Fixed Points of the Dickson Polynomials of the Second Kind

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 𝑝 be a prime, 𝑒≥1, 𝑞=𝑝^𝑒, and let 𝔽_𝑞 be the field of 𝑞 elements. Denote the greatest integer function of 𝑛 by ⌊𝑛⌋. For a fixed 𝑎∈𝔽_𝑞, the polynomials 𝐷_𝑘(𝑥,𝑎)=^⌊𝑘/2⌋_𝑗=1𝑘𝑗()𝑘−𝑗𝑘−𝑗−𝑎^𝑗𝑥^𝑘−2𝑗,𝐸_𝑘(𝑥,𝑎)=^⌊𝑘/2⌋_𝑗=1𝑗()𝑘−𝑗−𝑎^𝑗𝑥^𝑘−2𝑗 (1) are called the Dickson polynomials of the first and second kind, respectively.

It is well known (see [1]) that for any 𝑥∈𝔽_𝑞, there exists 𝑢∈𝔽_𝑞2 such that 𝑥=𝑢+𝑎𝑢⁻¹ and the Dickson polynomials may be expressed as 𝐷_𝑘(𝑥,𝑎)=𝑢^𝑘+𝑎^𝑘𝑢^−𝑘,𝐸_𝑘()=𝑢𝑥,𝑎^𝑘+1−𝑎^𝑘+1𝑢^−(𝑘+1)𝑢−𝑎𝑢⁻¹,if𝑢²(±√≠𝑎,𝑘+1)𝑎^𝑘,otherwise. (2) For 𝑎∈𝔽^∗_𝑞, denote 𝜂(𝑎)=1,if𝑎isasquare,−1,otherwise. (3)

According to [2], if 𝜂(𝑎)=1 and 𝑞 is odd, then the sign class of 𝑘 is defined to be the set 𝐴={(±𝑘₁,±𝑘₂,±𝑘₃)∈ℤ³} if 𝑘 satisfies the following congruences: 𝑘₁≡𝑘+1mod𝑝,𝑘₂≡𝑘+1mod𝑞−12,𝑘₃≡𝑘+1mod𝑞+12. (4) If 𝜂(𝑎)=−1 and 𝑞 is odd, then the sign class of 𝑘 is defined to be the set 𝐴={(±𝑘₁,±𝑘₂)∈ℤ²} if 𝑘 satisfies the following congruences: 𝑘₁≡𝑘+1mod𝑞−1,𝑘₂≡𝑘+1mod𝑞+1. (5)

If 𝐸_𝑘(𝑥,1) is a permutation over 𝔽_𝑞, then 𝑘 belongs to the class sign (2,2,2) (see Henderson and Matthews [3]). Moreover (see Cohen in [4]), these conditions are necessary for 𝑞∈{𝑝,𝑝²}, where 𝑝 is an odd prime. Henderson and Matthew extended this result to all squares 𝑎 such that 𝑞∈{𝑝,𝑝²}. Later they found new classes of permutation polynomials of the type 𝐸_𝑘(𝑥,1) over 𝔽_𝑞. They proved that for 𝑞=3^𝑒, 𝐸_𝑘(𝑥,1) permutes the elements of 𝔽_𝑞 if the sign class of 𝑘 contains one of the following triples:(i){2,10,10} for 𝑒=3; (ii){2,4,4} for 𝑒 odd; (iii){2,((3^𝑠−1)/2)⁻¹+1,((3^𝑡−1)/2)⁻¹+1}, where gcd(𝑠,𝑒)=gcd(𝑡,2𝑒)=1.

They also proved that if 𝑞=5^𝑒 and if 𝑘 belongs either to the sign class of {2,2,(𝑞−1)/4} or {2,2,2}, then 𝐸_𝑘(𝑥,1) permute the elements of 𝔽_𝑞. In general, for a large prime number 𝑝, few permutation polynomials of the form 𝐸_𝑘(𝑥,𝑎) have been identified.

For 𝑎∈{−1,0,1}, the cycle structure of 𝐷_𝑘(𝑥,𝑎) is well known. It was established by Ahmad in [5] for 𝑎=0 and later by Lidl and Mullen [6] for 𝑎∈{−1,1}. But the cycle structure of 𝐸_𝑘(𝑥,𝑎) remains unknown. We look partially to it by determining the number of fixed points of 𝐸_𝑘(𝑥,𝑎) over 𝔽_𝑞 for the Dickson polynomials of the second kind that are permutation polynomials over 𝔽_𝑞. For 𝑎=1, it is well known that the first and second Chebyshev polynomials 𝑇_𝑛(𝑥) and 𝑈_𝑛(𝑥), over the field ℤ_𝑝, are, respectively, conjugates of 𝐷_𝑘(𝑥,1) and 𝐸_𝑘(𝑥,1). Namely, we have 𝐷_𝑘(𝑥,1)=2𝑇_𝑘𝑥2,𝐸_𝑘(𝑥,1)=𝑈_𝑘𝑥2, (6) where 𝑇_𝑘(𝑥)((𝑥,𝑈=cos𝑘arccos))_𝑘((𝜃=)𝜃)cos))sin((𝑘+1(𝜃),().sin𝑘∈ℤ (7) We use these relations between Dickson polynomials and Chebyshev polynomials along with a new approach to define 𝑇_𝑛(𝑥) and 𝑈_𝑛(𝑥) over ℤ_𝑝 to give a new approach of the Dickson polynomials 𝐷_𝑘(𝑥,1) and 𝐸_𝑘(𝑥,1) on ℤ_𝑝 in Section 2. In Section 3, we study the number of fixed points of 𝐸_𝑘(𝑥,𝑎).

