# An Elementary Proof of the Twin Prime Conjecture

Berndt

(XXX; XXX)

###### Abstract

It’s well known that every prime number $p\geq 5$ has the form $6k-1$ or
$6k+1$. We’ll call $k$ the generator of $p$. Twin primes are distinghuished
due to a common generator for each pair. Therefore it makes sense to search
for the Twin Primes on the level of their generators. This paper present a new
approach to prove the Twin Prime Conjecture by a sieve method to extract all
Twin Primes on the level of the Twin Prime Generators. We define the
$\omega_{p_{n}}$–numbers $x$ as numbers for which holds that $6x-1$ and $6x+1$
are coprime to the prime $p_{n}$. By dint of the average distance
$\bar{\delta}(p_{n})$ between the $\omega_{p_{n}}$–numbers we can prove the
Twin Prime Conjecture indirectly.

GenselB. GenselSpittal an der Drau Carinthia University of Applied Sciences
Villacher Str. 1, A-9800, Spittal an der Drau, Austria<EMAIL_ADDRESS>
Release:

### Keywords

twin primes, primes, number theory - MSC2010: 11A41

## Notations

We’ll use the following notations:

$\begin{array}[]{llll}&\mathbb{N}\mbox{ the set of the positve integers},\\\
&\mathbb{P}\mbox{ the set of the primes},&\mathbb{P}^{*}\mbox{ primes}\geq
5,\\\ &\mathbb{P}_{-}=\\{p\in\mathbb{P}^{*}\mid
p\equiv-1(\operatorname{mod}6)\\}&\mbox{and
}\mathbb{E}_{-}=\\{n\in\mathbb{N}\mid 6n-1\in\mathbb{P}_{-}\\}\\\
&\mathbb{P}_{+}=\\{p\in\mathbb{P}^{*}\mid
p\equiv+1(\operatorname{mod}6)\\}&\mbox{and
}\mathbb{E}_{+}=\\{n\in\mathbb{N}\mid 6n+1\in\mathbb{P}_{+}\\}\\\ &&\mbox{and
}\mathbb{E}=\mathbb{E}_{-}\cap\mathbb{E}_{+}.\end{array}$

##  1 Introduction

The question on the infinity of the twin primes keeps busy many mathematicians
for a long time. 1919 V. Brun [3] had proved that the series of the inverted
twin primes converges while he had tried to prove the Twin Prime Conjecture.
Several authors worked on bounds for the length of prime gaps (see f.i. [4, 5,
6]). 2014 Y. Zhang [7] obtained a great attention with his proof that there
are infinitely many consecutive primes with a distance of $70,000,000$ at
most. With the project ”PolyMath8” this bound could be lessened down to $246$
respectively to $12$ assuming the validity of the Elliott–Halberstam
Conjecture [8].

We present in this paper another approach as in the most works on this topic.
We tranfer the looking for twin primes to the level of their generators
because each twin prime has a common generator.

##  2 Twin Prime Generators

It’s well known that every prime number $p\geq 5$ has the form $6k-1$ or
$6k+1$. We’ll call $k$ the generator of $p$. Twin primes are distinghuished
due to a common generator for each pair. Therefore it makes sense to search
for the twin primes on the level of their generators.

Let be

(1) $\kappa(p)=\left\\{\begin{array}[]{ll}\frac{p+1}{6}&\mbox{for
}p\in\mathbb{P}_{-}\\\ \frac{p-1}{6}&\mbox{for
}p\in\mathbb{P}_{+}\end{array}\right.$

an over $\mathbb{P}^{*}$ defined function, the generator of the pair
$(6\kappa(p)-1,6\kappa(p)+1)$.

A number $x$ is a member of $\mathbb{E}$ if $6x-1$ as well as $6x+1$ are
primes. This is true if the following statement holds.

