1 Introduction
Let 𝑝 be a prime number and 𝔽_{
q
} be a finite field with q = 𝑝^{
n
} elements and let
is an affine plane algebraic curve (over the finite field
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
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
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
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 socalled 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 𝑝cycles (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 𝑆_{2} property. More precisely: A subset 𝔖 of integers is an 𝑆_{2} set (or has the 𝑆_{2} 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 𝑆_{2}set, then its cardinality must be asymptotically equal to
The paper is organized as follows: In section 2 we give a brief exposition of some properties of 𝑝cycles. 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 𝑆_{2} 𝑝cycle. 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 𝑆_{2} 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 𝑆_{2} sets constructed in section 4.
2 𝑝Cycles
Let 𝑝 be a prime number and let = 𝑝^{𝑛} be a nonnegative power of 𝑝. The 𝑝adic expansion of a positive integer 𝑎 in the integer interval 𝐼_{𝑞−1}: = [1, 𝑞−1] is given by:
𝑎 = 𝑘_{0} + 𝑘_{1}𝑝 + 𝑘_{2}𝑝^{2} +…+ 𝑘_{𝑛1} 𝑝^{𝑛1},
where the numerals 0 ≤ 𝑘_{𝑗} < 𝑝 satisfies for j = 0, ··· ,n−1.
If each 𝑎 = 𝑘_{0} + 𝑘_{1}𝑝 + 𝑘_{2}𝑝^{2} +…+ 𝑘_{𝑛1} 𝑝^{𝑛1} ∈ 𝐼_{𝑞−1} is represented by the 𝑛tuple
𝑎^{⊓} = 𝑘_{𝑛1} + 𝑝𝑘_{0} + 𝑝^{2}𝑘_{1} + …+ 𝑝^{ n1 } 𝑘_{𝑛2} .
By ⊓^{𝑘} we will understand the iteration k times the cyclic numeralpermutation. The 𝑝adic period of 𝑎 is the small natural integer (𝑎) such that
A 𝑝cycle 𝔍 is an ordered set
Example 2.1 Let 𝑝 = 2 and 𝑛 = 4. The number 3 = 1∙2^{0} + 1∙2^{1} + 0∙2^{2} + 0∙2^{2} corresponds to the 4tuple (1,1,0,0) therefore 3^{⊓} = 0∙2^{0} + 1∙2^{1} + 0∙2^{2} = 6, 6^{⊓} = 12, 12^{⊓} = 9. Consequently (3,6,12,9) is a 2cycle 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 2cycle of length 7 and the element 3 ∈ [1,127] has period 7.
Since the process of determining 𝑝cycles in 𝐼_{𝑞−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 𝐼_{𝑞−1} as follows:
ρ: 𝐺 ⨯ 𝐼_{𝑞−1} → 𝐼_{𝑞−1}
(σ^{𝑘}, 𝑖) ↦ (𝑝^{𝑘} · 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 ∈ 𝐼_{𝑞−1}, the 𝑝cycle
Proof: Observe that if 𝑖 = 𝑖_{0} + 𝑖_{1}𝑝 + 𝑖_{2}𝑝^{2} +…+ 𝑖_{ n1 } 𝑝^{ n1 } is the 𝑝adic expansion of 𝑖, then
𝑖^{⊓} = 𝑖_{ n1 } + 𝑖_{0}𝑝 +𝑖_{1}𝑝^{2} +…+ 𝑖_{ n2 } 𝑝^{ n1 } .
Hence
𝑝𝑖 − 𝑖^{⊓} = +𝑖_{0}𝑝 + 𝑖_{1}𝑝^{2} +…+ 𝑖_{ n−1 } 𝑝^{ n } − (𝑖_{ m−1 } + 𝑖_{0}𝑝 + 𝑖_{1}𝑝^{2} +…+ 𝑖_{0n−2}𝑝^{n1})
= 𝑖_{ n−1 } (𝑞 − 1),
consequently 𝑖^{⊓} ≡ (mod (𝑞 − 1)).
Corollary 2.1 If q = p ^{ n } and ℑ is a 𝑝cycle, then 𝑙(ℑ)  𝑛.
Proof: We known that if G acts on a set 𝑆, then 𝐺_{𝑥} = {𝑔 ∈ 𝐺  𝑔· 𝑥 = 𝑥}
is a subgroup of 𝐺 and the cardinal number of the orbit
Proposition 2.1 Every 𝑝cycle has the form
ℑ = (i, pi ,..., p ^{ k } i, (p ^{ k+1 } i)_{ q−1 } ,..., (p ^{ℓ} i)_{ q−1 } )
where ℓ+1 is the length of ℑ and 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 𝑝cycle ℑ = (i, pi ,..., p ^{ k } i, (p^{ k+1 } i)_{ q1 } ,..., (p^{ℓ} i)_{ q1 } ), 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 𝑥^{𝑁}−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 𝑝cycles for 𝑝 = 2.3 and 𝑞 = 𝑝^{𝑛} for some values of 𝑛.
𝑞 = 2^{6} = 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) 
𝑞 = 3^{3} = 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) 
𝑞 = 3^{4} = 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 2cycle 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 𝑝cycle from any of its elements, we will say that an integer 𝑖 generates the 𝑝cycle ℑ if
ℑ = (i, 𝑝i ,..., 𝑝^{ k } i, ( 𝑝^{ k+1 } i)_{ q1 } ,..., (𝑝^{ℓ} i)_{ q1 } ),
and i < (𝑝^{ k+𝑗 } i)_{ q1 } for 𝑗 = 1,…, ℓ− 𝑘. In this case we write ℑ = ⟨ i ⟩.
