# New Identities Arising from the Circle Method with Applications to the Goldbach Conjecture

Bill Q Yue October

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# New Identities Arising from the Circle Method with Applications to the Goldbach Conjecture

Bill Q. Yue

October 16, 2025

Formerly Department of Mathematics, Western Michigan University, USA
Former Fellow of the Institute of Mathematics and its Applications (IMA)
Former Chartered Mathematician (CMath), Institute of Mathematics and its Applications (IMA)

Independent Researcher, Sarasota, FL USA
gestyue2@gmail.com

## Abstract

Using the identity $\text{GoldbachNumber}_3(2n + 1) = \sum_{p+q+r=2n+1} 1 = \int_0^1 \left( \sum_{3 \le p \le 2n+1} e^{2\pi ipz} \right)^3 e^{2\pi i(-2n-1)z} dz$, where $p, q, r$ are odd primes and $2n \ge 8$, Vinogradov's method has been highly successful in proving the Weak Goldbach Conjecture [2]. However, similar identities encounter significant challenges when addressing the minor arcs associated with the Goldbach Conjecture.

Through a different approach based on Ramanujan sums, we establish and prove new identities that link Goldbach numbers to specific function evaluations. These results may provide alternative perspectives for studying additive problems involving prime numbers.

Keywords: Goldbach Conjecture, Weak Goldbach Conjecture, circle method, prime numbers, analytic number theory, identities

# Introduction

## Background

In 1937, Vinogradov's method was highly successful in proving the Weak Goldbach Conjecture using the following identity:

$$ \text{GoldbachNumber}_3(2n+1) = \sum_{p+q+r=2n+1} 1 = \int_0^1 \left( \sum_{3 \le p \le 2n+1} e^{2\pi ipz} \right)^3 e^{2\pi i(-2n-1)z} dz = $$
$$ \int_0^1 \left( \sum_{3 \le p < 2n+1} e^{2\pi ipz} \right)^3 e^{2\pi i(-2n-1)z} dz, $$

where $2n \ge 8, p, q, r$ are odd primes.

**Remark:** We use $\sum_{3 \le p < 2n+1}$ rather than $\sum_{3 \le p < 2n+1}$ in the summation because this convention simplifies certain identities. This convention is applied to most summations throughout the paper.

Since then, many mathematicians have made considerable efforts to overcome the difficulties of dealing with minor arcs by using the following identity to approach the Goldbach Conjecture ($2n \ge 6$):

$$ \text{GoldbachNumber}_2(2n) = \sum_{p+q=2n} 1 = \int_0^1 \left( \sum_{3 \le p < 2n} e^{2\pi ipz} \right)^2 e^{2\pi i(-2n)z} dz = \int_0^1 \text{Goldbachz}(2n, z) dz, $$

where

$$ \text{Goldbachz}(2n, z) = \left( \sum_{3 \le p < 2n} e^{2\pi ipz} \right)^2 e^{2\pi i(-2n)z}. $$

It is perhaps surprising that, during almost a century, no other identities—capable of eliminating the difficulties involved in dealing with minor arcs—have been identified.

To address these challenges, we develop and prove various new identities. The key to these identities is Ramanujan's sum. These results were obtained gradually over time. From this process, we observed that "harder identities are sometimes easier to discover," while "easier identities are often harder to discover," and that "harder proofs may be easier to find," while "easier proofs may be harder to find."

In this paper, all symbolic and numerical calculations are carried out using Mathematica when the computational complexity exceeds human capability.

In this paper, all identities, theorems, lemmas, and corollaries—regardless of whether detailed proofs are provided—have been proved. On the other hand, none of the propositions have yet been proved. As a matter of fact:

**Proving one of these propositions could provide a possible pathway toward a proof of the Goldbach Conjecture.**

We have been working on all of these propositions and will share the progress. Where details of a proof are provided, the proof may not always be the most elegant; rather, the aim is to present the relevant techniques. Readers may choose among these techniques to derive alternative proofs, if they wish.

This paper is divided into three parts:

Part I focuses on the Goldbach Conjecture. In Theorem 51 of Part I, interesting results are presented that relate the distribution of odd prime numbers,
the Goldbach Conjecture, and the Twin Prime Conjecture. In Claim 57, also in Part I, we state a related and noteworthy observation concerning the number of odd primes less than $n$. This statement may offer additional insight into the distribution of odd primes and, in a broader sense, into analytic questions related to the Riemann Hypothesis. Further details about the Twin Prime Conjecture and the Prime Number Theorem—especially regarding the number of odd primes—are given in the papers by Yue [6, 7].

Part II is about the Weak Goldbach Conjecture, in which a relation between the Goldbach Conjecture and the Weak Goldbach Conjecture is examined. (There are many such relations, but we present only one.) As far as we know, no previous work has explored direct relations between the two conjectures. This relation may even suggest a way to approach a proof of the Goldbach Conjecture by contradiction. Several propositions are stated in Part II. The identities presented there may also offer a potential pathway to a new approach to the Weak Goldbach Conjecture, possibly strengthening it into the Weak Goldbach Theorem.

Part III considers possible research directions beyond the Goldbach Conjecture and the Weak Goldbach Conjecture.

In this paper, some identities are given in both complex and real forms, while others are provided only in one form. Regarding their structure, it is notable how simple and elegant some of them are. See the two examples below:

If we can prove one of the following, the Goldbach Conjecture would be proved. (See Theorem 30 and Theorem 51.)

$$ \sum_{1 \le j \le 2n} \left( \sum_{3 \le p < 2n} e^{2\pi ip \left( \frac{j}{2n} \right)} \right)^2 > 0 $$

$$ \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos \left( 2\pi p \left( \frac{j}{4n} \right) \right) \right]^2 (-1)^j > 0. $$

There may be different ways of proving the Goldbach Conjecture using different identities. The ultimate goal is to prove the asymptotic formula stated in Proposition 61.

For some propositions, the proof may require advanced techniques such as analytic number theory or probabilistic methods. For others, the proof may require only elementary (though not necessarily easy) approaches. Some propositions may also be particularly suitable for exploration using AI-based methods. Readers may pursue whichever approaches they find most suitable. It seems likely that mathematicians, amateurs, or AI systems could explore some of these propositions and potentially arrive at a proof of the Goldbach Conjecture.