######  Theorem 1.

A number $x$ is a member of $\mathbb{E}$ if and only if there is no
$p\in\mathbb{P}^{*}$ with $p<6x-1$ where one of the following congruences
holds:

(2) $\displaystyle x\equiv-\kappa(p)(\operatorname{mod}p)$ (3) $\displaystyle
x\equiv+\kappa(p)(\operatorname{mod}p)$

###### Proof.

  1. A.

$p\in\mathbb{P}_{-}$ , therefore is $p=6\kappa(p)-1$:

If (2) is true then there is an $n\in\mathbb{N}$ with

$\displaystyle x$ $\displaystyle=$
$\displaystyle-\kappa(p)+n\cdot(6\kappa(p)-1)$ $\displaystyle 6x$
$\displaystyle=$ $\displaystyle-6\kappa(p)+6n\cdot(6\kappa(p)-1)$
$\displaystyle 6x+1$ $\displaystyle=$
$\displaystyle-6\kappa(p)+6n\cdot(6\kappa(p)-1)+1$ $\displaystyle=$
$\displaystyle(6n-1)(6\kappa(p)-1)$ $\displaystyle\Rightarrow 6x+1$
$\displaystyle\equiv$ $\displaystyle 0(\operatorname{mod}p)\Rightarrow
x\notin\mathbb{E}$

For (3) the proof will be done with $6x-1$:

$\displaystyle 6x-1$ $\displaystyle=$ $\displaystyle
6\kappa(p)+6n\cdot(6\kappa(p)-1)-1$ $\displaystyle=$
$\displaystyle(6n+1)(6\kappa(p)-1)$ $\displaystyle\Rightarrow 6x-1$
$\displaystyle\equiv$ $\displaystyle 0(\operatorname{mod}p)\Rightarrow
x\notin\mathbb{E}$

  2. B.

$p\in\mathbb{P}_{+}$, therefore is $p=6\kappa(p)+1$: We go the same way with
(2) and $6x-1$ as well as (3) and $6x+1$:

$\displaystyle 6x-1$ $\displaystyle=$
$\displaystyle(6n-1)(6\kappa(p)+1)\Rightarrow 6x-1\equiv
0(\operatorname{mod}p)$ $\displaystyle 6x+1$ $\displaystyle=$
$\displaystyle(6n+1)(6\kappa(p)+1)\Rightarrow 6x+1\equiv
0(\operatorname{mod}p)$

With these it’s shown that $x\notin\mathbb{E}$ if the congruences (2) or (3)
are valid. They cannot be true both because they exclude each other.
If on the other hand $x\notin\mathbb{E}$, then is $6x-1$ or $6x+1$ no prime.
Let be $6x-1\equiv 0(\operatorname{mod}p)$ and $p\in\mathbb{P}_{-}$. Then we
have

$\displaystyle 6x-1$ $\displaystyle\equiv$ $\displaystyle
p(\operatorname{mod}p)$ $\displaystyle\equiv$
$\displaystyle(6\kappa(p)-1)(\operatorname{mod}p)$ $\displaystyle 6x$
$\displaystyle\equiv$ $\displaystyle 6\kappa(p)(\operatorname{mod}p)$
$\displaystyle x$ $\displaystyle\equiv$
$\displaystyle\kappa(p)(\operatorname{mod}p).$

For $p\in\mathbb{P}_{+}$ we have

$\displaystyle 6x-1$ $\displaystyle\equiv$
$\displaystyle-p(\operatorname{mod}p)$ $\displaystyle\equiv$
$\displaystyle-(6\kappa(p)+1)(\operatorname{mod}p)$ $\displaystyle 6x$
$\displaystyle\equiv$ $\displaystyle-6\kappa(p)(\operatorname{mod}p)$
$\displaystyle x$ $\displaystyle\equiv$
$\displaystyle-\kappa(p)(\operatorname{mod}p).$

The other both cases we can handle in the same way. Therefore either (2) or
(3) is valid if $x\notin\mathbb{E}$. ∎

If we consider that the least proper divisor of a number $6x-1$ or $6x+1$ is
less or equal to $\sqrt{6x+1}$ than $p$ in the congruences (2) and (3) can be
further limited by

$\hat{p}(x)=\operatorname{max}(p\in\mathbb{P}^{*}\mid p\leq\sqrt{6x+1}).$

Henceforth we will use the letter $p$ for a general prime number and $p_{n}$
if we describe an element of a sequence of primes. With $p_{n}$ as the $n$-the
prime number 111It is $p_{1}=2$ and $\pi(z)$ as the number of primes $\leq z$
we have with

(4) $\displaystyle x$ $\displaystyle\equiv$
$\displaystyle-\kappa(p_{n})(\operatorname{mod}p_{n})$ or $\displaystyle x$
$\displaystyle\equiv$ $\displaystyle+\kappa(p_{n})(\operatorname{mod}p_{n})$

for $3\leq n\leq\pi\left(\hat{p}(x)\right)$ a proofable system of criteria to
exclude a number $x\geq 4$ as not being a member of $\mathbb{E}$.

