Proof of the Twin Prime Conjecture (Together with the Proof of Polignac’S Conjecture for Cousin Primes) Marko Jankovic

Proof of the Twin Prime Conjecture (Together with the Proof of Polignac’s Conjecture for Cousin Primes) Marko Jankovic

To cite this version:

Marko Jankovic. Proof of the Twin Prime Conjecture (Together with the Proof of Polignac’s Conjec- ture for Cousin Primes). 2020. hal-02549967v2

HAL Id: hal-02549967 https://hal.archives-ouvertes.fr/hal-02549967v2 Preprint submitted on 27 May 2020 (v2), last revised 18 Aug 2021 (v12)

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Marko V. Jankovic

Department of Emergency Medicine, Bern University Hospital “Inselspital”, and ARTORG Centre for Biomedical Engineering Research, University of Bern, Switzerland

Abstract. In this paper proof of the twin prime conjecture is going to be presented. In order to do that, the basic formula for prime numbers was analyzed with the intention of finding out when this formula would produce a twin prime and when not. It will be shown that the number of twin primes is infinite. Originally very difficult problem (in observational space) has been transformed into a simpler one (in generative space) that can be solved. The same approach is used to prove the

Polignac’s conjecture for cousin crimes.

1 Introduction

In number theory, Polignac’s conjecture states: For any positive even number n, there are infinitely many prime gaps of size n. In other words: There are infinitely many cases of two consecutive prime numbers with the difference n [1]. For n = 2 it is known as twin prime conjecture.

Conditioned on the truth of the generalized Elliott-Halberstam Conjecture [2], in [3] it has been shown that there are infinitely many primes’ gaps that have value of at least 6. In this paper gap 2 is analyzed. Gap 4 can be analyzed in a very similar manner. The problem is

addressed in generative space, which means that prime numbers are not going to be analyzed directly, but rather their representatives that can be used to produce them.

Remark 1: Prime numbers 2 and 3 are in a sense special primes, since they do not share some of the common features of all other prime numbers. For instance, every prime number, apart from 2 and 3, can be expressed in the form 6l + 1 or 6l - 1, where l ϵ N. So, in this paper most of the time, prime numbers bigger than 3 are analyzed.

Remark 2: In this paper any infinite number series in the form c1 * l ± c2 is going to be called a thread, defined by number c1. Here c1 and c2 are constants that belong to the set of natural numbers (c2 can also be zero and usually is smaller than c1) and l represents an infinite series of consecutive natural numbers in the form (1, 2, 3, …).

2 Proof of the twin prime conjecture

It is well known that every two consecutive odd numbers (psn, pln), between two consecutive odd numbers divisible by 3 (e.g. 9 11 13 15, or 39 41 43 45), can be expressed as

푝푠 =6푘 − 1

푝푙 =6푘 + 1, 푘 휖푁.

Twin prime numbers are obtained in the case when both psn, and pln are prime numbers. If any of the psn or pln (or both) are a composite number, then we cannot have twin primes. So, the strategy is to check in which cases (for which k) it would not be possible to have twin primes.

We denote any composite number (that is represented as a product of prime numbers bigger than 3) with CPN5. A number in the form 6l + 1 is marked with mpl, while a number in the form 6s - 1 is marked with mps (l, s ϵ N). That means that any composite number CPN5 can be expressed in the form mpl x mpl, mps x mps or mpl x mps.

If psn represents a composite number, the following equation must hold

퐶푃푁5+1 푘 = 6

. (1)

Since CPN5 should be in the mps form, CPN5 can be generally expressed as a product mpl x mps, or

mpl = 6x + 1 and mps = 6y - 1(x, y ϵ N), which leads to

CPN5 = mpl x mps = 6(6xy - x + y) – 1, (2) or, due to symmetry

mpl = 6y + 1 and mps = 6x - 1, which leads to

CPN5 = mpl x mps = 6(6xy + x - y) – 1. (3)

If (2) or (3) is replaced in (1) forms of k that cannot produce the twin primes will be obtained. Those forms are expressed by the following equations

푘 = (6푥 −1)푦 + 푥 (4a)

푘 = (6푥 +1)푦 − 푥, (4b)

where x, y ϵ N. These equations are equivalent (they will produce the same numbers) and can be used interchangeably.

