A new enlightenment about gaps between primes

# A new enlightenment about gaps between primes

Belhaouari, Samir B.
Department of Computer science, INNOPOLIS University, Kazan, Russia
Department of Mathematics, University of Sharjah, UAE
email: samir.brahim@gmail.com
June 16, 2015
###### Abstract

The idea of generating prime numbers through sequence of sets of co-primes was the starting point of this paper that ends up by proving two conjectures, the existence of infinitely many twin primes and the Goldbach conjecture. The main idea of our approach is summarized on the creation and on the analyzing sequence of sets of distinct co-primes with the first primes, , and the important properties of the modulus linear combination of the co-prime sets, , that gives sets of even numbers . Furthermore, by generalizing our approach, the Polignac conjecture "the existence of infinitely many cousin primes, ," and the statement that "every even integer can be expressed as a difference of two primes," are derived as well.

Subject Classification: 11A41, 11L20 .

Keywords: Prime number, twin number, Goldbach’s conjecture.

## 1 Introduction

The proof of the existence of infinitely many twin primes has been one of the most difficult and the oldest question in number theory for several years. The twin prime conjecture states: There are infinitely many primes such that is also prime. Some attribute the conjecture to the Greek mathematician Euclid of Alexandria, who gave the oldest known proof that there exist an infinite number of primes.
In 1849 de Polignac made the more general conjecture that for every integer , there are infinitely many prime pairs such that . The twin prime conjecture is for .

Since every three consecutive numbers, , are not all co-primes with the number 3, then 2 and 3 are factors of . Subsequently, all twin primes are in the form of except for the first pair and also all cousin primes are in the form of and consecutively. On December 2011, in [2], the largest known twin prime pair, , were found with 200700 digits.

Similar to the distribution of prime numbers among positive integers, the number of twin primes less that naturally declines as increases. This brings up the question of how many pairs of twin primes are there?

Many research works have been devoted to provide a proof to the twin prime conjecture but it is still one of the open problems in mathematics and number theory. Some important result of twin primes and the gap between primes have been obtained. One of these results is Brun’s constant (B). In 1919 Brun, [3], proved that the sum of the reciprocals of the twin primes converge to a definite number called Brun’s constant and has the value of . The recently breakthrough, in 2013, of Yitang Zhang [5], where he proved that there are infinitely many pairs of primes with prime gap less than 70 million

 liminfn→∞(pn+1−pn)<7.107.

This proof shows that the gap between primes does not grow infinitely, and it is bounded by a definite number. This result was significantly improved by several researchers.

In May 2013, mathematicians had uncovered simple tweaks to Zhang’s argument that brought the bound below 60 million. A May 30 blog post by Scott Morrison of the Australian National University in Canberra ignited a firestorm of activity, as mathematicians vied to improve on this number, setting one record after another. By June 4, Terence Tao of the University of California, Los Angeles, a winner of the Fields Medal, mathematics’ highest honor, had created a Polymath project, an open, online collaboration to improve the bound that attracted dozens of participants. For weeks, the project moved forward at a breathless pace. At times, the bound was going down every thirty minutes, Tao recalled. By July 27, the team had succeeded in reducing the proven bound on prime gaps from 70 million to 4,680.

On November 19 by James Maynard [4], a postdoctoral researcher working on his own at the University of Montreal, has upped the ante. Just months after Zhang announced his result, Maynard has presented an independent proof that pushes the gap down to 600. A new Polymath project is in the planning stages, to try to combine the collaboration’s techniques with Maynard’s approach to push this bound even lower, for more details please visit the website WIRED, Nov. 2013.

Strong connection between the twin prime and Goldbach’s Conjectures was found along of our approach to prove the twin prime’s conjecture. The Goldbach conjecture, dating from Goldbach’s correspondence with Euler in 1742, that states: Every even integer greater than 2 is the sum of two prime numbers (not necessarily distinct). It has been verified that the even integers through have the stated property, but Goldbach’s conjecture remains unproved.