Proposition 2.2 Let 𝑞 = 𝑝^{𝑛} with 𝑛 = 𝑚·𝑙. If ℑ = ⟨i⟩, then 𝑙 = (ℑ) < 𝑛 if and only if i is a multiple of
Proof: By definition, ℓ = length(ℑ) if and only if, 𝑝^{ℓ} i= (𝑞  1) ℎ_{ℓ} + i
if and only if,
Proposition 2.3 The greatest integer i ∈ I _{𝑞−1} such that i generates a 𝑝cycle of length 𝑛, is i = 𝑝^{𝑛} ( 𝑝^{𝑛1} + 1). The corresponding 𝑝cycle is,
(𝑝^{𝑛} ( 𝑝^{𝑛1} + 1), 𝑝^{𝑛} ( 𝑝^{0} + 1), 𝑝^{𝑛} ( 𝑝 + 1), 𝑝^{𝑛} ( 𝑝^{2} + 1),…, 𝑝^{𝑛} ( 𝑝^{𝑛2} + 1)).
Proof: First observe that
𝑝i = 𝑝·(𝑝^{𝑛}  𝑝^{𝑛1}  1) = (𝑞  1)( 𝑝  1)  1 ≡ (𝑞  2)(mod 𝑞  1).
Hence 𝑝^{2} i ≡ 𝑞  𝑝  1 (mod 𝑞  1), so in general, 𝑝^{𝑘} i ≡ 𝑞  (𝑝^{𝑘1} + 1 (mod 𝑞  1) which leads to the desired expression. Secondly, each number i < 𝑗 < ??  1 can be rewritten as 𝑗 = (𝑞  1)  (𝑝^{𝑛1}  𝑘1) with 0 ≤ 𝑘 ≤ 𝑘 ≤ 𝑝^{𝑛1}  2. Now, the equation 𝑝𝑥  (𝑞  1)t = 𝑗 with 0 ≤ t ≤ 𝑝  1 has unique solution in I _{𝑞1} which implies that 𝑗 is the remainder modulus 𝑞  1 of some 𝑥 ( I _{𝑞1}.
Remark 2.2 Is know that, if 𝛼 = 𝛽^{ i } ∈ 𝔽_{ q } = 𝔽_{𝑝} (𝛽) and 𝑚_{𝛼} (𝑥) is its minimal polynomial, then
(Cf ^{10}, Theorem 4.1). Hence one can determine the number of 𝑝cycles of length 𝑑  𝑛 in I _{𝑞1}, such number is:
(here (·)is the Moebius function), as many as irreducible polynomials of degree 𝑑 in 𝔽_{𝑝} [𝑥 ].
3 𝑆 _{2}  Sets
A subset 𝔖 of integers is an 𝑆_{2}set if all the sums 𝑎 + 𝑎ʹ with 𝑎 ≠ 𝑎ʹ; 𝑎, 𝑎ʹ; ∈ 𝔖 are distinct. From now on, a set 𝔖 has the 𝑆_{2} property if 𝔖 is an 𝑆_{2}set. Similarly, a 𝑝cycle ℑ = (𝚤_{1},…, 𝚤_{𝑙}) is an 𝑆_{2} 𝑝cycle if the underlying set (𝚤_{1},…, 𝚤_{𝑙}) has the 𝑆_{2} property. In this section we give some criteria that allows us to decide when a 𝑝cycle ℑ = ⟨i⟩ has the 𝑆_{2} property. Also, we give conditions on the generators of a set of 𝑆_{2}cycles such that the union of these retain this property.
Theorem 3.1 If GCD (i,𝑞  1) = 1, then ℑ = ⟨i⟩ is an 𝑆_{2} 𝑝cycle.
Proof: Assume the contrary, so there exist integers 0 ≤ 𝑟, 𝑠, 𝑢, 𝑣 ≤ 𝑛 1such that
(𝑝^{𝑟} i) _{𝑞−1} +(𝑝^{𝑠} i)_{𝑞−1} = (𝑝^{𝑢} i)_{𝑞−1} + (𝑝^{𝑣} i)_{𝑞−1} 𝑟 < 𝑠, 𝑢< 𝑣, (1)
If 𝑟 = min{𝑟,u}, then (1) implies (𝑝^{𝑣𝑟} + 𝑝^{𝑢𝑟} − 𝑝^{𝑠𝑟}  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 3cycle generated by the divisor 11 of 242, (11,33,99,55,165) has the 𝑆_{2} property.
Corrollary 3.1 If 𝑞−1 is a prime number, then any 𝑝cycle is an 𝑆_{2}set.
Example 3.1 The Table 4 shows all the 2cycles of length 5 which, by 3.1 are 𝑆_{2}sets contained in the integer interval [1.31 ].
The next corollary give us information about the components of a 𝑝cycle 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 (𝑝^{𝑗}𝑖) _{𝑞−1} = 𝑝^{𝑗}𝑖  (𝑞−1) ℎ_{𝑗}.