## Notations
$(a, b) = \gcd(a, b)$, $n$ is an integer greater than or equal to $3$, $k$, $d$, $j$, $l$ and $m$ are positive integers, and $p$, $q$, $r$, $p_1$, $p_2$... and $P$ are odd prime numbers.

$|a| = \text{Abs}(a)$ for a number $a$.

$|S| = \text{Size of the set } S$.

$B(x) = O(A(x))$, where $A(x) > 0$, means there exists a positive real constant number $c$, such that $|B(x)| \le cA(x)$ for large $x$.

$B(x) \gg (A(x))$, where $A(x) > 0$ and $B(x) > 0$, means there exists a positive real constant number $c$, such that $B(x) \ge cA(x)$ for large $x$.

$B(x) = o(A(x))$, where $|A(x)| > 0$, means that $\lim_{x \to +\infty} |B(x)/A(x)| = 0$.

$\epsilon(A(x))$, where $|A(x)| > 0$, means that $\lim_{x \to +\infty} |\epsilon(A(x))/A(x)| = 0$.

$B(x) \sim A(x)$, where $A(x) > 0$ and $B(x) > 0$, means $B(x) = A(x) + \epsilon(A(x))$ for large $x$.

$B(x) \asymp A(x)$, where $A(x) > 0$ and $B(x) > 0$, means $B(x) \neq A(x) + \epsilon(A(x))$ for large $x$.

$\mu(m)$ is Möbius function, and $\phi(m)$ is Euler's totient function.

## Definitions

1) We define

$$ \pi_3(n) = \sum_{3 \le p \le 2n} 1 = \pi(n) - 1. $$

Note that

$$ \pi_3(2n) = \sum_{3 \le p \le 2n} 1 = \sum_{3 \le p < 2n} 1 = \pi(2n) - 1. $$

2) We define the following functions:

left

$$ S(n, z) = \sum_{3 \le p < n} e^{2\pi ipz} $$

$$ Scos(n, x) = \sum_{3 \le p < n} \cos(2\pi px) $$

$$ Ssin(n, x) = \sum_{3 \le p < n} \sin(2\pi px) $$

3) The Ramanujan's sum formula used in this paper is as follows

$$ c_m(k) = \mu\left(\frac{m}{(m, k)}\right) \frac{\phi(m)}{\phi\left(\frac{m}{(m, k)}\right)} $$

where $c_m(k) = \sum_{x \in R_m} e^{2\pi ikx}$, and $R_m$ is defined below in 4).

4) We define the following sets.

We define $R$ set of order $m$ $R_m$ as the set:

$$ R_m = \left\{ \frac{j}{m} \mid 1 \le j \le m, (j, m) = 1 \right\}. $$
Note that $R_1 = \{1\}$ and $|R_m| = \phi(m)$.

**Example 1.** Example for $m = 8$:

$$R_8 = \{1/8, 3/8, 5/8, 7/8\}.$$

We define $D_{2n}$ as the set:

$$D_{2n} = \{\frac{j}{2d} | 2d|2n, 1 \le j < 2d, (j, 2d) = 1\}$$

Note that

$$D_{2n} = \bigcup_{2d|2n} R_{2d}.$$

**Example 2.** Example for $2n = 2 \ast 6 = 12$: under the conditions of $2d|12$, the possible values for $2d$ are $2, 4, 6, 12$.

for $2d = 2$, $R_2 = \{1/2\}$,

for $2d = 4$, $R_4 = \{1/4, 3/4\}$,

for $2d = 6$, $R_6 = \{1/6, 5/6\}$,

for $2d = 12$, $R_{12} = \{1/12, 5/12, 7/12, 11/12\}$.

Thus, $D_{12} = \{1/2, 1/4, 3/4, 1/6, 5/6, 1/12, 5/12, 7/12, 11/12\}$.

We define $E_{2n}$ as the set:

$$E_{2n} = \{\frac{j}{4d} | 4d|2n, 1 \le j < 4d, (j, 4d) = 1\}$$

Note that

$$E_{2n} = \bigcup_{4d|2n} R_{4d}$$

So

$$E_{2n} = \{\} \text{ when } 2n = 4k + 2.$$

**Example 3.** Example for $2n = 2 \ast 6 = 12$: under the conditions of $4d|12$, the possible values for $4d$ can be $4, 12$.

for $4d = 4$, $R_4 = \{1/4, 3/4\}$,

for $4d = 12$, $R_{12} = \{1/12, 5/12, 7/12, 11/12\}$.

Thus, $E_{12} = \{1/4, 3/4, 1/12, 5/12, 7/12, 11/12\}$.

We define $V_m$ as the set:

$$V_m = \{\frac{j}{m} | 1 \le j \le m\}$$
hence for any function $f(z)$

$$ \sum_{z \in V_m} f(z) = \sum_{1 \le j \le m} f\left(\frac{j}{m}\right). $$

Note that

$$ |V_m| = m $$

and

$$ V_m = \bigcup_{d|m} R_d. $$

**Example 4.** Example for $2n = 2 \cdot 6 = 12$: under the conditions of $d|12$, the possible values for $d$ can be $1, 2, 3, 6, 12$.

for $d = 1, R_1 = \{1\}$,

for $d = 2, R_2 = \{1/2\}$,

for $d = 3, R_3 = \{1/3, 2/3\}$,

for $d = 4, R_4 = \{1/4, 3/4\}$

for $d = 6, R_6 = \{1/6, 5/6\}$

for $d = 12, R_{12} = \{1/12, 5/12, 7/12, 11/12\}$.

