A conjecture on permutation trinomials over finite fields of characteristic two

In this paper, by analyzing the quadratic factors of an 11-th degree polynomial over the finite field , a conjecture on permutation trinomials over [x] proposed very recently by Deng and Zheng is settled, where n=2m and m is a positive integer with (m,5)=1.

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1 Introduction

Let be a prime power and denote the finite field with elements. A polynomial over is called a permutation polynomial if the induced mapping from to itself is a bijection [15]. Permutation polynomials have been studied for several decades and have important applications in a wide range of areas such as coding theory [2, 9], combinatorial designs [4] and cryptography [18, 19].

The construction of permutation polynomials with a simple algebraic form is an interesting research problem and it has already attracted researchers’ much attention in recent years. By using certain techniques in dealing with equations or polynomials over finite fields, a number of permutation polynomials with a simple form have been obtained, the reader is referred to [3, 5, 6, 7, 8, 10, 12, 16, 23, 24, 26, 27] and the references therein. Motivated by the observation that more than half of the known permutation binomials and trinomials were constructed from Niho exponents, Li and Helleseth [12] aimed to investigate permutation trinomials over of the form

 f(x)=x+xs(2m−1)+1+xt(2m−1)+1, (1)

where , and , are two integers, and consequently, four classes of permutation trinomials over with the form (1) were obtained in [12] based on some subtle manipulation of solving equations with low degree over finite fields, and another two classes of such permutations were presented in [13] by virtue of the property of linear fractional polynomials over finite fields. Meanwhile, some similar and more general results on permutation trinomials over were also obtained in [5, 11]. For the permutation polynomials from Niho exponents, the reader is referred to [20, 21, 22] for some recent results and to a survey paper [14]. Very recently, followed the work of [12], by some delicate operation of solving equations with low degrees over finite fields, Deng and Zheng [1] presented two more classes of permutation trinomials over of the form (1), and proposed a conjecture on such a kind of permutation trinomials based on computer experiments. This paper is devoted to settle the conjecture proposed by Deng and Zheng in [1].

The remainder of this paper is organized as follows. Section 2 gives some notations and the conjecture proposed in [1]. Section 3 proves the conjecture by analyzing the quadratic factors of an -th degree polynomial over the finite field , and Section 4 concludes this paper.

2 A conjecture on permutation trinomials of the form (1)

A criterion for a polynomial of the form (1) to be a permutation polynomial had been characterized by the following lemma which was proved by Park and Lee [17], Wang [25] and Zieve [28].

Lemma 1.

([17, 25, 28]) Let be a prime power and . If such that , then permutes if and only if

(1) ;

(2) permutes the set of the -th roots of unity in .

From now on, let be a positive integer and denote the -th roots of unity in , i.e., the unit circle of by

 μ2m+1={x∈F2n|x2m+1=1}.
Conjecture 1.

([1]) Let and , then defined by (1) is a permutation polynomial over .

Note that if . Then, according to Lemma 1, to prove Conjecture 1, it suffices to show , or equivalently

 x11(1+x4+x10)2m−1 (2)

permutes the unit circle of . Observe that is irreducible of which implies that has solutions in if and only if . Thus, by , one gets for any . It is therefore (2) can be written as

 x11⋅1+x−4+x−101+x4+x10=x11+x7+xx10+x4+1. (3)

Then, to prove Conjecture 1, it suffices to show that (3) permutes the unit circle of if , i.e.,

 x11+x7+xx10+x4+1=t

has a unique solution in for any if , which is equivalent to proving that the equation

 x11+tx10+x7+tx4+x+t=0 (4)

has at most one solution in for any if .

3 Proof of Conjecture 1

This section presents the proof of Conjecture 1.

Lemma 2.

Let , where . If , where , is a factor of , then , must satisfy one of the following conditions:

(1) and ;

(2) and ;

(3) .

Proof.

Assume that can be factorized as

 F(x)=(x2+ax+b)(x9+9∑i=1cix9−i).

