Marko V Jankovic

Proof of Polignac's conjecture for gap equal to eight

In this paper proof of the Polignac's Conjecture for gap equal to eight is going to be presented. It will be shown that consecutive primes with gap eight could be obtained through two stage sieve process, and that will be used to prove that infinitely many primes with gap eight exist. The proof represents an simple extension of the recently presented proofs that infinitely many twin, cousin and sexy primes exist. The major contribution of this paper is presentation of all elementary modules that are necessary for the proof of Polgnac's conjecture in general case.

Here it is going to be proved that m in (2) is represented by threads defined by odd prime numbers. Now, the form of (2) for some values of i will be checked. Case i = 1: m = 3j + 1, Case i = 2: m = 5j + 2, Case i = 3: m = 7j + 3, Case i = 4: m = 9j + 4 = 3(3j + 1) + 1, Case i = 5: m = 11j + 5, Case i = 6: m = 13j + 6 , Case i = 7: m = 15y + 7 = 5(3j + 1) + 2, Case i = 8: m = 17j + 8,

Introduction

In number theory, Polignac's conjecture states: For any positive even number g, there are infinitely many prime gaps of size g. In other words: there are infinitely many cases of two consecutive prime numbers with the difference g [START_REF] De Polignac | Recherches nouvelles sur les nombres premiers[END_REF].

In [START_REF] Polymath | Variants of the Selberg sieve and bounded intervals containing many primes[END_REF] it has been shown that exists infinitely many consecutive prime numbers that have gaps that are not bigger than 246. Recently, the Polignac's conjecture was proved for gaps of the size 2 and 4 [START_REF] Jankovic | A Proof of the Twin Prime Conjecture (together with the Proof of Polignac's Conjecture for Cousin Primes)[END_REF], as well as for the gap equal to six [START_REF] Jankovic | A Proof of Polignac's Conjecture for gap equal to six[END_REF]. The problem was addressed in generative space, which means that prime numbers were not analyzed directly, but rather their representatives that can be used to produce them. This paper represents a simple extension of the previous work [START_REF] Jankovic | A Proof of Polignac's Conjecture for gap equal to six[END_REF]. Here, the gap equal to eight is going to be analyzed.

Basically, three groups of gaps bigger than 4 exist: the gaps of the size 6k, the gaps of the size 6k+2 and gaps of the size 6k+4, k ϵ N. In the text that follows we mark the prime numbers in the form 6k -1 as mps primes and prime numbers in the form 6k +1 as mpl primes, k ϵ N. The gaps of the size 6k could be related to the prime pairs in both (mps, mps) and (mpl, mpl) form. The gaps in the form 6k + 2 can only be related to the pairs of primes in (mps, mpl) form, while gaps in the form 6k + 4 can only be related to the pairs of primes in (mpl, mps) form. In other words there is not a single prime in mpl form that has consecutive prime that is 6k + 2 apart, and there is not a single prime in mps form that has consecutive prime 6k + 4 apart. It is trivial to show that by simple calculation.

Here it will be shown that exist an infinite number of primes with gaps of the size 8 (we mark them as g8-primes). It will be shown that primes with gap of the size 8 could be generated by two stage recursive type sieve process. This process will be compared to the other two stage recursion sieve process that leaves infinitely many numbers. Fact that sieve process that generate g8-primes leaves more numbers than the other sieve process will be used to prove that infinitely many consecutive primes with gap 8, exist.

Remark 1: In this paper any infinite series in the form c 1 •l ± c 2 is going to be called a thread defined by number c 1 (in literature these forms are known as linear factors -however, it seems that the term thread is probably better choice in this context). Here c 1 and c 2 are numbers that belong to the set of natural numbers (c 2 can also be zero and usually is smaller than c 1 ) and l represents an infinite series of consecutive natural numbers in the form (1, 2, 3, …).

Remark 2:

In the text that follows we will use fact that number of even/odd numbers is equal to the one half of the number of natural numbers (also the 1/3, 4/5 and other fractions of the number of natural numbers are used). This is not an usual way of discussing the number of numbers that are infinitely big, but it is quite suitable in this context. It is important to notice that there is no context in which is correct to state that number of natural numbers is equal to the number of odd or even numbers (or that number of natural numbers is equal to the number of numbers divisible by 3, or 5, and so on). What can be said is that it is possible to generate the same number of unique labels for odd or even numbers using the same number of unique labels for natural numbers. However, if we want to produce all even and odd labels at the same time, obviously, it is necessary to have two sets of natural numbers at the same time (it is necessary to clone the set of natural numbers), which means that the number of the natural numbers in that moment is two times bigger than the number of even or odd numbers. This analysis can also be made in quantum probabilistic context, but it is beyond the scope of this paper.