Thus, $V_{12} = \{1/12, 2/12, 3/12, 4/12, 5/12, 6/12, 7/12, 8/12, 9/12, 10/12, 11/12, 12/12\} =$

$$ \{1, 1/2, 1/3, 2/3, 1/4, 3/4, 1/6, 5/6, 1/12, 5/12, 7/12, 11/12\}. $$

We define $T_{2n}$ as the set:

$$ T_{2n} = \{ \frac{2j-1}{2n} \mid 1 \le j \le n \}. $$

Note that

$$ |T_{2n}| = n $$

and

$$ T_{2n} = \bigcup_{(2k+1)|2n} R_{2n/(2k+1)} \text{ where } k \ge 0. $$

**Example 5.** Example for $n = 12$: $2n = 24$.

$$ T_{24} = \{1/24, 1/8, 5/24, 7/24, 3/8, 11/24, 13/24, 5/8, 17/24, 19/24, 7/8, 23/24\}. $$

**Example 6.** Example for $n = 7$: $2n = 14$.
$$T_{14} = \{1/14, 3/14, 5/14, 1/2, 9/14, 11/14, 13/14\}.$$

We define

$$C_2 = \prod_{3 \le p} \left[ 1 - \frac{1}{(p-1)^2} \right] = 0.6601618158... - \text{Twin Prime Constant.}$$

# Part I

## Goldbach Conjecture

For $2n \ge 6$, we have

$$ \begin{aligned} \text{GoldbachNumber}_2(2n) &= \sum_{p+q=2n} 1 = \\ & \int_0^1 \left( \sum_{3 \le p \le 2n} e^{2\pi ipz} \right)^2 e^{2\pi i(-2n)z} dz = \\ & \int_0^1 \left( \sum_{3 \le p < 2n} e^{2\pi ipz} \right)^2 e^{2\pi i(-2n)z} dz. \end{aligned} $$

To prove Goldbach Conjecture we need and only need to prove $\text{GoldbachNumber}_2(2n) > 0$.

We define the following complex functions

$$ \text{Goldbachz}0(2n, z) = [S(2n, z)]^2 = \left( \sum_{3 \le p < 2n} e^{2\pi ipz} \right)^2 $$

and

$$ \text{Goldbachz}(2n, z) = \text{Goldbach0}(2n, z)e^{-2\pi i(2n)z}. $$

We define the following real functions

$$ \text{Goldbachx}0(2n, x) = [\text{Scos}(2n, x)]^2 - [\text{Ssin}(2n, x)]^2 = \left[ \sum_{3 \le p < 2n} \cos(2\pi px) \right]^2 - \left[ \sum_{3 \le p < 2n} \sin(2\pi px) \right]^2 $$

and

$$ \text{Goldbachx}(2n, x) = 4[\text{Scos}(2n, x)]^2 \cos(2\pi(2n)x). $$

Before presenting the details of the proof, we first state the following note-worthy identities.

The most striking and surprising identities are as follows:
$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{z \in V_{2n}} \text{Goldbachz0}(2n, z)}{2n} $$

$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{x \in V_{2n}} \text{Goldbachx0}(2n, x)}{2n} $$

The most elegant and simplest identity is as follows:

$$ \text{GoldbachNumber}_2(2n) = \frac{1}{2n} \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 (-1)^j. $$

# 1 The First Set of Identities

## 1.1 Identities and Examples

**Theorem 7.** *The First set of Identities*

Complex version:

$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{z \in D_{2n}} \text{Goldbachz0}(2n, z) - \frac{1}{2} \sum_{z \in E_{2n}} \text{Goldbachz0}(2n, z)}{n}. $$

Real version:

$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{x \in D_{2n}} \text{Goldbachx0}(2n, x) - \frac{1}{2} \sum_{x \in E_{2n}} \text{Goldbachx0}(2n, x)}{n}. $$

Please note that each of the following is an integer for any 2n, according to Ramanujan's sum.

$$ \sum_{z \in D_{2n}} \text{Goldbachz}(2n, z), \sum_{z \in E_{2n}} \text{Goldbachz}(2n, z), \sum_{x \in D_{2n}} \text{Goldbachx}(2n, x) \text{ and } \sum_{x \in E_{2n}} \text{Goldbachx}(2n, x). $$

**Example 8.** Example for $2n = 2 \cdot 6 = 12$:

Note that $\text{GoldbachNumber}_2(12) = 2$.

Complex version, by definition we have

$$ \text{Goldbachz0}(12, z) = [e^{2\pi i(3)z} + e^{2\pi i(5)z} + e^{2\pi i(7)z} + e^{2\pi i(11)z}]^2 $$

$$ D_{12} = \{1/2, 1/4, 3/4, 1/6, 5/6, 1/12, 5/12, 7/12, 11/12\} $$
$$E_{12} = \{1/4, 3/4, 1/12, 5/12, 7/12, 11/12\}.$$

By evaluating the functions, we obtain the following:

$$\frac{\sum_{z \in D_{12}} Goldbachz0(12, z) - \frac{1}{2} \sum_{z \in E_{12}} Goldbachz0(12, z)}{6} = \frac{9 - \frac{1}{2}(-6)}{6} = 2$$

$$\text{Thus, GoldbachNumber}_2(12) = \frac{\sum_{z \in D_{12}} Goldbachz0(12, z) - \frac{1}{2} \sum_{z \in E_{12}} Goldbachz0(12, z)}{6}.$$

Real version, by definition we have

$$\text{Goldbachx0}(12, x) = [\cos(2\pi(3)x) + \cos(2\pi(5)x) + \cos(2\pi(7)x) + \cos(2\pi(11)x)]^2 -$$

$$[\sin(2\pi(3)x) + \sin(2\pi(5)x) + \sin(2\pi(7)x) + \sin(2\pi(11)x)]^2$$

$$D_{12} = \{1/2, 1/4, 3/4, 1/6, 5/6, 1/12, 5/12, 7/12, 11/12\}$$

$$E_{12} = \{1/4, 3/4, 1/12, 5/12, 7/12, 11/12\}.$$

By evaluating the functions, we obtain the following:

$$\frac{\sum_{x \in D_{12}} Goldbachx0(12, x) - \frac{1}{2} \sum_{x \in E_{12}} Goldbachx0(12, x)}{6} = \frac{9 - \frac{1}{2}(-6)}{6} = 2$$

$$\text{Thus, GoldbachNumber}_2(12) = \frac{\sum_{x \in D_{12}} Goldbachx0(12, x) - \frac{1}{2} \sum_{x \in E_{12}} Goldbachx0(12, x)}{6}.$$