Several works on background information about Goldbach’s conjecture have been conducted previously. Namely Dickson [6] lists results obtained through the first decade of this century. Wang, [7], is a collection of this century’s most important papers on the Goldbach conjecture through the early 1980’s. Shanks, [8], is a marvelous book on conjectures in number theory, Goldbach’s among them, and on mathematical conjectures in general. Ribenboim, [9], is another treasure chest in number theory, with a section on Goldbach (pp. 229-235).

## 2 Abbreviation and Notation

Before starting the proof, some important definitions are needed to be introduced as it is indicated below

Let ,…, distinct ordered prime numbers, and for only notation reason we put .

For any two sets and of integer numbers, their sum and their difference is defined as follows

 U±V={x±y:x∈U,y∈V}.

For any set is multiplied by a value is defined as follows

 cU={c.x:x∈U}.

The difference between all elements of the set , is noted by

 ΔU={x−y:(x,y)∈U2}.

The set is set of elements in but not in .

The arithmetic sequence, with reason , will be noted by , where and are the initial and the final values of the sequence respectively, i.e.,

 [r0:r:rf]={r0,r0+r,r0+2r,...,rf}.

If the the reason of the arithmetic sequence is equal to one then it will be noted simply by , i.e.,

 [r0:rf]={r0,r0+1,r0+2,...,rf}.

The function gives the largest integer less than or equal to x.

The modulo operation is computing the remainder after division, Modulo will be noted as , i.e.,

 mmod(n)=m−n⌊mn⌋.

The set contains all co-prime numbers with all devisors of the integer and are less than .

The set contains all co-prime numbers with primes , , and less than .

The set is the set of co-primes with primes , and less then .

Let be the prime-counting function that gives the number of primes less or equal to .

Let be the co-prime-counting function that gives the number of co-primes with , and are less or equal to .

Let be the number of all positive integers less than which do not have a prime divisor which is less than and belongs to .

, unless it is defined differently.

.

The cardinality of a set is noted by

If the asymptomatic ration of two functions verify

 limn→∞∣∣∣f(n)g(n)∣∣∣=1,

then it will be noted by

 f∼g.

If there exist positive and such that

 |f(x)|≤M|g(x)| for x>δ,

then we write

 f(x)=O(g(x)) as x→∞.

The greatest common divisor pf two integers, n and m, will be noted by

 gcd(n,m).

## 3 Sketch of the proof

This section will summarized all the headlines of our approach, from the starting point, through important properties, till reaching the two conjectures. All headlines are briefed as follows

• Any odd number can be written under the form of , where .

• All co-primes numbers with 2 and 3 can be written under the form or , so the two forms can be merged under one form , where and the set .

• All co-primes number with 2, 3, and 5 can be written under the form of , where and .

• Theorem 3.1 generalizes this idea of finding the shape of all co-primes number with , as

 n=k(Πn+1i=1pi)+Vmod(∏n+1i=1Pi),

where the sets are generated by the following recurrence formula

 Vmod(∏n+1i=1Pi)=(Pn+1Vmod(∏ni=1Pi)+(n∏i=1Pi)[1:Pn+1−1])mod(∏n+1i=1Pi).
• We have noticed and proved that for all

 (V∏ni=1pi−V∏ni=1pi)mod(∏ni=1pi)=[0:2:n∏i=1pi−2].
• We have noticed also that

 V∏ni=1pi = {1,pn+1,...,(n∏i=1pi)−pn+1,(n∏i=1pi)−1} minV∏ni=1pi/{1} = pn+1 V∏ni=1pi⋂(1:pn+1) = ∅.
• The set contains only primes numbers.

• The asymptotic density of co-primes are found to be as follows

 π(x,n) ∼ ln(n)xln(x), π(x,n)x ∼ ln(n)ln(x), and card(V∏ni=1pi)∏ni=1pi = Πni=1(pi−1)Πni=1pi ∼ exp(∫n−dnnln(n)) ∼ Cln(n) ∼ Cπ(n)n.
• In order to investigate the difference between only primes number, the primes inside the will be considered as it is indicated in Lemma 3.5, which it gives extremely important properties as follows:

 (ΔV∏n+1i=1pi)mod(∏n+1i=1pi)=(ΔUk∏n+1i=1pi)mod(∏n+1−ki=1pi)+k−1∑j=0(n−j∏i=1pi)[0:pn+1−j−1],

and

 ΔUk,j∏n+1i=1pi⊇(ΔV∏n+1−ki=1pi)mod(∏n+1−ki=1pi)=[0:2:n−k∏i=1pi−2],

where .