Remark 3.2 Let us consider the 𝑆_{2} 3cycle of length 5, Ξ =(11,33,99,55,165) generated by 11, divisor of 242 = 3^{5} − 1 contained in the interval [1,242]. It is clear that its cardinality 5 is very small respect to
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:
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 2cycle Ξ, generated by 27, (27,54,108,216,177,99,198,141) is an 𝑆_{2} 2cycle. 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 𝑆_{2}set. Observe that 𝔖 as a subset of the interval [1,72 has cardinality closed to
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 2cycle generated by 9, which is {9,18,36,72,144,33,66,132}. This, as 𝑆_{2}set is very poor respect to
As we can see, there are examples of 𝑝cycles which have the 𝑆_{2} 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 𝑝cycle of length 𝑛=4t generated by
is an 𝑆_{2}set. (Here the hat means that the power 𝑝^{ t } was excluded.)
Proof: First observe that
⟨𝚤⟩ = (𝚤, 𝑝𝚤, 𝑝^{2}𝚤,…, 𝑝^{ 2t1 } 𝚤, (𝑝^{ 2t } 𝚤)_{𝑞−1}, (𝑝^{ 2t+1 } 𝚤)_{𝑞−1} ,…, (𝑝^{ 4t1 } 𝚤)_{𝑞−1})
where (𝑝^{𝑗}𝚤) _{𝑞−1} denotes the remainder of 𝑝^{𝑗}𝚤 modulus 𝑞 − 1.
As always, we will suppose that such a set is not an 𝑆_{2}set, therefore several cases may occur, namely:
Case 1. 𝑝^{𝑟}𝚤 + 𝑝^{𝑠}𝚤 = 𝑝^{𝑢}𝚤 +𝑝^{𝑣}?? with 𝑟, 𝑠, 𝑢, 𝑣 ≤2t − 1.
Case 2.
Case 3.
Case 4.
Case 5.
Case 6.
Our next goal is to determine the number (𝑝^{ 2t+𝑚 } 𝚤)_{q−1}. For this end, let us denote by 𝚤_{𝑘,} and θ _{𝑚} the following natural numbers:
𝚤_{𝑘,𝑙}: = 1 + 𝑝 + ... +
With above notation, the number (𝑝^{ 2t+𝑚 } 𝚤)_{𝑞−1} can be written as follows:
As an illustration we consider the case 6. Let us suppose that
For instance, if 0 ≤ 𝑟 < 𝑢 ≤ 2t − 𝑚 < 2t + 𝑚_{𝑠} < 2t + 𝑚_{𝑣} with 0 ≤ 𝑚_{𝑠} ≤ t − 1< 𝑚_{𝑣}, then with the above notation we have:
Now, if 𝑟 = 0 after cancel common terms, we get the following equality
which implies that 𝑝1.
On the other hand, 𝑟 > 0 implies
if 𝑚_{𝑠} < 𝑚_{𝑣} − t and
if 𝑚_{𝑠} ≥ _{𝑣} − 𝑡
If 𝑚_{𝑠} < 𝑚_{𝑣} − 𝑡, then 𝑟 > 𝑚_{𝑠} + 1 implies 𝑝1; 𝑟 = 𝑚_{𝑠} + 1, implies 𝑟 = 𝑚 > 𝑡 which is absurd and finally 𝑟 = 𝑚_{𝑠} + 1 implies
Example 3.3 If 𝑝 = 2 and ?? = 1, then 𝑛 = 8, 𝚤 = 27, Proposition (2.2) says that the 2cycle (27,54,108,216,177,99,198,141) is an 𝑆_{2}set contained in [1,256].
Proposition 3.2 If 𝑖 < 𝑝, then the 𝑝cycle ⟨𝑖⟩ is an 𝑆_{2} 𝑝cycle.
Proof: If there are different elements 𝑟, 𝑠, 𝑢, 𝑣 such that
𝑝^{𝑟}𝑖 + 𝑝^{𝑠}𝑖 = 𝑝^{𝑢}𝑖 +𝑝^{𝑣}𝑖,
then we would have 𝑝1.
Corollary 3.4 Let 𝑖, 𝑗 ∈ I _{𝑞−1}. Then if 𝑖 < 𝑝 and , (𝑗, 𝑞−1) = 1, the 𝑝cycle Ξ generated by 𝑖𝑗 is an 𝑆_{2} 𝑝cycle.
Proof: If not, we should have 𝑝^{𝑟}𝑖𝑗 + 𝑝^{𝑠}𝑖𝑗 = 𝑝^{𝑢}𝑖𝑗 + 𝑝^{𝑢}𝑖𝑗 (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 𝑝cycle generated by 𝑝 +1 is an 𝑆_{2} 𝑝cycle.
Proof: Observe that such 𝑝cycle is
(𝑝 + 1, 𝑝^{2} + 𝑝,…, 𝑝^{𝑛2} + 𝑝 ^{𝑛3}, 𝑝 ^{𝑛1} + 𝑝 ^{𝑛2}, 𝑝^{𝑛1} + 1)
First, let us suppose that there exist different integers 1 ≤ 𝑟, 𝑠, 𝑢, 𝑣 ≤ 𝑛 − 1 such that
𝑝^{𝑛𝑟}+𝑝^{𝑛𝑟1} + 𝑝^{𝑛𝑠} + 𝑝^{𝑛𝑠1} = 𝑝^{𝑛𝑢} + 𝑝^{𝑛𝑢1} + 𝑝^{𝑛𝑣} + 𝑝^{𝑛𝑣1},
then if 𝑠 = max {𝑟, 𝑠, 𝑢, 𝑣} we have 𝑝1. On the other hand, if there exist different integers 1 ≤ 𝑟, 𝑠, 𝑢 ≤ 𝑛 − 1 such that
𝑝^{𝑛𝑟} + 𝑝^{𝑛??1} + 𝑝^{𝑛𝑠}+𝑝^{𝑛𝑠1} = 𝑝^{𝑛𝑢} + 𝑝^{𝑛𝑢1} + 𝑝^{𝑛1} + 1, (3)
then since
𝑝^{𝑛} + 𝑝^{𝑛1} (𝑞  1) = 𝑝^{𝑛1} + 1,
(3) can be written as
𝑝^{𝑛𝑟} + 𝑝^{𝑛𝑟1} + 𝑝^{𝑛𝑠}+ 𝑝^{𝑛𝑠1} = 𝑝^{𝑛𝑢} + 𝑝^{𝑛𝑢1} + 𝑝^{𝑛 +}𝑝^{𝑛1} (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 𝑆_{2} 𝑝cycles is an 𝑆_{2}set. A first approach is given in the following proposition:
Proposition 3.4 If 1 < ?? < 𝑝, then the set
𝔖: ={⟨1⟩ ⋃ ⟨𝑗⟩}
is an 𝑆_{2}set.
