`Journal of Applied MathematicsVolume 2013 (2013), Article ID 472350, 7 pageshttp://dx.doi.org/10.1155/2013/472350`
Research Article

## Fixed Points of the Dickson Polynomials of the Second Kind

Department of Mathematical Sciences, United Arab Emirates University, P.O. Box 17551, Al Ain, Abu Dhabi 17551, UAE

Received 12 December 2012; Revised 6 March 2013; Accepted 7 March 2013

Copyright © 2013 Adama Diene and Mohamed A. Salim. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

The permutation behavior of Dickson polynomials of the first kind has been extensively studied, while such behavior for Dickson polynomials of the second kind is less known. Necessary and sufficient conditions for a polynomial of the second kind to be a permutation over some finite fields have been established by Cohen, Matthew, and Henderson. We introduce a new way to define these polynomials and determine the number of their fixed points.

#### 1. Introduction

Let be a prime, , , and let be the field of elements. Denote the greatest integer function of by . For a fixed , the polynomials are called the Dickson polynomials of the first and second kind, respectively.

It is well known (see [1]) that for any , there exists such that and the Dickson polynomials may be expressed as For , denote

According to [2], if and is odd, then the sign class of is defined to be the set if satisfies the following congruences: If and is odd, then the sign class of is defined to be the set if satisfies the following congruences:

If is a permutation over , then belongs to the class sign (see Henderson and Matthews [3]). Moreover (see Cohen in [4]), these conditions are necessary for , where is an odd prime. Henderson and Matthew extended this result to all squares such that . Later they found new classes of permutation polynomials of the type over . They proved that for , permutes the elements of if the sign class of contains one of the following triples:(i) for ; (ii) for odd; (iii), where .

They also proved that if and if belongs either to the sign class of or , then permute the elements of . In general, for a large prime number , few permutation polynomials of the form have been identified.

For , the cycle structure of is well known. It was established by Ahmad in [5] for and later by Lidl and Mullen [6] for . But the cycle structure of remains unknown. We look partially to it by determining the number of fixed points of over for the Dickson polynomials of the second kind that are permutation polynomials over . For , it is well known that the first and second Chebyshev polynomials and , over the field , are, respectively, conjugates of and . Namely, we have where We use these relations between Dickson polynomials and Chebyshev polynomials along with a new approach to define and over to give a new approach of the Dickson polynomials and on in Section 2. In Section 3, we study the number of fixed points of .

#### 2. The Dickson Polynomials and

Now we present a theoretical approach of the family of Dickson polynomials and on , which are essentially the first and the second Chebyshev polynomials and over . We will provide a better way to look at these polynomials which can help to find many known properties. Let us recall that the first and the second Chebyshev polynomials defined in (7) can also be, respectively, defined by the following linear recurrence relations: Obviously, , and they can be derived as conjugates of and , respectively. This implies that and , as permutations of , have the same number of cycles with the same length when written as a product of disjoint cycles, likewise for and . Therefore to understand the permutation structure of and , it suffices to know that of and .

Put . The structure of depends on whether is a square in or not. For a given ring , we denote their group of units by .

Lemma 1. Let . If is a square in , that is, for some , then there exists a group homomorphism , such that for any , the following equation holds, where and .

Proof. If , then is a square in . Let be its roots in . Then , so . If , then can be expressed as , where . Let be the mapping from to defined by Clearly, is a surjective group homomorphism from to . If , then . Furthermore, and it has order . If then Therefore, and satisfy the recurrence relations where and is the order of . That is, and satisfy the recurrence (8), which means that is the first coordinate of and Notice that is uniquely determined by up to a sign and we have where is the order of .

Remark 2. Let FC be the set of all the first coordinates of the elements in . FC contains elements including and . Those elements are in for which is a square in . Moreover, FC is invariant under the actions of and . For a given element in FC, we can find an element in FC such that is in . The mapping which sends to is an isomorphism from to . Therefore, the order of is simply determined by the least common multiple of the order of and . However, since in , , the kernel is cyclic. For any element in , if , a generator of the group , then is a generator of .

Lemma 3. Let . If is a square in , that is, for some , then there exists a group homomorphism , such that where and .

Proof. In this case, is not a square in , and . Therefore, . Again we can express every element as , where . Let be a map, defined by . Clearly, is a surjective group homomorphism from to . The inverse of is and for each element if , then Therefore, and satisfy the recurrence relations where and is the order of . That is, is the first coordinate of and is uniquely determined by up to a sign. is the second Chebyshev polynomial. In this case, is a cyclic group of order , which also implies that where is the order of .

Remark 4. Since is a cyclic group, a good way to find a generator is to first find the generator of and then take the power. If we let again FC to be the set of all the first coordinates of the elements of , then it contains elements including and. The includes all the elements in for which is a square in . It is invariant under the actions of and .