**Example 9.** Example for $2n = 2 * 15 = 30$:

Note that $GoldbachNumber_2(30) = 6$ and

$$\text{Goldbachz0}(30, z) = [e^{2\pi i(3)z} + e^{2\pi i(5)z} + e^{2\pi i(7)z} + e^{2\pi i(11)z} + e^{2\pi i(13)z} + e^{2\pi i(17)z} + e^{2\pi i(19)z} + e^{2\pi i(23)z} + e^{2\pi i(29)z}]^2$$

$$\text{Goldbachx0}(30, x) = [\cos(2\pi(3)x) + \cos(2\pi(5)x) + \cos(2\pi(7)x) + \cos(2\pi(11)x) + \cos(2\pi(13)x)$$
$$ 
\begin{aligned}
& +\cos(2\pi(17)x) + \cos(2\pi(19)x) + \cos(2\pi(23)x) + \cos(2\pi(29)x)]^2 - \\
& [\sin(2\pi(3)x) + \sin(2\pi(5)x) + \sin(2\pi(7)x) + \sin(2\pi(11)x) + \cos(2\pi(13)x) \\
& +\sin(2\pi(17)x) + \sin(2\pi(19)x) + \sin(2\pi(23)x) + \sin(2\pi(29)x)]^2
\end{aligned}
$$

$$ D_{30} = \{1/2, 1/6, 5/6, 1/10, 3/10, 7/10, 9/10, 1/30, 7/30, 11/30, 13/30, 17/30, 19/30, 23/30, 29/30\}. $$

$$ E_{30} = \{\} $$

Complex version, by evaluating the functions we obtain the following:

$$ \frac{\sum_{z \in D_{30}} \text{Goldbachz0}(30, z) - \frac{1}{2} \sum_{z \in E_{30}} \text{Goldbachz0}(30, z)}{15} = \frac{90 - \frac{1}{2}(0)}{15} = 6 $$

$$ \text{Thus, GoldbachNumber}_2(30) = \frac{\sum_{z \in D_{30}} \text{Goldbachz0}(30, z) - \frac{1}{2} \sum_{z \in E_{30}} \text{Goldbachz0}(30, z)}{15}. $$

Real version, by evaluating the functions, we obtain the following:

$$ \frac{\sum_{x \in D_{30}} \text{Goldbachx0}(2n, x) - \frac{1}{2} \sum_{x \in E_{30}} \text{Goldbachx0}(2n, x)}{15} = \frac{90 - \frac{1}{2}(0)}{15} = 6 $$

$$ \text{Thus, GoldbachNumber}_2(30) = \frac{\sum_{x \in D_{30}} \text{Goldbachx0}(30, x) - \frac{1}{2} \sum_{x \in E_{30}} \text{Goldbachx0}(30, x)}{15}. $$

## 1.2 Proof of the Identities

We prove complex version first, in other words we prove the following first.

$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{z \in D_{2n}} \text{Goldbachz0}(2n, z) - \frac{1}{2} \sum_{z \in E_{2n}} \text{Goldbachz0}(2n, z)}{n}. $$

Recall that

$$ \text{Goldbachz}(2n, z) = \text{Goldbachz0}(2n, z)e^{2\pi i(-2n)z} = $$
$$[S(2n, z)]^2 e^{2\pi i(-2n)z} =$$

$$[\sum_{3 \le p < 2n} e^{2\pi i p z}]^2 e^{2\pi i(-2n)z} =$$

$$[\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} e^{2\pi i(p+q)z}] e^{2\pi i(-2n)z} =$$

$$\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} e^{2\pi i(p+q-2n)z}.$$

**Lemma 10.** There are exactly $n$ integer values for $|(p+q-2n)|$.

Proof.

We first observe the possible integer values for $(p+q-2n)$.

For $(p+q-2n) < 0$, we have $6-2n \le (p+q-2n) \le -2$, hence there are $(-2-(6-2n))/2+1 = n-3$ values.

For $(p+q-2n) = 0$, there is one value.

For $0 < (p+q-2n)$, since the largest possible $p$ is $(2n-1)$, the largest possible $q$ is $(2n-1)$, we have $2 \le (p+q-2n) \le ((2n-1)+(2n-1)-2n) = 2n-2$, hence there are $((2n-2)-2)/2 + 1 = n-1$ values.

Note that $(n-1) > (n-3)$, thus as we put above together, $|(p+q-2n)|$ has exactly $(n-1)+1=n$ integer values. These values are 0, 2, 4, ..., $2n-2$.

Remark: $(p+q-2n)$ has $(n-3)+1+(n-1) = 2n-3$ integer values while $|(p+q-2n)|$ has $n$ integer values.

Next, we let $k$ be an integer with $2k \ge 2$. Recall that

$$R_{2k} = \{ \frac{j}{2k} \mid 1 \le j < 2k, (j, 2k) = 1 \}.$$

**Lemma 11.** The following is true for any integers $k$ and $m$, where $2k \ge 2$, $m \ge 2$.

$$\sum_{z \in R_{2k}} e^{2\pi i(m)z} = \sum_{z \in R_{2k}} e^{2\pi i(-m)z}.$$

Proof.

The proof is left to the reader.

**Lemma 12.** The following holds for any integers $k$, where $2k \ge 2$.

$$\sum_{z \in R_{2k}} \text{Goldbachz}(2n, z) = \sum_{z \in R_{2k}} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} e^{2\pi i(p+q-2n)z} = \\ \sum_{z \in R_{2k}} \sum_{0 \le m \le n-1} a_m e^{2\pi i(m)z}.$$

where $a_m$, $0 \le m \le n-1$ are $n$ variables with integer values.
Proof.

By Lemma 11 we have the following:

$$ \sum_{z \in R_{2k}} e^{2\pi i(m)z} = \sum_{z \in R_{2k}} e^{2\pi i(-m)z} $$

and by Lemma 10 we know that $|(p+q-2n)|$ has exactly $n$ integer values, which are 0, 2, 4, ..., $2n-2$. We define

$a_0$ as the coefficient of $e^{2\pi i(0)z}$ in the summation of $\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} e^{2\pi i|p+q-2n|z}$,

$a_1$ as the coefficient of $e^{2\pi i(2)z}$ in the summation of $\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} e^{2\pi i|p+q-2n|z}$,

...