Proof: Assume that 𝔖 is not an 𝑆_{2}set. Then, there exist ??, 𝑏, 𝑐, 𝑑 ∈ {1, 𝑗} and integer numbers 0 ≤ 𝑟, 𝑠, 𝑢, 𝑣 ≤ 𝑛  1 such that,
𝑎𝑝^{𝑟} + 𝑏𝑝^{𝑠} = 𝑐𝑝^{𝑢} + 𝑑𝑝^{𝑣}, (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 𝑝𝑎 or 𝑝𝑐 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 𝑝𝑑. The case δ = 𝑢 = 𝑣 leads to 𝑎𝑝^{ 𝑟δ } + 𝑎𝑝^{ 𝑠δ } = 𝑐 + 𝑑, which is again a contradiction. On the other hand, if δ = 𝑟 = 𝑢 = 𝑣 we obtain 𝑎 + 𝑎𝑝^{ 𝑠δ } = 𝑐 + 𝑑, now after to consider the different possibilities for 𝑎, 𝑐, 𝑑, we conclude that 𝑝1 or 𝑝𝑗.
Case 3: 𝑎 ≠ 𝑏 and 𝑐 ≠ 𝑑. If 𝑟, 𝑠, 𝑢, 𝑣 are distinct and δ = 𝑟, 𝑠, 𝑢, or 𝑣 then 𝑝𝑎, 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^{ 𝑣δ }  𝑑𝑝^{ 𝑠δ } 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^{ 𝑟δ } + 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 𝑆_{2}sets 𝑆 ⊂ [1, 𝑀 = 3^{𝑛} − 1] for different values of 𝑛.
As we can see, the first two sets have a reasonable cardinality in respect to
Proposition 3.5 If 1 < 𝑗 < 𝑘 < 𝑝 and 𝑘 + 𝑗 ≠ 𝑝 + 1, then the set
𝔖:= {⟨1⟩ ∪·⟨ 𝑗 ⟩ ∪ ⟨ 𝑘 ⟩}
is an 𝑆_{2}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,
𝑎𝑝^{𝑟} + 𝑏𝑝^{𝑠} = 𝑐𝑝^{𝑢} + 𝑑𝑝^{𝑣}, (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 𝑝^{δ} in (5) we obtain 𝑝𝑎 or 𝑝𝑐 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 (5) we obtain 𝑝𝑎; if δ = 𝑢 or 𝑣, then 𝑝𝑐 or 𝑝𝑑. The case δ = 𝑢 = 𝑣 leads to 𝑎𝑝^{𝑟δ} + 𝑎𝑝^{𝑠δ} = 𝑐 + 𝑑, which is again a contradiction. On the other hand, if δ = 𝑟 = 𝑢 = 𝑣 we obtain 𝑎 + 𝑎𝑝^{𝑠δ} = 𝑐 + 𝑑. Now after to consider the different possibilities for 𝑎, 𝑐, 𝑑, we obtain the following equalities:
𝑝^{𝑠δ} + 1 = 𝑗 + 𝑘, + 𝑗??^{𝑠δ} = 𝑘 +1, 𝑘 + 𝑘𝑝^{𝑠δ} = 𝑗 + 1. (6)
But 𝑠δ ≥ 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 𝑝𝑎, 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 = 𝑑𝑝^{𝑣δ} − 𝑏𝑝^{𝑠δ} 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 𝑎𝑝^{𝑟δ} + 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 𝑆_{2} sets obtained for different values of 𝑀 = 𝑝^{𝑛}1
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 𝑆_{2}set.
Example 3.6 If we take 𝑝 = 7 and 𝑛 = 2, we obtain the Table 7.
𝑀 = 𝑝^{𝑛1} 

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 < 𝑎_{1} < 𝑎_{2} < 𝑎_{3} < 𝑝, 𝑎_{𝑖} + 𝑎_{𝑗} ≠ 𝑝 + 1 and the set {1, 𝑎_{1}, 𝑎_{2}, 𝑎_{3}} is an 𝑆_{2}set, then the set
𝔖: = { ⟨1⟩ ∪ ⟨𝑎_{1}⟩ ∪ ⟨𝑎_{2}⟩ ∪ ⟨𝑎_{3}⟩ }
is an 𝑆_{2}set.
Proof: By Proposition (3.5) in order to prove our assertion, we only have to analyze the equation
𝑝^{𝑟} + 𝑎_{1}𝑝^{𝑠} = 𝑎_{2}𝑝^{𝑢} + 𝑎_{3}𝑝^{𝑣}, (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 + 𝑎_{1} − 𝑎_{2} = 𝑎_{3}𝑝^{ 𝑣δ } which is a contradiction.
