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## Print version ISSN 0121-1935

### rev. cienc. vol.21 no.1 Cali Jan./June 2017

Artículo de investigación

𝑝-cycles, S 2 -sets and Curves with Many Points

Álvaro Garzón R.1

1Departamento de Matemáticas, Universidad del Valle, Santiago de Cali - Colombia alvaro.garzon@correounivalle.edu.co

Abstract

We construct S 2-sets contained in the integer interval I q-1 = [1,q - 1] with q = 𝑝 n , 𝑝 a prime number and n ∈ Z+, by using the 𝑝-adic expansion of integers. Such sets come from considering 𝑝-cycles of length n. We give some criteria in particular cases which allow us to glue them to obtain good S 2-sets. After that we construct algebraic curves over the finite field 𝔽 q with many rational points via minimal (𝔽𝑝, 𝔽𝑝)-polynomials whose exponent set is an S 2-set.

Key words: 𝑝-adic expansion; S 2-sets; finite field; Kummer cover; rational points

1 Introduction

Let 𝑝 be a prime number and 𝔽 q be a finite field with q = 𝑝 n elements and let Fq¯ be an algebraic closure of 𝔽 q . Given a polynomial ƒ(𝑥, 𝑦) ∈ 𝔽 q [ 𝑥, 𝑦 ] which is irreducible over Fq¯ , the set

𝕮ƒ =(α, β)  F¯q × F¯q such that ƒ (α, β) = 0

is an affine plane algebraic curve (over the finite field F¯q ) and the points P = (α, β) ∈ ∈ ƒ such that (α, β) ∈ 𝔽q x 𝔽q are called rational points over 𝔽q.

In 1940 A. Weil proved the Riemann hypothesis for curves over finite fields. As an immediate corollary he obtained an upper bound for the number of rational points on a geometrically irreducible nonsingular curve ℭ of genus g(ℭ) over a finite field of cardinality q, namely

#𝕮(Fq)  q + 1 + 2𝑔(𝕮)q, (*)

Where ℭ(𝔽 q ) denotes the set of rational points of the curve ℭ. If the cardinality of the finite field is not a square, the upper bound above was improved by Serre 1 substituting 2𝑔q by its integer part [ 2𝑔q ]. If the cardinality is a square, (say q = 𝑟2), then the ℭ curve is called maximal over F𝑟2 If #ℭ(𝔽q) attains the Weil’s upper bound; i.e., #𝕰(Fq) = r2 + 1 + 2r𝒈(𝕰) .

The interest in curves over finite fields with many rational points with respect to their genera (i.e., with # ℭ.(𝔽 q )) close to known upper bounds; e.g., see tables in 2 and 3 was greatly renewed after algebraic geometry codes (AG codes) were introduced by Goppa in 4. Many constructions of curves over finite fields are often performed by using special polynomials (𝑥) ∈ 𝔽 q [ 𝑥 ]. The essential properties of (𝑥) are sometimes of the following form:

Property I. One has that (𝔽q) ( 𝔽𝑝, and for most elements α ∈ 𝔽q, α is a simple root of 𝑝(𝑥) - 𝑝(α).

Property II. The set  =γ F¯q;p(x)-γ has multiple root in F¯q  has low cardinality, and one has a nice description of the multiplicities of the roots.

Polynomials satisfying property p (𝔽 q ) ⊆ 𝔽 p are known as (𝔽 p , 𝔽 p )-polynomials. A particular case of (𝔽 q ,𝔽 p )-polynomials, which are studied in detail in 5, are the so-called minimal (𝔽 q ,𝔽 p )-polynomials; i.e., (𝔽 q ,𝔽 p )-polynomials whose degree is ≤ 𝑞 − 1 and whose exponent set is characterized as being the 𝑝-cycles (also introduced in 5) of the integer interval [1, 𝑞 − 1]. We point out that in many cases the exponent set 𝔖 of a minimal (𝔽 q , 𝔽 p )-polynomial has a nice property that has been extensively studied in number theory, namely the 𝑆2 property. More precisely: A subset 𝔖 of integers is an 𝑆2 -set (or has the 𝑆2 property) if all the sums 𝑎 + 𝑎ʹ with 𝑎 ≠ 𝑎ʹ; a, a' ∈ 𝔖 are distinct.

It is known that if 𝔖 ⊆ [1, 𝑀] is an 𝑆2-set, then its cardinality must be asymptotically equal to M , see 6. As we remarked above in many cases the 𝑝-cycles of the integer interval [ 1, 𝑞 − 1] are 𝑆2-sets, unfortunately its cardinality which is a divisor of is almost always small respect to q-1 . The goal in this work is: firstly, provide some criteria to glue a set of 𝑆2 𝑝-cyles; i.e., 𝑝-cyles such that the underlying set has the 𝑆2 property, to obtain 𝑆2-sets whose cardinality is close to q-1 ; secondly, use some of these 𝑆2 sets obtained to construct algebraic curves ℭ over the finite field 𝔽 q whose set of rational points ℭ(F𝑞) has cardinality close to know upper bounds. Again, we refer to 2 and 3.

The paper is organized as follows: In section 2 we give a brief exposition of some properties of 𝑝-cycles. We show how these can be constructed from a generator element. Section 3 contains some results which allow us to decide when a number 𝑖 ∈ 𝑞−1: = [1, 𝑞 − 1] generates an 𝑆2 𝑝-cycle. We give some criteria to glue some of them and then, obtain sets with a good cardinality. Many examples are included. Section 4 is devoted to study a few particular cases to obtain good 𝑆2 sets. Finally, in section 5 we construct Kummer covers of the projective line over finite fields with many rational points. The idea comes from 7 and that is the construction of rational functions 𝜇(x) ∈ 𝔽 q (x) having the value 1 for many elements 𝛼 ∈ 𝔽 q , such rational functions in our case are induced by (𝔽 p ,𝔽 p )-polynomials whose exponent sets are 𝑆2 sets constructed in section 4.

2 𝑝-Cycles

Let 𝑝 be a prime number and let = 𝑝𝑛 be a non-negative power of 𝑝. The 𝑝-adic expansion of a positive integer 𝑎 in the integer interval 𝐼𝑞−1: = [1, 𝑞−1] is given by:

𝑎 = 𝑘0 + 𝑘1𝑝 + 𝑘2𝑝2 +…+ 𝑘𝑛-1 𝑝𝑛-1,

where the numerals 0 ≤ 𝑘𝑗 < 𝑝 satisfies for j = 0, ··· ,n−1.

If each 𝑎 = 𝑘0 + 𝑘1𝑝 + 𝑘2𝑝2 +…+ 𝑘𝑛-1 𝑝𝑛-1 ∈ 𝐼𝑞−1 is represented by the 𝑛-tuple (k0, k1,...,kn-1)𝐅pn , then we denote by 𝑎, the integer number obtained after applying the cyclic numeral-permutation:

𝑎 = 𝑘𝑛-1 + 𝑝𝑘0 + 𝑝2𝑘1 + …+ 𝑝 n-1 𝑘𝑛-2 .

By ⊓𝑘 we will understand the iteration k times the cyclic numeral-permutation. The 𝑝-adic period of 𝑎 is the small natural integer (𝑎) such that aι(a) = a and it is clear that the period of an integer number 𝑎 depends of 𝑝 and n.

A 𝑝-cycle 𝔍 is an ordered set I = a, a, a2, ...,aι(a)-1 , we will denote by 𝑙(𝔍), the length of 𝔍 and is defined by 𝑙(𝔍) = 𝑙(𝑎), (see 5 for more details).

Example 2.1 Let 𝑝 = 2 and 𝑛 = 4. The number 3 = 1∙20 + 1∙21 + 0∙22 + 0∙22 corresponds to the 4-tuple (1,1,0,0) therefore 3 = 0∙20 + 1∙21 + 0∙22 = 6, 6 = 12, 12 = 9. Consequently (3,6,12,9) is a 2-cycle of length 4. The element 3 ∈ [1,15] has period 4. On the other hand, if 𝑛 = 7, then the ordered set (3,6,12,24,48,96,65) is a 2-cycle of length 7 and the element 3 ∈ [1,127] has period 7.

Since the process of determining 𝑝-cycles in 𝐼𝑞−1 play an important role in this work, we will study some properties in detail.

Let G = 〈𝜎〉 be the cyclic group of order 𝑛. The group G acts on the set 𝐼𝑞−1 as follows:

ρ: 𝐺 ⨯ 𝐼𝑞−1 → 𝐼𝑞−1

𝑘, 𝑖) ↦ (𝑝𝑘 · 𝑖) 𝑞−1 𝑘 = 0,1,…, 𝑛 −1

where (𝑎) 𝑞−1 is a representative for the residual class of 𝑎 modulus 𝑞−1.

Theorem 2.1 For each 𝑖 ∈ 𝐼𝑞−1, the 𝑝-cycle i, i, i2, ..., il(i)-1 is the orbit of 𝑖 with respect to the action ρ above.

Proof: Observe that if 𝑖 = 𝑖0 + 𝑖1𝑝 + 𝑖2𝑝2 +…+ 𝑖 n-1 𝑝 n-1 is the 𝑝-adic expansion of 𝑖, then

𝑖 = 𝑖 n-1 + 𝑖0𝑝 +𝑖1𝑝2 +…+ 𝑖 n-2 𝑝 n-1 .