$a_{(n-1)}$ as the coefficient of $e^{2\pi i(2n-2)z}$ in the summation of $\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} e^{2\pi i|p+q-2n|z}$.

This completes the proof.

Note that $a_0 = \text{GoldbachNumber}_2(2n)$, and $\sum_{0 \le m \le n-1} a_m = [\pi_3(2n)]^2$.

**Lemma 13.** A linear system with $n$ variables $\{a_0, a_1, a_2, ..., a_{n-1}\}$ can be expressed as follows.

$$ \begin{bmatrix} b_{11} & b_{12} & \dots & b_{1n} \\ b_{21} & b_{22} & \dots & b_{2n} \\ b_{31} & b_{32} & \dots & b_{3n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1} & b_{n2} & \dots & b_{nn} \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ \vdots \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} \sum_{z \in R_2} \text{Goldbachz}(2n, z) \\ \sum_{z \in R_4} \text{Goldbachz}(2n, z) \\ \sum_{z \in R_6} \text{Goldbachz}(2n, z) \\ \vdots \\ \sum_{z \in R_{2n}} \text{Goldbachz}(2n, z) \end{bmatrix} $$

where $b_{st} = \mu(\frac{s}{(s,t-1)}) \frac{\phi(2s)}{\phi(\frac{s}{(s,t-1)})}$, $1 \le s \le n$, $1 \le t \le n$.

Proof.

Note that for any $2k \ge 2$

$$ \sum_{z \in R_{2k}} \text{Goldbachz}(2n, z) = \sum_{z \in R_{2k}} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} e^{2\pi i(p+q-2n)z} = \\ \sum_{z \in R_{2k}} \sum_{0 \le m \le n-1} a_m e^{2\pi i(m)z}. $$

Rewrite the following equation

$$ \sum_{z \in R_{2k}} \sum_{0 \le m \le n-1} a_m e^{2\pi i(m)z} = \sum_{z \in R_{2k}} \text{Goldbachz}(2n, z) $$

as

$$ \sum_{0 \le m \le n-1} (\sum_{z \in R_{2k}} e^{2\pi i(m)z}) a_m = \sum_{z \in R_{2k}} \text{Goldbachz}(2n, z), $$

let $2k$ run over the range of $(2, 2n)$ we have
$$ (\sum_{z \in R_2} e^{2\pi i(0)z})a_0 + (\sum_{z \in R_2} e^{2\pi i(2)z})a_1 + \dots (\sum_{z \in R_2} e^{2\pi i((2n-2))z})a_{n-1} = \sum_{z \in R_2} \text{Goldbachz}(2n, z) \\ (\sum_{z \in R_4} e^{2\pi i(0)z})a_0 + (\sum_{z \in R_4} e^{2\pi i(2)z})a_1 + \dots (\sum_{z \in R_4} e^{2\pi i((2n-2))z})a_{n-1} = \sum_{z \in R_4} \text{Goldbachz}(2n, z) \\ (\sum_{z \in R_6} e^{2\pi i(0)z})a_0 + (\sum_{z \in R_6} e^{2\pi i(2)z})a_1 + \dots (\sum_{z \in R_6} e^{2\pi i((2n-2))z})a_{n-1} = \sum_{z \in R_6} \text{Goldbachz}(2n, z) \\ (\sum_{z \in R_{2n}} e^{2\pi i(0)z})a_0 + (\sum_{z \in R_{2n}} e^{2\pi i(2)z})a_1 + \dots (\sum_{z \in R_{2n}} e^{2\pi i((2n-2))z})a_{n-1} = \sum_{z \in R_{2n}} \text{Goldbachz}(2n, z) $$

which is

$$ \begin{bmatrix} \sum_{z \in R_2} e^{2\pi i(0)z} & \sum_{z \in R_2} e^{2\pi i(2)z} & \dots & \sum_{z \in R_2} e^{2\pi i((2n-2))z} \\ \sum_{z \in R_4} e^{2\pi i(0)z} & \sum_{z \in R_4} e^{2\pi i(2)z} & \dots & \sum_{z \in R_4} e^{2\pi i((2n-2))z} \\ \sum_{z \in R_6} e^{2\pi i(0)z} & \sum_{z \in R_6} e^{2\pi i(2)z} & \dots & \sum_{z \in R_6} e^{2\pi i((2n-2))z} \\ \vdots & \vdots & \ddots & \vdots \\ \sum_{z \in R_{2n}} e^{2\pi i(0)z} & \sum_{z \in R_{2n}} e^{2\pi i(2)z} & \dots & \sum_{z \in R_{2n}} e^{2\pi i((2n-2))z} \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ \vdots \\ a_{n-1} \end{bmatrix} = $$ $$ \begin{bmatrix} \sum_{z \in R_2} \text{Goldbachz}(2n, z) \\ \sum_{z \in R_4} \text{Goldbachz}(2n, z) \\ \sum_{z \in R_6} \text{Goldbachz}(2n, z) \\ \vdots \\ \sum_{z \in R_{2n}} \text{Goldbachz}(2n, z) \end{bmatrix} $$

By using Ramanujan's sum, we have $\sum_{z \in R_{2s}} e^{2\pi i(2t-2)z} = \mu(\frac{2s}{(2s,2t-2)}) \frac{\phi(2s)}{\phi(\frac{2s}{(2s,2t-2)})}$, $1 \le s \le n, 1 \le t \le n$.

Note that $\mu(\frac{2s}{(2s,2t-2)}) \frac{\phi(2s)}{\phi(\frac{2s}{(2s,2t-2)})} = \mu(\frac{s}{(s,t-1)}) \frac{\phi(2s)}{\phi(\frac{s}{(s,t-1)})}$, $1 \le s \le n, 1 \le t \le n$.

Let $b_{st} = \mu(\frac{s}{(s,t-1)}) \frac{\phi(2s)}{\phi(\frac{s}{(s,t-1)})}$, $1 \le s \le n, 1 \le t \le n$, that completes the proof of Lemma 13.