2. If ≠ δ = 𝑟 = 𝑠 ≠ 𝑢, we obtain 1 + 𝑎_{1} = 𝑎_{2}𝑝^{ 𝑢δ } + 𝑎_{3}??^{ 𝑣δ } 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 𝑎_{1}, 𝑎_{2}, 𝑎_{3} 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
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 𝑆_{2} 𝑝cycles ⟨𝑖⟩, ⟨𝑗⟩, ⟨𝑘⟩ such that, the resulting set 𝔖: = {⟨𝑖⟩ ∪ ⟨𝑗⟩ ∪ ⟨𝑘⟩} maintains the 𝑆_{2} 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 𝑝cycles (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 𝑆_{2} property. Now, observe that for this purpose it is easier to establish conditions for which 𝔖 does not have the 𝑆_{2} 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 𝑆_{2} set if there exist distinct _{1}, 𝑎_{2}, 𝑎_{3}, 𝑎_{4} ( 𝔖 such that 𝑎_{1} + 𝑎_{2} = 𝑎_{3} + 𝑎_{4}, 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 = (𝑝^{2} − 1)𝑚
12) 𝑝𝑖 − 𝑝𝑗 − (𝑝 − 1)𝑘 = (𝑝^{2} − 1)𝑚
13) 𝑝𝑖 + 𝑗 − (𝑝 + 1)𝑘 = (𝑝^{2} − 1)𝑚
14) (𝑝 + 1)𝑖 − 𝑝𝑗 − ??𝑘 = (𝑝^{2} − 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:
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 = (𝑝^{2} − 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= (𝑝^{2}−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 𝑆_{2} set for 2 ≤ 𝑘 ≤ 6. The following Table contains all the 𝑆_{2} sets which were obtained joining three 7cycles of length 2 whose generators satisfy the condition 1 = 𝑗 < 𝑘 < 𝑝 < 𝑖.
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 lefthand column can be put together with exactly one generator 𝑖 in the second column to obtain an 𝑆_{2} set 𝔖 = {𝑗, 𝑝𝑗, 𝑘, 𝑝𝑘, 𝑖, (𝑝𝑖)_{48}}. for example, line 4 says that the sets 𝔖_{1} ={2,6,9,14,15,42}, 𝔖_{2} ={2,6,11,14,29,42}, 𝔖_{3} = {2,6,14,17,23,42}, 𝔖_{4} = {2,6,14,20,42,44}, 𝔖_{5} = {2,6,14,27,42,45}, 𝔖_{6} = {2,6,9,14,33,39,42}, and 𝔖_{7} = {2,6,14,41,42,47} are 𝑆_{2} sets.
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 + 𝑝^{2} +…+ 𝑝^{𝑡1} + 𝑝^{𝑡} + 𝑝^{𝑡+1} +…+ 𝑝^{2𝑡}.
With this assumption, we have the following proposition:
Proposition 4.2.1 The set 𝔖 = {⟨1⟩, ⟨𝚤⟩} is an 𝑆_{2} set.
Proof: As always, we will assume that 𝔖 = {⟨1⟩, ⟨𝚤⟩} has not the 𝑆_{2} property. If this occurs, then at least one of following equalities is satisfied. For simplicity we write in this prove 𝑝^{𝑘}𝑖 instead of (𝑝^{𝑘}𝑖) _{𝑞−1}.
1) 𝑝^{𝑟} + 𝑝^{𝑠} = 𝑝^{𝑢} + 𝑝^{𝑣}𝚤
2) 𝑝^{𝑟} + 𝑝^{𝑠} = 𝑝^{𝑢} + (𝑝^{2𝑡 +𝑚𝑣}𝚤)
3) 𝑝^{𝑟} + 𝑝^{𝑠} = 𝑝^{𝑢}𝚤 + 𝑝^{𝑣}𝚤
4) 𝑝^{𝑟} + 𝑝^{𝑠} = 𝑝^{𝑢}𝚤 + (𝑝^{2𝑡 +𝑚𝑣}𝚤)
5) 𝑝^{𝑟} + 𝑝^{𝑠} = (𝑝^{2𝑡 +𝑚𝑢}𝚤) + (𝑝^{2𝑡 +𝑚𝑣}𝚤)
6) 𝑝^{𝑟} + 𝑝^{𝑠}𝚤 = 𝑝^{𝑢}𝚤 + 𝑝^{𝑣}𝚤
7) 𝑝^{𝑟} + 𝑝^{𝑠}𝚤 = 𝑝^{𝑢}𝚤 + (𝑝^{2𝑡 +𝑚𝑣}𝚤)
8) 𝑝^{𝑟} + 𝑝^{𝑠}𝚤 = (𝑝^{2𝑡 +𝑚𝑢}𝚤) + (𝑝^{2𝑡 +𝑚𝑣}𝚤)
9) 𝑝^{𝑟} + 𝑝^{2𝑡+𝑚??}𝚤) = (𝑝^{2𝑡 +𝑚𝑢}𝚤) + (𝑝^{2𝑡 +𝑚𝑣}𝚤)
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 + 𝑚_{𝑣} = 𝑟 < 𝑠 with 0 ≤ 𝑚_{𝑣} ≤ 𝑡 − 1, then by (2) the equation defined by item 4, can be rewritten as
𝑝^{𝑟} + 𝑝^{𝑠} = 𝑝^{𝑢}𝚤 + θ_{𝑚𝑣} + ^{2𝑡 +𝑚𝑣} 𝚤_{𝑚𝑣+1,𝑡}(8)
If 𝑢 = 0, after we cancel common powers in (8) we obtain:
𝑝^{𝑟} · (sum of powers of 𝑝) = 2 + 2𝑝 + 2𝑝^{2}+… + 2𝑝^{𝑚𝑣} + 𝑝^{𝑚𝑣+1} +…+ 𝑝^{2𝑡}.