• We can choose the integer such that all elements inside of are primes. Since the range of set is large and

 limk→∞minkUk,j∏n+1i=1pi/{1} = +∞ limk→∞maxkUk,j∏n+1i=1pi/{1} = +∞, and ΔUk,j∏n+1i=1pi = [0:2:n−k∏i=1pi−2],

it leads to the infinity existence of twin prime number and to the fact that every even integer greater than 2 can be expressed as the difference of two primes

 liminfn→∞(pn−pn+1)=2.
• Since there is no prime between two primes that defer by 4, we can conclude the existence of infinitely many cousin primes, , i.e.,

 limn→∞∑k>n1{pk−pk+1=4}≠0.
• we can prove that if every even integer greater than 2 can be expressed as the difference of two primes then every even integer greater than 2 can be expressed as the sum of two primes.

• Other methods are highlighted during the proofs.

The initial idea of these above points comes from the way of generating sets of co-prime numbers recursively by adding one prime at each step as it is indicated by the following theorem

###### Theorem 3.1.

Any co-prime number, , with the first prime numbers can be expressed according to the following equation

 n=(m∏i=1Pi)k+l,

where , and a set of co-prime numbers with the first prime numbers, .

The set of co-prime numbers, , can be generated recursively as follows:

 Vmod(∏m+1i=1Pi)=(Pm+1Vmod(∏mi=1Pi)+(m∏i=1Pi)[1:Pm+1−1])mod(∏m+1i=1Pi).

Concluding that any number is a prime number.

Some of important properties of set of co-prime numbers that will be very often used along this paper are highlighted by the following lemma

###### Lemma 3.2.

i. For all co-primes and , we have the following two properties:

 y=(x[1:y−1])mod(y)=[1:y−1]mod(y)
 ∀z1∈[1:y−1],∃!z2∈[1:y−1],z2≠z1, such % that z1=(z2x)mod(y)

ii. If is a co-prime number with , , we have the following properties:

 Vmod(∏mi=1Pi)=(yVmod(∏mi=1Pi))mod(∏mi=1Pi).
 ∀z1∈Vmod(∏mi=1Pi),∃!z2∈Vmod(∏mi=1Pi),z2≠z1, such that z1=(z2x)mod(∏mi=1Pi).

The main ideas of our proofs are on creating and on analysing of set sequence of distinct co-primes with the first primes, , by using Theorem 3.1 and the important properties of the linear combination of the co-prime sets, , that give sets of even numbers . The next three lemmas will explain our approach gradually.

First of all, some properties of the difference between sets of co-prime numbers will be pointed out and a recursive formula between these sets will be expressed as well.

###### Lemma 3.3.

For all and for all , we have the following equalities

1. The difference co-primes of the sets generate compact subsets of even numbers

 (ΔV∏ni=1pi)mod(∏ni=1pi) = (V∏ni=1pi−V∏ni=1pi)mod(∏ni=1pi) = [0:2:n∏i=1pi−2].
2. ,
where the set is the set of co-primes number with , for

The recursive relation between sets can de deducted, for , as follows

 wk+1=wkpn−k+1+(n+1∏i=n−k+2pi)[1:pn−k+1−1].

As initial investigation inside the set , a new question is raised about finding subsets that verified the first part of lemma 3.3, the next lemma gives interesting result

###### Lemma 3.4.

For all bigger than 1, let’s define the following sets

 U∏n+1i=1pi={x∈V∏n+1i=1pi:x∈[0,4n∏i=1pi]},

then

 (ΔU∏n+1i=1pi)mod(∏n+1i=1pi)⊇(ΔV∏ni=1pi)mod(∏ni=1pi),

and

 (ΔU∏n+1i=1pi)⊇[0:2:(n∏i=1pi)−2]

In order to explain the idea on how to prove that the difference or summation between all prime numbers is equals to the set of all even integers, the following lemma, a generalization and improved version of lemma 3.4, will help us to limit the difference only between primes numbers.