2 Proof of Polignac's conjecture for gap equal to eight

As it was already explained in the introduction part, if two consecutive prime numbers have the gap of the size 8 it is clear that smaller of those numbers has to be in mps form while the bigger one has to be in mpl form. Here, we are going to present a two stage process that can be used for generation of the smaller of the g8-prime pair that is in mps form.

In the first stage prime numbers are going to be produced by removal of all composite numbers from the set of natural numbers.

In the second stage, we are going to remove the following numbers:

-Number 2 (since it cannot generate g8-prime pair).

-All numbers in mps form that represent a smaller numbers in twin primes pairs.

-A quarter of sexy primes (or more precisely one half of mps sexy primes). In order to understand why is it so, we are going to consider 5 consecutive numbers 6k-1, 6k+1, 6k+3, 6(k+1)-1, 6(k+1)+1. We can see that 6k-1 and 6(k+1)+1 can create a g8-primes only in the case when they are both primes and when 6k+1 and 6(k+1)-1 are not primes (6k+3 is obviously composite number divisible by 3). Here, we are going to analyze three situations. When 6k+1 and 6(k+1)-1 are both primes, and in the case when only 6k+1 is prime, removal of twin primes will remove mps primes that cannot have g8-prime pair. However, when 6k+1 is not prime and 6(k+1)-1 is prime, it is also necessary to remove smaller number of sexy prime pair (6k-1, 6(k+1)-1) since it also cannot create a g8-prime pair.

-A half of the number of twin primes and quarter of the number of sexy primes from mpl primes is going to be removed in such a way that Dirichlet's theorem for arithmetic progressions [START_REF] Dirichlet | Bewies des Satzes, dass jede unbregenzte arithmetische Progression, deren erstes Glied und Differenz ganze Zahlen ohne gemeinschaftluchen Factor sind, unendlich viele Primzahlen enthält[END_REF] still holds after removal of all those numbers that are previously mentioned.

-All prime numbers in mpl form that are left.

-All prime numbers in mps form that have a bigger odd g8-neighbor (odd number that is by 8 bigger than the prime of interest) that is a composite number.

At the end, only the prime numbers in the mps form, that represent the smaller number of a g8prime pair, are going to stay. Their number is a half of the number of g8-primes. It is going to be shown that that number is infinite.

STAGE 1

Prime numbers can be obtained in the following way:

First, we remove all even numbers (except 2) from the set of natural numbers. Then, it is necessary to remove the composite odd numbers from the rest of the numbers. In order to do that, the formula for the composite odd numbers is going to be analyzed. It is well known that odd numbers bigger than 1, here denoted by a, can be represented by the following formula

a = 2n +1,
where n ϵ N. It is not difficult to prove that all composite odd numbers a c can be represented by the following formula

a c =2 (2 i j+i+ j)+1=2((2 j+1)i+ j )+1. (1)
where i, j ϵ N. It is simple to conclude that all odd composite numbers could be represented by product (2i + 1)(2j + 1), where i, j ϵ N. If it is checked how that formula looks like for the odd numbers, after simple calculation, equation ( 1) is obtained. This calculation is presented here. The form 2m + 1, m ϵ N will represent odd numbers that are composite. Then the following equation

holds 2 m+1=(2 i+1)(2 j+1) .
From that, it follows that m must be in the form

m = 2ij + i + j = (2i + 1) j + i. (2)
When all numbers represented by m are removed from the set of odd natural numbers bigger than 1, only the numbers that represent odd prime numbers are going to stay. In other words, only odd numbers that cannot be represented by (1) will stay. This process is equivalent to the sieve of Sundaram [START_REF] Aiyar | Sundaram's Sieve for prime numbers[END_REF].

Let us denote the numbers used for the generation of odd prime numbers with m2 (here we ignore number 2). Those are the numbers that are left after the implementation of Sundaram sieve. The number of those numbers that are smaller than some natural number n, is equivalent to the number of prime numbers smaller than n. If we denote with π(n) number of primes smaller than n, than the following equation holds [START_REF] Rosser | Approximate formulas for some functions of prime numbers[END_REF] π( n)≈ n ln (n)

.