2. The Dickson Polynomials 𝐷_𝑘(𝑥,1) and 𝐸_𝑘(𝑥,1)

Now we present a theoretical approach of the family of Dickson polynomials 𝐷_𝑘(𝑥,1) and 𝐸_𝑘(𝑥,1) on ℤ_𝑝, which are essentially the first and the second Chebyshev polynomials 𝑇_𝑛(𝑥) and 𝑈_𝑛(𝑥) over ℤ_𝑝. 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: 𝑇₀(𝑥)=1,𝑇₁(𝑥)𝑇=𝑥,_𝑘(𝑥)=2𝑥𝑇_𝑘−1(𝑥)−𝑇_𝑘−2(𝑥),𝑈₀(𝑥)=1,𝑈₁(𝑥)𝑈=2𝑥,_𝑘(𝑥)=2𝑥𝑈_𝑘−1(𝑥)−𝑈_𝑘−2(𝑥). (8)(9) Obviously, 𝐷_𝑘(𝑥,1)=2𝑇_𝑘(𝑥/2), 𝐸_𝑘(𝑥,1)=𝑈_𝑘(𝑥/2) and they can be derived as conjugates of 𝑇_𝑘(𝑥) and 𝑈_𝑘(𝑥), respectively. This implies that 𝐷_𝑘(𝑥,1) and 𝑇_𝑘(𝑥), as permutations of ℤ_𝑝, have the same number of cycles with the same length when written as a product of disjoint cycles, likewise for 𝐸_𝑘(𝑥,1) and 𝑈_𝑘(𝑥). Therefore to understand the permutation structure of 𝐷_𝑘(𝑥,1) and 𝐸_𝑘(𝑥,1), it suffices to know that of 𝑇_𝑘(𝑥) and 𝑈_𝑘(𝑥).

Put ℝ_𝑝=ℤ_𝑝[𝑧]/(𝑧²+1). The structure of ℝ_𝑝 depends on whether −1 is a square in ℤ_𝑝 or not. For a given ring 𝑅, we denote their group of units by 𝑈(𝑅).

Lemma 1. Let 𝑝≡1(mod4). If 1−𝑥² is a square in ℤ_𝑝, that is, 1−𝑥²=𝑦² for some 𝑦∈ℤ_𝑝, then there exists a group homomorphism 𝑁∶𝑈(ℝ_𝑝)→𝑈(ℤ_𝑝), such that for any 𝑥+𝑦𝑧∈Ker(𝑁), the following equation 𝑥+𝑦𝑧^𝑘=𝑃_𝑘𝑥,𝑦+𝑄_𝑘𝑥,𝑦𝑧 (10) holds, where 𝑇_𝑘(𝑥)=𝑃_𝑘(𝑥,𝑦) and 𝑈_𝑘(𝑥)=(1/𝑦)𝑄_𝑘(𝑥,𝑦).

Proof. If 𝑝≡1(mod4), then −1 is a square in ℤ_𝑝. Let ±𝑤 be its roots in ℤ_𝑝. Then ℝ_𝑝≅ℤ_𝑝⊕ℤ_𝑝, so |𝑈(ℝ_𝑝)|=(𝑝−1)². If 𝑋∈ℝ_𝑝, then 𝑋 can be expressed as 𝑋=𝑎+𝑏𝑧, where 𝑎,𝑏∈ℤ_𝑝. Let 𝑁 be the mapping from ℝ_𝑝 to ℤ_𝑝 defined by 𝑁(𝑎+𝑏𝑧)=𝑎²+𝑏². (11) Clearly, 𝑁 is a surjective group homomorphism from 𝑈(ℝ_𝑝) to 𝑈(ℤ_𝑝). If 𝑥+𝑦𝑧∈Ker(𝑁), then (𝑥+𝑦𝑧)⁻¹=𝑥−𝑦𝑧. Furthermore, Ker(𝑁)≅𝑈(ℤ_𝑝) and it has order 𝑝−1. If 𝑥+𝑦𝑧^𝑛=𝑃_𝑛𝑥,𝑦+𝑄_𝑛𝑥,𝑦𝑧,𝑥+𝑦𝑧∈Ker(𝑁), (12) then 𝑥+𝑦𝑧^𝑛+1=𝑃_𝑛𝑥,𝑦+𝑄_𝑛𝑧=𝑥,𝑦𝑥+𝑦𝑧𝑥𝑃_𝑛𝑥,𝑦−𝑦𝑄_𝑛+𝑥,𝑦𝑥𝑄_𝑛𝑥,𝑦+𝑦𝑃_𝑛𝑥,𝑦𝑧. (13) Therefore, 𝑃_𝑛(𝑥,𝑦) and 𝑄_𝑛(𝑥,𝑦) satisfy the recurrence relations 𝑃_𝑛+1𝑥,𝑦=𝑥+𝑦𝑥𝑃_𝑛𝑥,𝑦−𝑦𝑄_𝑛,𝑄𝑥,𝑦_𝑛+1𝑥,𝑦=𝑥𝑄_𝑛𝑥,𝑦+𝑦𝑃_𝑛𝑥,𝑦, (14) where 𝑃₁(𝑥)=𝑥,𝑃₂(𝑥)=2𝑥²𝑃−1,_𝑝(𝑥)=𝑥,𝑃_𝑘+𝑛(𝑥)=𝑃_𝑛(𝑥),1𝑦𝑄₁(𝑥)1=1,𝑦𝑄₂(𝑥)1=2𝑥,𝑦𝑄_𝑝(𝑥)1=1,𝑦𝑄_𝑘+𝑛(𝑥)=𝑄_𝑛(𝑥), (15) and 𝑘 is the order of 𝑥+𝑦𝑧. That is, 𝑃_𝑛(𝑥,𝑦) and 𝑄_𝑛(𝑥,𝑦)/𝑦 satisfy the recurrence (8), which means that 𝑇_𝑛(𝑥)=𝑃_𝑛(𝑥,𝑦) is the first coordinate of (𝑥+𝑦𝑧)^𝑛 and 𝑈_𝑛(𝑥)=𝑄_𝑛𝑥,𝑦𝑦. (16) Notice that 𝑄_𝑛(𝑥,𝑦) is uniquely determined by 𝑃_𝑛(𝑥,𝑦) up to a sign and we have 𝑇₁(𝑥)=𝑥,𝑇₂(𝑥)=2𝑥²𝑇−1,_𝑝(𝑥)=𝑥,𝑇_𝑘+𝑛(𝑥)=𝑇_𝑛(𝑥),𝑈₁(𝑥)=1,𝑈₂(𝑥)𝑈=2𝑥,_𝑝(𝑥)=1,𝑈_𝑘+𝑛(𝑥)=𝑈_𝑛(𝑥), (17) where 𝑘 is the order of 𝑥+𝑦𝑧.