**Lemma 14.** We define the following function

For integer $s, h(s) = \begin{cases} 1 & \text{if } s \text{ is odd} \\ \frac{1}{2} & \text{if } s \text{ is even} \end{cases}$

Then a linear system with $n$ variables $\{a_0, a_1, a_2, ..., a_{n-1}\}$ can be expressed as follows.

$$ \begin{bmatrix} g_{11} & g_{12} & \dots & g_{1n} \\ g_{21} & g_{22} & \dots & g_{2n} \\ g_{31} & g_{32} & \dots & g_{3n} \\ \vdots & \vdots & \ddots & \vdots \\ g_{n1} & g_{n2} & \dots & g_{nn} \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ \vdots \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} h(1) \sum_{z \in R_2} \text{Goldbachz}(2n, z) \\ h(2) \sum_{z \in R_4} \text{Goldbachz}(2n, z) \\ h(3) \sum_{z \in R_6} \text{Goldbachz}(2n, z) \\ \vdots \\ h(n) \sum_{z \in R_{2n}} \text{Goldbachz}(2n, z) \end{bmatrix} $$

where $g_{st} = \mu(\frac{s}{(s,t-1)}) \frac{\phi(s)}{\phi(\frac{s}{(s,t-1)})}$, $1 \le s \le n, 1 \le t \le n$.

Proof.

Note that $\phi(2s)h(s) = \phi(s)$, linear system in Lemma 13 is
$$
\begin{bmatrix} b_{11} & b_{12} & \dots & b_{1n} \\ b_{21} & b_{22} & \dots & b_{2n} \\ b_{31} & b_{32} & \dots & b_{3n} \\ \vdots & \vdots & \ddots & \vdots \\ b_{n1} & b_{n2} & \dots & b_{nn} \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ \vdots \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} \sum_{z \in R_2} \text{Goldbachz}(2n, z) \\ \sum_{z \in R_4} \text{Goldbachz}(2n, z) \\ \sum_{z \in R_6} \text{Goldbachz}(2n, z) \\ \vdots \\ \sum_{z \in R_{2n}} \text{Goldbachz}(2n, z) \end{bmatrix}
$$

$$
\text{where } b_{st} = \mu\left(\frac{s}{(s,t-1)}\right) \frac{\phi(2s)}{\phi\left(\frac{s}{(s,t-1)}\right)}, \quad 1 \le s \le n, 1 \le t \le n.
$$

Multiply both sides on the left by the diagonal matrix

$$
\begin{bmatrix} h(1) & 0 & \dots & 0 \\ 0 & h(2) & \dots & 0 \\ 0 & 0 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & h(n) \end{bmatrix} \quad \text{then we have the following linear system}
$$

$$
\begin{bmatrix} g_{11} & g_{12} & \dots & g_{1n} \\ g_{21} & g_{22} & \dots & g_{2n} \\ g_{31} & g_{32} & \dots & g_{3n} \\ \vdots & \vdots & \ddots & \vdots \\ g_{n1} & g_{n2} & \dots & g_{nn} \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ \vdots \\ a_{n-1} \end{bmatrix} = \begin{bmatrix} h(1) \sum_{z \in R_2} \text{Goldbachz}(2n, z) \\ h(2) \sum_{z \in R_4} \text{Goldbachz}(2n, z) \\ h(3) \sum_{z \in R_6} \text{Goldbachz}(2n, z) \\ \vdots \\ h(n) \sum_{z \in R_{2n}} \text{Goldbachz}(2n, z) \end{bmatrix}
$$

$$
\text{where } g_{st} = \mu\left(\frac{s}{(s,t-1)}\right) \frac{\phi(s)}{\phi\left(\frac{s}{(s,t-1)}\right)}, \quad 1 \le s \le n, 1 \le t \le n.
$$

This completes the proof.

$$
\text{Let } G = \begin{bmatrix} g_{11} & g_{12} & \dots & g_{1n} \\ g_{21} & g_{22} & \dots & g_{2n} \\ g_{31} & g_{32} & \dots & g_{3n} \\ \vdots & \vdots & \ddots & \vdots \\ g_{n1} & g_{n2} & \dots & g_{nn} \end{bmatrix}, \text{ where } g_{st} = \mu\left(\frac{s}{(s,t-1)}\right) \frac{\phi(s)}{\phi\left(\frac{s}{(s,t-1)}\right)}, \quad 1 \le s \le n,
$$

$1 \le t \le n$, we have the following Lemma.

**Lemma 15.** The following holds:

$$
\mathrm{Det}(G) = (-1)^{n+1}n! \neq 0. \text{ So } G^{-1} \text{ exists, where } G^{-1} \text{ is inverse of } G.
$$

To keep the paper concise, the proof is omitted.

Note that to compute $a_0 = \text{GoldbachNumber}_2(2n)$, so we only need the first
row of $G^{-1}$; the other rows are irrelevant.

**Lemma 16.** The following holds:

$$
\text{The first row of } G^{-1} = [u(1, n)^{\frac{1}{n}}, u(2, n)^{\frac{1}{n}}, u(3, n)^{\frac{1}{n}}, \dots, u(n, n)^{\frac{1}{n}}],
$$

$$
\text{where } u(j, n) = \begin{cases} 1 & \text{if } j \mid n \\ 0 & \text{if } j \nmid n \end{cases}.
$$

To keep the paper concise, the proof is omitted.

Finally, we have

$a_0 = \text{GoldbachNumber}_2(2n)$

$$
= [u(1, n)^{\frac{1}{n}}, u(2, n)^{\frac{1}{n}}, u(3, n)^{\frac{1}{n}}, \dots, u(n, n)^{\frac{1}{n}}]
\begin{bmatrix}
h(1) \sum_{z \in R_2} \text{Goldbachz}(2n, z) \\
h(2) \sum_{z \in R_4} \text{Goldbachz}(2n, z) \\
h(3) \sum_{z \in R_6} \text{Goldbachz}(2n, z) \\
\vdots \\
h(n) \sum_{z \in R_{2n}} \text{Goldbachz}(2n, z)
\end{bmatrix}
$$
$$= \sum_{j|n} \frac{h(j) \sum_{z \in R_{2j}} Goldbachz(2n, z)}{n}.$$

**Lemma 17.** The following holds:

$$a_0 = GoldbachNumber_2(2n)$$

$$= \sum_{j|n} \frac{h(j) \sum_{z \in R_{2j}} Goldbachz(2n, z)}{n}$$

$$= \frac{\sum_{z \in D_{2n}} Goldbachz0(2n, z) - \frac{1}{2} \sum_{z \in E_{2n}} Goldbachz0(2n, z)}{n}.$$

Proof is left for the reader.