Hence

𝑝𝑖 − 𝑖 = +𝑖0𝑝 + 𝑖1𝑝2 +…+ 𝑖 n−1 𝑝 n − (𝑖 m−1 + 𝑖0𝑝 + 𝑖1𝑝2 +…+ 𝑖0n−2𝑝n-1)

= 𝑖 n−1 (𝑞 − 1),

consequently 𝑖 ≡ (mod (𝑞 − 1)).

Corollary 2.1 If q = p n andis a 𝑝-cycle, then 𝑙(ℑ) | 𝑛.

Proof: We known that if G acts on a set 𝑆, then 𝐺𝑥 = {𝑔 ∈ 𝐺 | 𝑔· 𝑥 = 𝑥}

is a subgroup of 𝐺 and the cardinal number of the orbit x¯ = 𝑔 · x 𝑔𝐺 of 𝑥, is (𝐺: 𝐺𝑥) the index of 𝐺𝑥 in 𝐺. 8, II,4.3.

Proposition 2.1 Every 𝑝-cycle has the form

ℑ = (i, pi ,..., p k i, (p k+1 i) q−1 ,..., (p i) q−1 )

where ℓ+1 is the length ofand k > 0 is the smallest integer satisfying p k i < 𝑞−1 ≤ p+1 k+1 i.

Proof: Since

p k+𝑗 i = 𝑟 (𝑞−1) + (p k+𝑗 i) 𝑞−1,

then

p k+𝑗+1 i = 𝑝(p k+𝑗 i) = (𝑟𝑝 + 𝑚)(𝑞−1) + (𝑝(p k+𝑗 i)𝑞−1)𝑞−1

where

(p k+𝑗 i) 𝑞−1 = 𝑚 (𝑞 − 1) + ((p k+𝑗 i) 𝑞−1)−1.

Hence,

(p k+𝑗+1 i)𝑞−1 = ((p k+𝑗 i) 𝑞−1)−1.

Remark 2.1 In accordance with Proposition (2.1), one can see that a 𝑝-cycle ℑ = (i, pi ,..., p k i, (p k+1 i) q-1 ,..., (p i) q-1 ), is nothing but the cyclotomic coset of 𝑖 mod(𝑞 − 1). Cyclotomic cosets mod(𝑁) play an important role in the factorization in 𝔽 p [x] of the polynomial 𝑥𝑁−1 and consequently in coding theory. We refer to 9 for details.

Example 2.2 In the following Tables 1, 2 and 3 we exhibit the different 𝑝-cycles for 𝑝 = 2.3 and 𝑞 = 𝑝𝑛 for some values of 𝑛.

Table 1  2-cycles of length 1, 2, 3 and 6.

 𝑞 = 26 = 64 (63) (21,42) (9,18,36) (27,54,45) (1,2,4,8,16,32) (3,6,12,24,48,33) (5,10,20,40,17,34 (7,14,28,56,49,35) (11,22,44,25,50,37) (13,26,52,41,19,38) (15,30,60,57,51,39) (23,46,29,58,43,53) (31,62,61,59,55,47)

Table 2 3-cycles of length 1 and 3.

 𝑞 = 33 = 27 (13) (26) (1,3,9) (2,6,18) (4,12,10) (5,15,19) (7,21,11) (8,24,20) (14,16,22) (17,25,23)

Table 3 3-cycles of length 1, 2 and 4.

 𝑞 = 34 = 81 (40) (80) (10,30) (20,60) (50,70) (1,3,9,27) (2,6,18,54) (4,12,36,28) (5,15,45,55) ((7,21,63,29) (8,24,72,56) (11,33,19,57) (13,39,37,31) (14,42,46,58) (16,48,64,32) (17,51,73,59) (22,66,38,34) (23,6947,61) (25,75,65,35) (26,78,74,62) (41,43,49,67) (44,52,76,68) (53,79,77,71)

Observe for example, that for 𝑞 = 64 not all 2-cycle has length 6. The following proposition says when these situations occur. Before that we introduce a convenient notation. Although it is true that we can obtain a 𝑝-cycle from any of its elements, we will say that an integer 𝑖 generates the 𝑝-cycle ℑ if

ℑ = (i, 𝑝i ,..., 𝑝 k i, ( 𝑝 k+1 i) q-1 ,..., (𝑝 i) q-1 ),

and i < (𝑝 k+𝑗 i) q-1 for 𝑗 = 1,…, ℓ− 𝑘. In this case we write ℑ = ⟨ i ⟩.

Proposition 2.2 Let 𝑞 = 𝑝𝑛 with 𝑛 = 𝑚·𝑙. If ℑ = ⟨i⟩, then 𝑙 = (ℑ) < 𝑛 if and only if i is a multiple of

j=0m-1 pl.j

Proof: By definition, ℓ = length(ℑ) if and only if, 𝑝 i= (𝑞 - 1) ℎ + i

if and only if, i =j=0m-1 pl.jhl .

Proposition 2.3 The greatest integer iI 𝑞−1 such that i generates a 𝑝-cycle of length 𝑛, is i = 𝑝𝑛 -( 𝑝𝑛-1 + 1). The corresponding 𝑝-cycle is,

(𝑝𝑛 -( 𝑝𝑛-1 + 1), 𝑝𝑛 -( 𝑝0 + 1), 𝑝𝑛 -( 𝑝 + 1), 𝑝𝑛 -( 𝑝2 + 1),…, 𝑝𝑛 -( 𝑝𝑛-2 + 1)).

Proof: First observe that

𝑝i = 𝑝·(𝑝𝑛 - 𝑝𝑛-1 - 1) = (𝑞 - 1)( 𝑝 - 1) - 1 ≡ (𝑞 - 2)(mod 𝑞 - 1).

Hence 𝑝2 i ≡ 𝑞 - 𝑝 - 1 (mod 𝑞 - 1), so in general, 𝑝𝑘 i ≡ 𝑞 - (𝑝𝑘-1 + 1 (mod 𝑞 - 1) which leads to the desired expression. Secondly, each number i < 𝑗 < ?? - 1 can be rewritten as 𝑗 = (𝑞 - 1) - (𝑝𝑛-1 - 𝑘-1) with 0 ≤ 𝑘 ≤ 𝑘 ≤ 𝑝𝑛-1 - 2. Now, the equation 𝑝𝑥 - (𝑞 - 1)t = 𝑗 with 0 ≤ t ≤ 𝑝 - 1 has unique solution in I 𝑞-1 which implies that 𝑗 is the remainder modulus 𝑞 - 1 of some 𝑥 ( I 𝑞-1.

Remark 2.2 Is know that, if 𝛼 = 𝛽 i ∈ 𝔽 q = 𝔽𝑝 (𝛽) and 𝑚𝛼 (𝑥) is its minimal polynomial, then

mα(x) = ti(x-βt) ,

(Cf 10, Theorem 4.1). Hence one can determine the number of 𝑝-cycles of length 𝑑 | 𝑛 in I 𝑞-1, such number is:

Nq(d) = 1dd'dμ(d')qdd'

(here (·)is the Moebius function), as many as irreducible polynomials of degree 𝑑 in 𝔽𝑝 [𝑥 ].

3 𝑆 2 - Sets

A subset 𝔖 of integers is an 𝑆2-set if all the sums 𝑎 + 𝑎ʹ with 𝑎 ≠ 𝑎ʹ; 𝑎, 𝑎ʹ; ∈ 𝔖 are distinct. From now on, a set 𝔖 has the 𝑆2 property if 𝔖 is an 𝑆2-set. Similarly, a 𝑝-cycle ℑ = (𝚤1,…, 𝚤𝑙) is an 𝑆2 𝑝-cycle if the underlying set (𝚤1,…, 𝚤𝑙) has the 𝑆2 property. In this section we give some criteria that allows us to decide when a 𝑝-cycle ℑ = ⟨i⟩ has the 𝑆2 property. Also, we give conditions on the generators of a set of 𝑆2-cycles such that the union of these retain this property.

Theorem 3.1 If GCD (i,𝑞 - 1) = 1, then ℑ = ⟨i⟩ is an 𝑆2 𝑝-cycle.

Proof: Assume the contrary, so there exist integers 0 ≤ 𝑟, 𝑠, 𝑢, 𝑣 ≤ 𝑛 -1such that

(𝑝𝑟 i) 𝑞−1 +(𝑝𝑠 i)𝑞−1 = (𝑝𝑢 i)𝑞−1 + (𝑝𝑣 i)𝑞−1 𝑟 < 𝑠, 𝑢< 𝑣, (1)

If 𝑟 = min{𝑟,u}, then (1) implies (𝑝𝑣-𝑟 + 𝑝𝑢-𝑟 − 𝑝𝑠-𝑟 - 1) ≡ (mod 𝑞−1) which is absurd.

Remark 3.1 The reciprocal of Theorem 5.1 is false, in fact if 𝑝 = 3 and = 5, then is easily proved that the 3-cycle generated by the divisor 11 of 242, (11,33,99,55,165) has the 𝑆2 property.

Corrollary 3.1 If 𝑞−1 is a prime number, then any 𝑝-cycle is an 𝑆2-set.

Example 3.1 The Table 4 shows all the 2-cycles of length 5 which, by 3.1 are 𝑆2-sets contained in the integer interval [1.31 ].