Remark 2. Let FC be the set of all the first coordinates of the elements in Ker(𝑁). FC contains (𝑝+1)/2=(𝑝−1)/2+1 elements including 0 and 1. Those elements are 𝑥 in ℤ_𝑝 for which 1−𝑥² is a square in ℤ_𝑝. Moreover, FC is invariant under the actions of 𝑇_𝑛 and 𝑈_𝑛. For a given element 𝑢 in FC, we can find an element 𝑣 in FC such that 𝑢+𝑣𝑧 is in Ker(𝑁). The mapping which sends 𝑢+𝑣𝑧 to 𝑢+𝑣𝑤𝑢−𝑣𝑤 is an isomorphism from Ker(𝑁) to 𝑈(ℤ_𝑝). Therefore, the order of 𝑢+𝑣𝑧 is simply determined by the least common multiple of the order of 𝑢+𝑣𝑤 and 𝑢−𝑣𝑤. However, since in ℤ_𝑝, (𝑢+𝑣𝑤)(𝑢−𝑣𝑤)=1, the kernel Ker(𝑁) is cyclic. For any element 𝑢+𝑣𝑧 in Ker(𝑁), if 𝑢+𝑣𝑤=𝑔, a generator of the group 𝑈(ℤ_𝑝), then 𝑢+𝑣𝑧 is a generator of Ker(𝑁).

Lemma 3. Let 𝑝≡3(mod4). If 1−𝑥² is a square in ℤ_𝑝, that is, 1−𝑥²=𝑦² for some 𝑦∈ℤ_𝑝, then there exists a group homomorphism 𝑁∶𝑈(ℝ_𝑝)→𝑈(ℤ_𝑝), such that 𝑥+𝑦𝑧^𝑛=𝑃_𝑛𝑥,𝑦+𝑄_𝑛𝑥,𝑦𝑧,𝑥+𝑦𝑧∈Ker(𝑁), (18) where 𝑇_𝑛(𝑥)=𝑃_𝑛(𝑥,𝑦) and 𝑈_𝑛(𝑥)=𝑄_𝑛(𝑥,𝑦)/𝑦.

Proof. In this case, −1 is not a square in ℤ_𝑝, and |ℝ_𝑝|=𝑝². Therefore, |𝑈(ℝ_𝑝)|=𝑝²−1. Again we can express every element 𝑋∈ℝ_𝑝 as 𝑋=𝑎+𝑏𝑧, where 𝑎,𝑏∈ℤ_𝑝. Let 𝑁∶ℝ_𝑝→ℤ_𝑝 be a map, defined by 𝑁(𝑎+𝑏𝑧)=𝑎²+𝑏². Clearly, 𝑁 is a surjective group homomorphism from 𝑈(ℝ_𝑝) to 𝑈(ℤ_𝑝). The inverse of 𝑥+𝑦𝑧∈Ker(𝑁) is (𝑥+𝑦𝑧)⁻¹=𝑥−𝑦𝑧 and for each element 𝑥+𝑦𝑧∈Ker(𝑁) if (𝑥+𝑦𝑧)^𝑛=𝑃_𝑛(𝑥,𝑦)+𝑄_𝑛(𝑥,𝑦), then 𝑥+𝑦𝑧^𝑛+1=𝑃_𝑛𝑥,𝑦+𝑄_𝑛𝑧=𝑥,𝑦𝑥+𝑦𝑧𝑥𝑃_𝑛𝑥,𝑦−𝑦𝑄_𝑛+𝑥,𝑦𝑥𝑄_𝑛𝑥,𝑦+𝑦𝑃_𝑛𝑥,𝑦𝑧. (19) Therefore, 𝑃_𝑛(𝑥,𝑦) and 𝑄_𝑛(𝑥,𝑦) satisfy the recurrence relations 𝑃_𝑛+1𝑥,𝑦=𝑥,𝑦𝑥𝑃_𝑛𝑥,𝑦−𝑦𝑄_𝑛,𝑄𝑥,𝑦_𝑛+1𝑥,𝑦=𝑥𝑄_𝑛𝑥,𝑦+𝑦𝑃_𝑛𝑥,𝑦, (20) where 𝑃₁(𝑥)=𝑥,𝑃₂(𝑥)=2𝑥²−1,𝑃_𝑝+1(𝑥)𝑃=1,_𝑝+2(𝑥)=𝑥,𝑃_𝑘−𝑛(𝑥)=𝑃_𝑛(𝑥),1𝑦𝑄₁(𝑥)1=1,𝑦𝑄₂(𝑥)1=2𝑥,𝑦𝑄_𝑝+2(𝑥)1=1,𝑦𝑄_𝑝+3(𝑥)1=2𝑥,𝑦𝑄_𝑘+𝑛(𝑥)=𝑄_𝑛(𝑥), (21) and 𝑘 is the order of 𝑥+𝑦𝑧. That is, 𝑇_𝑛(𝑥)=𝑃_𝑛(𝑥,𝑦) is the first coordinate of (𝑥+𝑦𝑧)^𝑛 and 𝑄_𝑛(𝑥,𝑦) is uniquely determined by 𝑃_𝑛(𝑥,𝑦) up to a sign. 𝑈_𝑛=𝑄_𝑛(𝑥,𝑦)/𝑦 is the second Chebyshev polynomial. In this case, Ker(𝑁) is a cyclic group of order 𝑝+1, which also implies that 𝑇_𝑝+1(𝑥)=1,𝑇_𝑝+2(𝑥)=𝑥,𝑇_𝑘−𝑛(𝑥)=𝑇_𝑛(𝑥),𝑈_𝑝+2(𝑥)=1,𝑇_𝑝+3(𝑥)=2𝑥,𝑇_𝑘−𝑛(𝑥)=𝑇_𝑛(𝑥), (22) where 𝑘 is the order of 𝑥+𝑦𝑧.

