Two classes of permutation trinomials with Niho exponents over finite fields with even characteristic

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Abstract

In this paper, we consider two classes of permutation trinomials with Niho-type exponents over the finite field F22m. Some sufficient conditions are obtained to characterize the coefficients of the permutation trinomials. Our numerical result suggests that those sufficient conditions for one class are also necessary.

Introduction

Let Fq be the finite field with q elements. A polynomial fFq[x] is called a permutation polynomial of Fq if its associated polynomial mapping f:cf(c) from Fq to itself is a bijection. Permutation polynomials over finite fields have wide applications in coding theory, cryptography, and combinatorial design theory [21], and we refer the readers to [10], [19], [29] for more details of the recent advances and contributions to the area.

Permutation polynomials with a few terms are of great interest for their simple algebraic forms and some additional properties [3], [4], [5], [6], [7], [9], [11], [12], [14], [15], [16], [17], [18], [23], [24], [25], [27], [30], [31], [32], [33]. In particular, permutation polynomials having the form xrf(x(q1)/) were intensively studied [22], [26], [28], [34]. Permutation trinomials of the formf(x)=x+axs(q1)+1+bxt(q1)+1Fq2[x], where s and t are two integers, have attracted much attention in recent years [5], [6], [8], [9], [11], [12], [15], [17], [18], [24], [25]. Note that s, t should be interpreted as modulo q+1. For instance, let q=2m, (s,t)=(12,34)=(2m1+1,2m2+1). Given (s,t), finding conditions on a, b that are sufficient and necessary for f to be a permutation polynomial of Fq2 is a hard question, in most known cases the coefficients are assumed to be trivial, that is, (a,b)=(1,1), and a number of possible pairs (s,t) with trivial coefficients such that f in (1) are permutations have been determined. A few years ago, Hou determined all the coefficients a, bFq2 making f in (1) a permutation polynomial in the case (s,t)=(1,2) [8], [9]. To the best of our knowledge, this excellent work is the first instance that all possible coefficients for f in (1) to be a permutation polynomial are completely determined. Recently, Tu and Zeng determined all possible coefficients a, bFq2 making f in (1) permutations for the case (s,t)=(12,12) [24]. They transformed the problem to that of investigating the number of solutions in the unit circle by using the well-known result (Lemma 2.1 in Section 2). Then they can find the sufficient conditions. By a known result on the Kloosterman sum, they showed that the sufficient conditions are also necessary. In a recent work [25], Tu et al. investigated the case (s,t)=(2,q) for even q. They found sufficient conditions on the coefficients a, b for f in (1) to be a permutation polynomial of Fq2 and due to some numerical results they conjectured that the sufficient conditions are also necessary. The conjecture has been confirmed by Bartoli in [3] and later by Hou in [11] utilizing the Hasse-Weil bound via different approaches. In [24], Tu and Zeng considered the case (s,t)=(14,34) with q even, and a, bFq2. They proved that f(x)=x+axs(q1)+1+bxt(q1)+1 is a permutation of Fq2 if b=a2q and x3+x+a1q0 for all xFq. Very recently, Hou showed that the sufficient condition for the case (s,t)=(14,34) found in [24] is also necessary using a strategy similar to that of [11] through quite involved computations [12].

Let n=2m be a positive integer. The purpose of this paper is to find more new permutation trinomials f(x) as in (1) over F2n with general coefficients a and b. Recently, Wang, Zhang and Zha studied the permutation behavior of x+x2m+xd over F2n for m even and d=i=0n22ik with k being an odd positive integer satisfying gcd(k,m2)=1 [30, Section 3]. Inspired by this work, we investigate a family of infinite classes of permutation trinomials with the form f(x)=x+ax2m+bxd defined on F2n, however, for m being any positive integer, d=i=0n22ik with k being an odd positive integer satisfying gcd(k,m)=1, and a, bF2m. We find that the exponents of the family of these infinite classes are of Niho-type, and then with Lemma 2.1 we need to analyze some affine equations over F2m with degree 2k. We prove that the conditions on a and b are sufficient, and the numerical experiments indicate that they are also necessary. On the other hand, inspired by a recent work of Li and Helleseth [18], we consider the case of (s,t)=(2k2k1,12k1), where k is a positive integer satisfying 1km1 and gcd(2k1,2m+1)=1. Assuming that a, bF2m, we obtain some sufficient conditions on a and b such that f(x)=x+axs(2m1)+1+bxt(2m1)+1 is a permutation of F2n.

The remainder of this paper is organized as follows. In Section 2, we introduce some basic concepts and investigate the conditions on a, b that are sufficient and necessary for the affine equation x2k+ax+b=0 to have exactly two solutions (and one solution, respectively) in the unit circle, where a, bF2m, and k satisfies gcd(k,m)=1. In Section 3, we propose two classes of permutation trinomials having the form as in (1). Section 4 concludes the paper.

Section snippets

Preliminaries

Let m and n be two positive integers satisfying m|n, we use Trmn() to denote the trace function form F2n to F2m, i.e.,Trmn(x)=x+x2m+x22m++x2(n/m1)m.

The following result was discovered independently by several authors. It is also worth pointing out that it can also be derived as a special case of the AGW criterion ([1, Lemma 1]).

Lemma 2.1

[2], [22], [26], [28], [34] Let ℓ, r>0 with |(q1), and let h(x)Fq[x]. Then f(x)=xrh(x(q1)/) permutes Fq if and only if both

  • (1)

    gcd(r,(q1)/)=1 and

  • (2)

    xrh(x)(q1)/

Two classes of permutation trinomials from Niho exponents

In this section, we consider two classes of permutation trinomials from Niho exponents with the form (1) over Fq2 for q even. It can be easily seen that if f(x)=x+axs(q1)+1+bxt(q1)+1Fq2[x] is a permutation of Fq2, where ab0, then a+b+10. Thus, the assumption that a+b+10 is needed when we consider the permutation property of f. We have the following theorem.

Theorem 3.1

Let k be an odd positive integer, and m, n and d be positive integers such that n=2m, 1kn1, gcd(m,k)=1 and d=i=0m2ik. Let a, bF2m

Conclusion

In this paper, we investigate the possible coefficients a, bF2m, for which (1) is a permutation over F2n, in the two cases: 1) (s,t)=(1,d), where d=2k(2mk1)(2k1)(2m1), and k is an odd integer such that gcd(k,m)=1 and 1k2m1; 2) (s,t)=(2k2k1,12k1), where k is a positive integer with k<m such that gcd(2k1,2m+1)=1. The conditions in Theorem 3.1 are proved to be sufficient, and numerical result suggests that they are also necessary. It is a question worthy to be investigated in the

CRediT authorship contribution statement

Lijing Zheng: Conceptualization, Methodology, Software, Investigation, Writing - original draft. Haibin Kan: Resources, Supervision. Jie Peng: Resources, Writing - review & editing, Data curation.

Acknowledgements

The authors would like to thank Professor Xiang-dong Hou for his helpful suggestions which improved this paper. The authors thank the anonymous reviewers for their valuable suggestions which significantly improved both the quality and the presentation of this paper. This research is supported by National Key R & D Program of China (No. 2019YFB2101703) and National Natural Science Foundation of China (Grant Nos. 61672166, 11701488, 61972258 and U19A2066), and Scientific Research Fund of Hunan

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