##  3 The Twin Sieve

The congruences in (4) can be combined in the following way:

(5) $x^{2}\equiv\kappa(p_{n})^{2}(\operatorname{mod}p_{n})\mbox{ for }3\leq
n\leq\pi\left(\hat{p}(x)\right)$

because if $x\equiv\pm\kappa(p_{n})(\operatorname{mod}p_{n})$ then there is a
number $t$ with $x=\pm\kappa(p_{n})+tp_{n}$. Squared this produces
$x^{2}=\kappa(p_{n})^{2}+p_{n}(t^{2}p_{n}\pm 2t\kappa(p_{n}))$ and we get
$x^{2}\equiv\kappa(p_{n})^{2}(\operatorname{mod}p_{n})$. This results in a
system of sieves with sieve functions $\psi(x,p_{n})$ for which hold for
$3\leq n\leq\pi\left(\hat{p}(x)\right)$

$\displaystyle x^{2}-\kappa(p_{n})^{2}$ $\displaystyle\equiv$
$\displaystyle\psi(x,p_{n})(\operatorname{mod}p_{n})\mbox{ respectively}$ (6)
$\displaystyle\psi(x,p_{n})$ $\displaystyle=$
$\displaystyle\left(x^{2}-\kappa(p_{n})^{2}\right)\operatorname{Mod}p_{n}.$

Obviously $\psi(x,p)$ is a periodical function in $x$ with a period length of
$p$. We’ll call the sieve represented by $\psi(x,p_{n})$ as
$\textbf{{S}}_{n}$. For the system of the sieves
$\textbf{{S}}_{3}\times\ldots\times\textbf{{S}}_{n}$ we’ll build the aggregate
sieve functions

$\displaystyle\Psi(x,p_{n})$ $\displaystyle=$
$\displaystyle\prod_{i=3}^{n}\frac{\psi(x,p_{i})}{p_{i}}$ (7) and
$\displaystyle\hat{\Psi}(x)$ $\displaystyle=$
$\displaystyle\Psi(x,\hat{p}(x)).$

Because the value set of $\psi(x,p)$ consists of positive integers between $0$
and $p-1$, $\Psi(x,p)$ and $\hat{\Psi}(x)$ have rational values between $0$
and $<1$.

A number $x$ will be “sieved” by $\textbf{{S}}_{n}$ if and only if
$\psi(x,p_{n})=0$. With (7) in this case also is $\Psi(x,p_{n})=0$. In
contrast to the sieve of ERATOSTHENES in our sieve the exclusion of a number
$x$ will be not controlled by $x\operatorname{Mod}p=0$, but by
$\left(x^{2}-\kappa(p)^{2}\right)\operatorname{Mod}p=0$.

Let be

(8) $O_{n}=\operatorname{min}(x\in\mathbb{N}\mid\hat{p}(x)=p_{n}).$

For $x\geq O_{n}$ “works” the sieve $\textbf{{S}}_{n}$, i.e. $O_{n}$ is the
origin of the sieve $\textbf{{S}}_{n}$. Every sieve has up from $O_{n}$ in
every $\psi$–period just $p_{n}-2$ positions with $\psi(x,p_{n})\neq 0$ and
two positions with $\psi(x,p_{n})=0$, once if (2) and on the other hand if (3)
is valid. We speak about $a$– and $b$–bars of the sieve $\textbf{{S}}_{n}$.
From (2) and (3) it is easy to see that the distance between an $a$– and a
$b$–bar is $2\kappa(p_{n})$.

It is $p_{n}\leq\hat{p}(x)\leq\sqrt{6x+1}$ and therefore $p_{n}^{2}\leq 6x+1$.
Then

(9) $O_{n}=\frac{p_{n}^{2}-1}{6}$

is the least number which meets this relation. It is easy to prove that for
every prime $p$ holds that $p^{2}-1$ is an integer divisible by $6$.