Remark 4. Since Ker(𝑁) is a cyclic group, a good way to find a generator is to first find the generator of 𝑈(ℝ_𝑝) and then take the (𝑝−1)th power. If we let again FC to be the set of all the first coordinates of the elements of Ker(𝑁), then it contains ((𝑝+1)/2)+1 elements including 0 and1. The Ker(𝑁) includes all the elements 𝑥 in ℤ_𝑝 for which 1−𝑥² is a square in ℤ_𝑝. It is invariant under the actions of 𝑇_𝑛 and 𝑈_𝑛.

Finally, we assume that 1−𝑥² is not a square in ℤ_𝑝. Put ℤ_𝑝=ℤ_𝑝𝑦𝑦²−1−𝑥². (23) Clearly, ℤ_𝑝 is a field of 𝑝² elements and ℤ_𝑝 can be embedded in ℤ_𝑝 in the conventional way. For convenience, we will treat ℤ_𝑝 as a subset of ℤ_𝑝 according to the standard embedding. Here, −1 becomes naturally a square in ℤ_𝑝, and ±𝑦 are the two square roots of (1−𝑥²) in ℤ_𝑝. The group of units 𝑈(ℤ_𝑝) of ℤ_𝑝 is a cyclic group of order 𝑝²−1. Let ℍ_𝑝=ℤ_𝑝[𝑧]/(𝑧²+1). Then ℍ_𝑝 is a ring but not a field. Let 𝕊_𝑝 denote the subset of all elements in ℍ_𝑝 of the form of 𝑎+𝑏𝑦𝑧 where 𝑎,𝑏∈ℤ_𝑝. Clearly, 𝕊_𝑝 is a subring of ℍ_𝑝.

Lemma 5. Let 𝑝≡1(mod4). There exists a group homomorphism 𝑁∶𝑈(𝕊_𝑝)→𝑈(ℤ_𝑝), such that 𝑥+𝑦𝑧^𝑛=𝑃_𝑛𝑥,𝑦+𝑄_𝑛𝑥,𝑦𝑧,𝑥+𝑦𝑧∈Ker(𝑁), (24) where 𝑇_𝑛(𝑥)=𝑃_𝑛(𝑥,𝑦) and 𝑈_𝑛(𝑥)=𝑄_𝑛(𝑥,𝑦)/𝑦.