Expanding the right hand side of and comparing the coefficients of where gives

 c1 = a+t, c2 = b+a2+at, c3 = a3+a2t+bt, c4 = 1+b2+a2b+a4+a3t, c5 = a+ab2+a5+a2bt+b2t+a4t,

and comparing the coefficients of for gives

 c9 = tb, c8 = b+atb2, c7 = bt+ab+a2tb3, c6 = b2+a2b+a3tb4, c5 = b4t+b2t+a2bt+a4t+a3bb5, c4 = b3+a2b2+ab4t+ab2t+a5t+a4bb6.

Then, according to the values of and , one gets

 {(1+b2+a2b+a4+a3t)b6=b3+a2b2+a4b+(ab4+ab2+a5)t,(a+ab2+a5+a2bt+b2t+a4t)b5=a3b+(b4+b2+a2b+a4)t,

i.e.,

 (ab4+ab2+a5+a3b6)t=b6+b8+a2b7+a4b6+b3+a2b2+a4b, (5) (a2b6+b7+a4b5+b4+b2+a2b+a4)t=ab5+ab7+a5b5+a3b. (6)

In the following we shall consider three cases to prove Lemma 2.

Case 1. If , i.e., since . Then by (5), one obtains that , i.e., . Replacing by gives

 b5+b7+a4b5+b2+a2b+b4+b2=0.

Thus, in this case we have

 ab3+b2+b+a2=0, (7) b6+(a4+1)b4+b3+a2=0. (8)

Case 2. If , by (6) one gets , which implies , and one then has

 b2+b3+a2b2+a=0, (9) a2b6+b5+b4+b2+a4=0 (10)

due to .

Case 3. If and . Then, by (5) and (6), one gets

 b6+b8+a2b7+a4b6+b3+a2b2+a4bab4+ab2+a5+a3b6=ab5+ab7+a5b5+a3ba2b6+b7+a4b5+b4+b2+a2b+a4,

which can be written as

 a10+(b4+b2+1)a6+(b5+b)a4+(b7+b)a2+b10+b8+b7+b5+b3+b2+1=0

by a detailed calculation. This completes the proof. ∎

To prove Conjecture 1, we need to show that in Lemma 2 cannot have two solutions in for any , i.e., cannot have a quadratic factor satisfying have two solutions in . Observe that if are two solutions to , then

 x1+x2=a,x1x2=b.

Moreover, one has

 x2m1+x2m2=a2m=1x1+1x2=x1+x2x1x2=ab,

i.e., . This implies that if is a factor of satisfying has two solutions in , then it must have . Actually this fact has been found in [22] and the number of solutions in to has also been characterized there. We provide the proof of the relation here to make the paper self-contained.

Due to this fact, we further consider the conditions in Lemma 2.

Lemma 3.

Let , where . If , where and , is a quadratic factor of , then must satisfy

 v5+(u4+u2)v3+v2+(u6+u4)v+u10+u4+1=0, (11)

where and .

Proof.

According to Lemma 2, we can discuss the three cases in Lemma 2 as follows:

1. Taking -th power on both sides of (7) gives

 a2mb3⋅2m+b2⋅2m+b2m+a2⋅2m=0. (12)

On the other hand, by and , one gets , and then

 a2mb3⋅2m=ab⋅1b3=ab4,b2⋅2m=1b2,b2m=1b,a2⋅2m=a2b2.

Then, (12) can be written as

 ab4+1b2+1b+a2b2=1b4(a+b2+b3+a2b2)=0. (13)

Combining (7) and (13) gives

 (ab3+a2)b=a+a2b2,

i.e.,

 ab(b+1)=(b+1)4.

If , then (7) leads to since , a contradiction to (8). Thus, one gets and then . Substituting it into (7), one obtains

 b2(b+1)3+b2+b+(b+1)6b2=0,

which can be reduced to , i.e., . Then, and . Thus, (8) is reduced to

 1+(b8+1)b4+1+b4=0,

i.e., , a contradiction to . Then, the first case in Lemma 2 cannot happen if .