######  Theorem 2.

Every sieve $\textbf{{S}}_{n}$ with $n\geq 3$ starts at its origin $O_{n}$
with a sieve bar and we have $\psi(O_{n},p_{n})=0$.

###### Proof.

We substitute $p_{n}$ by $6\kappa(p_{n})\pm 1$. With this and (9) holds

$\displaystyle O_{n}$ $\displaystyle=$
$\displaystyle\frac{\left(6\kappa(p_{n})\pm 1\right)^{2}-1}{6}$
$\displaystyle=$ $\displaystyle\frac{6\kappa(p_{n})\left(6\kappa(p_{n})\pm
2\right)}{6}$ $\displaystyle=$
$\displaystyle\kappa(p_{n})\left(6\kappa(p_{n})\pm 1\right)\pm\kappa(p_{n})$
$\displaystyle=$ $\displaystyle\kappa(p_{n})\cdot p_{n}\pm\kappa(p_{n})$
$\displaystyle\equiv$
$\displaystyle\pm\kappa(p_{n})(\operatorname{mod}p_{n})\rightarrow\psi(O_{n},p_{n})=0.$

∎

We see that $\textbf{{S}}_{n}$ starts for $p_{n}\in\mathbb{P}_{-}$ with an
$a$–bar (2) and in the other case with a $b$–bar(3).

For every $x\geq O_{n}$ the local position in the sieve $\textbf{{S}}_{n}$
relative to the phase start 222For $p_{n}\in\mathbb{P}_{-}$ the phase start is
$O_{n}$ and else it is $O_{n}-2\kappa(p_{n})$. can be determined by the
position function $\tau(x,p_{n})$:

$\displaystyle x+\kappa(p_{n})$ $\displaystyle\equiv$
$\displaystyle\tau(x,p_{n})(\operatorname{mod}p_{n})\mbox{ respectively}$ (10)
$\displaystyle\tau(x,p_{n})$ $\displaystyle=$
$\displaystyle\left(x+\kappa(p_{n})\right)\operatorname{Mod}p_{n}.$

Between the sieve function $\psi(x,p)$ and the position function $\tau(x,p)$
there is the following relationship:

(11) $\displaystyle\psi(x,p)$ $\displaystyle=$
$\displaystyle\tau(x,p)\cdot\left(x-\kappa(p)\right)\operatorname{Mod}p$
$\displaystyle=$
$\displaystyle\tau(x,p)\cdot\left(\tau(x,p)-2\kappa(p)\right)\operatorname{Mod}p.$

Obviously is $\psi(x,p)=0$ if and only if $\tau(x,p)=0$ ($a$–bar) or
$\tau(x,p)=2\kappa(p)$ ($b$–bar).

##  4 The Permeability of the Sieves
$\textbf{{S}}_{3}\times\ldots\times\textbf{{S}}_{n}$

For every $x$ in the interval

(12) $\mathcal{A}_{n}:=\left[O_{n},O_{n+1}-1\right]$

$\hat{p}(x)$ persists constant on the value $p_{n}$. The length of this
interval 333 Really is
$\mathcal{A}_{n}:=\left[O_{n},O_{n+1}-1\right]\cap\mathbb{N}$. Henceforth all
intervals will be defined as sections of the number line. will be denoted as
$d_{n}$. It is depending on the distance between successive primes. Since they
can only be even, we have with $a=2,4,6,\ldots$

(13) $\displaystyle d_{n}$ $\displaystyle=$
$\displaystyle\frac{(p_{n}+a)^{2}-1}{6}-\frac{p_{n}^{2}-1}{6}$
$\displaystyle=$ $\displaystyle\frac{2ap_{n}+a^{2}}{6}$ $\displaystyle=$
$\displaystyle\frac{a}{3}(p_{n}+\frac{a}{2})$ $\displaystyle\geq$
$\displaystyle\frac{2}{3}(p_{n}+1).$

On the other hand it results because of $p_{n+1}<2p_{n}$ (see [2], p. 188)

