiplicative inequivalent to known permutations. Further, they are also multiplicative inequivalent. Now let us discuss the first two classes of permutation trinomials with trivial coefficients. We need to show that they are multiplicative inequivalent to all the known classes. To this end, we used a Magma program to confirm this conclusion for at least one small field F2m , where 4 ≤ m ≤ 10. Consequently, these two classes of permutation trinomials are also new. Further, these two classes are also multiplicative inequivalent to each other. 5. Conclusions Permutation binomials and permutation trinomials over finite fields are both interesting and important in theory and in many applications. In this paper, we find a new result about permutation binomials and construct several new classes of permutation trinomials, and some of them are generalizations of known ones. However, from computer experiments, we found that there should be more classes of permutation binomials and permutation trinomials. A complete determination of all permutation binomials or all permutation trinomials over finite fields seems to be out of reach for the time bing. The constructed functions here may lay a good foundation for the further research. At last, we would like to mention that the constructed functions have many applications. For instances, they can be employed in linear codes [6] and cyclic codes [12], they can also be used to construct highly nonlinear functions such as bent and semi-bent functions. For more details, please refer to the last paragraph in [13] and the references therein.

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