1. Introduction

2. Notation

3. Lemma

Proof. 

a . Let ${\left\{p_n\right\}}_1^{\infty }$ denote the odd prime sequence. Let

$$\begin{array}{c}\hfill h_n=(p_n-1)/2,\end{array}$$

(1)

and we call $h_n$ as the generator of odd prime numbers. So we have

$$\begin{array}{c}\hfill H={\left\{h_n\right\}}_1^{\infty }=\{1,2,3,5,6,8,9,11,14\cdots \}.\end{array}$$

Similarly, let $c_n$ be the generator of odd composite numbers. We have

$$\begin{array}{c}\hfill C={\left\{c_n\right\}}_1^{\infty }=\{4,7,10,12,13,16,17\cdots \}.\end{array}$$

Let $o_n$ be the generator of odd numbers, where the odd numbers do not contain 1. We have

$$\begin{array}{c}\hfill O=H+C=\{1,2,3,4,5,6,7\cdots \}.\end{array}$$

b. Let us construct a generating function for $a_r$ , the number of partitions of the positive integer $r(r⩾2)$ which into two generator of odd prime numbers.

Let

$$\begin{array}{c}\hfill H\left(x\right)=x^1+x^2+x^3+x^5+x^6+\cdots +x^{h_n}+x^{h_{n+1}}+\cdots .\end{array}$$

(2)

So, the generating function $Y\left(x\right)$ as follows:

$$\begin{array}{cc}\hfill Y\left(x\right)=H{\left(x\right)}^2=& (x^1+x^2+x^3+x^5+x^6+\cdots +x^{h_n}+x^{h_{n+1}}+\cdots )\hfill \\ \hfill & ·(x^1+x^2+x^3+x^5+x^6+\cdots +x^{h_n}+x^{h_{n+1}}+\cdots ).\hfill \end{array}$$

(3)

c. Obviously, it is only necessary to prove that $a_r⩾1(r⩾2)$, and the lemma is proved.

Since the value of $a_r$ cannot be obtained by the theory of generating functions, and Euclid’s proof demonstrates that the set of prime numbers is countable infinite [5].

So, let us prove it by mathematical induction.

STEP 1: Easy to know

$$\begin{array}{c}\hfill a_1=0,a_2=1,a_3=2,a_4=3,a_5=2,\dots .\end{array}$$

Thus, $a_t⩾1(2⩽t⩽5)$, When $r=h_4=5$.

STEP 2: Let us assume $a_t⩾1(2⩽t⩽h_n)$ is true for $r=h_n(n⩾5)$.

STEP 3: Let’s consider the statement with $r=h_{n+1}$.

For every $n⩾1$, there is some prime number p with $n<p⩽2n$. It was first proved for all n by Pafnuty Chebyshev in 1850. [1]. In other words, it has

$$\begin{array}{c}\hfill 13⩽p_n<p_{n+1}⩽2p_n(n⩾5).\end{array}$$

(4)

Thus, we have

$$\begin{array}{c}\hfill 6⩽h_n<h_{n+1}⩽2h_n(n⩾5).\end{array}$$

(5)

So let

$$\begin{array}{c}\hfill h_{n+1}=h_n+k(1⩽k⩽h_n).\end{array}$$

(6)

Obviously, $h_n+1,h_n+2,\cdots ,h_n+k-1$ are the generator of odd composite numbers. Let

$$\begin{array}{c}\hfill D=\{h_n+1,h_n+2,\cdots ,h_n+k-1\}.\end{array}$$

(7)

Thus

$$\begin{array}{c}\hfill C^{h_{n+1}-1}=\{4,7,10,\cdots ,h_n-1,h_n+1,\cdots ,h_n+k-1\}.\end{array}$$

(8)

Consider the formal power series as follows:

$$\begin{array}{cc}\hfill H^{h_{n+1}}\left(x\right)=& x^1+x^2+x^3+x^5+\cdots +x_n^h+x^{h_{n+1}}\hfill \\ \hfill =& H^{h_n}\left(x\right)+x^{h_{n+1}}\hfill \end{array}$$

$$\begin{array}{cc}\hfill D\left(x\right)=& x^{h_n+1}+x^{h_n+2}+x^{h_n+3}+\cdots +x^{h_n+k-1}\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill C^{h_{n+1}-1}\left(x\right)=& x^4+x^7+x^{10}+\cdots +x^{h_n-1}+x^{h_n+1}+\cdots +x^{h_n+k-1}\hfill \\ \hfill =& C^{h_n-1}\left(x\right)+D\left(x\right)\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill O^{h_{n+1}}\left(x\right)=& x^1+x^2+x^3+\cdots +x^{h_n}+\cdots +x^{h_{n+1}}\hfill \\ \hfill =& O^{h_n}\left(x\right)+D\left(x\right)+x^{h_{n+1}}.\hfill \end{array}$$

(12)