Using similar procedure, we can see that pln, that represents CPN5 number in the mpl form, will correspond to the composite number in the case

퐶푃푁5−1 푘 = . 6

(5)

In this case, CPN5 can be expressed in the form mpl1 x mpl2, or as mps1 x mps2. Two possibilities exist (x, y ϵ N):

mpl 1= 6x + 1 and mpl2 = 6y + 1, which leads to

CPN5 = mpl1 x mpl2 = 6(6xy +x + y) + 1, (6) or,

mps1 = 6x - 1 and mps2 = 6y - 1, which leads to

CPN5 = mps1 x mps2 = 6(6xy - x - y) + 1. (7)

When (6) and (7) are replaced in (5), together with (4a) the forms for all k that cannot produce a twin prime pair, are obtained. Those forms are expressed by the following equation

(6푥 −1)푦 + 푥 푘 = (6푥 −1)푦 − 푥, (8) (6푥 +1)푦 + 푥

where x, y ϵ N. This is a sufficient and necessary condition for k, so that it cannot be used for the generation of twin primes. In other words, at least one of the twin odds generated by a k in any of the forms (8) will be a composite number, and if any of the odds generated by a k is a composite number, then k must be in one of the forms (8). Alternatively, it is possible to use the equation (4b) instead of (4a). In that case a different set of equations that produce the same numbers as the equation (8), is obtained. Here, a list of the k s(first 7) that cannot be presented in the form (8) and that generate all twin primes bigger than 3 and smaller than 100, is presented.

k 1 2 3 5 7 10 12

Twin prime 1 5 11 17 29 41 59 71

Twin prime 2 7 13 19 31 43 61 73

In order to prove that there are infinitely many twin prime pairs we need to prove that infinitely many natural numbers that cannot be expressed in the form (8) exists. First, the form of (8) for some values of x will be checked.

Case x=1: k = 5y - 1, k = 5y + 1, k = 7y + 1,

Case x=2: k = 11y - 2, k = 11y + 2, k = 13y + 2,

Case x=3: k = 17y - 3, k = 17y + 3, k = 19y + 3,

Case x=4: k = 23y - 4, k = 23y + 4, k = 25y + 4 = 5(5y +1) – 1,

Case x=5: k = 29y - 5, k = 29y + 5, k = 31y + 5,

Case x=6: k = 35y – 6 = 7(5y – 1) + 1, k = 35y + 6 = 5(7y + 1) +1, k = 37y + 6,

Case x=7: k = 41y - 7, k = 41y + 7, k = 43y + 7,

Case x=8: k = 47y – 8, k = 47y + 8, k = 49y + 8 = 7(7y + 1) +1.

It can be seen that k is represented by the threads that are defined by prime numbers bigger than 3. From examples (cases x = 4, x = 6 and x = 8), it can be seen that if (6x - 1) or (6x +

1) represent a composite number, k that is represented by that number also has representation by one of the prime factors of that composite number. This can be proved easily in the general case, by direct calculation, using representations similar to (2) and (3).

Here, only one case is going to be analyzed. All other cases can be analyzed analogously. In this case, assume that

(6푥 −1) = (6푙 +1)(6푠 −1) where (l, s ϵ N), and that leads to

푥 =6푙푠 − 푙 + 푠.

Considering that and using the following representation of k that includes the form (6x - 1)

푘 = (6푥 −1)푦 + 푥, the simple calculations leads to

푘 = (6푙 +1)(6푠 −1)푦 +6푙푠 − 푙 + 푠 = (6푙 +1)(6푠 −1)푦 + 푠(6푙 +1) − 푙, or

푘 = (6푙 +1)(6푠 −1)푦 + 푠 − 푙 which means

푘 = (6푙 +1)푓 − 푙

and that represents the already existing form of the representation of k for the factor (6l+1), where

푓 = (6푠 −1)푦 + 푠.

