Sequences in finite fields yielding divisors of Mersenne, Fermat and Lehmer numbers, II

: Let ρ be an odd prime ≥ 11 . In Part I, starting from an M -cycle in a finite field F ρ , we have established how the divisors of Mersenne, Fermat and Lehmer numbers arise. The converse question is taken up in this Part with the introduction of an arithmetic function and the notion of a split-associated prime


Introduction
Numbers of the forms 2 n − 1, 2 n + 1 and 2 2 n + 1 are referred to as Mersenne, Lehmer and Fermat numbers, respectively.The main purpose of this study is to establish the algebraic principle upon which the factors of these numbers arise.In Part I, [9], the author has introduced the polynomial sequences {F k (x)}, {G k (x)} and {H k (x)} over Z defined as follows: In [9], the equivalence of the following statements have been proved: The concept of satellite polynomial has been introduced.
(i) A polynomial p(x) ∈ Z[x] is said to be a satellite polynomial for G j (x) if p(x) | G j (x) but p(x) ̸ ∈ {G k (x)}.
(ii) A polynomial q(x) ∈ Z[x] is said to be a satellite polynomial for H j (x) if q(x) | H j (x) but q(x) ̸ ∈ {H k (x)}.
With ρ a prime, the values assumed by the sequences in the field F ρ have been considered, leading to the sequences {M (t)}, {θ t,k } and {ψ t,k }, respectively.Let ρ be an odd prime ≥ 11.Let M (t) ∈ F ρ − {0, ±1, ±2} such that M 2 k ̸ = 2, 3 for all k in the cycle Let ω be the smallest positive integer such that ψ t,ω = 0. Then it has been proved in Part I [9], that ω ≥ n and 2ω + 1 | 2 n − 1 or 2 n + 1.It has also been proved that n | 1  2 Φ(2ω + 1).In Part I, starting from an M -cycle in F ρ , we have established how the divisors of Mersenne, Fermat and Lehmer numbers arise.The converse question is taken up in this part.One may refer to Brent, Crandall, Dilcher and van Halewyn [1], Brillhart and Johnson [2], Brillhart [3], Gostin [5], Kang [7], Kravitz [8] and Ribenboim [10] for several results on the factors of Mersenne and Fermat numbers.Starting with such a factor, how to find an odd prime ρ and the M -cycle in F ρ contributing the factor under consideration?This is the focus of attention in this part.
First we develop the necessary preliminaries and then proceed to settle the converse question.The main results of this study are contained in Theorems 3.2, 3.3, 4.2, 5.10, 6.2, Corollary 6.3 and Theorem 7.3.A summary of results is furnished in Section 8.
2 Classification of M -cycles in the field F ρ Consider an M -cycle of length n in the field F ρ denoted by Let ω be the corresponding pivotal position in C 1 (t).In view of [9,Corollary 5.3], M 1 ,M 2 ,. ..,M n are roots of the polynomial H ω (x) over F ρ .We have the following: Definition 2.1 (Types of M -cycles).We say that the M -cycle of length n in F ρ is of type I if the M -cycle contributes all the roots of H ω (x) and of type II if the elements of the M -cycle form a proper subset of the set of roots of H ω (x).i.e., the M -cycle is of type I if ω = n and of type II if ω > n.
Example 2.1.Consider the field F 43 .Let ( p q ) denote the Jacobi symbol.We obtain ( 6 43 ) = 1.It is noted that 36 2 = 1296 ≡ 6 (mod 43).Hence 36 2 − 2 ≡ 4 (mod 43).Choosing M (t) = 4, we get the cycle 4 → 14 where } and they are non-trivial polynomials over Z.Since the degree of p(x) is less than the degree of H ω (x), not all the roots of H ω (x) in F ρ can be roots of p(x).Therefore, there is at least one root of H ω (x) in F ρ which also satisfies H d (x).Since ω = n, all the roots of H ω (x) are the elements of an M -cycle.Hence H d (M (t)) = 0 for some M (t) ∈ F ρ .However, since H d (x) is an element of the sequence {H k (x)}, from [9, Theorem 6.1] it is seen that all the other elements in the M -cycle also satisfy H d (x).Consequently, we have the degree of H d (x) equals the degree of H ω (x).This implies that degree of p(x) = 0, which is a contradiction.
Remark 2.2.The converse of the above theorem, however, does not hold as noted in Example 2.2.
Proof.By assumption, ω = n.Hence, the polynomial H ω (x) has degree n.The coefficients of x n and x n−1 in H ω (x) are both 1.Since M 1 , M 2 , . . ., M n exhaust all the roots of H ω (x), it follows from the theory of equations that n i=1 M i ≡ −1 (mod ρ).
3 Classification of prime numbers in relation to G(x) and H(x)-sequences A result provided by [9, Theorems 2.12 and 2.14] is the distinction between odd primes and odd composite numbers in the context of the polynomial sequences {G k (x)} and {H k (x)}.While considering the polynomials G m (x) and H m (x), we have to distinguish between the following two cases: (i) 2m + 1 is a composite number, and (ii) 2m + 1 is a prime number.
The first case has already been considered in [9, Section 2].Now let us consider the case when 2m + 1 is a prime number.A few definitions become imperative in this context.The concept of a satellite polynomial has been introduced in [9, Section 2, Subsection 7].By [9, Definition 3.9], when an M -cycle is considered in the field F ρ , we refer to ρ as the background prime.It has been proved in [9,Theorem 6.1] that the elements of an M -cycle are the roots of some H(x)-polynomial.In this regard, we have the following classification of prime numbers.Definition 3.1 (Split-associated prime).If 2m + 1 is a prime and if H m (x) has a satellite polynomial ∈ F ρ [x] whose roots are in F ρ for some background prime ρ, then 2m + 1 is called a split-associated prime.
If 2m + 1 is a prime and if H m (x) has a satellite polynomial ∈ F ρ [x], then it follows that all the proper factors of H m (x) are satellite polynomials.Definition 3.2 (Non-split-associated prime).If 2m + 1 is a prime and if H m (x) does not have a satellite polynomial ∈ F ρ [x] where ρ is a background prime for 2m + 1, then 2m + 1 is called a non-split-associated prime.