Proof. In this case, −1 is a square in ℤ_𝑝 and the equation 𝑎²+𝑏²(1−𝑥²)=0 has no nonzero solutions for 𝑎,𝑏∈ℤ_𝑝. Therefore, 𝕊_𝑝 is a field.
Let 𝑁 be the mapping from 𝑈(𝕊_𝑝) to 𝑈(ℤ_𝑝), defined by 𝑁𝑎+𝑏𝑦𝑧=𝑎²+𝑏²1−𝑥². (25) Clearly, 𝑁 is a surjective group homomorphism from 𝑈(𝕊_𝑝) to 𝑈(ℤ_𝑝). Moreover, Ker(𝑁) is a cyclic group of order 𝑝+1, and for every element 𝑥+𝑦𝑧∈Ker(𝑁), (𝑥+𝑦𝑧)⁻¹=𝑥−𝑦𝑧 and for each element of the form 𝑥+𝑦𝑧∈Ker(𝑁) if 𝑥+𝑦𝑧^𝑛=𝑃_𝑛𝑥,𝑦+𝑄_𝑛𝑥,𝑦, (26) then 𝑥+𝑦𝑧^𝑛+1=𝑃_𝑛𝑥,y+𝑄_𝑛𝑧=𝑥,𝑦𝑥+𝑦𝑧𝑥𝑃_𝑛𝑥,𝑦−𝑦𝑄_𝑛+𝑥,𝑦𝑥𝑄_𝑛𝑥,𝑦+𝑦𝑃_𝑛𝑥,𝑦𝑧. (27) Therefore, 𝑃_𝑛(𝑥,𝑦) and 𝑄_𝑛(𝑥,𝑦) satisfy the recurrence relations 𝑃_𝑛+1𝑥,𝑦=𝑥,𝑦𝑥𝑃_𝑛𝑥,𝑦−𝑦𝑄_𝑛,𝑄𝑥,𝑦_𝑛+1𝑥,𝑦=𝑥𝑄_𝑛𝑥,𝑦+𝑦𝑃_𝑛𝑥,𝑦, (28) where 𝑃₁(𝑥)=𝑥,𝑃₂(𝑥)=2𝑥²−1,𝑃_𝑝+1(𝑥)𝑃=1,_𝑝+2(𝑥)=𝑥,𝑃_𝑘−𝑛(𝑥)=𝑃_𝑛(𝑥),1𝑦𝑄₁(𝑥)1=1,𝑦𝑄₂(𝑥)1=2𝑥,𝑦𝑄_𝑝+2(𝑥)1=1,𝑦𝑄_𝑝+3(𝑥)1=2𝑥,𝑦𝑄_𝑘+𝑛(𝑥)=𝑄_𝑛(𝑥), (29) and 𝑘 is the order of 𝑥+𝑦𝑧. That is, 𝑇_𝑛(𝑥)=𝑃_𝑛(𝑥,𝑦) is the first coordinate of (𝑥+𝑦𝑧)^𝑛 and 𝑄_𝑛(𝑥,𝑦) is uniquely determined by 𝑃_𝑛(𝑥,𝑦) up to a sign. 𝑈_𝑛(𝑥)=𝑄_𝑛(𝑥,𝑦)/𝑦 is the second Chebyshev polynomial. Since Ker(𝑁) is a cyclic group of order 𝑝+1, we have 𝑇_𝑝+1(𝑥)=1,𝑇_𝑝+2(𝑥)=𝑥,𝑇_𝑘−𝑛(𝑥)=𝑇_𝑛(𝑥),𝑈_𝑝+2(𝑥)=1,𝑇_𝑝+3(𝑥)=2𝑥,𝑈_𝑘−𝑛(𝑥)=𝑈_𝑛(𝑥), (30) where 𝑘 is the order of 𝑥+𝑦𝑧.

Remark 6. We can find a generator of Ker(𝑁) by raising to the (𝑝−1)th a generator of 𝑈(𝕊_𝑝). If 𝐹_𝐶𝐾𝑝 denotes the set of all the first coordinates of the elements of Ker(𝑁), then it contains (𝑝+1)/2+1 elements including ±1, but not 0. 𝐹_𝐶𝐾𝑝 without the elements ±1 consists of all the elements 𝑥 in ℤ_𝑝 for which 1−𝑥² is not a square in ℤ_𝑝.

Lemma 7. Let 𝑝≡3(mod4). There exists a group homomorphism 𝑁∶𝑈(ℝ_𝑝)→𝑈(ℤ_𝑝), such that 𝑇_𝑛(𝑥)=𝑃_𝑛(𝑥,𝑦) and 𝑈_𝑛(𝑥)=𝑄_𝑛(𝑥,𝑦)/𝑦, whenever 𝑥+𝑦𝑧∈Ker(𝑁) and (𝑥+𝑦𝑧)^𝑛=𝑃_𝑛(𝑥,𝑦)+𝑄_𝑛(𝑥,𝑦)𝑧.

Proof. Here, we know that −1 is not a square in ℤ_𝑝 and 𝑎²+𝑏²(1−𝑥²)=0 has nonzero solutions for 𝑎 and 𝑏. The total number of solutions is 2𝑝−1. Therefore, 𝕊_𝑝 is not a field.
Let 𝑁∶𝕊_𝑝→ℤ^∗_𝑝 be a map, defined by 𝑁𝑎+𝑏𝑦𝑧=𝑎²+𝑏²1−𝑥². (31) Clearly, 𝑁 is a surjective group homomorphism from 𝑈(𝕊_𝑝) to 𝑈(ℤ_𝑝). Also Ker(𝑁) is a cyclic group of order 𝑝−1 and for every element 𝑥+𝑦𝑧∈Ker(𝑁), (𝑥+𝑦𝑧)⁻¹=𝑥−𝑦𝑧. Furthermore, if 𝑥+𝑦𝑧^𝑛=𝑃_𝑛𝑥,𝑦+𝑄_𝑛𝑥,𝑦𝑥+𝑦𝑧∈Ker(𝑁), (32) then 𝑥+𝑦𝑧^𝑛+1=𝑃_𝑛x,𝑦+𝑄_𝑛𝑧=𝑥,𝑦𝑥+𝑦𝑧𝑥𝑃_𝑛𝑥,𝑦−𝑦𝑄_𝑛+𝑥,𝑦𝑥𝑄_𝑛𝑥,𝑦+𝑦𝑃_𝑛𝑥,𝑦𝑧. (33) Therefore, 𝑃_𝑛(𝑥,𝑦) and 𝑄_𝑛(𝑥,𝑦) satisfy the recurrence relations 𝑃_𝑛+1𝑥,𝑦=𝑥,𝑦𝑥𝑃_𝑛𝑥,𝑦−𝑦𝑄_𝑛,𝑄𝑥,𝑦_𝑛+1𝑥,𝑦=𝑥𝑄_𝑛𝑥,𝑦+𝑦𝑃_𝑛𝑥,𝑦, (34) where 𝑃₁(𝑥)=𝑥,𝑃₂(𝑥)=2𝑥²−1,𝑃_𝑝+1(𝑥)𝑃=1,_𝑝+2(𝑥)=𝑥,𝑃_𝑘−𝑛(𝑥)=𝑃_𝑛(𝑥),1𝑦𝑄₁(𝑥)1=1,𝑦𝑄₂(𝑥)1=2𝑥,𝑦𝑄_𝑝+2(𝑥)1=1,𝑦𝑄_𝑝+3(𝑥)1=2𝑥,𝑦𝑄_𝑘+𝑛(𝑥)=𝑄_𝑛(𝑥), (35) and 𝑘 is the order of 𝑥+𝑦𝑧. That is, 𝑇_𝑝−1(𝑥)=1,𝑇_𝑝(𝑥)=𝑥,𝑇_𝑘−𝑛(𝑥)=𝑇_𝑛(𝑥),𝑈_𝑝+2(𝑥)=1,𝑈_𝑝+3(𝑥)=2𝑥,𝑈_𝑘−𝑛(𝑥)=𝑈_𝑛(𝑥), (36) where 𝑘 is the order of 𝑥+𝑦𝑧.