Table 4. 𝑆2 2-cycles of length 5.

 1,2,4,8,16) (3,6,12,24,17) (5,10,20,9,18) 7,14,28,25,19) (11,22,13,26,21) (15,30,29,27,23)

The next corollary give us information about the components of a 𝑝-cycle when 𝑖 | 𝑞−1. More precisely.

Corrollary 3.2 Let 𝑖 | 𝑞−1 and Ξ = 〈 i 〉, then all its components are multiples of 𝑖.

Proof: By Proposition (2.1), each component of Ξ has the form (𝑝𝑗𝑖) 𝑞−1 = 𝑝𝑗𝑖 - (𝑞−1) ℎ𝑗.

Remark 3.2 Let us consider the 𝑆2 3-cycle of length 5, Ξ =(11,33,99,55,165) generated by 11, divisor of 242 = 35 − 1 contained in the interval [1,242]. It is clear that its cardinality 5 is very small respect to 242  15 even so in accordance with the previous corollary, if we cancel the common factor 11 of each member of the 3-cycle Ξ, we obtain the set 𝔖 ={1,3,9,5,15} which is again an 𝑆2-set although it is not the underlying set of a 3-cycle. The important fact here is that the cardinality of 𝔖 is 5 and now we have a nice 𝑆2-set in the interval [1.15] whose cardinality is now very good respect to 15  4 . We can obtain each member of this set as follows.

Corollary 3.3 Let 𝑖 ∈ [1, 𝑞−1], 𝑑 = 𝐺𝐶D (𝑖, 𝑞−1) and Ξ = 〈 i 〉, The set obtained by canceling the common factor 𝑑 to each component of Ξ has the for:

𝔖=id,...,pkid,pk+1dq-1d,...,pn-1idq-1d

Proof: It is clear from the uniqueness of the residue.

Example 3.2 If 𝑝 = 2, 𝑛 = 8 and i = 27 ∈ [1,255], then the 2-cycle Ξ, generated by 27, (27,54,108,216,177,99,198,141) is an 𝑆2 2-cycle. Now since (27.255) = 3, then the set 𝔖={9,18,36,72,59,33,66,47} obtained canceling the common factor 3 of each component of Ξ, is an 𝑆2-set. Observe that 𝔖 as a subset of the interval [1,72 has cardinality closed to 72<9 .

On the other hand, since 255/3 = 85 and the first four terms 9,18,36,72 do not exceed 85, but 2 72 ⨯ 72 = 144 ⨯ 1 + 59; 2 ⨯ 59 = 108 ⨯ 1+33 and 2 ⨯ 66 = 112 = 85 ⨯ 1+ 47, then we can generate the new set from 9 taking remainder mod 85.

We emphasize that the set 𝔖 = {9,18,36,72,59,33,66,47} is not the underlying set of the 2-cycle generated by 9, which is {9,18,36,72,144,33,66,132}. This, as 𝑆2-set is very poor respect to 25515 , but by Corollary (3.2), induces a new 𝑆2-set contained in the interval [1.48] namely, {3,6,12,24,48,11,22,44}. Again, this is not the underlying set of the 2-cycle generated by 3, which is {3,6,12,24,48,96192,129}. That is again an 𝑆2-set of small cardinality respect to 255 but again induces a new 𝑆2-set {1,2,4,8,16,32,64,43} ⊂ [1.64]. Clearly this is not the underlying set of the 2-canonical cycle {1,2,4,8,16,32,64,128}.

As we can see, there are examples of 𝑝-cycles which have the 𝑆2 property and whose generator 𝑖 is neither prime with 𝑞 - 1 nor divisor of 𝑞 − 1. The following result provides a concrete example.

Proposition 3.1 Let t be an integer number and 𝑛=4t, then the underlying set of the 𝑝-cycle of length 𝑛=4t generated by

ι:=pμμ=02t - pt =1+p+p2+...+pt-1+pt+pt+1+...+p2t

is an 𝑆2-set. (Here the hat means that the power 𝑝 t was excluded.)

Proof: First observe that

⟨𝚤⟩ = (𝚤, 𝑝𝚤, 𝑝2𝚤,…, 𝑝 2t-1 𝚤, (𝑝 2t 𝚤)𝑞−1, (𝑝 2t+1 𝚤)𝑞−1 ,…, (𝑝 4t-1 𝚤)𝑞−1)

where (𝑝𝑗𝚤) 𝑞−1 denotes the remainder of 𝑝𝑗𝚤 modulus 𝑞 − 1.

As always, we will suppose that such a set is not an 𝑆2-set, therefore several cases may occur, namely:

Case 1. 𝑝𝑟𝚤 + 𝑝𝑠𝚤 = 𝑝𝑢𝚤 +𝑝𝑣?? with 𝑟, 𝑠, 𝑢, 𝑣 ≤2t − 1.

Case 2. prι+psι=puι+p2t+mvιq-1 with 𝑟, 𝑠, 𝑢, 𝑣 ≤ 2t − 1.

Case 3. pr𝜄+ps𝜄=(p2t+mv𝜄)q-1 with 𝑟, 𝑠, 𝑢 ≤ 2t − 1.

Case 4. pr𝜄+(p2t+ms𝜄)q-1=(p2t+mu𝜄)q-1 +(p2t+mv𝜄)q-1 with 𝑟, 𝑠 ≤2t − 1.

Case 5. (p2t+mr𝜄)q-1+(p2t+ms𝜄)q-1=(p2t+mu𝜄)q-1+(p2t+mv𝜄)q-1 .

Case 6. pr𝜄+(p2t+ms𝜄)q-1=pu𝜄+(p2t+mv𝜄)q-1 with 𝑟, 𝑢 ≤ 2t − 1.

Our next goal is to determine the number (𝑝 2t+𝑚 𝚤)q−1. For this end, let us denote by 𝚤𝑘, and θ 𝑚 the following natural numbers:

𝚤𝑘,𝑙: = 1 + 𝑝 + ... + pi^ + .. + 𝑝 2t- and θ 𝑚: = 1 + 𝑝 + ... + 𝑝𝑚.

With above notation, the number (𝑝 2t+𝑚 𝚤)𝑞−1 can be written as follows:

(p2t+m𝜄)q-1=θm+p2t+m𝜄m+1t if 0 mt-1,𝜄2t-m,m-t+p2t-mθ2t-m-1 if tm2t-1. (2)

As an illustration we consider the case 6. Let us suppose that

pr𝜄+(p2t+ms𝜄)q-1=pu𝜄+(p2t+mv𝜄)q-1

For instance, if 0 ≤ 𝑟 < 𝑢 ≤ 2t − 𝑚 < 2t + 𝑚𝑠 < 2t + 𝑚𝑣 with 0 ≤ 𝑚𝑠t − 1< 𝑚𝑣, then with the above notation we have:

prι+θms+p2t+msιms+1,t=puι+ι2t-mv,mv-t+p2t+mvθ2t-mv-1,

Now, if 𝑟 = 0 after cancel common terms, we get the following equality

(pmv-t+...+p2t)+(1+p+...+pms)+p2t+msιms+1,t=puι+p2t+mvθ2t-mv-1,

which implies that 𝑝|1.

On the other hand, 𝑟 > 0 implies

prι+p2t+msιms+1,t=puι+(pms+1+...+pmv-t^+...+pmv)+p2t+mvθ2t-mv-1,

if 𝑚𝑠 < 𝑚𝑣t and

prι+pmv-t+p2t+msιms+1,t=puι+(pms+1+...+pmv)+p2t+mvθ2t-mv-1

if 𝑚𝑠𝑣 − 𝑡

If 𝑚𝑠 < 𝑚𝑣 − 𝑡, then 𝑟 > 𝑚𝑠 + 1 implies 𝑝|1; 𝑟 = 𝑚𝑠 + 1, implies 𝑟 = 𝑚 > 𝑡 which is absurd and finally 𝑟 = 𝑚𝑠 + 1 implies pms+1|2(1+p+...+pms) . Similar arguments led us to contradictions when we consider the case 𝑚𝑠 ≥ − 𝑡; we omit details.

Example 3.3 If 𝑝 = 2 and ?? = 1, then 𝑛 = 8, 𝚤 = 27, Proposition (2.2) says that the 2-cycle (27,54,108,216,177,99,198,141) is an 𝑆2-set contained in [1,256].

Proposition 3.2 If 𝑖 < 𝑝, then the 𝑝-cycle ⟨𝑖⟩ is an 𝑆2 𝑝-cycle.

Proof: If there are different elements 𝑟, 𝑠, 𝑢, 𝑣 such that

𝑝𝑟𝑖 + 𝑝𝑠𝑖 = 𝑝𝑢𝑖 +𝑝𝑣𝑖,

then we would have 𝑝|1.

Corollary 3.4 Let 𝑖, 𝑗 ∈ I 𝑞−1. Then if 𝑖 < 𝑝 and , (𝑗, 𝑞−1) = 1, the 𝑝-cycle Ξ generated by 𝑖𝑗 is an 𝑆2 𝑝-cycle.

Proof: If not, we should have 𝑝𝑟𝑖𝑗 + 𝑝𝑠𝑖𝑗 = 𝑝𝑢𝑖𝑗 + 𝑝𝑢𝑖𝑗 (mod 𝑞−1) for some integers 𝑟, 𝑠, 𝑢, 𝑣, but this congruence contradicts Proposition (3.2).