Definition 3.3 (Universal satellite polynomial).
A satellite polynomial of H m (x), when it exists, is said to be universal if it is the same irrespective of the background prime under consideration.
Example 3.1.The polynomial p(x) = x 4 − x 3 − 4x 2 + 4x + 1 is a factor of H 7 (x), whatever be the background prime, and it is not in {H k (x)}.Hence, p(x) is a universal satellite polynomial for H 7 (x).As a result, H 15 (x) factorizes as in F 61 and F 311 , respectively.Thus, the coefficients of the factors of H 15 (x) depend on the concerned background primes.Consequently, the satellite polynomials of H 15 (x) are local, whenever they exist in some field F ρ .It follows that 31 is a split-associated prime.
Theorem 3.1.A necessary condition for a prime p to be split-associated is that is composite, but not conversely.
That the converse does not hold is illustrated by the following example.Consider the prime p = 37.A background prime for p is 149.The following M -cycle exists in For this cycle, ω = 18.Since ω = n, it follows that p is non-split-associated.However, , then all the resulting factors of H ω (x) are of equal degree.
Proof.If there are two factors of Since the divisibility by 2ω + 1 is associated with the smallest n occurring as an exponent in 2 n − 1 or 2 n + 1, it follows that n 1 = n 2 .A similar argument applies to the case of the occurrence of several satellite polynomials as factors of Employing a similar argument as in the above theorem, we have the following two theorems.Theorem 3.3.Let ρ and ρ ′ be two background primes for a prime 2ω + 1.If H ω (x) splits into satellite polynomials, then the satellite polynomials of , then all the resulting factors of H ω (x) are local satellite polynomials.
Proof.By Theorem 3.2, the resulting factors of H ω (x) are of equal degree.Let the degree of any such polynomial be α and let the number of such factors be s.Then the roots of H ω (x) in F ρ form s number of M -cycles of length α each.Suppose these M -cycles are: For each i = 1, 2, . . .s, let S i,j denote the elementary symmetric functions formed by the elements of the i-th M -cycle, where j = 1, 2, . . ., α.Then Since the S i,j 's in (3.1) depend on F ρ [x], each factor of H ω (x) is a local satellite polynomial.
Corollary 3.1.For given H ω (x), the degree α of any aforesaid satellite polynomial of H ω (x) is unique.

An arithmetic function
Introduction of a new arithmetic function becomes necessary for our study.
Definition 4.1 (Odd part of a natural number).Define δ : N → N as follows: Given a natural number n, we can write n = 2 λ m, where λ ≥ 0 and m is an odd integer ≥ 1.We define provided m is an odd integer.We refer to δ(n) as the odd part of n.
We have the following theorem.
Next we prove a crucial identity involving the arithmetic function δ.The result is contained in the following theorem.
Corollary 4.1.There do not exist two distinct primes for which both corresponding odd parts are identical.

