The Generating Functions and Goldbach Conjecture

1. Introduction

In 1742, Goldbach proposed two conjectures: the binary Goldbach conjecture, which states that every even integer greater than 2 can be expressed as the sum of two primes, and the ternary Goldbach conjecture, which states that every odd integer greater than 5 can be expressed as the sum of three primes [7].

In 2013, Helfgott successfully proved the ternary Goldbach conjecture [6]. The binary Goldbach conjecture is still unresolved.

For the binary Goldbach conjecture, by the sieve method, since Brun first proved $9+9$ in 1920 [3]. Currently the best result is due to Jingrun Chen [4], who proved $1+2$ in 1973, which prompts that every sufficiently large even integer can be written as the sum of a prime and product of at most two primes. Obviously, it is difficult to solve Goldbach conjecture by studying prime numbers directly.

The Goldbach conjecture is actually a special type of integer partitions. G.E. Andrews and K.Eriksson use generating functions as products to find the number of partitions of n into multiple parts [2]. Therefore, in this paper we use the theory of generating functions to prove that there exist two positive integers $h_i$ and $h_j$, such that $h_i+h_j=r(r⩾2)$, where $h_n=(p_n-1)/2(p_n⩾3)$. Then we complete the proof of Goldbach conjecture.

2. Notation

$p_n,p_i,p_j$: odd prime numbers.

$h_n,h_i,h_j$: the generator of odd prime numbers, positive integers.

$c_n$: the generator of odd composite numbers, positive integers.

$o_n$: the generator of odd numbers, positive integers.

$n,r,s,t$: positive integers.

$a_r,a_t,a_r^{\prime }$: nonnegative integers.

${\left\{p_n\right\}}_1^{\infty },{\left\{h_n\right\}}_1^{\infty },{\left\{c_n\right\}}_1^{\infty }$: infinite sequence.

$P,H,C,D,O$:set.

$C^{h_{n+1}-1}$:a set containing only $c_i(c_i⩽h_{n+1}-1)$.

$Y\left(x\right),H\left(x\right),D\left(x\right),C\left(x\right),O\left(x\right)$: formal power series.

$Y^{h_n}\left(x\right)$: a formal power series containing only $a_r{x}^r(1⩽r⩽h_n)$ ; others are similar.

3. Lemma

Lemma 1.

If $p_n$ denote the n-th odd prime, then there exist two positive integers $h_i$ and $h_j$, such that $h_i+h_j=r(r⩾2)$, where $h_n=(p_n-1)/2$.

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 $a_t(h_n+1⩽t⩽h_n+k)$.

		${\mathit{a}}_{\mathit{t}}$	${\mathit{x}}^{{\mathit{h}}_{\mathit{n}}+1}$	${\mathit{x}}^{{\mathit{h}}_{\mathit{n}}+2}$	${\mathit{x}}^{{\mathit{h}}_{\mathit{n}}+3}$	⋯	${\mathit{x}}^{{\mathit{h}}_{\mathit{n}}+\mathit{k}}$
	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

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

Theorem 1.

There exist two odd primes, $p_i$ and $p_j$, such that $p_i+p_j=2r+2(r⩾2).$

Proof. 

Consider the sequence

$$\begin{array}{c}\hfill \{2r+2,r⩾2\}=\{6,8,10,12,\dots ,2r+2,\dots \}.\end{array}$$

With help of Lemma 1, we have

$$\begin{array}{c}\hfill 2r+2=2(h_i+h_j)+2=(2h_i+1)+(2h_j+1).\end{array}$$

(21)

Since $p_n=2h_n+1$, so we have

$$\begin{array}{c}\hfill 2r+2=p_i+p_j.\end{array}$$

(22)

Hence there exist two odd primes $p_i$ and $p_j$ such that

$$\begin{array}{c}\hfill p_i+p_j=2r+2(r⩾2),\end{array}$$

(23)

which proves the theorem. □

Thus, With help of Theorem 1, for the binary Goldbach conjecture, We have

$$\begin{array}{c}\hfill 2+2=4,p_i+p_j=2r+2(r⩾2).\end{array}$$

(24)

For the ternary Goldbach conjecture, We have

$$\begin{array}{c}\hfill 2+2+3=7,p_i+p_j+3=2r+2+3=2r+5(r⩾2).\end{array}$$

(25)

This proves the Goldbach conjecture.