Proposition 3.3 If 𝑛 is odd, then the 𝑝-cycle generated by 𝑝 +1 is an 𝑆2 𝑝-cycle.

Proof: Observe that such 𝑝-cycle is

(𝑝 + 1, 𝑝2 + 𝑝,…, 𝑝𝑛-2 + 𝑝 𝑛-3, 𝑝 𝑛-1 + 𝑝 𝑛-2, 𝑝𝑛-1 + 1)

First, let us suppose that there exist different integers 1 ≤ 𝑟, 𝑠, 𝑢, 𝑣 ≤ 𝑛 − 1 such that

𝑝𝑛-𝑟+𝑝𝑛-𝑟-1 + 𝑝𝑛-𝑠 + 𝑝𝑛-𝑠-1 = 𝑝𝑛-𝑢 + 𝑝𝑛-𝑢-1 + 𝑝𝑛-𝑣 + 𝑝𝑛-𝑣-1,

then if 𝑠 = max {𝑟, 𝑠, 𝑢, 𝑣} we have 𝑝|1. On the other hand, if there exist different integers 1 ≤ 𝑟, 𝑠, 𝑢 ≤ 𝑛 − 1 such that

𝑝𝑛-𝑟 + 𝑝𝑛-??-1 + 𝑝𝑛-𝑠+𝑝𝑛-𝑠-1 = 𝑝𝑛-𝑢 + 𝑝𝑛-𝑢-1 + 𝑝𝑛-1 + 1, (3)

then since

𝑝𝑛 + 𝑝𝑛-1 -(𝑞 - 1) = 𝑝𝑛-1 + 1,

(3) can be written as

𝑝𝑛-𝑟 + 𝑝𝑛-𝑟-1 + 𝑝𝑛-𝑠+ 𝑝𝑛-𝑠-1 = 𝑝𝑛-𝑢 + 𝑝𝑛-𝑢-1 + 𝑝𝑛 +𝑝𝑛-1 -(𝑞 - 1),

this implies that 𝑝 + 1 divides 𝑞 - 1 which is absurd.

Now we are interested in finding conditions to decide when the union of 𝑆2 𝑝-cycles is an 𝑆2-set. A first approach is given in the following proposition:

Proposition 3.4 If 1 < ?? < 𝑝, then the set

𝔖: ={⟨1⟩ ⋃ ⟨𝑗⟩}

is an 𝑆2-set.

Proof: Assume that 𝔖 is not an 𝑆2-set. Then, there exist ??, 𝑏, 𝑐, 𝑑 ∈ {1, 𝑗} and integer numbers 0 ≤ 𝑟, 𝑠, 𝑢, 𝑣 ≤ 𝑛 - 1 such that,

𝑎𝑝𝑟 + 𝑏𝑝𝑠 = 𝑐𝑝𝑢 + 𝑑𝑝𝑣, (4)

with 𝑟 ≠ 𝑠 if 𝑎 = 𝑏 and 𝑢 ≠ 𝑣 if 𝑐 = 𝑑. The proof is somewhat technical and therefore divided into several cases. To begin suppose that δ = min {𝑟, 𝑠, 𝑢, 𝑣}.

Case 1: If 𝑎 = 𝑏 and 𝑐 = 𝑑, then after canceling 𝑝 δ in (4) we obtain 𝑝|𝑎 or 𝑝|𝑐 which is a contradiction.

Case 2: 𝑎 = 𝑏 and 𝑐 ≠ 𝑑. In this case, if δ = 𝑟 or 𝑠, then after canceling 𝑝 δ in (4) we obtain 𝑝 | 𝑎; if δ = 𝑢 or 𝑣, then 𝑝| or 𝑝|𝑑. The case δ = 𝑢 = 𝑣 leads to 𝑎𝑝 𝑟-δ + 𝑎𝑝 𝑠-δ = 𝑐 + 𝑑, which is again a contradiction. On the other hand, if δ = 𝑟 = 𝑢 = 𝑣 we obtain 𝑎 + 𝑎𝑝 𝑠-δ = 𝑐 + 𝑑, now after to consider the different possibilities for 𝑎, 𝑐, 𝑑, we conclude that 𝑝|1 or 𝑝|𝑗.

Case 3: 𝑎 ≠ 𝑏 and 𝑐 ≠ 𝑑. If 𝑟, 𝑠, 𝑢, 𝑣 are distinct and δ = 𝑟, 𝑠, 𝑢, or 𝑣 then 𝑝|𝑎, 𝑏, 𝑐 or 𝑑 which is a contradiction. If two exponents are equal, for example, 𝑟 = 𝑢 and δ = 𝑟 = 𝑢, then we have 𝑎 - 𝑐 = 𝑑𝑝 𝑣-δ - 𝑑𝑝 𝑠-δ which is absurd. Finally, if three exponents are equal, for example, 𝑢 = 𝑣 = 𝑠 and δ = 𝑢 = 𝑣 = 𝑠, then equation (4) is nothing else but 𝑎𝑝 𝑟-δ + 𝑏 = 𝑐 + 𝑑. Now again after consider all the possibilities for 𝑎, 𝑏, 𝑐, 𝑑, we obtain the same equations and conclusions as in the Case 2.

Example 3.4 If we take 𝑝 = 3, then in this case 𝑗 = 2. We show in the following Table some 𝑆2-sets 𝑆 ⊂ [1, 𝑀 = 3𝑛 − 1] for different values of 𝑛.

Table 5 2-sets as union of two 3-cycles.

 𝑛 𝑀 -sets 2 8 2.82 1,2,3,6 3 26 5.09 1,2,3,6,9,18 4 80 8.9 1,2,3,6,9,18,27,54 5 242 15.55 1,2,3,6,9,18,27,54,81,162

As we can see, the first two sets have a reasonable cardinality in respect to M , while the latter two do not. However, it should be noted that since any subset of 𝑆2-set retains the 𝑆2 property, the set obtained for 𝑀 = 80 is a nice 𝑆2-set, if it is considered as subset of the integer interval [1.54], likewise the sets {1,2,3,6,9,18,27,54,81} ( [1,81] and {1,2,3,6,9,18,27} ( [1.27] have also a nice cardinality with respect to 9 and 5 respectively.

Proposition 3.5 If 1 < 𝑗 < 𝑘 < 𝑝 and 𝑘 + 𝑗 ≠ 𝑝 + 1, then the set

𝔖:= {⟨1⟩ ∪·⟨ 𝑗 ⟩ ∪ ⟨ 𝑘 ⟩}

is an 𝑆2-set.

Proof: Using the notation as in the proof of Proposition (3.4), let us assume that there exist 𝑎, 𝑏, 𝑐, 𝑑 ( {1, 𝑗, 𝑘} and integer numbers 0 ≤ 𝑟, 𝑠, 𝑢, 𝑣 ≤ 𝑛 − 1 such that,

𝑎𝑝𝑟 + 𝑏𝑝𝑠 = 𝑐𝑝𝑢 + 𝑑𝑝𝑣, (5)

with 𝑟 ≠ 𝑠 if 𝑎 = 𝑏 and 𝑢 ≠ 𝑣 if 𝑐 = 𝑑. We distinguish three cases:

Case 1. If 𝑎 = 𝑏 and 𝑐 = 𝑑, then after canceling 𝑝δ in (5) we obtain 𝑝|𝑎 or 𝑝|𝑐 which is a contradiction.

Case 2. 𝑎 = 𝑏 and 𝑐 ≠ 𝑑. In this case, if δ = 𝑟 or 𝑠, then after canceling 𝑝δ in (5) we obtain 𝑝|𝑎; if δ = 𝑢 or 𝑣, then 𝑝|𝑐 or 𝑝|𝑑. The case δ = 𝑢 = 𝑣 leads to 𝑎𝑝𝑟-δ + 𝑎𝑝𝑠-δ = 𝑐 + 𝑑, which is again a contradiction. On the other hand, if δ = 𝑟 = 𝑢 = 𝑣 we obtain 𝑎 + 𝑎𝑝𝑠-δ = 𝑐 + 𝑑. Now after to consider the different possibilities for 𝑎, 𝑐, 𝑑, we obtain the following equalities:

𝑝𝑠-δ + 1 = 𝑗 + 𝑘, + 𝑗??𝑠-δ = 𝑘 +1, 𝑘 + 𝑘𝑝𝑠-δ = 𝑗 + 1. (6)

But 𝑠-δ ≥ 1 implies that none of the above equations are satisfied.

Case 3. 𝑎 ≠ and 𝑐 ≠ 𝑑. If 𝑟, 𝑠, 𝑢, 𝑣 are distinct and δ = 𝑟, 𝑠, 𝑢 or 𝑣, then 𝑝|𝑎, 𝑏, 𝑐 or which is a contradiction. If two exponents are equal, for example 𝑟 = 𝑢 and δ = 𝑟 = 𝑢, then we have 𝑎 − 𝑐 = 𝑑𝑝𝑣-δ − 𝑏𝑝𝑠-δ which is absurd. Finally, if three exponents are equal, for example 𝑢= 𝑣 = 𝑠 and δ = 𝑢= 𝑣 = 𝑠, then equation (5) is nothing else but 𝑎𝑝𝑟-δ + 𝑏 = 𝑐 + 𝑑. Now again after consider all the possibilities for 𝑎, 𝑏, 𝑐, 𝑑, we obtain the same equations and conclusions as (6).