Relationship between the pivotal position and the background prime
The concept of pivotal position has been introduced in [9, Definition 5.4].Suppose ω is the pivotal position in the ψ t,k -sequence for an M (t)-cycle in F ρ [x].In [9, Theorem 5.6], we have proved that the middlemost positions in each compartment of a(M (t)) are occupied by the values of 2 M (t−1) and 0 in the first and second rows, respectively and these values are not attained at any other places in the concerned compartment.Given t, it has been proved in [9,Corollary 5.4] that the period of the cyclic sequence θ t,k (resp.ψ t,k ) as a function of k is 2ω + 1.On the basis of these results, we establish a relationship between ω and ρ.First we develop the preliminaries.

The linkage with arithmetic progressions
The role of arithmetic progressions has been brought out in [9,Theorem 2.13].Further linkage of the pivotal position in the ψ t,k -sequence with certain arithmetic progressions is considered in the sequel.Definition 5.1 (Fundamental arithmetic progression associated with a natural number).Suppose n ∈ N .An arithmetic progression with first term n and common difference 2n + 1 is called the fundamental arithmetic progression associated with n and is denoted by S(n), i.e., S(n) = {n, 3n + 1, 5n + 2, . . .}.

Divisibility among odd numbers is transformed into an equivalent problem of arithmetic progressions as follows:
Theorem 5.1.Suppose j, m ∈ N with j < m.Then 2j + 1 | 2m + 1 if and only if S(m) ⊂ S(j).

Root points
We consider the positions in the second row of the matrix a(M (t)) at which the ψ t,k -sequence attains a zero with k ≥ ρ.First we have the following theorem.
Theorem 5.2.Suppose a polynomial H(x)-sequence attains a root at ω in the second row of the principal compartment C 1 (t) of the matrix a(M (t)).Then the positions in the second row of a(M (t)) at which the polynomial attains a root are precisely the elements of S(ω).
Proof.When the ψ t,k -sequence attains a zero at k = ω, by [9, Corollary 5.4] the next zero of the sequence occurs at k = 3ω + 1.The proof is completed by a repetition of this argument.
Let us find the smallest non-negative integer h such that whenever the ψ t,k -sequence contains a root of H ω (x) in the field F ρ , it contains all the roots of H ω (x) in F ρ at ψ t,k when k takes the value of ρ + h.Definition 5.2 (Root point).Let ρ be a given odd prime.Suppose ω ∈ N .The root point of ρ with respect to ω, if it exists, is denoted by r(ρ, ω) and is defined as the least natural number ρ + h where h is the least non-negative integer such that H ω (x) attains all of its ω roots in F ρ at ψ t,k with k = ρ + h. (5.1) If the root point of ρ with respect to ω does not exist, then we say that r(ρ, ω) is undefined.
Proof.From [9, Corollary 5.4], it is seen that the roots of H ω (x) are attained in the ψ t,k -sequence at regular intervals of 2ω + 1. Hence the result.

Properties of the root points
We consider the derivation of a formula for r(ρ, ω), when it exists, in terms of ρ and ω.For this purpose we establish certain properties of r(ρ, ω).
Next we establish an important condition for the polynomial H ω (x) to attain roots in two different finite fields.
Let s ∈ N such that C s+1 (t, ρ, ω) is the compartment in a(t) corresponding to which ρ is the index of the cell.The index of the last cell in the completed compartments before the occurrence of ρ as the index of a cell is s(2ω + 1) − 1.The cells with indices ρ and ρ + h occur in the immediate next compartment.
Let s ′ ∈ N such that C s ′ +1 (t, ρ, ω) is the compartment in a(t) corresponding to which ρ ′ is the index of a cell.The cells with indices ρ ′ and ρ ′ + h ′ occur in the immediate next compartment.
Let us consider the forward movement along the ψ t,k -sequence from the s-th compartment to (s ′ + 1)-st compartment.Since ρ ′ ≡ ρ (mod 2ω + 1), the number of cells from the leftmost cell in the (s + 1)-st compartment to the cell with index ρ is the same as the number of cells from the leftmost cell in the (s ′ + 1)-st compartment to the cell with index ρ ′ .Since a root of H ω (x) occurs after h positions from ρ in the ψ t,k -sequence and the roots of H ω (x) are attained in the ψ t,k -sequence at regular intervals of 2ω + 1, it follows that a root of H ω (x) occurs after h positions from ρ ′ in the ψ t,k -sequence.Since h ′ is the least non-negative integer such that H ω (x) attains all of its ω roots in F ρ ′ at ψ t,k where k = ρ ′ + h ′ , it follows that h ′ ≤ h.Similarly, considering the backward movement along the ψ t,k -sequence from the (s ′ + 1)-st compartment to (s + 1)-st compartment, it is seen that h ≤ h ′ .Hence we obtain h ′ = h.Consequently, r(ρ ′ , ω) − r(ρ, ω) = ρ ′ − ρ.

