On the Proof of Goldbach's Conjecture and the Structure of Prime Number Gaps

On the Proof of Goldbach’s Conjecture and the Structure of Prime Number Gaps Zaroual Ayouba,∗, Lamnii Abdellaha aUniversity of Abdelmalek Essaadi, Laboratory ISI, ENSATe, Tetouan, Morocco. Abstract This paper investigates the intricate structure of prime numbers and offers insights into Goldbach’s Conjecture, a long-standing and captivating problem in number theory. It is conjectured that any even integer larger than 2 can be expressed as the sum of two prime numbers. We begin by exploring the construction of specific prime number sequences and propose general formulas that describe their distribution and gaps. Through an algebraic and probabilistic approach, we delve into patterns of prime pairs, including twin primes, and apply these findings to verify the conjecture using newly defined sets, {uk(n)}∞ n=0. Finally, we present a novel framework for proving Goldbach’s Conjecture and discuss the broader implications for future research in prime number theory. Keywords: Prime numbers, Goldbach’s Conjecture, twin Prime, Number theory 1. Introduction The Goldbach Conjecture, proposed in 1742 by Christian Goldbach, remains one of the most intriguing and longstanding unsolved problems in number theory. It postulates that every even integer greater than 2 can be expressed as the sum of two prime numbers. Despite its apparent simplicity, the conjecture has withstood centuries of mathematical scrutiny. Over the years, notable progress has been made in understanding its structure. In 1937, Vinogradov proved that every sufficiently large odd integer can be expressed as the sum of three primes, laying the groundwork for the ternary version of Goldbach’s Conjecture [1]. This result was further cemented when Helfgott completed the proof for all cases in 2013 [2]. In another breakthrough, Chen Jingrun showed in 1973 that every sufficiently large even number can be written as the sum of a prime and a semiprime (a number with at most two prime factors), a result now known as Chen’s Theorem [3]. These advances illustrate the deep connection between the additive properties of prime numbers and their distribution. Modern research continues to explore this relationship by investigating conjectures such as the Elliott–Halberstam Conjecture [4] and by extending Goldbach’s ideas to more general algebraic frameworks [5]. In this paper, we take a fresh approach to Goldbach’s Conjecture by analyzing the structural properties of prime numbers through algebraic and probabilistic methods. We introduce a new class of sequences {uk(n)}∞ n=0 that generate subsets of integers ⋆This work was supported by the University of Abdelmalek Essaadi. ∗Corresponding author with controlled divisibility properties. These sequences allow us to constructively define prime-like sets and study their interactions, including prime gaps and twin primes. We leverage these results to propose a novel method of verifying Goldbach’s Conjecture by establishing a correspondence between sums of primes and the structured elements of these sequences. In doing so, we also highlight asymptotic results and probabilistic arguments that reinforce the plausibility of the conjecture for all even integers. The remainder of this paper is organized as follows: Section 2 introduces the algebraic construction of prime-related sequences and their properties. Section 3 explores the distribution and density of prime numbers via these sequences. In Section 4, we examine prime gaps and their statistical behavior. Section 5 presents a constructive proof of the Goldbach Conjecture using our sequence-based framework. Finally, Section 6 summarizes our findings and outlines directions for future research. 