(14) $\displaystyle d_{n}$ $\displaystyle=$
$\displaystyle\frac{p_{n+1}^{2}-1}{6}-\frac{p_{n}^{2}-1}{6}$ $\displaystyle=$
$\displaystyle\frac{p_{n+1}^{2}-p_{n}^{2}}{6}$ $\displaystyle=$
$\displaystyle\frac{(p_{n+1}+p_{n})(p_{n+1}-p_{n})}{6}$ $\displaystyle<$
$\displaystyle\frac{3p_{n}\cdot p_{n}}{6}=\frac{p_{n}^{2}}{2}.$

The congruences from (10)

(15) $x+\kappa(p_{i})\equiv\tau(x,p_{i})(\operatorname{mod}p_{i}),\quad 3\leq
i\leq n$

meet the requirements of the Chinese Remainder Theorem (see [1], p. 89).
Therefore it is modulo $5\cdot 7\cdot\ldots\cdot p_{n}$ uniquely resolvable.
With

(16) $p_{n}\sharp_{5}:=\prod_{i=3}^{n}p_{i}=5\cdot 7\cdot\ldots\cdot p_{n}$

it’s $(\operatorname{mod}p_{n}\sharp_{5})$ 444It is
$p_{n}\sharp_{5}=\frac{p_{n}\sharp}{6}$, with the primorial $p_{n}\sharp$.
uniquely resolvable. Therefore the sieves
$\textbf{{S}}_{3}\times\ldots\times\textbf{{S}}_{n}$ have the period length
$p_{n}\sharp_{5}$ and for the aggregate sieve function holds:

$\Psi(x+a\cdot p_{n}\sharp_{5},p_{n})=\Psi(x,p_{n})\mid a\in\mathbb{N}.$

######  Definition 4.1.

A positive integer $x$ will be called an “$\omega_{p}$–number” if both $6x-1$
and $6x+1$ are coprime 555 Then is $\gcd(36x^{2}-1,p\sharp_{5})=1.$ to $p$. In
this case is $\Psi(x,p)>0$.

Let be

$\mathcal{P}_{n}:=\left[O_{n},O_{n}+p_{n}\sharp_{5}-1\right]$

the interval of the period of the sieves
$\textbf{{S}}_{3}\times\ldots\times\textbf{{S}}_{n}$. We’ll denote it
henceforth as period section. Evidently is
$\mathcal{A}_{n}\subset\mathcal{P}_{n}$ for all $n\geq 3$.

The values of the function $\tau(x,p)$ are the numbers $0,1,\ldots,p-1$. Two
of them result in the exlcuding of $x$ and $p-2$ don’t. Therefore by working
of the sieves $\textbf{{S}}_{3}\times\ldots\times\textbf{{S}}_{n}$ we have

(17) $\varphi(p_{n})=\prod_{i=3}^{n}(p_{i}-2)$

$\omega_{p_{n}}$–numbers in $\mathcal{P}_{n}$. If a lot of them are in
$\mathcal{A}_{n}$, they are members of $\mathbb{E}$ because the sieves
$\textbf{{S}}_{3}\times\ldots\times\textbf{{S}}_{n}$ here are working only.
The relation between (17) and the period length of (15) results in

(18)
$\eta(p_{n})=\frac{\varphi(p_{n})}{p_{n}\sharp_{5}}=\prod_{i=3}^{n}\frac{p_{i}-2}{p_{i}},$

as a measure of the mean “permeability” of working of the sieves
$\textbf{{S}}_{3}\times\ldots\times\textbf{{S}}_{n}$ or as the density of the
$\omega_{p_{n}}$–numbers in $\mathcal{P}_{n}$. Obviously $\eta(p)$ is a strong
monotonously decreasing function. Its inversion

(19) $\bar{\delta}(p)=\frac{1}{\eta(p)}$

discribes the average distance between the $\omega_{p}$–numbers in their
period section.