Example 3.5 In the followingTable

we show some 𝑆2 -sets obtained for different values of 𝑀 = 𝑝𝑛-1

Table 6 Examples based in Proposition 3.5.

𝑝 𝑛 𝑀 𝑆 2 -sets
5 2 24 4.89 1,2,3,5,10,15
7 2 48 6.9 1,2,3,7,14,21
1,2,4,7,14,28
1,2,5,7,14,35
1,3,4,7,21,28
1,3,6,7,21,42
1,4,5,7,28,35
1,4,6,7,28,42
1,5,6,7,35,42

With a few modifications at the proof of Proposition (3.5), we have:

Corollary 3.5 If 𝑗 < 𝑘 < 𝑙 < 𝑝 and 𝑘 + 𝑙 ≠ 𝑝𝑗 + 𝑗, then the set

𝑆: = { ⟨𝑗⟩ ∪ ⟨𝑘⟩ ∪ ⟨𝑙⟩ }

is an 𝑆2-set.

Example 3.6 If we take 𝑝 = 7 and 𝑛 = 2, we obtain the Table 7.

Table 7 Joining 3 7-cycles of length 2.

𝑀 = 𝑝𝑛-1 𝑆 2 -sets
48 6.9 2,3,4,14,21,28
2,3,5,14,21,35
2,3,6,14,21,42
2,4,5,14,28,35
2,4,6,14,28,42
2,5,6,14,35,42
3,4,5,21,28,35
3,4,6,21,28,42
4,5,6,28,35,42

Proposition 3.6 If 1 < 𝑎1 < 𝑎2 < 𝑎3 < 𝑝, 𝑎𝑖 + 𝑎𝑗 ≠ 𝑝 + 1 and the set {1, 𝑎1, 𝑎2, 𝑎3} is an 𝑆2-set, then the set

𝔖: = { ⟨1⟩ ∪ ⟨𝑎1⟩ ∪ ⟨𝑎2⟩ ∪ ⟨𝑎3⟩ }

is an 𝑆2-set.

Proof: By Proposition (3.5) in order to prove our assertion, we only have to analyze the equation

𝑝𝑟 + 𝑎1𝑝𝑠 = 𝑎2𝑝𝑢 + 𝑎3𝑝𝑣, (7)

with 0 < 𝑟, 𝑠, 𝑢, 𝑣 < 𝑛. As in the proof of Proposition (3.5) let 𝛿 = min {𝑟, 𝑠, 𝑢, 𝑣}.

1. If δ = 𝑟 = 𝑠 = 𝑢 ≠ 𝑣, then equation (7) becomes 0 < 1 + 𝑎1 − 𝑎2 = 𝑎3𝑝 𝑣-δ which is a contradiction.

2. If ≠ δ = 𝑟 = 𝑠 ≠ 𝑢, we obtain 1 + 𝑎1 = 𝑎2𝑝 𝑢-δ + 𝑎3?? 𝑣-δ which is absurd.

3. The cases δ = 𝑟 with δ ≠ 𝑠, 𝑢, 𝑣 and δ = 𝑟 = 𝑠 = 𝑢 = 𝑣 are trivial.

Example 3.7 If 𝑝 = 11 and 𝑛 = 2, there exist 36 possibilities for 𝑎1, 𝑎2, 𝑎3 which satisfies the hypothesis of the Proposition (3.6), we exhibit in the Table 8 only some of such sets. Before this note that although this example did not give sets whose cardinality is near to 12011 , we point out that each of these sets provides examples of 𝑆2 sets for any values less than 𝑀 =120. In some cases, we find good examples.

Table 8 Joining 4 11-cycles of length 2.

 1,2,3,5,11,22,33,55 1,2,3,6,11,22,33,66 1,2,3,7,11,22,33,77 1,2,3,8,11,22,33,88 1,3,5,10,11,33,55,110 1,3,6,7,11,33,66,77 1,3,6,10,11,33,66,110 1,3,7,8,11,33,77,88 1,3,7,10,11,33,77,110 1,4,6,7,11,44,66,77 1,4,6,10,11,44,66,110 1,4,7,9,11,44,77,99 1,4,9,10,11,44,99,110 1,5,9,10,11,55,99,110 1,6,7,8,11,66,77,88 1,6,9,10,11,66,99,110 1,7,8,9,11,77,88,99 1,7,9,10,11,77,99,110

4 Particular Examples

So far, we have provided some criteria that allow us to glue 𝑆2 𝑝-cycles ⟨𝑖⟩, ⟨𝑗⟩, ⟨𝑘⟩ such that, the resulting set 𝔖: = {⟨𝑖⟩ ∪ ⟨𝑗⟩ ∪ ⟨𝑘⟩} maintains the 𝑆2 property, the condition on their generators 𝑖, 𝑗, 𝑘 is 1 < 𝑖 < 𝑗 < 𝑘 < 𝑝. In this section we will try to go a little further.

4.1 The Case 𝑞 = 𝑝 2

Let us suppose that 1 ≤ 𝑗 < 𝑘 < 𝑝 < 𝑖 are the generators of the 𝑝-cycles (𝑗, 𝑝𝑗), (𝑘, 𝑝𝑘) and (𝑖, (𝑝𝑖) 𝑞−1). As always, we want to establish conditions on their generators so that the resulting set 𝔖: = {⟨𝑖⟩ ∪ ⟨𝑗⟩ ∪ ⟨𝑘⟩} has the 𝑆2 property. Now, observe that for this purpose it is easier to establish conditions for which 𝔖 does not have the 𝑆2 property. Now, observe that for this purpose it is easier to establish conditions for which 𝔖 does not have the S 2 property. In fact, 𝔖: = {𝑗, 𝑝𝑗, 𝑘, 𝑝𝑘, 𝑖 (𝑝𝑖) 𝑞−1} is not an 𝑆2 set if there exist distinct 1, 𝑎2, 𝑎3, 𝑎4 ( 𝔖 such that 𝑎1 + 𝑎2 = 𝑎3 + 𝑎4, then it is evident that this equality leads us to consider a large number of equations. Fortunately many of these equations are not possible, for example it is impossible that 𝑘 + 𝑗 = 𝑝𝑘 + 𝑝𝑗 or 𝑘 + 𝑗 = 𝑝𝑘 +(𝑝𝑖)𝑞−1 or 𝑘 + 𝑗 = 𝑖 +𝑝𝑗. After check all the possibilities, we must consider only the following equations:

1) 𝑖 + (𝑝 + 1) 𝑗 + 𝑘 = 0

2) 𝑖 − (𝑝 + 1) 𝑗 + 𝑝𝑘 = 0

3) 𝑖 − (𝑝 − 1)𝑗 − 𝑝𝑘 = 0

4) 𝑖 + 𝑝𝑗 − (𝑝 − 1)𝑘 = 0

5) 𝑖 − 𝑗 − (𝑝 − 1)𝑘 = 0

6) 𝑖 - 𝑗 = (𝑝 + 1)𝑚

7) 𝑖 + 𝑗 − (𝑝 + 1)𝑘 = 0

8) 𝑖 + 𝑗 = (𝑝 + 1)𝑚

9) 𝑖 + 𝑝𝑗 − (𝑝 + 1)𝑘 = 0

10) 𝑖 − 𝑘 = (𝑝 − 1)𝑚

11) 𝑝𝑖 − (𝑝 − 1)𝑗 - 𝑝𝑘 = (𝑝2 − 1)𝑚

12) 𝑝𝑖 − 𝑝𝑗 − (𝑝 − 1)𝑘 = (𝑝2 − 1)𝑚

13) 𝑝𝑖 + 𝑗 − (𝑝 + 1)𝑘 = (𝑝2 − 1)𝑚

14) (𝑝 + 1)𝑖 − 𝑝𝑗 − ??𝑘 = (𝑝2 − 1)𝑚

Equation 6 corresponds to case 𝑗 + (𝑝𝑖) 𝑞−1 = 𝑖 + 𝑝𝑗, equation 11 corresponds to case 𝑗 + (𝑝𝑖)𝑞−1 = 𝑝𝑗 + 𝑝𝑘 and so on. To illustrate, we consider 𝑝 = 7 and we will to analyze the equations 3, 5, 6, 9, 10 and 12.