2. The Structure of Prime Numbers In this section, we examine the construction and properties of prime numbers using algebraic methods. We define new sequences {uk(n)}∞ n=0 that serve as a basis for exploring the distribution and behavior of prime numbers. Through these sequences, we uncover patterns and relationships that will be crucial for understanding the gaps between primes and eventually for approaching Goldbach’s conjecture. We begin by defining the structure of these sequences and provide examples to illustrate their application. Now let us define Ukas the set of positive integers ≥2 that are not divisible by any of the primes p0, p1, . . . , pk−1. Thus, for n∈Uk, we have gcd(n, p0·p1···pk−1) = 1. Additionally: gcd(p0·p1···pk−1·m+n, p0·p1···pk−1) = 1 for m∈N,(1) which implies p0·p1···pk−1·m+n∈Uk. The set Ukconsists of elements from Uk−1that are not divisible by pk−1. The number of elements ≤1 + p0·p1···pk−1of Uk−1that are not divisible by pk−1 is given by: X 1<n≤1+p0·p1···pk−1 gcd(n,p0·p1···pk−1)=1 1 = ϕ(p0·p1···pk−1),(2) where ϕis Euler’s totient function. Using Equation 2, the set Ukcan be defined as: Uk={uk(n)|n∈N},(1) where: uk(n) = uk(n−Mk) + Nk,(2) with Nk= k−1 Y i=0 ui(0), 2 and Mk=ϕ(Nk) = k−1 Y i=0 ui(0) −1, subject to the initial condition: u0(n) = n+ 2. The first terms of Ukare given by: uk(n) for 0 ≤n < k−1 Y i=0 (ui(0) −1).(3) These terms represent the first elements of Uk−1that are not divisible by uk−1(0). Example 2.1. For a more in-depth explanation of the structure of the set Uk, we provide some examples. 1. U0:{2,3,4,5,6,7,8,9, . . .} u1(n) = u1(n−1) + 2, u1(0) = 3, U1={3,5,7,9,11, . . .} 2. U1:{3,5,7,9,11, . . .} u2(n) = u2(n−2) + 6, u2(0) = 5, u2(1) = 7, U2={5,7,11,13,17,19,23,25, . . .} 3. U2:{5,7,11,13,17,19,23,25, . . .} u3(n) = u3(n−8) + 30, u3(0) = 7, u3(1) = 11, u3(2) = 13 ··· In the remainder of this section, and to simplify the calculations, we propose an iterative version of the sequence {uk(n)}∞ n=0. It is already established that the recursive formula for {uk(n)}∞ n=0 is given by: uk(n) = uk(n−Mk) + Nk, We can apply this recursively jtimes: uk(n−Mk) = uk(n−2Mk) + Nk, uk(n−2Mk) = uk(n−3Mk) + Nk, 3 . . . uk(n−jMk) = uk(n−(j+ 1)Mk) + Nk. Continuing this process leads us to express uk(n) in terms of ukat a reduced argument: Thus, we obtain: uk(n) = uk(n−jMk) + jNk.(4) Remark 2.2. For n−jMkto be minimal, jmust be the quotient of the division of nby Mk: j=n Mkand n−jMk=nmod Mk. By 2.2, Equation 4 can be written as: uk(n) = Nkn Mk+uk(nmod Mk).(5) Theorem 2.3. The sequence {uk}∞ k=0 provides a mathematical formulation of the Sieve of Eratosthenes algorithm (see [9]). Furthermore, the sequence of prime numbers can be derived as pk=uk(0) = min(Uk), where pkrepresents the k-th prime number. Proof. The set Ukconsists of positive integers ≥2 not divisible by p0, p1, . . . , pk−1. The smallest such integer is pk, as it is the first prime not divisible by any of the previous primes. Hence, pkis the first element of Uk. Now, we present an important lemma for the proof of the main result of this work. Lemma 2.4. For all x, y ∈Uk, the product x·yalso belongs to Uk. That is, if x and yare elements of Uk, then x·y∈Uk. Proof. By definition, the set Ukconsists of all positive integers ≥2 that are not divisible by any of the primes p0, p1, . . . , pk−1. Suppose x∈Ukand y∈Uk. This means that xand yare not