Attainment of roots of H(x)-polynomial
With the necessary tools having been constructed, we prove a result on the attainment of roots of H(x)-polynomial in the field F ρ with ρ ≥ 11.We have to distinguish between two cases: Theorem 5.8.If 3 | 2ω + 1, then a necessary condition for H ω (x) to attain all of its roots in a field ( This implies that Hence the theorem.Theorem 5.9.Suppose 3 ∤ 2ω + 1 and H ω (x) attains all of its roots in a field F ρ .Suppose 2ω ′ + 1 = 3(2ω + 1).If H ω ′ (x) attains all of its roots in some field F ρ ′ then 2ω + 1 | δ(ρ − 1) or δ(ρ + 1).
Thus the theorem holds for H ω (x).
Combining Theorems 5.8 and 5.9, we obtain the following condition for the attainment of all the roots of a H(x)-polynomial.

Full complement of the roots of polynomials of H(x)-sequence
The results in this section come as offshoots of Theorem 4.2.We find an answer to the converse problem emanating from the result contained in Theorem 5.10.

Existence of non-singular M -cycles in the field F ρ
Let ρ be a given prime ≥ 11.In the ordered pair (n, ω) with n, ω ∈ N , let n denote the length of an M -cycle in a field F ρ and ω denote the pivotal position in C 1 (t) at which the ψ t,k -sequence attains a zero.We have established in Section 5 that 2ω + 1 | δ(ρ − 1) or δ(ρ + 1).Because of this property, the M -cycles in F ρ can be put in the following two disjoint classes: (i) M -cycles associated with δ(ρ − 1), and (ii) M -cycles associated with δ(ρ + 1).
In Definition 3.9 of [9], the concept of background prime for the M -cycle has been introduced.Definition 6.1 (Minimum background prime).Given 2ω + 1 ∈ N , the least among all the background primes of 2ω + 1 is called the minimum background prime for 2ω + 1 .
Certain primes are named after Sophie Germain (1776-1831).Definition 6.2 (Sophie Germain and safe primes).A prime p is said to be Sophie Germain if 2p + 1 is also a prime (see for e.g., Ribenboim [10], Roberts [11] and Shanks [12]).An odd prime p is called a safe prime if is also a prime.
Large Sophie German primes were determined by Dubner [4].The distribution of these primes was studied by Yates in [13].
An important property possessed by any odd prime ρ in respect of the roots of H(x)polynomials is obtained in the following theorem.Theorem 6.2 (Existence of full complement of the roots of H ω (x)).If ρ is an odd prime ≥ 11 and if 2ω + 1 | δ(ρ − 1) or δ(ρ + 1), then the polynomial H ω (x) attains all of its roots in F ρ .
Proof.We give a proof by induction on ω.First we prove the result for the minimum background prime for 2ω + 1 and then extend it to a general background prime for 2ω + 1.
Assume the theorem for all satellite polynomials of H(x) of degree < m and for all fields F ρ where ρ is the minimum background prime for 2ω + 1.Now consider H m (x).Let ρ be the minimum background prime for 2m + 1.By Theorem 6.1, it follows that H(x) has at least one root in F ρ .Starting from this root, we obtain an M -cycle in F ρ of length ≥ 2. We have to consider separately two cases viz.when 2m + 1 is: (i) a prime, and (ii) a composite number.
Case (i).2m + 1 is a prime.We have to consider two sub-cases.
• Sub-case (i) (A).m is a prime.In this case m is a Sophie Germain prime.It is seen that H m (x) has no satellite polynomial.The result in this case follows from [9, Theorems 2.12 and 6.1].
-Sub-case (i) (B) (I).2m + 1 is a non-split-associated prime.In this case, H m (x) has no satellite polynomial and so it attains all of its ω roots in F ρ .
By the induction assumption, µ(x) has all of its roots in F ρ .Since each root of µ(x) is also a root of H m (x), putting together the roots of λ(x) and µ(x) we obtain all the m roots of H m (x) in F ρ .
Case (ii).2m + 1 is composite.In this case, there exists a prime divisor 2j + 1 of 2m + 1.By [9, Theorems 2.12], H j (x) is a divisor of H m (x).Since j < m, by Case (i), H j (x) attains all of its roots in F ρ .Now Hm(x) H j (x) , being a satellite polynomial of H m (x) of degree m − j < m, attains all of its roots in F ρ which in turn implies the result for H m (x).
Thus the theorem holds in all the sub-cases, in respect of the field F ρ where ρ is the minimum background prime for 2ω + 1.We now extend the proof to all the fields F η where η is any other background prime for 2ω + 1 implying η > ρ.
The elements in F ρ which form M -cycles are linked to the quadratic residues in F ρ .If 2ω + 1 is associated with δ(ρ − 1) in F ρ and δ(η − 1) in F η , then considering the number of quadratic residues associated with the odd parts in the concerned fields, it is seen that at least as much quadratic residues are associated with δ(η − 1) in F η as are associated with δ(ρ − 1) in F ρ .A similar result holds if 2ω + 1 is associated with δ(ρ − 1) and δ(η + 1) or δ(ρ + 1) and δ(η − 1), or δ(ρ + 1) and δ(η + 1) in the concerned fields.So δ(η − 1) (or δ(η + 1), as the case may be) is associated with at least ω roots of H(x) in F η .Since the number of roots of H(x) cannot exceed ω, it follows that H(x) has all of its roots in F η .Corollary 6.2.Given an odd prime ρ ≥ 11, the polynomials Hδ(ρ−1)−1 Proof.Follows from Corollary 6.1 and Theorem 6.2.