The Table 9 contains the solutions:

Table 9 solutions

Equation Solutions
(3) 𝑖 − (𝑝 − 1)𝑗 - 𝑝𝑘 = 0 (1,2,20);(1,3,27);(1,4,34);(1,5,39);
(1,6,48)
(2,3,33);(2,4,40);(2,5,47)
(3,4,46)
(5) 𝑖 − 𝑗 − (𝑝 − 1)𝑘 = 0 (1,2,13);(1,3,19);(1,4,25);(1,5,31)
(1,6,37)
(2,3,20);(2,4,26);(2,5,32);(2,6,38).
(3,4,27);(3,5,33);(3,6,39).
(4,5,34);(4,6,40).
(5,6,41).
(6) 𝑖 - 𝑗 = (𝑝 + 1)𝑚 (1,𝑘,9,1) (1, 𝑘,17,2); (1, 𝑘,25,3);
(1,,33,4);(1,𝑘,41,5).
(2,𝑘,10,1);(2,𝑘,18,2);(2,𝑘,26,3);
(2,,34,4);(2,𝑘,42,5).
(3,𝑘,11,1);(3,𝑘,19,2);(3,𝑘,27,3);
(3,,35,4);(3,𝑘,43,5).
(4,𝑘,12,1);(4,𝑘,20,2);(4,𝑘,28,3);
(4,,36,4);(4,𝑘,44,5).
(5,𝑘,13,1);(5,𝑘,21,2);(5,𝑘,29,3);
(5,,37,4);(5,𝑘,45,5).
(6,𝑘,14,1);(6,𝑘,22,2);(6,𝑘,30,3);
(6,,38,4);(6,𝑘,46,5).
(9) 𝑖 + 𝑝𝑗 − (𝑝 + 1)𝑘 = 0 (1,3,11); (1,4,17);(1,5,23);(1,6,29).
(2,4,10); (2,5,16);(2,6,22).
(3,5,9);(3,6,15).
(4,6,8).
(10) 𝑖 − 𝑘 = (𝑝 − 1)𝑚 (𝑗,3,9,1)
(𝑗,4,10,1).
(𝑗,5,11,1); (𝑗,5,17,2).
(𝑗,6,12,1);(𝑗,6,18,2).
(12) 𝑝𝑖 - 𝑝𝑗 − (𝑝 − 1)𝑘 = (𝑝2 − 1)𝑚 (1,2,37,5); (1,3,31,4); (1,4,25,3).
(1,5,19,2);(1,6,13,1).
(2,3,32,4);(2,4,26,3); (2,5,20,2).
(2,4,27,3).

Note that equations containing (𝑝𝑖) 𝑞−1 such that (6), (10) and (12), have as solutions quadruples (𝑗,,,). The value 𝑚 appears because 𝑖 > 𝑝 = 7 and therefore, 𝑝𝑖= (𝑝2−1)+(𝑝𝑖)𝑞−1 = 48𝑚 + (7𝑖). The other equations have as solutions triples (𝑗, 𝑘, 𝑖). The first solution of the equation (6) for example, says that the set 𝔖 = {1, 7, k, 7k, 9, 15} is not an 𝑆2 set for 2 ≤ 𝑘 ≤ 6. The following Table contains all the 𝑆2 sets which were obtained joining three 7-cycles of length 2 whose generators satisfy the condition 1 = 𝑗 < 𝑘 < 𝑝 < 𝑖.

Table 10 𝑆2 sets as union of three 7-cycles of length 2.

 1,2,7,11,14,29 1,4,7,11,28,29 1,5,7,26,35,38 1,2,7,12,14,36 1,4,7,13,28,43 1,5,7,27,35,45 1,2,7,14,27,45 1,4,7,18,28,30 1,5,7,34,35,46 1,3,7,10,21,22 1,4,7,19,28,37 1,6,7,10,22,42 1,3,7,12,21,36 1,4,7,26,28,38 1,6,7,11,29,42 1,3,7,13,21,43 1,4,7,27,28,45 1,6,7,20,42,44 1,3,7,18,21,30 1,5,7,10,22,35 1,6,7,26,38,42 1,3,7,20,21,44 1,5,7,12,35,36 1,6,7,27,42,45 1,3,7,21,26,38 1,5,7,18,30,35 1,6,7,34,42,46 1,3,7,21,34,46 1,5,7,20,35,44

The next Table should be read as follows: The two generators 𝑗, 𝑘 that appear in the left-hand column can be put together with exactly one generator 𝑖 in the second column to obtain an 𝑆2 set 𝔖 = {𝑗, 𝑝𝑗, 𝑘, 𝑝𝑘, 𝑖, (𝑝𝑖)48}. for example, line 4 says that the sets 𝔖1 ={2,6,9,14,15,42}, 𝔖2 ={2,6,11,14,29,42}, 𝔖3 = {2,6,14,17,23,42}, 𝔖4 = {2,6,14,20,42,44}, 𝔖5 = {2,6,14,27,42,45}, 𝔖6 = {2,6,9,14,33,39,42}, and 𝔖7 = {2,6,14,41,42,47} are 𝑆2 sets.

Table 11 generators 𝑗, 𝑘 that appear in the left-hand column can be put together with exactly one generator 𝑖 in the second column to obtain an 𝑆2 set 𝔖 = {𝑗, 𝑝𝑗, 𝑘, 𝑝𝑘, 𝑖, (𝑝𝑖)48

 𝑗, 𝑘 𝑖 2,3 12,17,25,41 2,4 9,11,13,17,19,25,27,33,41 2,5 9,12,19,25,27,33 2,6 9,11,17,20,27,33,41 3,4 5,6,13,17,18,25,26,33,41 3,5 6,10,12,18,20,26,34,41 3,6 10,13,17,20,25,26,34,41 4,5 6,9,18,19,25,26,33,41 4,6 9,11,13,17,19,25,27,33,34,41 5,6 9,10,19,20,25,26,27,33

4.2 The Case 𝑞 = 𝑝4𝑡

With the notations as in Proposition 3.1, let 𝑡 be an integer number, 𝑛 = 4𝑡 and

𝚤:= 1 + 𝑝 + 𝑝2 +…+ 𝑝𝑡-1 + 𝑝𝑡 + 𝑝𝑡+1 +…+ 𝑝2𝑡.

With this assumption, we have the following proposition:

Proposition 4.2.1 The set 𝔖 = {⟨1⟩, ⟨𝚤⟩} is an 𝑆2 set.

Proof: As always, we will assume that 𝔖 = {⟨1⟩, ⟨𝚤⟩} has not the 𝑆2 property. If this occurs, then at least one of following equalities is satisfied. For simplicity we write in this prove 𝑝𝑘𝑖 instead of (𝑝𝑘𝑖) 𝑞−1.

1) 𝑝𝑟 + 𝑝𝑠 = 𝑝𝑢 + 𝑝𝑣𝚤

2) 𝑝𝑟 + 𝑝𝑠 = 𝑝𝑢 + (𝑝2𝑡 +𝑚𝑣𝚤)

3) 𝑝𝑟 + 𝑝𝑠 = 𝑝𝑢𝚤 + 𝑝𝑣𝚤

4) 𝑝𝑟 + 𝑝𝑠 = 𝑝𝑢𝚤 + (𝑝2𝑡 +𝑚𝑣𝚤)

5) 𝑝𝑟 + 𝑝𝑠 = (𝑝2𝑡 +𝑚𝑢𝚤) + (𝑝2𝑡 +𝑚𝑣𝚤)

6) 𝑝𝑟 + 𝑝𝑠𝚤 = 𝑝𝑢𝚤 + 𝑝𝑣𝚤

7) 𝑝𝑟 + 𝑝𝑠𝚤 = 𝑝𝑢𝚤 + (𝑝2𝑡 +𝑚𝑣𝚤)

8) 𝑝𝑟 + 𝑝𝑠𝚤 = (𝑝2𝑡 +𝑚𝑢𝚤) + (𝑝2𝑡 +𝑚𝑣𝚤)

9) 𝑝𝑟 + 𝑝2𝑡+𝑚??𝚤) = (𝑝2𝑡 +𝑚𝑢𝚤) + (𝑝2𝑡 +𝑚𝑣𝚤)

We must show that if one of these equations is satisfied, then we obtain a contradiction. We give the proof only for some particular cases of equations 4.7 and 9.

For equation 4, let us suppose that 0 ≤ 𝑢 < 2𝑡 + 𝑚𝑣 = 𝑟 < 𝑠 with 0 ≤ 𝑚𝑣 ≤ 𝑡 − 1, then by (2) the equation defined by item 4, can be rewritten as

𝑝𝑟 + 𝑝𝑠 = 𝑝𝑢𝚤 + θ𝑚𝑣 + 2𝑡 +𝑚𝑣 𝚤𝑚𝑣+1,𝑡(8)

If 𝑢 = 0, after we cancel common powers in (8) we obtain:

𝑝𝑟 · (sum of powers of 𝑝) = 2 + 2𝑝 + 2𝑝2+… + 2𝑝𝑚𝑣 + 𝑝𝑚𝑣+1 +…+ 𝑝2𝑡.

These equalities imply that 𝑝𝑚𝑣+1 | 20𝑚𝑣 which is absurd. On the other hand, 0 < 𝑢 carries us again to the absurd divisibility relation 𝑝𝑢 | 0𝑚𝑣.

Now, suppose that the equation defined by item 7 holds; i.e.,

𝑝𝑟 + 𝑝𝑠𝚤 = 𝑝𝑢𝚤 + (𝑝2𝑡 +𝑚𝑣 𝚤) (9)

Note that implicitly 0 ≤ 𝑠, 𝑢 ≤ 2𝑡 −1. Moreover, we have assumed that 0 < 𝑟 ≤ 4 𝑡 −1 and 0 ≤ 𝑠 < 𝑢 ≤ 2𝑡 −1 < 2𝑡 + 𝑚𝑣 with 0 ≤ 𝑚𝑣 ≤ 𝑡 −1.

