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 . 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  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  aimed to investigate permutation trinomials over of the form
where , and , are two integers, and consequently, four classes of permutation trinomials over with the form (1) were obtained in  based on some subtle manipulation of solving equations with low degree over finite fields, and another two classes of such permutations were presented in  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 . Very recently, followed the work of , by some delicate operation of solving equations with low degrees over finite fields, Deng and Zheng  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 .
2 A conjecture on permutation trinomials of the form (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
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
has a unique solution in for any if , which is equivalent to proving that the equation
has at most one solution in for any if .
3 Proof of Conjecture 1
This section presents the proof of Conjecture 1.
Let , where . If , where , is a factor of , then , must satisfy one of the following conditions:
(1) and ;
(2) and ;
Assume that can be factorized as
Expanding the right hand side of and comparing the coefficients of where gives
and comparing the coefficients of for gives
Then, according to the values of and , one gets
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
Thus, in this case we have
Case 2. If , by (6) one gets , which implies , and one then has
due to .
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
Moreover, one has
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  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.
Let , where . If , where and , is a quadratic factor of , then must satisfy
where and .
Taking -th power on both sides of (7) gives
On the other hand, by and , one gets , and then
Taking -th power on both sides of (9) gives
Again by and , one has
Then, (14) can be written as
This together with (9) implies that
since . Substituting it into (9) gives
which is equivalent to
By , one gets , , which implies
Let and , then we have
Thus, by Lemma 2 (3), one obtains
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
Notice that is irreducible over and , then all the roots of are conjugate over and lie in [15, Thm. 2.14]. Define
The polynomial can be factorized over as follows:
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
which implies that
Note that . Then, , and , i.e., for any and any . This implies the proof. ∎
has no solution if .
According to Lemma 4, it suffices to show that
cannot hold for , with .
Suppose that , then taking -th power on both sides of (17) gives
Again by , i.e., , one has
then , which leads to
a contradiction to and . This completes the proof. ∎
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  was settled.
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