divisible by any of the primes p0, p1, . . . , pk−1. Assume, for the sake of contradiction, that x·y /∈Uk. Then x·ymust be divisible by at least one prime piwhere i∈ {0,1, . . . , k −1}. This implies that pi|x·y. Since piis a prime number, it follows that pimust divide either xor y. However, by the assumption that x∈Ukand y∈Uk, neither xnor yis divisible by pi. This contradiction arises from the properties of prime numbers. Therefore, the initial assumption that x·y /∈Ukis false. Hence, x·y∈Uk. 4 Theorem 2.5. Each set Ukis divided into two parts: one that is a multiple of pk and one that is not a multiple of pk. Thus, we have: Uk=Uk+1 ∪(pk·Uk)∪{pk}(6) Proof. Let y∈Uk. By Lemma 2.4, we know that pk·y∈Uk, which implies that pk·Uk⊂Uk. Since pk/∈pk·Uk, we can define Uk+1 as follows: Uk+1 =Uk\(pk·Uk)∪{pk}. The set Uk+1 consists of elements of Ukthat are not divisible by pk, while also ensuring that the elements of Ukare not divisible by p0, p1, . . . , pk−1. Therefore, we can express Ukas: Uk=Uk+1 ∪(pk·Uk)∪{pk}. 3. The sequence Ukand the distribution of prime numbers In this section, we analyze how the sequence {uk}∞ k=0 is related to the distribution of prime numbers. To achieve this, we employ fundamental results, such as Mertens’ formula and key theorems linked to the Riemann Hypothesis. These results will allow us to approximate indices and establish connections with classical prime-counting functions (see [10] for an introduction and [11, 12] for deeper results in analytic number theory). Prime-Counting Function We begin by recalling the prime-counting function, π(x), which represents the number of primes less than or equal to x: π(x) = {p≤x|p∈P},(3) where Pdenotes the set of all prime numbers. Remark 3.1. Since each set Ukexcludes multiples of p0, p1, . . . , pk−1, the smallest composite number appearing in Ukis uk(0)2=p2 k. This observation plays a crucial role in many subsequent counting arguments. To facilitate our calculations, we define ikas the smallest integer nsuch that uk(n) = p2 k. Thus, we obtain: uk(ik) = p2 k=Nk·ik Mk+ukikmod Mk, where Nk= k−1 Y i=0 ui(0), Mk=ϕ(Nk) = k−1 Y i=0ui(0) −1. Since uk(ikmod Mk)< Nk, there exists some integer ℓksuch that uk(ikmod Mk) = Nk−ℓk. 5 Thus, we obtain: ik Mk=p2 k−Nk+ℓk Nk . Rearranging the terms ik Mk=ik Mk−ε1 k=p2 k−Nk+ℓk Nk ,0≤ε1 k<1. Then ik=Mkp2 k Nk+Mk−1 + ℓk Nk+ε1 k. By defining εk=−1 + ℓk Nk+ε1 k,we deduce ik=Mkp2 k Nk+εkMk,|εk|<1. Since ikrepresents the number of primes in the interval [pk, p2 k), we obtain: π(p2 k)−π(pk) = ik=Mkp2 k Nk +εkMk,|εk|<1, =p2 kY p<pk1−1 p+εkMk. From this, we deduce: εk=π(p2 k)−π(pk) Mk−p2 k Nk . As k→ ∞, we observe that εk→0 since Mk≫π(p2 k)−π(pk) and Nk≫p2 k(For instance, numerical approximations yield ε20 ≈5.562 ×10−26). On the other hand,we establish the following asymptotic relation: |εk| ∼ p2 k Nk as k→ ∞. Using Mertens’ formula (see [6]): Y p≤x1−1 p=e−γ log x1 + O1 log x, we obtain:  εkMk 2k+1 ∼ Mkp2 k Nk 2k+1 ∼ p2 ke−γ 2k+1 log pk→0, since 2k+1 log pk≫p2 ke−γ. Then εkMk=O(2k+1). Hence π(p2 k)−π(pk) = p2 kY p<pk1−1 p+O(2π(pk)). By substituting x=p2 kinto the classical formula of the Sieve of Eratosthenes (see [14]), we obtain: π(x)−π(√x) = xY p<√x1−1 p+O2π(√x). 