This completes the proof of Theorem 7 of the complex version.

To prove Theorem 7 for the real version, we only need to prove

$$\sum_{z \in D_{2n}} Goldbachz0(2n, z) = \sum_{x \in D_{2n}} Goldbachx0(2n, x)$$

and

$$\sum_{z \in E_{2n}} Goldbachz0(2n, z) = \sum_{x \in E_{2n}} Goldbachx0(2n, x).$$

Note that

$$D_{2n} = \bigcup_{2d|2n} R_{2d} = \bigcup_{j|n} R_{2j}$$

and

$$E_{2n} = \bigcup_{4d|2n} R_{4d} = \bigcup_{2j|n} R_{2(2j)}.$$

So we only need to prove the following holds for all $R_{2j}$ where $j|n$:

$$\sum_{j|n, z \in R_{2j}} Goldbachz0(2n, z) = \sum_{j|n, x \in R_{2j}} Goldbachx0(2n, x). \quad (1.1)$$

Note that $e^{iw} = \cos(w) + i\sin(w)$, we have

$$\begin{aligned} Goldbachz0(2n, z) &= \left( \sum_{3 \le p < 2n} e^{2\pi ipz} \right)^2 = \left( \sum_{3 \le p < 2n} e^{2\pi ipx} \right)^2 \\ &= \left[ \sum_{3 \le p < 2n} \cos(2\pi px) + i\sin(2\pi px) \right]^2 \end{aligned}$$
$$
\begin{align*}
&= [\sum_{3 \le p < 2n} \cos(2\pi px)]^2 + 2i \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\sin(2\pi qx) - [\sum_{3 \le p < 2n} \sin(2\pi px)]^2 \\
&= \{[\sum_{3 \le p < 2n} \cos(2\pi px)]^2 - [\sum_{3 \le p < 2n} \sin(2\pi px)]^2\} + 2i \{\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\sin(2\pi qx)\} \\
&= \{[S\cos(2n, x)]^2 - [S\sin(2n, x)]^2\} + 2i \{\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\sin(2\pi qx)\} \\
&= \operatorname{Goldbachx0}(2n, x) + 2i \{\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\sin(2\pi qx)\}.
\end{align*}
$$

Hence we have

$$
\sum_{j|n,z \in R_{2j}} \text{Goldbach}0(2n, z) = \sum_{j|n,x \in R_{2j}} \text{Goldbach}0(2n, x) + 2i \sum_{j|n,z \in R_{2j}} \{ \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\sin(2\pi qx) \}.
$$

By Ramanujan's sum we immediately obtain that $\sum_{z \in R_k} \text{Goldbachz}(2n, z)$ is an integer. Hence, the following must be true. Readers can prove it directly as well.

$$
2i \sum_{j|n,z \in R_{2j}} \left\{ \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px) \sin(2\pi qx) \right\} = 0.
$$

Thus, we have

$$
\sum_{j|n,z \in R_{2j}} \text{Goldbach}0(2n,z) = \sum_{j|n,x \in R_{2j}} \text{Goldbach}0(2n,x).
$$

This completes the proof of Theorem 7 for the real version.

1.3 Propositions

**Proposition 18.** One of the following statements holds:

$$
\sum_{z \in D_{2n}} \text{Goldbach}0(2n, z) - \frac{1}{2} \sum_{z \in E_{2n}} \text{Goldbach}0(2n, z) > 0
$$

or

$$
\sum_{z \in D_{2n}} \text{Goldbach}0(2n, z) - \frac{1}{2} \sum_{z \in E_{2n}} \text{Goldbach}0(2n, z) > 0.
$$
# 2 The Second Set of Identities

## 2.1 Identities and Examples

**Theorem 19.** For any $2m > 2n$, we have

$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{z \in D_{2m}} \text{Goldbachz}(2n, z) - \frac{1}{2} \sum_{z \in E_{2m}} \text{Goldbachz}(2n, z)}{m} $$

**Example 20.** Example for $2n = 12$, $n = 6$, $m = n+1 = 7$, and $2m = 2*7 = 14$.

Note that $\text{GoldbachNumber}_2(12) = 2$.

By definition, we have

$$ \text{Goldbachz}(12, z) = [e^{2\pi i(3)z} + e^{2\pi i(5)z} + e^{2\pi i(7)z} + e^{2\pi i(11)z}]^2 e^{2\pi i(-12)z} $$

$$ D_{14} = \{1/2, 1/14, 3/14, 5/14, 9/14, 11/14, 13/14\} $$

$$ E_{14} = \{\} $$

By evaluating the functions, we obtain the following:

$$ \frac{\sum_{z \in D_{14}} \text{Goldbachz}(12, z) - \frac{1}{2} \sum_{z \in E_{14}} \text{Goldbachz}(12, z)}{7} = \frac{14 - \frac{1}{2}(0)}{7} = 2 $$

$$ \text{Thus, } \text{GoldbachNumber}_2(12) = \frac{\sum_{z \in D_{14}} \text{Goldbachz}(12, z) - \frac{1}{2} \sum_{z \in E_{14}} \text{Goldbachz}(12, z)}{7} $$

**Example 21.** Example for $2n = 12$, $n = 6$, $m = n + 2 = 8$, $2m = 2*8 = 16$.

Note that $\text{GoldbachNumber}_2(12) = 2$.