Thus, we have

$$\begin{array}{cc}\hfill & Y^{h_{n+1}}\left(x\right)\hfill \\ \hfill & =H^{h_{n+1}}{\left(x\right)}^2\hfill \\ \hfill & =O^{h_{n+1}}{\left(x\right)}^2-C^{h_{n+1}-1}{\left(x\right)}^2-2C^{h_{n+1}-1}\left(x\right)·H^{h_{n+1}}\left(x\right)\hfill \\ \hfill & =O^{h_{n+1}}{\left(x\right)}^2-{(C^{h_n-1}\left(x\right)+D\left(x\right))}^2-2(C^{h_n-1}\left(x\right)+D\left(x\right))·(H^{h_n}\left(x\right)+x^{h_{n+1}})\hfill \\ \hfill & =O^{h_{n+1}}{\left(x\right)}^2-C^{h_n-1}{\left(x\right)}^2-2C^{h_n-1}\left(x\right)·D\left(x\right)-D{\left(x\right)}^2-2C^{h_n-1}\left(x\right)·H^{h_n}\left(x\right)\hfill \\ \hfill & -2D\left(x\right)·H^{h_n}\left(x\right)-2C^{h_n-1}\left(x\right)·x^{h_{n+1}}-2D\left(x\right)·x^{h_{n+1}}\hfill \\ \hfill & =O^{h_{n+1}}{\left(x\right)}^2-O^{h_n}{\left(x\right)}^2+H^{h_n}{\left(x\right)}^2-2D\left(x\right)·O^{h_n}\left(x\right)-2C^{h_{n+1}-1}\left(x\right)·x^{h_{n+1}}-D{\left(x\right)}^2.\hfill \end{array}$$

Obviously,

$$\begin{array}{cc}\hfill O^{h_{n+1}}{\left(x\right)}^2-O^{h_n}{\left(x\right)}^2& =2x^{h_n+1}+4x^{h_n+2}+\cdots +2k{x}^{h_n+k}+\cdots +x^{2(h_n+k)}.\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill H^{h_n}{\left(x\right)}^2& =a_1{x}^1+a_2{x}^2+a_3{x}^3+\cdots +a_{h_n}{x}^{h_n}+\cdots +x^{2h_n}.\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill D\left(x\right)·O^{h_n}\left(x\right)& =x^{h_n+2}+2x^{h_n+3}+\cdots +(k-1)x^{h_n+k}+\cdots +x^{2h_n+k-1}.\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill C^{h_{n+1}-1}\left(x\right)·x^{h_{n+1}}& =x^{h_n+k+4}+x^{h_n+k+7}+\cdots +x^{2h_n+2k-1}.\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill D{\left(x\right)}^2& =x^{2h_n+2}+2x^{2h_n+3}+3x^{2h_n+4}+\cdots +x^{2h_n+2k-2}.\hfill \end{array}$$

(17)

The above data are sorted out in the following table:

Table 1. The values of $a_t(h_n+1⩽t⩽h_n+k)$.

Table 1. The values of $a_t(h_n+1⩽t⩽h_n+k)$.

		${\mathit{a}}_{\mathit{t}}$	${\mathit{x}}^{{\mathit{h}}_{\mathit{n}}+1}$	${\mathit{x}}^{{\mathit{h}}_{\mathit{n}}+2}$	${\mathit{x}}^{{\mathit{h}}_{\mathit{n}}+3}$	⋯	${\mathit{x}}^{{\mathit{h}}_{\mathit{n}}+\mathit{k}}$
	a	$O^{h_{n+1}}{\left(x\right)}^2-O^{h_n}{\left(x\right)}^2$	2	4	6	⋯	$2k$
	b	$H^{h_n}{\left(x\right)}^2$	⩾ 0	⩾ 0	⩾ 0	⋯	⩾ 0
	c	$D\left(x\right)·O^{h_n}\left(x\right)$	0	1	2	⋯	$k-1$
	d	$C^{h_{n+1}-1}\left(x\right)·x^{h_{n+1}}$	0	0	0	⋯	0
	e	$D{\left(x\right)}^2$	0	0	0	⋯	0
	$a+b-2c-2d-e$	$Y^{h_{n+1}}\left(x\right)$	⩾ 2	⩾ 2	⩾ 2	⋯	⩾ 2

As can be seen from the table, we have

$$\begin{array}{c}\hfill a_t⩾2(h_n<t⩽h_{n+1}).\end{array}$$

(18)

So $a_t⩾1(h_n<t⩽h_{n+1})$ is true. Therefore, by principle of mathematical induction, $a_r⩾1(r⩾2)$.

Thus,this indicates that there exist two positive integers $h_i$ and $h_j$ such that

$$\begin{array}{c}\hfill h_i+h_j=r(h_n=(p_n-1)/2,r⩾2).\end{array}$$

(19)

In addition, considering symmetry, let $a_r^{\prime }$ be the number of non repeating partitions of the integer r, so

$$\begin{array}{c}\hfill a_r^{\prime }=\left\{\begin{array}{cc}(a_r+1)/2,\hfill & a_r=2s-1(s⩾1)\hfill \\ a_r/2,\hfill & a_r=2s(s⩾1),\hfill \end{array}\right.\end{array}$$

(20)

which proves the lemma. □

4. Demonstration of Goldbach Conjecture

References

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2. G. E. Andrews, and K. Eriksson, Integer partitions, Cambridge University Press, 2004.
3. V. Brun, Le crible d’Eratosthene et le theoreme de Goldbach, Selsk, Skr, 3, 1920.
4. J. R. Chen, On the representation of a larger even integer as the sum of a prime and the product of at most two primes, Sci. Sinica, 16:157–176, 1973.
5. R. Garnier, and J. Taylor, Discrete Mathematics for new Technology 2nd edition, Institute of Physics Publishing, Bristol, 2002.
6. H. A. Helfgott. The ternary Goldbach conjecture is true. arXiv e-prints, December 2013. arXiv:1312.7748.
7. R. C. Vaughan, Goldbach’s Conjectures: A Historical Perspective, In Open problems in mathematics, pages 479–520, Springer, 2016.