From [START_REF] Rosser | Approximate formulas for some functions of prime numbers[END_REF] we also know that following holds

π( n)> n ln( n) , n⩾ 17. ( 3
)

STAGE 2

What was left after the first stage are prime numbers. With the exception of number 2, all other prime numbers are odd numbers. All odd primes can be expressed in the form 2n + 1, n ϵ N. It is simple to understand that their bigger odd g8-neighbor must be in the form 2n + 9, n ϵ N. Now, we should implement a second stage in which we are going to remove:

A. Number 2 (since 2 cannot make a g8-prime pair);

B. All twin primes in mps form, a quarter of sexy primes (smaller sexy numbers in mps form). Also, a number of primes in mpl form that is equal to the number that represents the sum of the half of the numbers of twin primes and quarter of the numbers of sexy primes, in such a way that Dirichlet's theorem for arithmetic progressions [START_REF] Dirichlet | Bewies des Satzes, dass jede unbregenzte arithmetische Progression, deren erstes Glied und Differenz ganze Zahlen ohne gemeinschaftluchen Factor sind, unendlich viele Primzahlen enthält[END_REF] still holds after removal of all those numbers that are mentioned-so number of numbers that is going to be left is the number of primes minus the number of twin primes minus one half of the number of sexy primes (number 2 is ignored, and that has no impact on the analysis that follows). It is not difficult to prove that number of numbers left, is infinite. It is not difficult to understand it having in mind that relative density of twin and sexy primes to the density of all primes is zero.

C. The rest of mpl primes -it is trivial to see that it can be done by one thread that is defined by 3so in this step it is going to be removed, one half of the numbers that are left after step B; D. All odd primes in the form 2m + 1 such that 2m + 9, m ϵ N represents a composite number (all primes whose bigger odd g8-neighbor is composite number). If we make the same analysis, like in the Stage 1, it is simple to understand that m must be in the form

m = 2ij + i + j -4 = (2i + 1) j + i -4. ( 4
)
All numbers (in observational space) that are going to stay must be numbers in mps form and they represent a smaller primes of the g8-prime pairs that is in mps form. What has to be noticed is that thread in (4) that is defined by prime number 3 (for i = 1) is not going to remove any number from the numbers left.

Let us mark the number of g8-primes with π G8 . Also, let us define the number of numbers that is left after two consecutive implementations of Sundaram sieve as pd8. The prime numbers that are obtained after removal of number 2, twin primes in mps form and a quarter of sexy primes (a half of sexy primes in mps form) and the corresponding number of mpl primes (equal to the sum of a half of the number of twin primes and a quarter of the number of the sexy primes) are going to be called g8-fs primes. The numbers obtained after recursive implementation of two Sundaram sieves (where the second Sundaram sieve is implemented on g8-fs primes) are going to be called g8double primes. The second stage sieve that is identical to the first stage sieve can be obtained if the g8-fs primes are lined up next to each other and then the numbers are removed from the exactly same positions like in the first stage -the second stage sieve is applied on the indexes of g8-fs primes. In that case it is easy to understand that the following equation would holds (n ϵ N)

pd8 ( n)≈ π (n)-π G2 (n)-0.5 π G6 (n) ln ( π( n)-π G2 ( n)-0.5 π G6 (n) ) , (5)
where pd8(n) represents the number of g8-double primes smaller than some natural number n (this number can be easily found out (by simple counting), once it is known which mpl numbers are removed in the process of obtaining the g8-fs primes). Here, it will be shown that number of g8double primes, marked as pd8, is smaller than the number π G8 /2. In order to understand why it is so, we are going to analyze ( 2) and ( 4) in more detail. (This is equivalent to the procedure already used in [3] [4].)

It is not difficult to be seen that m in ( 2) and ( 4) is represented by the threads that are defined by odd prime numbers. For details see Appendix A. Now we are going to compare stages 2 for generation of g8-double primes and mps g8-primes, step by step, for a few initial steps (analysis can be easily extended to any number of steps). Starting point for the second stage is the moment when the primes g8-fs are generated. Remove numbers defined by thread defined by 11 (obtained for i = 5) amount of numbers left is 9/10 of the numbers that are left after previous step It should be kept in mind that in the case of g8-double primes sieve is implemented on indexes of g8-fs prime numbers (indexes are consecutive numbers 1, 2, 3, … ).