2. Taking -th power on both sides of (9) gives

 b2⋅2m+b3⋅2m+a2⋅2mb2⋅2m+a2m=0. (14)

Again by and , one has

 b2⋅2m=1b2,b3⋅2m=1b3,a2⋅2mb2⋅2m=a2b4,a2⋅2m=a2b2.

Then, (14) can be written as

 1b2+1b3+a2b4+a2b2=1b4(b2+b+a2+a2b2)=0.

This together with (9) implies that

 (a2+a2b2)b=a2b2+a,i.e.,a=1b(b2+b+1) (15)

since . Substituting it into (9) gives

 b2+b3+1(b2+b+1)2+1b(b2+b+1)=0,

which is equivalent to

 (b+1)(b7+b5+b3+b+1)=0. (16)

Since and , then by (15) one has

 ab=a2m=1b2m(b2⋅2m+b2m+1)=1b−1(b−2+b−1+1)=b3b2+b+1,

which together with (15) implies that . If , then , a contradiction to (10). If , then by and (16) one gets , a contradiction to (15). Thus, the second case in Lemma 2 cannot happen either if .

3. By , one gets , , which implies

 a−1+a−2m=b+1a,a−1⋅a−2m=ba2.

Let and , then we have

• ;

• ;

• ;

• .

Thus, by Lemma 2 (3), one obtains

 v5+(u4+u2)v3+v2+(u6+u4)v+u10+u4+1=0.

This completes the proof. ∎

Notice that if (4) has two or more solutions in for some in , then must have a quadratic factor, say . If so, by Lemma 3, the coefficients , must satisfy the relation (11). Then, to prove Conjecture 1, we only need to show that (11) has no solution in if ,

To this end, define

 G(x,y)=x5+(y2+y)x3+x2+(y3+y2)x+y5+y2+1.

Notice that is irreducible over and , then all the roots of are conjugate over and lie in [15, Thm. 2.14]. Define

 H={x∈F25m|x5+x2+1=0}.
Lemma 4.

The polynomial can be factorized over as follows:

 G(x,y)=∏θ∈H(x+θ−1y+θ).
Proof.

Since all the roots of are distinct and lie in , thus to complete the proof, we only need to show that if for any . By a direct computation, if , then one gets

 (y2+y)x3 = θ2x5+θx4+(θ4+θ2)x3, (y3+y2)x = θ3x4+(θ4+θ2)x3+θ5x2+(θ6+θ4)x, y5+y2+1 = θ5x5+θ6x4+θ2x2+θ9x+θ10+θ4+1,

which implies that

 G(x,θx+θ2) = (θ5+θ2+1)x5+(θ6+θ3+θ)x4+(θ5+θ2+1)x2 +(θ9+θ6+θ4)x+(θ5+θ2+1)2.

Note that . Then, , and , i.e., for any and any . This implies the proof. ∎

Lemma 5.

has no solution if .

Proof.

According to Lemma 4, it suffices to show that

 x+θ−1y+θ=0 (17)

cannot hold for , with .

Suppose that , then taking -th power on both sides of (17) gives

 x+θ−2my+θ2m=0. (18)

Notice that . Otherwise we have , a contradiction to the facts that and satisfying . Then, by (17) and (18), one can obtain that

 ⎧⎪⎨⎪⎩x=θ+θ2m,y=θ+θ2mθ−1+θ−2m.

Again by , i.e., , one has

 (θ+θ2m)2m=θ+θ2m,

 θ∈F22m∩F25m=F2m,

a contradiction to and . This completes the proof. ∎

4 Conclusion

In this paper, by analyzing the possible quadratic factors of an -th degree polynomial over the finite field , a conjecture on permutation trinomials over proposed by Deng and Zheng in [1] was settled.

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