These equalities imply that 𝑝^{𝑚𝑣+1}  20_{𝑚𝑣} which is absurd. On the other hand, 0 < 𝑢 carries us again to the absurd divisibility relation 𝑝^{𝑢}  0_{𝑚𝑣}.
Now, suppose that the equation defined by item 7 holds; i.e.,
𝑝^{𝑟} + 𝑝^{𝑠}𝚤 = 𝑝^{𝑢}𝚤 + (𝑝^{2𝑡 +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 + 𝑚_{𝑣} with 0 ≤ 𝑚_{𝑣} ≤ 𝑡 −1.
Under these hypotheses and using (2), we have 𝑝  θ_{𝑚𝑣} which is absurd. On the other hand if 𝑡 ≤ 𝑚_{𝑣} ≤ 2𝑡 −1, then again by (2) we have:
𝑝^{𝑟} = 𝑝^{𝑠}𝚤 (𝑝^{𝑢𝑠} −1) + 𝑝^{2𝑡+𝑚𝑣} θ_{2𝑡−𝑚𝑣−1} + 𝚤_{2𝑡−𝑚𝑣+1, 𝑚𝑣−𝑡}. (10)
And therefore, if 𝑡 < 𝑚_{𝑣} it is clear that we have a contradiction, but if 𝑚_{𝑣} = 𝑡 (10) becomes, 𝑝^{𝑟} = 𝑝^{𝑠}𝚤 (𝑝^{𝑢𝑠} −1) + 𝑝^{2𝑡+𝑚𝑣} θ_{𝑡−1} + 𝚤_{𝑡,0} (11)
Now 𝑠 = 0 implies 𝑝  𝚤, while 𝑠 > 0 lead us to 𝑝  θ_{𝑡−1}, both facts being absurd.
Finally, to analyze the equation 𝑝^{𝑟} + (𝑝^{2𝑡+𝑚s}) = (𝑝^{2𝑡+𝑚𝑢}) + (𝑝^{2𝑡+𝑚𝑣}) defined by item 9, we will assume that 0 < 𝑟 < 4𝑡− 1 and 0 ≤ 𝑚_{𝑠} < 𝑚_{𝑢} < ?? ≤ 𝑚_{𝑣} ≤ 2𝑡− 1. Now again by (2), we have: 𝑝^{𝑟} + θ _{𝑚𝑠} + 𝑝^{2𝑡+𝑚s} = θ _{𝑚𝑢} + 𝑝^{2𝑡+𝑚𝑢} + _{2𝑡−𝑚𝑣, 𝑚𝑣−𝑡} + 𝑝^{2𝑡+𝑚𝑣} θ_{2𝑡−𝑚𝑣−1} (12)
Now this equality is the same as
𝑝^{𝑟} + 𝑝^{2𝑡+𝑚s} = 𝑝^{𝑚s+1}θ_{𝑚𝑢 −𝑚𝑠 −1} + 𝑝^{2𝑡+} + _{2𝑡−𝑚𝑣, 𝑚𝑣−𝑡} + 𝑝^{2𝑡+𝑚𝑣} θ_{2𝑡−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 𝑆_{2} set in the integer interval [1.255]. Note that we have actually provided a good example of an 𝑆_{2} set which has a nice property: For each element 𝑎_{𝑗} ∈ 𝔖 the subset S_{𝑎𝑗} = 𝔖 ∩ I_{𝑎𝑗} ⊂ 𝔖 is a good 𝑆_{2} set.
4.3 The Case 𝑞 = 3^{4}
The following Proposition provides a criterion to construct 𝑆_{2} sets in the integer interval [1.80] by joining 𝑆_{2} 3cycles of length 4. By Theorem (5.1) and Corollary (3.4) we have that the set of generators 𝑆_{2} 3cycles 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 𝑆_{2} set.
Proof: To begin with, observe that 𝔖 is not an 𝑆_{2} 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:
3^{𝑟} 𝑖 + 3^{𝑠} 𝑖 ≡ 3^{𝑢} + 3^{𝑣} (mod(80)) (14)
3^{𝑟} 𝑖 +3^{𝑣} ≡ 3^{𝑢} + 3^{𝑠} (mod(80)) (15)
3^{𝑟} 𝑖 +3^{𝑠} 𝑖 ≡ 3^{𝑢} + 3^{𝑣} (mod(80)) (16)
If 3^{𝑟} 𝑖 + 3^{𝑢} 𝑗 + 3^{𝑣} 𝑗 (mod (80)) and I =2_{𝚤}. By parity arguments, the congruence (14) is equivalent to:
Now since (3^{𝑣−𝑢} + 1)/4 is 1 or 7, then (3^{𝑠−𝑟} + 1)/2 must be equal to 5 and hence 52_{𝑘} +1 which is a contradiction.