Remark 8. As in Remark 6, finding a generator 𝑢+𝑣𝑧 for Ker(𝑁) is equivalent to finding a generator 𝑔 in 𝑈(𝕊_𝑝) such that 𝑔=𝑢+𝑣𝑤. If 𝐹_𝐶𝕂𝑝 denotes the set of all the first coordinates of the elements of Ker(𝑁), then it contains (𝑝+1)/2 elements including ±1, but not 0. 𝐹_𝐶𝕂𝑝 without the element ±1 consists of all the elements 𝑥 in ℤ_𝑝 for which 1−𝑥² is not a square in ℤ_𝑝.

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

Corollary 9. If 𝑝≡1(mod4), then the number of elements 𝑥∈ℤ_𝑝  for which 1−𝑥² is a square in ℤ_𝑝 is (𝑝+1)/2.
If 𝑝≡3(mod4), then the number of elements 𝑥∈ℤ_𝑝  for which 1−𝑥² is a square in ℤ_𝑝 is (𝑝+3)/2.

3. Number of Fixed Points of 𝐸_𝑘(𝑥,𝑎)

Since Dickson permutation polynomials are closed under composition of polynomials if and only if 𝑎=0,1, or −1 (see [7]), we will only focus on these 3 cases.

The case when 𝑎=0 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 𝐸_𝑘(𝑥,0) over 𝔽_𝑞 is given by gcd𝑞−1,𝑘−1+1. (37)

In the next two theorems, we treat the cases 𝑎=1 and 𝑎=−1. Their proofs follow the lines of the one of Lidl and Mullen [6] for the Dickson polynomials of the first kind.

Theorem 11. For 𝑥≠±2, the number of fixed points of 𝐸_𝑘(𝑥,1) over 𝔽_𝑞 is given by 12(()gcd𝑞+1,𝑘−1+gcdq+1,2k+3)+gcd𝑞−1,𝑘−1+gcd𝑞−1,2𝑘+3−gcd𝑞+1,𝑘+3−gcd𝑞−1,𝑘+3−𝜀₁, (38) where 𝜀₁=5,𝑖𝑓𝑝𝑎𝑛𝑑𝑛𝑎𝑟𝑒𝑜𝑑𝑑,8∣𝑞+1,𝑎𝑛𝑑8∣𝑛−1,1,𝑖𝑓𝑝𝑎𝑛𝑑𝑛𝑎𝑟𝑒𝑜𝑑𝑑,8∤𝑞+1,𝑎𝑛𝑑8∤𝑛−1,0,𝑖𝑓𝑝𝑖𝑠𝑜𝑑𝑑𝑎𝑛𝑑𝑛𝑎𝑟𝑒𝑒𝑣𝑒𝑛. (39)