It can be seen that all patterns for k can be represented by the thread defined by the prime number bigger than 3. Now, it is going to be proved that the number of natural numbers that cannot be represented by the models (8) is infinite.

In order to do it, a method similar to the sieve of Eratosthenes [4] is going to be used. When all numbers that can be represented in the form

5y − 1 and 5y + 1,

are removed from the set of natural numbers N, it can be seen that a ratio of r1 = 2/5 of all

natural numbers is removed. The ratio c1 = 1 − 2/5 = 3/5 of all natural numbers cannot be represented by those two patterns and they still contain some numbers that could be used for representation of twin primes.

What does that actually mean? The proper interpretation of this result is: All natural numbers can be represented by 5 threads: 5y, 5y-1, 5y-2, 5y-3 and 5y-4 (y ϵ N). It means that all natural numbers that cannot be represented by (8) can only belong to the threads that are in the form 5y, 5y - 3 or 5y - 2. That means that there are three threads that potentially contain natural numbers that cannot be represented by (8).

If in addition, the natural numbers in the form

7y + 1, are removed, then the ratio of removed numbers can be calculated by the following equation

(checking that every removed number is calculated only once; basically, the formula for

calculation of the probability of occurring of two events that are not mutually exclusive is applied: P(A ∪ B) = P(A) + P(B) − P(A ∩ B))

1 1 1 2 1 2 17 푟 = 푟 + − × 푟 = 푟 + (1− 푟 ) = + 1− = 7 7 7 5 7 5 35

The ratio of all natural numbers that can still be used for the generation of twin primes, is defined by

1 1 3×6 푐 =1− 푟 =1− 푟 − (1− 푟 ) = 1− × 푐 = . 7 7 5×7

Again – the proper interpretation of this result is: All natural numbers can be represented with 35 threads in the form 35y-i (y ϵ N, i ϵ {0, 1, 2, … , 34}). From previous step it is known that 14 threads defined by 5*(7y-j)-1 and 5*(7y-j) + 1, where y ϵ N, j ϵ {0, 1, 2, …, 6}, do not contain the numbers that can be used for the generation of twin primes. From the current step it is known that 5 threads defined by 7*(5y-j)+1, where y ϵ N, j ϵ {0, 1, 2, 3, 4}, do not contain any number that can be used for the generation of the twin primes. However, these formulas, from the first two steps, produce some threads that overlap:

35푦 +1=5(7푦 −0) +1=7(5푦 −0) + 1,

and

35푦 −6=5(7푦 −1) −1=7(5푦 −1) + 1.

That leave us with 18 threads that potentially contain numbers that cannot be represented by (8).

Now, we denote prime numbers bigger than 3 as p5, where p5(1) = 5, p5(2) = 7 and so on.

Also, we denote the numbers p5-1, if p5 is a prime number in the mpl form and p5-2, if p5 is a prime number in the mps form, with p5r .

After step n the ratio rn of all numbers removed and ratio the cn of all numbers still potentially available for the generation of twin primes, are obtained. In the step n + 1 we will have

푝5(푛 +1) − 푝5푟(푛 + 1) 푝5(푛 +1) − 푝5푟(푛 +1) 푟 = 푟 + − × 푟 . 푝5(푛 +1) 푝5(푛 +1)

After a few elementary calculations, the following equation is obtained

푝5(푛 +1) − 푝5푟(푛 + 1) 푟 = 푟 + × (1− 푟 ). 푝5(푛 +1)

Now, the following equation holds

푝5(푛 +1) − 푝5푟(푛 + 1) 푐 =1− 푟 =1− 푟 − × (1− 푟 ) 푝5(푛 +1)

푝5(푛 +1) − 푝5푟(푛 + 1) = 푐 − × 푐 푝5(푛 +1) or

푐 = 푝5푟(푛 +1) × 푐. (9)

Equation (9) can also be written in the following form

∏ 푝5푟(푗) 푐 = . ∏ 푝5(푗)

(10)

Hence, after n+1-st step in which a thread defined by the n+1-st p5 is removed, we know

풏ퟏ that ∏풋ퟏ 풑ퟓ풓(풋) threads that potentially contain numbers that can be used for the generation of twin primes exists.