Under these hypotheses and using (2), we have 𝑝 | θ𝑚𝑣 which is absurd. On the other hand if 𝑡 ≤ 𝑚𝑣 ≤ 2𝑡 −1, then again by (2) we have:

𝑝𝑟 = 𝑝𝑠𝚤 (𝑝𝑢-𝑠 −1) + 𝑝2𝑡+𝑚𝑣 θ2𝑡−𝑚𝑣−1 + 𝚤2𝑡−𝑚𝑣+1, 𝑚𝑣−𝑡. (10)

And therefore, if 𝑡 < 𝑚𝑣 it is clear that we have a contradiction, but if 𝑚𝑣 = 𝑡 (10) becomes, 𝑝𝑟 = 𝑝𝑠𝚤 (𝑝𝑢-𝑠 −1) + 𝑝2𝑡+𝑚𝑣 θ𝑡−1 + 𝚤𝑡,0 (11)

Now 𝑠 = 0 implies 𝑝 | 𝚤, while 𝑠 > 0 lead us to 𝑝 | θ𝑡−1, both facts being absurd.

Finally, to analyze the equation 𝑝𝑟 + (𝑝2𝑡+𝑚s) = (𝑝2𝑡+𝑚𝑢) + (𝑝2𝑡+𝑚𝑣) defined by item 9, we will assume that 0 < 𝑟 < 4𝑡− 1 and 0 ≤ 𝑚𝑠 < 𝑚𝑢 < ?? ≤ 𝑚𝑣 ≤ 2𝑡− 1. Now again by (2), we have: 𝑝𝑟 + θ 𝑚𝑠 + 𝑝2𝑡+𝑚s = θ 𝑚𝑢 + 𝑝2𝑡+𝑚𝑢 + 2𝑡−𝑚𝑣, 𝑚𝑣−𝑡 + 𝑝2𝑡+𝑚𝑣 θ2𝑡−𝑚𝑣−1 (12)

Now this equality is the same as

𝑝𝑟 + 𝑝2𝑡+𝑚s = 𝑝𝑚s+1θ𝑚𝑢 −𝑚𝑠 −1 + 𝑝2𝑡+ + 2𝑡−𝑚𝑣, 𝑚𝑣−𝑡 + 𝑝2𝑡+𝑚𝑣 θ2𝑡−𝑚𝑣−1 (13)

and consequently 𝑝 |2𝑡−𝑚𝑣,𝑚𝑣−𝑡 which is impossible

Example 4.2.1. Taking 𝑝 = 2 and 𝑡 = 1, we have that 𝚤 = 27 and by Propositions (4.2.1.) and (3.1) the set

𝔖 = {1,2,4,8,16,27,32,54,64,99,108,128,141,177,198,216}

is an 𝑆2 set in the integer interval [1.255]. Note that we have actually provided a good example of an 𝑆2 set which has a nice property: For each element 𝑎𝑗 ∈ 𝔖 the subset S𝑎𝑗 = 𝔖 ∩ I𝑎𝑗 ⊂ 𝔖 is a good 𝑆2 set.

4.3 The Case 𝑞 = 34

The following Proposition provides a criterion to construct 𝑆2 sets in the integer interval [1.80] by joining 𝑆2 3-cycles of length 4. By Theorem (5.1) and Corollary (3.4) we have that the set of generators 𝑆2 3-cycles of length 4 is Λ = {1,2,7,11,13,14,17,22,23,26,41,53} .

Proposition 4.3.1 Let 𝑖, 𝑗 ∈ Λ such that 𝑗 - 𝑖 ≡ 1 (mod 2), then 𝔖 = {⟨ 𝑖 ⟩ ⋃ ⟨ 𝑗 ⟩} is an 𝑆2 set.

Proof: To begin with, observe that 𝔖 is not an 𝑆2 set if and only if there exists (𝑟, 𝑠, 𝑢, 𝑣), with 𝑟 < 𝑠; < 𝑣 and 𝑟, 𝑠, 𝑢, 𝑣 ∈ {0,1,2,3} such that one of the following congruencies is satisfied:

3𝑟 𝑖 + 3𝑠 𝑖 ≡ 3𝑢 + 3𝑣 (mod(80)) (14)

3𝑟 𝑖 +3𝑣 ≡ 3𝑢 + 3𝑠 (mod(80)) (15)

3𝑟 𝑖 +3𝑠 𝑖 ≡ 3𝑢 + 3𝑣 (mod(80)) (16)

If 3𝑟 𝑖 + 3𝑢 𝑗 + 3𝑣 𝑗 (mod (80)) and I =2𝚤. By parity arguments, the congruence (14) is equivalent to:

20t =3r ι3s-r +12 -3u(2k+1) 3v-u+14

Now since (3𝑣−𝑢 + 1)/4 is 1 or 7, then (3𝑠−𝑟 + 1)/2 must be equal to 5 and hence 5|2𝑘 +1 which is a contradiction.

On the other hand if i = 2𝜄+1 and keeping in mind that 3𝜇 - 1 =2,8 or 26, then congruence (15) is equivalent to:

20t =3u k3v-u -12 -3r(2ι+1) ·2·

This equality implies that (3𝑣−𝑢 − 1)/2 must be even and hence equals to 4, consequently

10𝑡 = 3𝑢 · 2 · 𝑘−3(2+1),

Finally, congruence (16) is nothing but

80𝑡 = 3𝑟 i (1+3𝑠−𝑟− 3𝑣−𝑟) − 3𝑢 𝑗.

But this is impossible since the RHS is negative.

Example 4.3.1 By Proposition (4.3.1), we can put together the following 3-cycles generated by i and 𝑗 (Table).

Table 12 3-cycles generated by i and 𝑗

 𝑖 𝑗 1 2,14,22,26 2 7,11,13,17,23,41,53 7 14,22,26 11 14,22,26 13 14,22,26 14 17,23,41,53 17 22,26 22 23,41,53 23 26 26 41,53

The following Table contains some of these sets:

Table 13 Union of two 3-cycles of length 4.

 1,2,3,6,9,18,27,54 2,6,11,18,19,33,54,57 2,6,18,53,54,71,77,79 7,14,21,29,42,46,58,63 7,21,22,29,34,38,63,66 11,14,19,33,42,46,57,58 13,22,31,34,37,38,39,66 14,42,46,53,58,71,77,79 17,22,34,38,51,59,66,73 22,34,38,53,66,71,77,79 23,26,47,61,62,69,74,78 26,53,62,71,74,77,78,79

5 Curves with many rational points over finite fields

We consider the non-singular complete irreducible Kummer curve 𝔈 over 𝔽 q defined by the affine equation:

𝑦𝑟= (𝑥),

Where 𝑟| 𝑞 − 1 and the rational function 𝜇(x) ∈ 𝔽 q (𝑥) satisfies the following conditions:

1. 𝜇 is not the 𝑑-th power of an element 𝑣 ∈ 𝔽 q (𝑥) for any divisor 𝑑 > 1 of 𝑟;

2. 𝜇 = 1 on a substantial subset 𝖅𝜇 of ℙ1 (𝔽 q );

3. (𝑥) has many multiple zeros and poles.

For details about this conditions and proofs we refer 7 and 11. With above notation we have:

Proposition 5.1 (11, Proposition 2.1) The curve 𝔈 over the finite field 𝔽 q given by the Kummer equation 𝑦𝑟= 𝜇(𝑥), where 𝑟 divides 𝑞 − 1 and the rational function 𝜇(𝑥) is not the 𝑑-th power of an element 𝑣(𝑥) ∈ 𝔽 q (𝑥) for any divisor 𝑑 of 𝑟 with 𝑑 > 1, has the following properties:

1.If (μ)=1-indi·Pi is the divisor of 𝜇 with distinct 𝑃𝑖 ∈ ℙ1 (𝔽 q ) and there exists 𝑖 such that 𝑔cd(𝑟|𝑑𝑖|), then the genus 𝑔 of 𝔈 is given by

2𝑔(𝕰)-2=r·(n-2)-i=n𝑔dc(r,di ).

2. The set of 𝔽 q −rational points (essentially) satisfies |𝔈 (𝔽 q )| ≥ 𝑟 ∙ |𝖅𝜇|.

Proof: For details and proofs, we refer to the literature on algebraic function fields for instance, 12.

Now, let us explain briefly how construct such rational functions 𝜇: Let ℓ(𝑥) ∈ 𝔽𝑝 [𝑥 ], and we denote by 𝔙 the set { 𝛼∈ 𝔽 q ; ℓ(𝛼)= 0}

1. We split ℓ(𝑥) as ℓ(𝑥) = 𝑓 (𝑥) + 𝑔(𝑥) with 𝑓 (𝑥), 𝑔(𝑥) ∈ 𝔽𝑝 [ 𝑥]. We denote the zero sets (in 𝔽 q ) of 𝑓 (resp 𝑔) by 𝔙𝑓 (resp, 𝔙𝑔), then the rational function

μ(x):=-f(x)𝑔(x)

satisfies (𝛼) = 1 for 𝛼 ∈ 𝔙𝓁 \ (𝔙𝑓 ⋃ 𝔙𝑔).

2. Given (𝑥) ∈ 𝔽𝑝(𝑥), we will denote by ℛ (𝑓 (𝑥)) the remainder of the Euclidean division of 𝑓 (𝑥) by ℓ(x). In this way, we have (essentially):

(α)=0f(x)𝓡(f(x))=1

In accordance with above observation, we need consider polynomials ℓ(𝑥) ( 𝔽𝑝 [𝑥] having many roots in 𝔽𝑝.