Values of the fractions presented in the Table 1 are asymptotically correct (in the finite case those values are only approximately correct -for details see [START_REF] Jankovic | A Proof of the Twin Prime Conjecture (together with the Proof of Polignac's Conjecture for Cousin Primes)[END_REF] ). Here is important to understand that it is assumed, without proof, that Dirichlet's theorem on arithmetic progressions [START_REF] Dirichlet | Bewies des Satzes, dass jede unbregenzte arithmetische Progression, deren erstes Glied und Differenz ganze Zahlen ohne gemeinschaftluchen Factor sind, unendlich viele Primzahlen enthält[END_REF] holds on the subset of primes marked as g8-fs primes (in other words, after removal of twin primes in mps form and after removal of the quarter of the sexy primes (half of the sexy primes in mps form) from prime numbers, it is assumed that it is possible to select the same number of mpl primes that are going to be removed in such a way that Dirichlet's theorem on arithmetic progressions [START_REF] Dirichlet | Bewies des Satzes, dass jede unbregenzte arithmetische Progression, deren erstes Glied und Differenz ganze Zahlen ohne gemeinschaftluchen Factor sind, unendlich viele Primzahlen enthält[END_REF] still holds). For the arithmetic progressions that are defined by finite numbers this is a simple task, but it requires additional attention when we are speaking about progressions defined by infinite numbers.

From the table, it can be noticed that threads defined by the same number in first and second column will not remove the same percentage of numbers. The reason is obvious -consider for instance the thread defined by 3: in the first column it will remove 1/3 of the numbers left, but in the second column it will remove ½ of the numbers left, since the thread defined by 3 in stage 1 has already removed one third of the numbers (odd numbers divisible by 3 in observation space). So, only odd numbers (in observational space) that give residual 1 and -1 when they are divided by 3 are left, and there is asymptotically the same number of numbers that give residual -1 and numbers that give residual 1, when the number is divided by 3 (see [START_REF] Jankovic | A Proof of the Twin Prime Conjecture (together with the Proof of Polignac's Conjecture for Cousin Primes)[END_REF]Appendix C]). Same way of reasoning can be applied for all other threads defined by the same prime in different columns.

From Table 1 can be seen that in every step, except step 1, threads in the second column will leave bigger percentage of numbers than the corresponding threads in the first coulmn. This could be easily understood from the analysis that follows:

suppose that we have two natural numbers j, k such that j -1 ≥ k (j, k ϵ N ), then the following set of equations is trivially true

j+k-1⩾ 2k -j-k +1⩽ -2k jk-j-k +1⩽ jk-2k ( j -1)(k -1)⩽ ( j-2) k k-1 k ⩽ j-2 j-1
The equality sign holds only in the case j = k + 1. In the set of prime numbers there is only one case when j = k + 1 and that is in the case of primes of 2 and 3. In all other cases p(i) -p(i -1) > 1 , (i > 1, i ϵ N, p(i) is i-th prime number). So, in all cases i > 2

p(i-1)-1 p (i-1) < p (i)-2 p (i )-1 .
From Table 1 (or last equation) we can see that bigger number of numbers is left in every step of stage 2 then in the stage 1 (except 1 st step). From that, we can conclude that after every step bigger than 1, part of the numbers that is left in stage 2 is bigger than number of numbers left in the stage 1

(that is also noticeable if we consider amount of numbers left after removal of all numbers generated by threads that are defined by all prime numbers smaller than some natural number).

From previous analysis we can safely conclude that the following equation holds

π G8 > π G8 2 > pd8= lim n → ∞ pd8(n) .
Having in mind (3) and ( 5), we can say that for some n big enough the following inequality holds

pd8 ( n)> π( n)-π G2 ( n)-0.5 π G6 (n) ln ( π (n)-π G2 (n)-0.5 π G6 ( n) ) . ( 6)
(The n for which equations holds can be specified once the numbers in mpl form that are going to be eliminated are known -for instance it can be said that n ≥ 317 can be chosen, since 317 is the mpl twin primes, and a quarter of the sexy primes are eliminated from the prime numbers set, and if we assume that all corresponding mpl numbers (that "balance" a quarter of sexy primes that are numbers in mps form) that are going to be removed are all bigger than 317). It is not difficult to prove that the number of primes that are left when 2 and twin primes in mps form and quarter of sexy primes, and corresponding number of mpl primes are eliminated, is infinite. One way to understand it, is the fact that the relative density of twin primes and all primes is zero as well as the relative density of sexy primes and all primes is zero.