If the process is continued until all possible patterns (defined by (8)) related to all prime numbers bigger than 3 (and that is an infinite number) are removed, a number C can be defined by the following equation

퐶 = lim ∏ 푝5푟(푗). (11) →

It can easily be concluded that C is an infinite number. We know that C represents the number of threads that contain natural numbers that cannot be represented by (8). Since the set of natural numbers is closed for multiplication, it means that every one of those threads contains at least one number and that means that the number of natural numbers that cannot be represented by equation (8) is infinite.

That completes the proof that the number of twin primes is infinite.

3 Proof that the number of cousin primes is infinite

The cousin prime numbers are prime numbers with the gap 4. It is clear that cousin primes represent pairs of odd numbers that surround odd number divisible by 3 (e.g. (7 9 11), or

(13 15 17)). A pair can only represent a cousin primes if both of those numbers are primes.

So, if we denote the pair of odd numbers that surround the odd number divisible by 3 as pln

= 6k+1 and psm = 6(k+1)−1 = 6k+5, k ∈ N, these numbers can represent cousin primes only in the case when both pln and psm are prime numbers. If any of the pln or psm (or both) is a composite number, then we cannot have cousin primes.

The strategy is to check in which cases (for which k) it would not be possible to have cousin primes.

If psm represents a composite number the following equation must hold

6k + 5 = CPN5, where CPN5 represents a composite number in the mps form. After some elementary calculations, the following equation can be obtained

−5 + 퐶푃푁5 퐶푃푁5+1 푘 = = − 1. 6 6

(12)

Since CPN5 should be in the mps form, CPN5 can generally be expressed as a product mpl × mps, or

mpl = 6x + 1 and mps = 6y − 1(x, y ∈ N), which leads to

CPN5 = mpl × mps = 6(6xy − x + y) − 1, (13) or, due to symmetry

mpl = 6y + 1 and mps = 6x − 1, which leads to

CPN5 = mpl × mps = 6(6xy + x − y) − 1. (14)

Following a similar procedure, it can be seen that pln, that represents a CPN5 number in the mpl form, will correspond to the composite number in the case

퐶푃푁5−1 푘 = . 6

(15)

In this case, CPN5 can be expressed in the form mpl1 × mpl2 or as mps1 × mps2, and we will have two possibilities (x, y ∈ N)

mpl1 = 6x + 1 and mpl2 = 6y + 1, which leads to

CPN5 = mpl1 × mpl2 = 6(6xy + x + y) + 1, (16) or

mps1 = 6x − 1 and mps2 = 6y − 1, which leads to

CPN5 = mps1 × mps2 = 6(6xy − x − y) + 1. (17)

So, if we replace (13) (equivalently (14)) in (12), and (16) and (17) in (15) we obtain forms for all k that will not produce a cousin prime pair. Those forms are expressed by the following equation

(6푥 −1)푦 + 푥 −1 푘 = (6푥 −1)푦 − 푥 , (18) (6푥 +1)푦 + 푥 where x, y ∈ N. This is sufficient and necessary condition for k, so that it cannot be used for the generation of the cousin primes.

Now, using the same method like in the case of the twin prime conjecture, it can be proved that exists infinitely many natural numbers that cannot be presented in the form (18), and that completes the proof that the number of cousin primes is infinite.

References

[1] de Polignac, A. (1849). Recherches nouvelles sur les nombres premiers. Comptes

Rendus des S´eances de l’Acad´emie des Sciences.

[2] Neale, V. (2017). Closing the Gap. Oxford, U. K., Oxford University Press, p. 141-144.

[3] D.H.J. Polymath. Variants of the Selberg sieve, and bounded intervals containing many primes. Research in the Mathematical Sciences. 1(12). arXiv:1407.4897 (2014).

[4] Wells, D. (2005). Prime Numbers: The Most Mysterious Figures in Math. Hoboken, NJ:

John Wiley & Sons, Inc., p. 58-59.