On the other hand if i = 2𝜄+1 and keeping in mind that 3^{𝜇}  1 =2,8 or 26, then congruence (15) is equivalent to:
This equality implies that (3^{𝑣−𝑢} − 1)/2 must be even and hence equals to 4, consequently
10𝑡 = 3^{𝑢} · 2 · 𝑘−3(2+1),
which is again an contradiction.
Finally, congruence (16) is nothing but
80𝑡 = 3^{𝑟} i (1+3^{𝑠−𝑟}− 3^{𝑣−𝑟}) − 3^{𝑢} 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 3cycles generated by i and 𝑗 (Table).
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:
5 Curves with many rational points over finite fields
We consider the nonsingular complete irreducible Kummer curve 𝔈 over 𝔽_{ q } defined by the affine equation:
𝑦^{𝑟}= (𝑥),
Where 𝑟 𝑞 − 1 and the rational function 𝜇(x) ∈ 𝔽_{ q } (𝑥) satisfies the following conditions:
1. 𝜇 is not the 𝑑th power of an element 𝑣 ∈ 𝔽_{ q } (𝑥) for any divisor 𝑑 > 1 of 𝑟;
2. 𝜇 = 1 on a substantial subset 𝖅_{𝜇} of ℙ^{1} (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 𝑦^{𝑟}= 𝜇(𝑥), where 𝑟 divides 𝑞 − 1 and the rational function 𝜇(𝑥) is not the 𝑑th power of an element 𝑣(𝑥) ∈ 𝔽_{ q } (𝑥) for any divisor 𝑑 of 𝑟 with 𝑑 > 1, has the following properties:
1.If
2. The set of 𝔽_{ q } −rational points (essentially) satisfies 𝔈 (𝔽_{ q } ) ≥ 𝑟 ∙ 𝖅_{𝜇}.
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) ∈ 𝔽_{𝑝} [𝑥 ], and we denote by 𝔙_{ℓ} 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) ∈ 𝔽_{𝑝} [ 𝑥]. We denote the zero sets (in 𝔽_{ q } ) of 𝑓 (resp 𝑔) by 𝔙_{𝑓} (resp, 𝔙_{𝑔}), then the rational function
satisfies (𝛼) = 1 for 𝛼 ∈ 𝔙_{𝓁} \ (𝔙_{𝑓} ⋃ 𝔙_{𝑔}).
2. Given (𝑥) ∈ 𝔽_{𝑝}(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):
In accordance with above observation, we need consider polynomials ℓ(𝑥) ( 𝔽_{𝑝} [𝑥] having many roots in 𝔽_{𝑝}.
Definition 5.1 A polynomial (𝑥) ∈ 𝔽_{ q } [ 𝑥 ] is a restricted range polynomial if ??(𝛼) ∈ 𝔙 ⫋ 𝔽_{𝑝} for some proper subset of 𝔽_{𝑝} and for all 𝛼 ∈ 𝔽_{ q } . In particular, when 𝔙 = 𝔽_{𝑝}, we say that (𝑥) is a 𝔽_{ q } , 𝔽_{𝑝}polynomial.
A classic example of restricted range polynomial is the norm polynomial
Definition 5.2 A nonzero (𝔽_{ q } , 𝔽_{𝑝})−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 } , 𝔽_{𝑝})polynomial.
We recall briefly some properties of (𝔽_{ q } , 𝔽_{𝑝})polynomial. For proofs, we refer to ^{5} and ^{13}.
Proposition 5.2^{13} (𝔽_{ q } , 𝔽_{𝑝})polynomials are subjective.
Theorem 5.1^{5} (Characterization of (𝔽_{ q } , 𝔽_{𝑝}) polynomials) The exponent sets of the minimal (𝔽_{ q } , 𝔽_{𝑝})polynomials are the 𝑝cycles of set{0,…, 𝑞 − 1}. For each 𝑝cycle ℑ, all the minimal (𝔽_{ q } , 𝔽_{𝑝})polynomials with exponent set ℑ are:
where 𝑖 is an arbitrary but fixed representative of ℑ. In addition, we have all the different (𝔽_{ q } , 𝔽_{𝑝})polynomials of less or equal degree to 𝑞−1 by sums of polynomials ??_{ℑ} (𝑥, 𝛼) corresponding to different cycles.
Example 5.1 Using Theorem 5.1 and Example 2.2, we exhibit some (𝔽_{27}: 𝔽_{3}) polynomials.
Remark 5.1 Proposition 5.2 says that each (𝔽_{ q } , 𝔽_{𝑝})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 𝔽_{9} given by 𝑦^{2} = −(𝑥^{6}+ 𝑥^{5} + 𝑥^{4} + 2𝑥^{2}+ 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 3cycles of length 2, namely (1,3),(2,6) and (5,7), then set {1,3,5,7} which is not an 𝑆_{2} set! induces the 𝔽_{9},𝔽_{3}polynomial ℓ(𝑥) = 𝑥^{7} + 𝑥^{5} + 𝑥^{3} + 𝑥 which has its roots in 𝔽_{9}. We will take advantage of this fact to construct our curve. In general, given two coprimes polynomials ℓ(𝑥), 𝑓(𝑥) ∈ 𝔽_{𝑝}[ 𝑥] and 𝑟 𝑞 − 1, the Euclidean division
(𝑥)^{𝑟} = ℓ(𝑥) ∙ ℎ(𝑥) + ℛ_{ℓ}(𝑓(𝑥)^{𝑟}),
implies that for each 𝛼 ∈ 𝔽_{ q } root of ℓ(𝑥), ℛ_{ℓ} (𝑓(𝑥)^{𝑟})(𝛼) = 𝑓(𝛼)^{𝑟}; i.e. ℛ_{ℓ} (𝑓(𝑥)^{𝑟})(𝛼) is an 𝑟th power in 𝔽_{ q } , therefore the polynomial 𝑦^{𝑟} = ℛ_{ℓ} (𝑓(𝑥)^{𝑟})(𝛼) splits completely en 𝔽_{ q } [𝑦], this means many points. Now in our situation taking (𝑥) = (𝑥 −1) (𝑥 +1) (𝑥^{4} + 2𝑥 + 2) we have, ℛ_{ℓ} ((𝑥)^{2}) = −(𝑥^{6} + 𝑥^{5} + 𝑥^{4} + 2𝑥^{2} + 2), this carry us to our equation.