Having that in mind it it easy to show that following holds

lim n → ∞ π(n)-π G2 ( n)-0.5 π G6 ( n) ln ( π( n)-π G2 (n)-0.5 π G6 ( n) ) =∞ .
Then, the following equation holds

pd8= lim n → ∞ pd8 ( n)=∞ .
Now, we can safely conclude that the number of g8-primes is infinite. That concludes the proof.

Here we will state the following conjecture: for n big enough, number of g8-primes is given by the following equation

π G8 (n) ∼ 4C 2 ⋅ ( ( π (n) -π G2 (n)-0.5 π G6 ( n) ) ln ( π( n) -π G2 ( n)-0.5 π G6 (n) ) ) .
Constant C 2 s known as twin prime constant [START_REF] Hardy | Some Problems of 'Partitio Numerorum.' III. On the Expression of a Number as a Sum of Primes[END_REF]. The equation can be expressed by using only n, but it would be cumbersome. Why it is reasonable to make such conjecture can be understood form the procedure that is similar to one used in [START_REF] Jankovic | A Proof of the Twin Prime Conjecture (together with the Proof of Polignac's Conjecture for Cousin Primes)[END_REF]Appendix B]. If we mark the number of primes smaller than some natural number n with π(n) ≈ f (n), where function f (n) gives good estimation of the number of primes smaller than n, than π G8 (n), for n big enough, is given by the following equation

π G8 ( n) ∼ 4C 2 ⋅ ( f ( f (n) -π G2 ( n)-0.5 π G6 ( n) )) .
Here we can see that constant C 2 has a misleading name. It is connected with repeated implementation of a sieve that produces prime numbers which is also, but not exclusively, connected to the twin primes.

It can be seen that m is represented by the threads that are defined by odd prime numbers. From examples (cases i = 4, i = 7), it can be seen that if (2i + 1) represent a composite number, m that is represented by thread defined by that number also has a representation by the the thread defined by one of the prime factors of that composite number. That can be proved easily in the general case, by direct calculation, using representations similar to [START_REF] Polymath | Variants of the Selberg sieve and bounded intervals containing many primes[END_REF]. Here, that is going to be analyzed. Assume that 2i + 1 is a composite number, the following holds It is easy to understand that if the (2l+1) is a composite number, the proposed method has to be implemented again and again until the prime factor is achieved.)

In the same way this can be proved for (4).

4

 4

2i + 1 =

 1 (2l + 1)(2s + 1) where (l, s ϵ N). That leads to i = 2ls + l + s. The simple calculation leads to m = (2l +1) (2s + 1) j + 2ls + l + s = (2l + 1)(2s+1)j + s(2l + 1) + l or m = (2l+1)((2s+1)j + s) + l which means m = (2l + 1)f + l, and that represents the already exiting form of the representation of m for the factor (2l + 1), where f = (2s + 1)j + s.

Table 1

 1 Comparison of the stage 2 for generation of g8-double primes and mps g8-primes -

	threads defined by a few smallest primes
	Step	Stage 2 -g8-double primes	Step	Stage 2 -mps g8-primes
	1	Remove even numbers (except 2)	1	Remove numbers defined by thread
				defined by 3 (obtained for i = 1)
		amount of numbers left is 1/2		amount of numbers left is 1/2
	2	Remove numbers defined by thread	2	Remove numbers defined by thread
		defined by 3 (obtained for i = 1)		defined by 5 (obtained for i = 2)
		amount of numbers left is 2/3 of the		amount of numbers left is 3/4 of the
		numbers that are left after previous step		numbers that are left after previous step
	3	Remove numbers defined by thread	3	Remove numbers defined by thread
		defined by 5 (obtained for i = 2)		defined by 7 (obtained for i = 3)
		amount of numbers left is 4/5 of the		amount of numbers left is 5/6 of the
		numbers that are left after previous step		numbers that are left after previous step
	4	Remove numbers defined by thread
		defined by 7 (obtained for i = 3)
		amount of numbers left is 6/7 of the
		numbers that are left after previous step