######  Theorem 3.

The density of the $\omega_{p}$–numbers in their period section is lower
bounded by

$\eta(p)>\frac{3}{p}\quad\mbox{for }p\in\mathbb{P}^{*}.$

###### Proof.

Let be $\mathbb{Q}_{p}=\\{q\in\mathbb{P}^{*}\mid q\leq p\\}$ and
$\mathbb{U}_{p}=\\{u\equiv 1(\operatorname{mod}2)\mid 5\leq u\leq p\\}$.
Because all primes $>2$ are odd numbers it holds
$\mathbb{Q}_{p}\subset\mathbb{U}_{p}$ for $p>7$ 666For $p\leq 7$ is
$\mathbb{Q}_{p}=\mathbb{U}_{p}$.. All factors of $\eta(p)$ are less than $1$.
It results

$\eta(p)>\prod_{u\in\mathbb{U}_{p}}^{p}\frac{u-2}{u}=\frac{3}{5}\cdot\frac{5}{7}\cdot\frac{7}{9}\cdot\ldots\cdot\frac{p-4}{p-2}\cdot\frac{p-2}{p}=\frac{3}{p}.$

∎

By inversion of this relationship, we obtain for the average distance

(20) $\bar{\delta}(p)<\frac{p}{3}.$

Under consideration of (13) we obtain furthermore 777 We can even prove that
$\bar{\delta}(p)^{2}<p$ and $\bar{\delta}(p_{n})^{2}<d_{n}$ for
$p_{n}>200(n>45)$.

(21) $2\bar{\delta}(p_{n})<\frac{2p_{n}}{3}<\frac{2}{3}(p_{n}+1)\leq
d_{n}\rightarrow\bar{\delta}(p_{n})<\frac{d_{n}}{2}.$

This means that the avarage distance between $\omega_{p_{n}}$–numbers remains
ever less than the half of the length of $\mathcal{A}_{n}$, the interval where
$\omega_{p_{n}}$–numbers are twin prime generators.

##  5 The Sieve Process on Average

The intervals $\mathcal{A}_{n},n\geq 3$ defined by (12) cover the positive
integers $\geq 4$ gapless and densely. It is

$\mathbb{N}=\\{1,2,3\\}\cup\bigcup_{n=3}^{\infty}\mathcal{A}_{n}\mbox{ and
}\bigcap_{n=3}^{\infty}\mathcal{A}_{n}=\emptyset.$

They are the beginnings of the period sections $\mathcal{P}_{n}$ of the
$\omega_{p_{n}}$–numbers. Hereafter let’s say A-sections to the intervals
$\mathcal{A}_{n}$. Every $\omega_{p_{n}}$–number which lies in an A-section is
a twin prime generator (see above). In contrast to the A-sections the period
sections $\mathcal{P}_{n}$ overlap each other very densely. So the period
section $\mathcal{P}_{9}$ reachs over $1739$ A-sections up to the beginning of
the period section $\mathcal{P}_{1748}$ and the next $\mathcal{P}_{10}$ over
$7863$ A-sections up to the beginning of $\mathcal{P}_{7873}$.

######  Theorem 4.

Each origin $O_{n}$ cannot be located at the beginnning $O_{m}+a\cdot
p_{m}\sharp_{5}\mid a\in\mathbb{N}$ of any period of the sieves
$\textbf{{S}}_{3}\times\ldots\times\textbf{{S}}_{m}$ for $m<n$. Therefore it
holds for $m<n$

$O_{n}\not\equiv O_{m}(\operatorname{mod}p_{m}\sharp_{5}).$

###### Proof.

The equation

$\frac{p_{m}^{2}-1}{6}+a\cdot p_{m}\sharp_{5}=\frac{p_{n}^{2}-1}{6}\mbox{ and
thus }p_{m}^{2}+a\cdot p_{m}\sharp=p_{n}^{2}$

is for no primes $p_{m}<p_{n}$ solvable, because of $\gcd(p_{m},p_{n})=1$. ∎

Vice versa holds that every period section $\mathcal{P}_{n+1}$ starts always
inside of the previous period section $\mathcal{P}_{n}$ nearby to its origin
because (see (14) also)

$\displaystyle O_{n+1}$ $\displaystyle=$ $\displaystyle O_{n}+d_{n}\mbox{
and}$ $\displaystyle d_{n}$ $\displaystyle<$
$\displaystyle\frac{p_{n}^{2}}{2}\ll\frac{p_{n}\sharp_{5}}{2}.$