Proof. First we can notice that the only permutation polynomials of the form 𝐸_𝑘(𝑥,1) over 𝔽_𝑞 found are those for which 𝑞 is odd and 𝑘 belongs to the class sign (2,𝑎,𝑏), where 𝑎,𝑏 vary depending on if 𝑞=𝑝,  𝑝², 3^𝑒, or 5^𝑒. Let 𝛼∈𝔽_𝑞 be a fixed point of 𝐸_𝑘(𝑥,1). That is for all these permutations, we must have 𝑘+1≡2mod𝑝,𝑘+1≡−2mod𝑞−12. (40) Then we have 𝐸_𝑘(𝛼,1)=𝛼; that is, 𝑢^𝑘+1−𝑢^−(𝑘+1)𝑢−𝑢⁻¹1=𝑢+𝑢, (41) which leads to 𝑢^𝑘+1−1𝑢^𝑘+1−𝑢²+1𝑢²=0. (42) Hence, we get 𝑢^𝑘+3+1𝑢^𝑘−1−1=0. (43)
Let 𝑀={𝑢∈𝔽_𝑞2∶𝑢^𝑞−1=1or𝑢^𝑞+1=1} be the set of all solutions of the 𝑞 quadratic equations of the form 𝑥²−𝛼𝑥+1=0, with 𝛼∈𝔽_𝑞. Recall that if 𝜔 is a primitive element of 𝔽_𝑞2 and if 𝑀₁={𝜔^{(𝑞−1)𝑟}∶𝑟=0,1,…,𝑞} and 𝑀₂={𝜔^(𝑞+1)𝑠∶𝑠=0,1,…,𝑞−2}, then 𝑀=𝑀₁∪𝑀₂ and 𝑀₁∩𝑀₂={±1}. Also, 𝑢 is a solution of 𝑥²−𝛼𝑥+1=0 if and only if 𝑢⁻¹ is also a solution. So each fixed point 𝛼∈𝔽_𝑞 is associated with a pair (𝑢,𝑢⁻¹) in 𝑀. Therefore, the number of solutions is equal to half the sum of the solutions in both 𝑀₁ and 𝑀₂ excluding the common solutions.
Now 𝜔^{(𝑞−1)𝑟}∈𝑀₁ is a solution if and only if 𝜔^{(𝑞−1)𝑟(𝑘+3)}=−1  or  𝜔^{(𝑞−1)𝑟(𝑘−1)}=1.
Hence, we have ()𝑟().2𝑟𝑘+3≡0mod𝑞+1or𝑘−1≡0mod𝑞+1 (44) The numbers of solutions of these congruences in 𝑀₁ excluding the common solutions are, respectively, gcd(𝑞+1,2(𝑘+3))−gcd(𝑞+1,𝑘+3) and gcd(𝑞+1,𝑘−1)−2 if 𝑝 and 𝑛 are odd. Therefore, the number of fixed points associated to the elements in 𝑀₁ is 12(.gcd𝑞+1,2𝑘+3)−gcd𝑞+1,𝑘+3+gcd𝑞+1,𝑘−1−2 (45) Similarly, if 𝑝 and 𝑛 are odd, the number of fixed points associated to elements in 𝑀₂ excluding the common solutions is 12(.gcd𝑞−1,2𝑘+3)−gcd𝑞−1,𝑞+3−1+gcd𝑞−1,𝑘−1−2 (46) Using the same argument, the numbers of solutions of these congruences in 𝑀₁ excluding the common solutions are, respectively, gcd(𝑞+1,2(𝑘+3))−gcd(𝑞+1,𝑘+3) and gcd(𝑞+1,𝑘−1)−2 if 𝑝 and 𝑛 are even. Therefore, the number of fixed points associated to elements in 𝑀₁ is 12gcd𝑞+1,2(𝑘+3)−gcd𝑞+1,𝑘+3−1. (47) Similarly, if 𝑝 and 𝑛 are even, the number of fixed points associated to elements in 𝑀₂ excluding the common solutions is 12gcd𝑞−1,2(𝑘+3)−gcd𝑞−1,𝑞+3−1. (48) To complete the proof, we need to subtract the number of solutions when 𝜔^{(𝑞−1)𝑟(𝑘+3)}=−1 and𝜔^{(𝑞−1)𝑟(𝑘−1)}=1 at the same time. These are given by [7] and are equal to 8,ifpand𝑛areodd,8∣𝑞+1,and8∣𝑛−1,0,otherwise. (49) Combine all the results above to conclude the answer. The case where 𝑞 is even is pointless since in this case no permutation has been found.

Theorem 12. The number of fixed points of 𝐸_𝑘(𝑥,−1) over 𝔽_𝑞 is given by 12𝑎₁⋅gcd𝑘+3,2𝑞+1+𝑎₂⋅gcd𝑘+3,𝑞−1+𝑎₃⋅gcd𝑘−1,2𝑞+1+gcd𝑘−1,𝑞−1−𝜀₋₁, (50) where 𝑎₁=1,𝑖𝑓𝜐₂𝑞+1=𝜐₂(),𝑎𝑘+30,𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒,₂=1,𝑖𝑓𝜐₂(𝑘+3)≤𝜐₂𝑎𝑞−1−1,0,𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒,₃=1,𝑖𝑓𝜐₂𝑞+1<𝜐₂(),𝜀𝑘−10,𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒,₋₁=2,𝑖𝑓𝑞≡1(mod4),0,𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒. (51)

Proof. First of all, notice that for all permutations 𝐸_𝑘(𝑥,−1) over 𝔽_𝑞 found so far, 𝑘 is an odd integer. Let 𝛼∈𝔽_𝑞 be a fixed point of the permutation 𝐸_𝑘(𝑥,−1) over 𝔽_𝑞. Then, 𝐸_𝑘(𝛼,−1)=𝛼; that is, 𝐸_𝑘()=𝑢𝑥,−1^𝑘+1−𝑢^−(𝑘+1)𝑢+𝑢⁻¹1=𝑢−𝑢, (52) which implies 𝑢^𝑘+1−𝑢^−(𝑘+1)=𝑢²−𝑢⁻². (53) Hence, we have 𝑢^𝑘+3+1𝑢^𝑘−1−1=0. (54) Let 𝜔 be a primitive element of 𝔽_𝑞2. So we get 𝑀₁(−1)={𝜔^{(𝑞−1)𝑟/2}∶𝑟=1,3,...,2𝑞+1}, 𝑀₂(−1)={𝜔^(𝑞+1)𝑠∶𝑠=0,1,....,𝑞−2}, and 𝑀(−1)=𝑀₁(−1)∪𝑀₂(−1). For 𝑖=0,1, let 𝑢_𝑖=𝜔^{(𝑞2−1)(1+2𝑖)/4}. From [6], we have 𝑀₁(−1)∩𝑀₂(−1)={𝑢₁,𝑢₂} if 𝑞≡1(mod4) and 𝑀₁(−1)∩𝑀₂(−1)=∅ if 𝑞≡3(mod4). Therefore, as it was mentioned in [6], if we defined 𝑁₃(−1)=𝑢₁,𝑢₂(),𝑁if𝑞≡1mod4₃(−1)=∅if𝑞≡3(mod4) (55) and if 𝑁₁(−1)=𝑀₁(−1)⧵𝑁₃(),𝑁−1₂(−1)=𝑀₂(−1)⧵𝑁₃(−1), (56) then 𝑀(−1)=𝑁₁(−1)∪𝑁₂(−1)∪𝑁₃(−1) (57) is a disjoint union.
Finally, from again [6], for 𝑢∈𝔽_𝑞, if 𝑢=𝑢⁻¹, then  𝑢∈𝑁₃(−1). Now a solution 𝑢 of (𝑢^𝑘+3+1)(𝑢^𝑘−1−1)=0 is a solution of both (𝑢^𝑘+3+1) and (𝑢^𝑘−1−1) if and only if 𝑢∈𝑁₃(−1). Therefore, the number of solutions of (54) is the sum of the solutions in 𝑁₁(−1),𝑁₂(−1), and 𝑁₃(−1).
Let 𝜐_𝑝(𝑚) denote the highest power of 𝑝 dividing 𝑚 if 𝑚≠0 and set 𝜐_𝑝(0)=∞. An element 𝑢∈𝑀₁(−1) is a solution of 𝑢^𝑘+3+1=0 if and only if 𝑟(𝑘+3)≡𝑞+1mod2𝑞+1, (58) which has solutions if and only if gcd(𝑘+3,2(𝑞+1)) divides (𝑞+1) if and only if 𝜐₂(𝑘+3)≤𝜐₂(𝑞+1).
Let 𝑑=gcd(𝑘+3,2(𝑞+1)). It is easy to see that 𝜐₂(𝑑)=min𝜐₂(k+3),𝜐₂2𝑞+1=𝜐₂(𝑘+3). (59) Also let 𝛼,𝛽∈ℤ, such that 𝛼(𝑘+3)+2𝛽(𝑞+1)=𝑑. Then the solutions of (58) are 𝛼𝑟=𝑞+1𝑑+2𝑞+1𝑑𝑖,(𝑖=0,…,𝑑−1). (60) Using the same argument as in [6], 𝛼 should be odd, because otherwise 𝜐₂(𝑑)≥min𝜐₂(𝑘+3)+1,𝜐₂2𝑞+1+1>𝜐₂(𝑘+3). (61) Now 𝜐₂(𝑞+1/𝑑)=𝜐₂(𝑞+1)−𝜐₂(𝑑)=𝜐₂(𝑞+1)−𝜐₂(𝑘+1). Therefore, (𝑞+1) divides 𝑑 if and only if 𝜐₂(𝑞+1)=𝜐₂(𝑘+3) and (58) has an odd solution 𝑟 if and only if 𝜐₂(𝑞+1)=𝜐₂(𝑘+3). In this case, we have exactly gcd(𝑘+3,2(𝑞+1)) solutions.
An element 𝑢∈𝑀₂(−1) is a solution of 𝑢^𝑘+3+1=0 if and only if 𝜔^{(𝑞+1)𝑠𝑘+3}=−1, (62) if and only if ((𝜔^(𝑞+1)𝑠)^𝑘+3)²=1(mod(𝑞²−1)), if and only if (𝜔^{(𝑞+1)𝑠(𝑘+3)})²=1(mod𝑞²−1), if and only if 2𝑞+1𝑠(𝑘+3)=0mod𝑞²−1, (63) if and only 2𝑠(𝑘+3)=𝑞−1(mod(𝑞−1)), and if and only if 𝑠(𝑘+3)=𝑞−12mod𝑞−1. (64) The last equation has a solution if and only if gcd(𝑘+3,𝑞−1)) divides (𝑞−1)/2, which is equivalent to the condition 𝜐₂(𝑘+3)≤𝜐₂(𝑞−1)−1. Then the total number of those solutions is gcd(𝑘+3,𝑞−1).
An element 𝑢∈𝑀₁(−1) is a solution of 𝑢^𝑘−1−1=0 if and only if 𝜔^{(𝑘−1)(𝑞−1)𝑟/2}=1mod𝑞²−1 (65) if and only if 𝑟(𝑘−1)≡0mod2𝑞+1, (66) which has gcd((𝑘−1),2(𝑞+1)) solutions if it is solvable.
If 𝑑=gcd(𝑘−1,2(𝑞+1)), then 𝜐₂(𝑑)=min𝜐₂(𝑘−1),𝜐₂2𝑞+1=𝜐₂2𝑞+1, (67) which implies that 2(𝑞+1)/𝑑 is odd. But this can occur if and only if 𝜐₂(𝑞+1)<𝜐₂(𝑘−1), and the solutions of 8 in this case are 2𝑟=𝑞+1𝑑𝑖,(𝑖=0,…,𝑑−1). (68) An element 𝑢∈𝑀₂(−1) is a solution of 𝑢^𝑘−1−1=0 if and only if (𝜔^(𝑞+1)𝑠)^𝑘−1=1, if and only if 𝑠(𝑘−1)≡𝑞−1mod𝑞−1, (69) if and only if 𝑠(𝑘−1)≡0mod𝑞−1, (70) which has gcd(𝑘−1,𝑞−1) solutions.
To complete the proof, notice that by a similar method to that of the case 𝑎=1, 𝑢 is a solution of 𝑢^𝑘+3+1=0 (resp., 𝑢^𝑘−1−1=0) if and only if −𝑢⁻¹ is also a solution, and the set of solutions of (58) and (66) on 𝑁₃(−1) is the set of all solutions of 𝑢^𝑘+3+1=0 and 𝑢^𝑘−1−1=0 on 𝑁₃(−1), which is empty if 𝑞≡3(mod4) and is equal to {𝑢₁,𝑢₂} if 𝑞≡1(mod4).