Example 5.3 The curve ℭ over 𝔽_{27} given by 𝑦^{2} = −(𝑥^{4} + 2𝑥^{3} + 𝑥^{2} + 1)^{3} has 𝑔(ℭ) = 1 and #ℭ (𝔽_{27}) = 38.
By Proposition 3.4, the union of the underlying sets of the 3cycles (1,3,9) and (2,6,18) is an 𝑆_{2} set. This set induces the (𝔽_{27}:𝔽_{3})polynomial ℓ(𝑥) = 𝑥^{18} + 𝑥^{9} + 𝑥^{6} + 𝑥^{3} + 𝑥^{2} + 𝑥 which induces the rational function 𝜇(𝑥) = −𝑥^{2} (𝑥+1)^{2} (𝑥^{4} + 2𝑥^{3} + 𝑥^{2} + 1)^{3} and consequently the algebraic curve ℭ over 𝔽_{27} defined by the equation:
𝑦^{2} = −(𝑥^{4} + 2𝑥^{3} + 𝑥^{2} + 1)^{3}.
(Here we split ℓ(𝑥) as ℓ(𝑥) = ℓ_{1}(𝑥) + ℓ_{2}(𝑥) where ℓ_{1}(𝑥) = 𝑥 + 𝑥^{2} and ℓ_{2}(𝑥) = 𝑥^{3} + 𝑥^{6} + 𝑥^{9} + 𝑥^{18} = 𝑥^{3} (𝑥+1)^{3} (𝑥^{4} +2𝑥^{3} + 𝑥^{2} + 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), 𝑥^{27}  𝑥) = 𝑥 + 𝑥^{2} + 𝑥^{4} + 2𝑥^{5} + 𝑥^{6} +2𝑥^{7} + 𝑥^{9}, 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 𝑆_{2} 3cycle of length 3, (17,25,23) induces the minimal (𝔽_{27}:𝔽_{3})polynomial ℓ(𝑥) = 𝑥^{17} + 𝑥^{25} + 𝑥^{23} which has 9 roots in 𝔽_{27}, namely the zeros of the polynomial 𝑓(𝑥) = 𝑥 + 𝑥^{7} + 𝑥^{9} = 𝑥(1 + 𝑥)(2 + 𝑥)(2+ 𝑥^{2} + 𝑥^{3})(1+ 2𝑥^{2} + 𝑥^{3}). If we split ℓ(𝑥) as ℓ(𝑥) = 𝑥^{17} + 𝑥^{23} (𝑥^{2} + 1) and consider the rational function 𝜇(𝑥) := −𝑥^{6} (𝑥^{2} + 1), then for each α root of 𝑓(𝑥), 𝜇(𝛼) = 1.
Now, the curve ℭ over 𝔽_{27} defined by the Kummer equation
𝑦^{26} = (𝑥): = −𝑥^{6} (𝑥^{2} + 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 ℭ_{1} and ℭ_{2} over 𝔽_{49} with (ℭ_{1}) = 1 and (ℭ_{2}) = 3.
In Table 9, we exhibit some 𝑆_{2} sets 𝔖 ⊂ I_{48} which was obtained as union of 3 7cycles of length 2. For this example we consider the set 𝔖 ={1,2,7,11,14,29} = { ⟨1⟩ ∪⟨2⟩ ∪ ⟨11⟩}. This set induces the (𝔽_{49}:𝔽_{7})polynomial ℓ(𝑥) = 𝑥 + 𝑥^{2} + 𝑥^{7} + 𝑥^{11} + 𝑥^{14} + 𝑥^{29}). By Proposition 5.2 there exist a subset 𝔙 ⊂ 𝔽_{49} such that ℓ(𝔙) = 1 and after some computations we obtain that 𝔙 is the zero set of the polynomial 𝑙(𝑥) = 1+ 𝑥+ 𝑥^{2} + 6𝑥^{3} + 𝑥^{4}. 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+ 4𝑥^{2} + 𝑥^{4}) + (6𝑥^{3}), the polynomial 1+ 𝑥+ 4𝑥^{2} + 𝑥^{4} can be factored as 1+ 𝑥+ 4𝑥^{2} + 𝑥^{4} = (5 + 𝑥)^{3} (6 + 𝑥) and therefore we consider the Kummer cover defined by the equation
This algebraic function field has genus 𝑔 = 𝑟 − with 𝑑= 𝐺𝐶𝘋(𝑟,3), this gives, for 𝑟 =2, 𝑔(ℭ_{1}) = 1 and for 𝑟 =4, 𝑔(ℭ_{2}) = 3, the rational points satisfies #ℭ_{1} = 64 and #ℭ_{2} = 92 (see tables in ^{3}).