6 4. Gaps between primes In this section, we explore the gaps between consecutive primes in the sequence Uk. These gaps, defined as the difference ∆k,n, provide insights into the distribution of prime numbers. We will derive formulas for these gaps and analyze their properties. Understanding these gaps is crucial for examining the clustering and density of primes. This exploration aids in advancing conjectures related to prime number behavior. Let’s define the gaps between two elements of Ukas: ∆k,n =uk(n+ 1) −uk(n) =Nkn+ 1 Mk+uk((n+ 1) mod Mk)−Nkn Mk−uk(nmod Mk) =Nkn+ 1 Mk−n Mk+uk((n+ 1) mod Mk)−uk(nmod Mk) To simplify and make it more manageable for the rest of this section, we will now explicitly express the ∆k,n formula. ∆k,n =                uk(1) −uk(0) if nmod Mk= 0 uk(2) −uk(1) if nmod Mk= 1 . . .. . . uk(Mk−1) −uk(Mk−2) if nmod Mk=Mk−2 Nk+ (uk(0) −uk(Mk−1)) if nmod Mk=Mk−1 Hence, we have: Mk−1 X j=0 ∆k,j =Nk On the other hand, we can express Nkand Mkas: Nk= Rk X i=1 2iαk,i αk,i ∈Nand Rk≫kand Mk= N X i=1 αk,i, where αk,i represents the number of times ∆k,n = 2ioccurs when 0 ≤n≤Mk−1, defined as: For i= 1: αk,1=X 1≤n<Nk gcd(n,Nk)=1 gcd(n−2,Nk)=1 1 For i > 1: αk,i =X 1≤n<Nk gcd(n,Nk)=1 gcd(n−2i,Nk)=1 gcd(n−2j,Nk)=1 1≤j<i 1 7 Remark 4.1. As known, the unique triplet (3,5,7) violates the rule that there cannot be a triplet of prime numbers (p, p+2, p+4), since one of the three numbers p,p+ 2, or p+ 4 must be divisible by 3. Therefore, we have αk,1=αk,2. This observation confirms the theory presented in [15, page 205], which discusses the properties of twin primes and prime triplets. Definition 4.2. Define π2n(x) as the counting function for consecutive prime pairs with a difference of 2n, given by π2n(x) = #{pk≤x|pk+1 =pk+ 2n, k ∈N}. The following result provides a probabilistic verification of the conjecture on the existence of twin prime numbers. Theorem 4.3. As k→ ∞, the probability of finding twin primes in the interval {pk≤p < p2 k}is very high. Specifically, the following asymptotic approximation holds: π2(p2 k)−π2(pk)∼ik αk,1 Mk .(4) Furthermore, we have the following expression : αk,1= k−1 Y j=1 (pj−2), k ≥2.(5) Proof. We begin by proving the formula (5) as it will be used to establish the formula (4). Let n=p×m, where p∈P,m∈N, and p∤m. We define the function F(n) as F(n) = X 1≤i<n gcd(i,n)=1 gcd(i−2,n)=1 1. This function will be simplified based on the values of p, leading to the desired result, namely the expression for αk,1. Case 1: p > 2 Express ias i≡r(mod p), where 0 ≤r < p. According to the Chinese Remainder Theorem (see [7]), there is an isomorphism between Z/(p·m)Zand Z/pZ×Z/mZ. Using this, we can rewrite the sum as: F(p·m) = p−1 X r=1 r=2 X 0≤l<m gcd(l,m)=1 gcd(l−2,m)=1 1. Since there are p−2 valid values for r, the expression simplifies to: F(p·m) = (p−2) ·F(m). Case 2: p= 2 Similarly, we have: F(2 ·m) = 1 X r=1 X 0≤l<m gcd(l,m)=1 gcd(l−2,m)=1 1. 8 Since there is only 1 valid value for r, the result is: F(2 ·m) = F(m). Finally, we conclude: F(p×m) = (F(m) if p= 2, (p−2)F(m) if p > 2. By replacing p×mwith the product Nk, we obtain: αk,1=F(Nk) = F k−1 Y j=1 pj!= k−1 Y j=1 (pj−2). We now consider the probability that ∆k,n = 2 is uniform for 0 ≤n < Mk. Let Xbe a random variable representing the length of the sequence [∆k,n 0≤n < X]. The probability P(X=m) with m∈Nrepresents the probability of finding ∆k,n = 2 in the sequence [∆k,n 0≤n < m]. We have: P(X=m) = 1 −1−αk,1 Mkm . We know that the first non-prime number in Ukis p2 kwith an index equal to ik. Thus, P(X=ik) = 1 −1−αk,1 Mkik . Using a logarithmic approximation, this can be expressed as: P(X=ik)=1−eik·ln1−αk,1 Mk. Approximating further: P(X=ik)≈1−e−ik·αk,1 Mk. Substituting αk,1 Mk: P(X=ik)≈1−e−Mkp2 k Nk·αk,1 Mk. Therefore, P(X=ik)≈1−e−Qk−1 j=1 1−2 pj·p2 k 2. By Mertens’ Formula (see [6]), we have: Y p≤x1−1 p=e−γ log(x)1 + O1 log(x), x ≥2. Taking logarithms: log Y p≤x1−1 p!=X p≤x log 1−1 p. 9