Finally, we assume that is not a square in . Put Clearly, is a field of elements and can be embedded in in the conventional way. For convenience, we will treat as a subset of according to the standard embedding. Here, becomes naturally a square in , and are the two square roots of in . The group of units of is a cyclic group of order . Let . Then is a ring but not a field. Let denote the subset of all elements in of the form of where . Clearly, is a subring of .

Lemma 5. Let . There exists a group homomorphism , such that where and .

Proof. In this case, is a square in and the equation has no nonzero solutions for . Therefore, is a field.
Let be the mapping from to , defined by Clearly, is a surjective group homomorphism from to . Moreover, is a cyclic group of order , and for every element , and for each element of the form if then Therefore, and satisfy the recurrence relations where and is the order of . That is, is the first coordinate of and is uniquely determined by up to a sign. is the second Chebyshev polynomial. Since is a cyclic group of order , we have where is the order of .

Remark 6. We can find a generator of by raising to the a generator of . If denotes the set of all the first coordinates of the elements of , then it contains elements including , but not . without the elements consists of all the elements in for which is not a square in .

Lemma 7. Let . There exists a group homomorphism , such that and , whenever and .

Proof. Here, we know that is not a square in and has nonzero solutions for and . The total number of solutions is . Therefore, is not a field.
Let be a map, defined by Clearly, is a surjective group homomorphism from to . Also is a cyclic group of order and for every element , . Furthermore, if then Therefore, and satisfy the recurrence relations where and is the order of . That is, where is the order of .

Remark 8. As in Remark 6, finding a generator for is equivalent to finding a generator in such that . If denotes the set of all the first coordinates of the elements of , then it contains elements including ±1, but not . without the element ±1 consists of all the elements in for which is not a square in .

As corollaries of the previous discussion, we obtain the following.

Corollary 9. If , then the number of elements   for which is a square in is .
If , then the number of elements   for which is a square in is .

#### 3. Number of Fixed Points of

Since Dickson permutation polynomials are closed under composition of polynomials if and only if , or (see [7]), we will only focus on these cases.

The case when was proven by Ahmad [5]. His proof can also be found in [6]. The theorem is formulated as follows.

Theorem 10. The number of fixed points of over is given by

In the next two theorems, we treat the cases and . Their proofs follow the lines of the one of Lidl and Mullen [6] for the Dickson polynomials of the first kind.

Theorem 11. For , the number of fixed points of over is given by where

Proof. First we can notice that the only permutation polynomials of the form over found are those for which is odd and belongs to the class sign , where vary depending on if ,  , , or . Let be a fixed point of . That is for all these permutations, we must have Then we have ; that is, which leads to Hence, we get
Let be the set of all solutions of the quadratic equations of the form , with . Recall that if is a primitive element of and if and , then and . Also, is a solution of if and only if is also a solution. So each fixed point is associated with a pair in . Therefore, the number of solutions is equal to half the sum of the solutions in both and excluding the common solutions.
Now is a solution if and only if   or  .
Hence, we have The numbers of solutions of these congruences in excluding the common solutions are, respectively, and if and are odd. Therefore, the number of fixed points associated to the elements in is Similarly, if and are odd, the number of fixed points associated to elements in excluding the common solutions is Using the same argument, the numbers of solutions of these congruences in excluding the common solutions are, respectively, and if and are even. Therefore, the number of fixed points associated to elements in is Similarly, if and are even, the number of fixed points associated to elements in excluding the common solutions is To complete the proof, we need to subtract the number of solutions when and at the same time. These are given by [7] and are equal to Combine all the results above to conclude the answer. The case where is even is pointless since in this case no permutation has been found.

Theorem 12. The number of fixed points of over is given by where

Proof. First of all, notice that for all permutations over found so far, is an odd integer. Let be a fixed point of the permutation over . Then, ; that is, which implies Hence, we have Let be a primitive element of . So we get , , and . For , let . From [6], we have if and if . Therefore, as it was mentioned in [6], if we defined and if then is a disjoint union.
Finally, from again [6], for , if , then  . Now a solution of is a solution of both and if and only if . Therefore, the number of solutions of (54) is the sum of the solutions in , and .
Let denote the highest power of dividing if and set . An element is a solution of if and only if which has solutions if and only if divides if and only if .
Let . It is easy to see that Also let , such that . Then the solutions of (58) are Using the same argument as in [6], should be odd, because otherwise Now . Therefore, divides if and only if and (58) has an odd solution if and only if . In this case, we have exactly solutions.
An element is a solution of if and only if if and only if , if and only if , if and only if if and only , and if and only if The last equation has a solution if and only if divides , which is equivalent to the condition . Then the total number of those solutions is .
An element is a solution of if and only if if and only if which has solutions if it is solvable.
If , then which implies that is odd. But this can occur if and only if , and the solutions of 8 in this case are An element is a solution of if and only if , if and only if if and only if which has solutions.
To complete the proof, notice that by a similar method to that of the case , is a solution of (resp., ) if and only if is also a solution, and the set of solutions of (58) and (66) on is the set of all solutions of and on , which is empty if and is equal to if .

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