###### Lemma 3.5.

For all integer , it exists , such that for all integer , the difference between co-primes numbers, , can be expressed recursively in function of difference between subsets of as it is indicated below

 (ΔV∏n+1i=1pi)mod(∏n+1i=1pi)=(ΔUk,j∏n+1i=1pi)mod(∏n+1−ki=1pi)+k−1∑j=0(n−j∏i=1pi)[0:pn+1−j−1],

where

To extend this result and omit the modulo from the right side of the previous equation i.e., from the set , we enlarge the definition of the set as

 Uk,j∏n+1i=1pi=V∏n+1i=1pi⋂[jn+1−k∏i=1pi:(j+2)n+1−k∏i=1pi],

then it leads, under the condition that to

• For all integers , we have

 ΔUk,j∏n+1i=1pi⊇(ΔV∏n+1−ki=1pi)mod(∏n+1−ki=1pi)=[0:2:(n+1−k∏i=1pi)−2].

The threshold will be the smallest integer verified the condition, i.e.,

 k0=min{k:Πk−2i=1pi>pn+2}.

An important result will be needed in order to prove the lemma 3.5, evaluation the density of co-primes with , and are less or equal to , noted . RiChard Warlimont, [10], and D. A. Golden, [11], have worked on the estimation of co-primes density under certain conditions, our following lemma will express an asymptotic density of co-primes with consecutive primes.

###### Lemma 3.6.

Let be a set of primes and we denote by the number of all positive integers less than which do not have a prime divisor which is less than and belongs to .
Put , , and let .

Then we have the following statements

• For any positive integers we have

 Ψ(x,y,Q)=Card{i∈(y+kPQ,x+kPQ]|gcd(i,PQ)=1}.
• For any positive integers we have

 Ψ(x,y,Q)=Ψ(x+kPQ,y,Q)−Ψ(y+kPQ,y,Q)+y,

and if and then

 Ψ(x,y,Q)+Ψ(PQ−x,y,Q)=Ψ(PQ,y,Q)+1=PQ∏pi∈Q(1−1pi)+1.
• Uniformly in we have

 Ψ(x,y,Q)=(x−y)∏pi∈Q(1−1pi)(1+O(R(y)))+y.
• If and are big enough and then

 Ψ(x,y,Q)∼(x−y)ln(y)ln(x)+y,

and if then

 Ψ(x,y,Q)∼π(x,y/ln(y))∼xln(y)ln(x).

The coming sections will be devoted to prove the Theorem 3.1 and the Lemma 3.2 at first step of our approach as an introduction to the proof of the lemma 3.3, 3.4, and 3.5 respectively.

Finally the proof of the two conjectures with their extension will be at the end of this paper.

## 4 Proof of Theorem 3.1

Let and be two positive integers. If is not divisible by , then needs to be a multiple of added to another positive integer , where is smaller than ; under the condition that and are co-primes, the positive integer is co-prime with . This idea leads to the generation of a large prime number from a sequence of previous prime numbers. This idea can be written in mathematical form as follows:

 n=ak+b,where b

if and are co–primes than and are also co-prime numbers.

We will generalize this idea to generate co-prime numbers recursively by including one prime at each step.

Let be a set of co-primes with the first prime numbers, so can be expressed as:

 ω(m)=(m∏i=1pi).k+Vmod(∏mi=1pi),

A recursive question will be raised in how to find the set from the in order to extract all elements of . To do that, let suppose is co-prime also with and by definition it is co-prime with , so it exists two integers, and such that

 {ω(m+1)=(∏mi=1pi).k+Vmod(∏mi=1pi)ω(m+1)=(pm+1).k′+[1:(pm+1−1)], (1)

Let suppose the integer as a co-prime with all , , and is co-prime with , by using Lemma 3.2 the previous system of equations can be formulated as follows:

 {ω(m+1)=(∏mi=1pi).k+Vmod(∏mi=1pi)x1ω(m+1)=(pm+1).k′+[1:(pm+1−1)]x2, (2)

Without losing generality let and so can be expressed by

 (3)

The set of co-primes can be expressed as given below:

 (4)

Let and .