Consideration of the converse question
Given an M -cycle in a field F ρ , it follows from [9, Theorem 6.1] and Theorem 5.10 that the corresponding ψ t,k -sequence attains a zero at ω in C 1 (t) such that 2ω + 1 divides one of δ(ρ − 1), δ(ρ + 1).The converse question is: Given a divisor 2ω + 1 of δ(ρ − 1) or δ(ρ + 1), is there an M -cycle in F ρ for which the corresponding ψ t,k -sequence attains a zero at ω in C 1 (t)?The answer is in the affirmative.From Theorems 6.1 and 6.2 and Corollary 6.2, we are led to the following corollary.Corollary 6.3.Given an odd prime ρ ≥ 11 and any divisor 2ω + 1 of δ(ρ − 1) or δ(ρ + 1), the polynomial H ω (x) attains all of its roots in F ρ .
This establishes the sufficiency of the condition in Theorem 5.10.2. The former sequence does not contribute a root of any polynomial in the H(x)-sequence.In contrast, as seen in the course of the proof of Theorem 6.2, the latter sequence contributes a root of H 1 (x).Further, if 2ω + 1 is divisible by 3, the latter sequence also satisfies H ω (x) along with some other polynomial ∈ H(x)-sequence and a satellite polynomial of H ω (x).Because of this distinction, we refer to the former as singular whereas the latter is said to be non-singular.
7 M -cycles in the field F ρ from the divisors of Mersenne and Lehmer numbers Previously we have seen the role of arithmetic progressions in the method of cyclic sequences in [9, Section 2] and Section 5 of this Part II.Yet another role of arithmetic progressions is brought out in this section.

Problem to be considered
In Part I, starting from an M -cycle in a field F ρ , we have seen how the divisors of Mersenne and Lehmer numbers are obtained as established in [9,Theorem 8.2].In the other direction, now we take up the following question: Starting from a given divisor of 2 n − 1 or 2 n + 1, is it possible to obtain a field F ρ which contains an M -cycle of length n such that the corresponding ψ t,k -sequence attains a zero at ω in the compartment C 1 (t)?
To answer this question, we consider an application of Dirichlet's theorem on primes in arithmetic progression.Dirichlet (see for e.g., Hardy and Wright [6]) proved the following theorem: If a is positive and a and b have no common divisor except 1, then there are infinitely many primes of the form an + b.
Employing the above result, we have the following theorem.
1, there exists at least one root of H m (x) in F ρ.By [9, Theorem 6.1], starting from one root ofH m (x), we obtain a non-trivial M -cycle in F ρ all of whose elements are roots of H m (x).Let us form a polynomial λ(x), using each one of these elements as a root exactly once.By Theorem 3.4, λ(x) is a local satellite polynomial of degree ≥ 4 for H m