Let be

$\displaystyle\mathcal{P}_{n}^{k}$ $\displaystyle:=$
$\displaystyle\left[O_{n}+k\cdot p_{n}\sharp_{5},O_{n}+(k+1)\cdot
p_{n}\sharp_{5}-1\right]\mbox{ and}$ $\displaystyle\mathcal{A}_{n}^{+}$
$\displaystyle:=$
$\displaystyle\left[O_{n}+p_{n+1}\sharp_{5},O_{n+1}-1+p_{n+1}\sharp_{5}\right].$

With these we can show the recursive structure of the period sections

(22)
$\mathcal{P}_{n+1}=\mathcal{P}_{n}\setminus\mathcal{A}_{n}\cup\left(\bigcup_{k=1}^{p_{n+1}-1}\mathcal{P}_{n}^{k}\right)\cup\mathcal{A}_{n}^{+}.$

We can clearly see that the period section $\mathcal{P}_{n}$ overlaps
$\mathcal{P}_{n+1}$ up to the end of $\mathcal{P}_{n}\setminus\mathcal{A}_{n}$
and $\mathcal{P}_{n+1}$ has much space for A–sections $\mathcal{A}_{t}$ with
$t\geq n+1$.
The one consequence of this dense overlapping of the period sections is that a
plurality of the $\omega_{p_{m}}$–gaps from the period section
$\mathcal{P}_{m}$ persist constant as also $\omega_{p_{n}}$–gaps for $n>m$ but
in a shifted position relative to their origin $O_{n}$ (see Theorem 4 and
(22)). On the other hand this dense overlapping guarantees that extreme
anomalies of the distribution of the $\omega_{p_{n}}$–numbers cannot occur.

For the quantity of the $\omega_{p_{n}}$–numbers in $\mathcal{P}_{n}$ is
corresponding with (17)

$\varphi(p_{n})=\varphi(p_{n-1})\cdot(p_{n}-2).$

In $\mathcal{P}_{n}$ the $\varphi(p_{n})$ $\omega_{p_{n}}$–numbers are spread
888 It is easy to prove that the $\omega_{p}$–numbers in their period section
are symmetrically distributed around $\frac{p\sharp_{5}}{2}$ and
$p\sharp_{5}$. Nevertheless the distribution is non-uniform.  over
$p_{n}\sharp_{5}$ positions. According to (22) we have
$p_{n+1}\cdot\varphi(p_{n})$ $\omega_{p_{n}}$–numbers in $\mathcal{P}_{n+1}$.
In comparison between them and the $\omega_{p_{n+1}}$–numbers resulting from
the working of the sieve $\textbf{{S}}_{n+1}$ we see

(23)
$p_{n+1}\cdot\varphi(p_{n})-\varphi(p_{n+1})=p_{n+1}\cdot\varphi(p_{n})-(p_{n+1}-2)\cdot\varphi(p_{n})=2\varphi(p_{n}).$

We loose by the working of $\textbf{{S}}_{n+1}$ in the period section
$\mathcal{P}_{n+1}$ just $2\varphi(p_{n})$ potential generators of twin
primes. In other words, the sieve $\textbf{{S}}_{n+1}$ has $2\varphi(p_{n})$
“beating bars” in $\mathcal{P}_{n+1}$. At these positions $x$ holds

(24) $\Psi(x,p_{n})>0\mbox{ and }\psi(x,p_{n+1})=0.$

Only the beating bars let grow the gaps by exclusion of the
$\omega_{p_{n}}$–number between two $\omega_{p_{n}}$–gaps to one
$\omega_{p_{n+1}}$–gap. By the working of the sieve $\textbf{{S}}_{n+1}$ we
obtain the following sieve balance “on average”:

> The distances between the $\omega_{p_{n}}$–numbers persist unchanged at
> $\bar{\delta}(p_{n})$ on average except of those $2\varphi(p_{n})$
> $\omega_{p_{n}}$–numbers which are met by the beating bars of the sieve
> $\textbf{{S}}_{n+1}$. Thereby a distance $D$ occurs between the adjacent
> $\omega_{p_{n+1}}$–numbers on average:

$\displaystyle p_{n+1}\sharp_{5}$ $\displaystyle=$ $\displaystyle\mbox{changed
}+\mbox{unchanged}$ $\displaystyle=$ $\displaystyle 2\varphi(p_{n})\cdot
D+\left(\varphi(p_{n+1})-2\varphi(p_{n})\right)\cdot\bar{\delta}(p_{n})$
$\displaystyle=$ $\displaystyle 2\varphi(p_{n})\cdot
D+\left(\varphi(p_{n})(p_{n+1}-2)-2\varphi(p_{n})\right)\cdot\frac{p_{n}\sharp_{5}}{\varphi(p_{n})}$
$\displaystyle=$ $\displaystyle 2\varphi(p_{n})\cdot
D+p_{n+1}\sharp_{5}-4p_{n}\sharp_{5}$ and therefore $\displaystyle 0$
$\displaystyle=$ $\displaystyle 2\varphi(p_{n})\cdot D-4p_{n}\sharp_{5}\mbox{
and hence}$ $\displaystyle D$ $\displaystyle=$
$\displaystyle\frac{2p_{n}\sharp_{5}}{\varphi(p_{n})}=2\bar{\delta}(p_{n}).$

Therefore even the gaps between the $\omega_{p_{n+1}}$–numbers
($\omega_{p_{n+1}}$–gaps) which result from the beating bars persist less than
$d_{n+1}$ on average because

(25)
$D=2\bar{\delta}(p_{n})<\frac{2p_{n}}{3}<\frac{2p_{n+1}}{3}<\frac{2}{3}(p_{n+1}+1)\leq
d_{n+1}.$

##  6 Proof of the Twin Prime Conjecture

###### Proof.

The proof will be done indirectly. We assume that there is only a finite
number of twin primes and therefore there is only a finite number of twin
prime generators. Let be $y_{o}$ the greatest one. It lies in the A-section
$\mathcal{A}_{n_{o}}$ with $n_{o}=\pi\left(\hat{p}(y_{o})\right)$, the
beginning of the period section $\mathcal{P}_{n_{o}}$. In the subsequent
A-sections $\mathcal{A}_{t}$ with $t>n_{o}$ consequently there cannot be any
twin prime generators and therefore no $\omega_{p_{t}}$–numbers. But then we
have $\omega_{p_{t}}$–gaps with lengths $>d_{t}$ in all (infinitely many)
period sections $\mathcal{P}_{t}$ for $t>n_{o}$.

Because

  * •

all period sections $\mathcal{P}_{t}$ are very densely overlapped and
therefore extreme anomalies of the distribution of the
$\omega_{p_{t}}$–numbers cannot occur,

  * •

the average distances between the $\omega_{p_{t}}$–numbers are less than
$\frac{d_{t}}{2}$,

  * •

and even the $\omega_{p_{t}}$–gaps which are generated by beating bars of the
sieves $\textbf{{S}}_{t}$ are less than $d_{t}$ on average,

therefore it is not possible to have for all $t>n_{o}$ only period sections
$\mathcal{P}_{t}$ with $\omega_{p_{t}}$–gaps at their beginnings which are all
greater than $d_{t}$.

Therefore the proof assumtion cannot be valid and thus the Twin Prime
Conjecture must be true.

∎

## References

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  * [2] Ribenboim, P.: Die Welt der Primzahlen , Springer, 2011.
  * [3] Brun, V.: La série $\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\ldots$ où les dénominateurs sont ”nombres premiers jumeaux” est convergente où finie. Bull. Sci. Math. (2) 43 (1919), 100-104 and 124-128.
  * [4] Bombieri, E., Davenport, H.: Small differences between prime numbers. Proc. Roy. Soc. (A) 293 (1966), 1-18
  * [5] Goldston, D.A., Motohashi, Y., Pintz, J., Yıldırım, C.Y.: Small gaps between primes exist. In: Proceedings of the Japan Academy, Series A 82, 2006, 61-65
  * [6] Green, B., Tao, T.: The primes contain arbitrarily long arithmetic progressions. Annals of Math. (2) 167 (2008), 481-547
  * [7] Zhang, Y.: Bounded gaps between primes. Annals of Math. 179 (2014), 1121-1174
  * [8] Maynard, J.: Small gaps between primes. Annals of Math., Second Series, 181 (1), 383-413 (2015)