To calculate the equation 4 needs to be solved, by subtraction their two equations we can find

 m∏i=1pi.(k−l1)=(pm+1).(k′−l2),

where and can be calculated as:

 {(k−l1)=ϑ.pm+1(k′−l2)=ϑ.(∏mi=1pi),

where is any value inside .

By isolating the two variables and , we have:

 {k=ϑ.pm+1+l1.k′=ϑ.(∏mi=1pi)+l2 ,

Substituting the value of in system (3), we have:

 ⎧⎪⎨⎪⎩ω(m+1)=ϑ(∏m+1i=1pi)+l1(∏mi=1pi)+Vmod(∏mi=1pi)ω(m+1)=ϑ(∏m+1i=1pi)+l2(pm+1)+[1:(pm+1−1)],

By substitution the values of and in the above equations we conclude

 ⎧⎪⎨⎪⎩ω(m+1)=ϑ.(∏m+1i=1pi)+([1:(pm+1−1)].(∏mi=1pi))+(pm+1).Vmod(∏mi=1pi)\parω(m+1)=ϑ.(∏m+1i=1pi)+([1:(pm+1−1)].(∏mi=1pi))+(pm+1).Vmod(∏mi=1pi),

where both equations are equivalent due to the similarity of property of the modulo of the last term.

The general formula of that are co-primes with the first prime numbers will be as follows:

 ω(m+1)=ϑ.(m+1∏i=1pi)+([1:(pm+1−1)].(m∏i=1pi))+(pm+1).Vmod(∏mi=1pi),

We can general also the set of co-primes with the first prime numbers in order to get the following:

 ω(m+1)=ϑ.(m+1∏i=1pi)+Vmod(∏m+1i=1pi) ,

where

## 5 Proof of Lemma 3.2

### 5.1 Proof of Lemma 3.2 part (i)

We will start proving the first part of the Lemma 3.2 with following equality:

 (m∏i=1pi).[1:(pm+1−1)]mod(pm+1)=[1:(pm+1−1)]mod(pm+1) .

Since is co-prime with , there exists , where can be written as:

, where .

Then it implies that

 (k.pm+1+r)[1,2,3,........(pm+1−1)]pm+1=[r,2r,3r,........(pm+1−1).r]pm+1.

By contradiction we can prove that

 [ir]mod(pm+1)≠0 (5)

If , then it exists such that which it means that or are multiple of , where it contradicts of the fact that and are both less than .

To finish the proof of Lemma 3.2 part (i) and first part of (ii), we still need to show by contraction the following:

 [ir]mod(pm+1)=[jr]mod(pm+1),  ∀ i≠j∈{1,2...,pm+1−1} (6)

Let’s suppose that it exists two integers and such that:

 {ir=k′.pm+1+r0,  k′∈Zjr=k′′.pm+1+r0,  k′′∈Z .

By subtracting the two equation, we have that

 (i−j)r=(k′−k′′).pm+1,

which it contradicts with the equation 5 due to the facts that

 r ∈ [1:pm+1−1] (i−j)mod(pm+1) ≤ pm+1−1

From the equations 5 and 6, The following equation can be deducted easily

 [r 2r 3r ........(pm+1−1)r]mod(pm+1)=[1:(pm+1−1)]mod(pm+1),

then it implies that

 (m∏i=1pi).[1:(pm+1−1)]mod(pm+1)=[1:(pm+1−1)]mod(pm+1).

Since for all all two co-primes, and , we can write , where , then we can Generalize our result as follows

 (k1[1:(k2−1)])mod(k2)=[1:(k2−1)]mod(k2).

### 5.2 Proof of Lemma 3.2 part (ii)

Here we will prove the second part of Lemma 3.2

 pm+1.Vmod(∏mi=1pi)=Vmod(∏mi=1pi),

For all ; the cadinality of the set can be calculated as follows

 Card(Vmod(∏mi=1pi))=m∏i=2(pi−1).

Let , so

 Vmod(∏mi=1pi)={r1,r2,r3,......,rM}.

Now, we will test first the non existing of such that:

 ri:ripm+1=k(m∏i=1pi),

if the last equation is true, then for all we have

 ripj∈N,

which it implies that

 ri=k′(m∏i=1pi),

this statement contradict with the fact that , i.e.,

 (ri)mod(