Definition 5.1 A polynomial (𝑥) ∈ 𝔽 q [ 𝑥 ] is a restricted range polynomial if ??(𝛼) ∈ 𝔙 ⫋ 𝔽𝑝 for some proper subset of 𝔽𝑝 and for all 𝛼 ∈ 𝔽 q . In particular, when 𝔙 = 𝔽𝑝, we say that (𝑥) is a 𝔽 q , 𝔽𝑝-polynomial.

A classic example of restricted range polynomial is the norm polynomial

N𝔽q/𝔽p(x)=xpn-1...p+1  𝔽q[x].

Definition 5.2 A nonzero (𝔽 q , 𝔽𝑝)−polynomial 𝑓(𝑥) ∈ 𝔽 q [𝑥] will be called minimal, if deg (𝑓(𝑥)) ≤ 𝑞 − 1 and none its proper partial sums is a (𝔽 q , 𝔽𝑝)-polynomial.

We recall briefly some properties of (𝔽 q , 𝔽𝑝)-polynomial. For proofs, we refer to 5 and 13.

Proposition 5.213 (𝔽 q , 𝔽𝑝)-polynomials are subjective.

Theorem 5.15 (Characterization of (𝔽 q , 𝔽𝑝) -polynomials) The exponent sets of the minimal (𝔽 q , 𝔽𝑝)-polynomials are the 𝑝-cycles of set{0,…, 𝑞 − 1}. For each 𝑝-cycle ℑ, all the minimal (𝔽 q , 𝔽𝑝)-polynomials with exponent set ℑ are:

f𝕴(x,α)=k=0o(𝕴)-1 αpk xk     αFpo(𝕴)*

where 𝑖 is an arbitrary but fixed representative of ℑ. In addition, we have all the different (𝔽 q , 𝔽𝑝)-polynomials of less or equal degree to 𝑞−1 by sums of polynomials ?? (𝑥, 𝛼) corresponding to different cycles.

Example 5.1 Using Theorem 5.1 and Example 2.2, we exhibit some (𝔽27: 𝔽3) -polynomials.

Remark 5.1 Proposition 5.2 says that each (𝔽 q , 𝔽𝑝)-polynomial is subjective, hence, we can use this fact to construct appropriate rational functions which leads us to obtain curves over the finite field 𝔽 q with good parameters. The following examples explain how. The reader who is not familiar with the concepts of algebraic function fields (i.e., algebraic curves) as genus, rational points etc is referred to 12.

Example 5.2 The curve ℭ over 𝔽9 given by 𝑦2 = −(𝑥6+ 𝑥5 + 𝑥4 + 2𝑥2+ 2) has 𝑔(ℭ)=2 and 20 rational points. The best value possible, cf 2.

The affirmation is clear. Let us briefly explain how we obtained this equation. Observe that there are 3-cycles of length 2, namely (1,3),(2,6) and (5,7), then set {1,3,5,7} which is not an 𝑆2 set! induces the 𝔽9,𝔽3-polynomial ℓ(𝑥) = 𝑥7 + 𝑥5 + 𝑥3 + 𝑥 which has its roots in 𝔽9. We will take advantage of this fact to construct our curve. In general, given two co-primes polynomials ℓ(𝑥), 𝑓(𝑥) ∈ 𝔽𝑝[ 𝑥] and 𝑟 |𝑞 − 1, the Euclidean division

(𝑥)𝑟 = ℓ(𝑥) ∙ ℎ(𝑥) + ℛ(𝑓(𝑥)𝑟),

implies that for each 𝛼 ∈ 𝔽 q root of ℓ(𝑥), ℛ (𝑓(𝑥)𝑟)(𝛼) = 𝑓(𝛼)𝑟; i.e. ℛ (𝑓(𝑥)𝑟)(𝛼) is an 𝑟-th power in 𝔽 q , therefore the polynomial 𝑦𝑟 = ℛ (𝑓(𝑥)𝑟)(𝛼) splits completely en 𝔽 q [𝑦], this means many points. Now in our situation taking (𝑥) = (𝑥 −1) (𝑥 +1) (𝑥4 + 2𝑥 + 2) we have, ℛ ((𝑥)2) = −(𝑥6 + 𝑥5 + 𝑥4 + 2𝑥2 + 2), this carry us to our equation.

Example 5.3 The curve ℭ over 𝔽27 given by 𝑦2 = −(𝑥4 + 2𝑥3 + 𝑥2 + 1)3 has 𝑔(ℭ) = 1 and #ℭ (𝔽27) = 38.

By Proposition 3.4, the union of the underlying sets of the 3-cycles (1,3,9) and (2,6,18) is an 𝑆2 set. This set induces the (𝔽27:𝔽3)-polynomial ℓ(𝑥) = 𝑥18 + 𝑥9 + 𝑥6 + 𝑥3 + 𝑥2 + 𝑥 which induces the rational function 𝜇(𝑥) = −𝑥2 (𝑥+1)2 (𝑥4 + 2𝑥3 + 𝑥2 + 1)3 and consequently the algebraic curve ℭ over 𝔽27 defined by the equation:

𝑦2 = −(𝑥4 + 2𝑥3 + 𝑥2 + 1)3.

(Here we split ℓ(𝑥) as ℓ(𝑥) = ℓ1(𝑥) + ℓ2(𝑥) where ℓ1(𝑥) = 𝑥 + 𝑥2 and ℓ2(𝑥) = 𝑥3 + 𝑥6 + 𝑥9 + 𝑥18 = 𝑥3 (𝑥+1)3 (𝑥4 +2𝑥3 + 𝑥2 + 1). Observe that the places corresponding to 𝑥 = 0 and 𝑥 = −1 are unramified.)

It is easy to see that the curve ℭ has genus 𝑔 = 1. For the number of rational points (or places of degree one in the language of algebraic function fields), observe that (ℓ(𝑥), 𝑥27 - 𝑥) = 𝑥 + 𝑥2 + 𝑥4 + 2𝑥5 + 𝑥6 +2𝑥7 + 𝑥9, therefore the curve ℭ has at least 2 ∙ #ℓ−1(0) = 2 ∙ 9 = 18. We use computer program Mathematica to complete the determination of the rational points, we refer to 11, Remark 2.2 for details.

Example 5.4 The 𝑆2 3-cycle of length 3, (17,25,23) induces the minimal (𝔽27:𝔽3)-polynomial ℓ(𝑥) = 𝑥17 + 𝑥25 + 𝑥23 which has 9 roots in 𝔽27, namely the zeros of the polynomial 𝑓(𝑥) = 𝑥 + 𝑥7 + 𝑥9 = 𝑥(1 + 𝑥)(2 + 𝑥)(2+ 𝑥2 + 𝑥3)(1+ 2𝑥2 + 𝑥3). If we split ℓ(𝑥) as ℓ(𝑥) = 𝑥17 + 𝑥23 (𝑥2 + 1) and consider the rational function 𝜇(𝑥) := −𝑥6 (𝑥2 + 1), then for each α root of 𝑓(𝑥), 𝜇(𝛼) = 1.

Now, the curve ℭ over 𝔽27 defined by the Kummer equation

𝑦26 = (𝑥): = −𝑥6 (𝑥2 + 1),

has genus 𝑔 = 24 and 208 rational points. This is the best value know for (𝑞, 𝑔)=(27,24) (see 2).

Example 5.5 We will construct here two maximal curves ℭ1 and ℭ2 over 𝔽49 with (ℭ1) = 1 and (ℭ2) = 3.

In Table 9, we exhibit some 𝑆2 sets 𝔖 ⊂ I48 which was obtained as union of 3 7-cycles of length 2. For this example we consider the set 𝔖 ={1,2,7,11,14,29} = { ⟨1⟩ ∪⟨2⟩ ∪ ⟨11⟩}. This set induces the (𝔽49:𝔽7)-polynomial ℓ(𝑥) = 𝑥 + 𝑥2 + 𝑥7 + 𝑥11 + 𝑥14 + 𝑥29). By Proposition 5.2 there exist a subset 𝔙 ⊂ 𝔽49 such that ℓ(𝔙) = 1 and after some computations we obtain that 𝔙 is the zero set of the polynomial 𝑙(𝑥) = 1+ 𝑥+ 𝑥2 + 6𝑥3 + 𝑥4. We use this polynomial to construct our curves instead of ℓ(x), the reason is that ℓ(x) has high degree compared to |𝔙|. Now we split 𝑙(𝑥) as 𝑙(𝑥) = (1+ 𝑥+ 4𝑥2 + 𝑥4) + (6𝑥3), the polynomial 1+ 𝑥+ 4𝑥2 + 𝑥4 can be factored as 1+ 𝑥+ 4𝑥2 + 𝑥4 = (5 + 𝑥)3 (6 + 𝑥) and therefore we consider the Kummer cover defined by the equation

yr=x3(5+x)3(6+x)  r4.

This algebraic function field has genus 𝑔 = 𝑟 − with 𝑑= 𝐺𝐶𝘋(𝑟,3), this gives, for 𝑟 =2, 𝑔(ℭ1) = 1 and for 𝑟 =4, 𝑔(ℭ2) = 3, the rational points satisfies #ℭ1 = 64 and #ℭ2 = 92 (see tables in 3).

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Received: December 16, 2016; Accepted: June 13, 2017 This is an open-access article distributed under the terms of the Creative Commons Attribution License