By definition, we have

$$ \text{Goldbachz}(12, z) = [e^{2\pi i(3)z} + e^{2\pi i(5)z} + e^{2\pi i(7)z} + e^{2\pi i(11)z}]^2 e^{2\pi i(-12)z} $$

$$ D_{16} = \{1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, 1/16, 3/16, 5/16, 7/16, 9/16, 11/16, 13/16, 15/16\} $$
$E_{16} = \{1/4, 3/4, 1/8, 3/8, 5/8, 7/8, 1/16, 3/16, 5/16, 7/16, 9/16, 11/16, 13/16, 15/16\}$.

By evaluating the functions, we obtain the following:

$$ \frac{\sum_{z \in D_{16}} Goldbachz(12, z) - \frac{1}{2} \sum_{z \in E_{16}} Goldbachz(12, z)}{8} = \frac{16 - \frac{1}{2}(0)}{8} = 2 $$

$$ \text{Thus, } GoldbachNumber_2(12) = \frac{\sum_{z \in D_{16}} Goldbachz(12, z) - \frac{1}{2} \sum_{z \in E_{16}} Goldbachz(12, z)}{8}. $$

**Example 22.** Example for $2n = 12$, $n = 6$, $m = n + 4 = 10$, $2m = 2*10 = 20$.

Note that $GoldbachNumber_2(12) = 2$.

By definition, we have

$$ Goldbachz(12, z) = [e^{2\pi i(3)z} + e^{2\pi i(5)z} + e^{2\pi i(7)z} + e^{2\pi i(11)z}]^2 e^{2\pi i(-12)z} $$

$$ D_{20} = \{1/2, 1/4, 3/4, 1/10, 3/10, 7/10, 9/10, 1/20, 3/20, 7/20, 9/20, 11/20, 13/20, 17/20, 19/20\} $$

$$ E_{20} = \{1/4, 3/4, 1/20, 3/20, 7/20, 9/20, 11/20, 13/20, 17/20, 19/20\}. $$

By evaluating the functions, we obtain the following:

$$ \frac{\sum_{z \in D_{20}} Goldbachz(12, z) - \frac{1}{2} \sum_{z \in E_{20}} Goldbachz(12, z)}{10} = \frac{25 - \frac{1}{2}(10)}{10} = 2 $$

$$ \text{Thus, } GoldbachNumber_2(12) = \frac{\sum_{z \in D_{20}} Goldbachz(12, z) - \frac{1}{2} \sum_{z \in E_{20}} Goldbachz(12, z)}{10}. $$

## 2.2 Proof of the Identities

*Proof.* By following arguments similar to those used in the proof of 7, we can prove Theorem 19. □
## 2.3 Propositions

**Proposition 23.** For any $2n \ge 6$, there exists at least one $2m > 2n$ such that

$$ \sum_{z \in D_{2m}} \text{Goldbachz}(2n, z) - \frac{1}{2} \sum_{z \in E_{2m}} \text{Goldbachz}(2n, z) > 0. $$

# 3 The Third Set of Identities

## 3.1 Identities and Examples

**Theorem 24.** For any $P > 3(2n)$

$$ \text{GoldbachNumber}_2(2n) = \frac{4 \sum_{x \in T_{2P}} [\text{Scos}(2n, x)]^2 \cos(2\pi(2n)x)}{P} $$

$$ = \frac{4 \sum_{x \in T_{2P}} [\sum_{3 \le p < 2n} \cos(2\pi px)]^2 \cos(2\pi(2n)x)}{P}. $$

**Example 25.** Example for $2n = 8$.

We take $P = 29 > 3*8 = 24$, $2P = 58$, we have $\text{GoldbachNumber}_2(8) = 2$.

$$ \frac{4 \sum_{x \in T_{2P}} [\sum_{3 \le p \le 2n} \cos(2\pi px)]^2 \cos(2\pi(2n)x)}{P} = \frac{4 \sum_{x \in T_{58}} [\sum_{3 \le p < 8} \cos(2\pi px)]^2 \cos(2\pi(2n)x)}{29}. $$

$T_{58} = \{1/58, 3/58, 5/58, 7/58, 9/58, 11/58, 13/58, 15/58, 17/58, 19/58, 21/58, 23/58, 25/58, 27/58, 1/2, 31/58, 33/58, 35/58, 37/58, 39/58, 41/58, 43/58, 45/58, 47/58, 49/58, 51/58, 53/58, 55/58, 57/58\}.$

$$ [\sum_{3 \le p < 8} \cos(2\pi px)]^2 \cos(2\pi(2n)x) = $$

$$ [\cos(2\pi(3)x) + \cos(2\pi(5)x) + \cos(2\pi(7)x)]^2 \cos(2\pi(8)x). $$

$$ \sum_{x \in T_{58}} [\cos(2\pi(3)x) + \cos(2\pi(5)x) + \cos(2\pi(7)x)]^2 \cos(2\pi(8)x) = $$
$$23 - 17 \cos\left(\frac{\pi}{29}\right) + 17 \cos\left(\frac{2\pi}{29}\right) - 17 \cos\left(\frac{3\pi}{29}\right) + 17 \cos\left(\frac{4\pi}{29}\right) - 17 \cos\left(\frac{5\pi}{29}\right) + \\ 17 \cos\left(\frac{6\pi}{29}\right) - 17 \cos\left(\frac{7\pi}{29}\right) + 17 \sin\left(\frac{\pi}{58}\right) - 17 \sin\left(\frac{3\pi}{58}\right) + \\ 17 \sin\left(\frac{5\pi}{58}\right) - 17 \sin\left(\frac{7\pi}{58}\right) + 17 \sin\left(\frac{9\pi}{58}\right) - 17 \sin\left(\frac{11\pi}{58}\right) + 17 \sin\left(\frac{13\pi}{58}\right) = \frac{29}{2}.$$

Thus,

$$\frac{4 \sum_{x \in T_{2P}} \left[ \sum_{3 \le p \le 2n} \cos(2\pi px) \right]^2 \cos(2\pi(2n)x)}{P} = \\ \frac{4 \sum_{x \in T_{58}} \left[ \sum_{3 \le p < 8} \cos(2\pi px) \right]^2 \cos(2\pi(2n)x)}{29} = \frac{4 * \frac{29}{2}}{29} = 2.$$

## 3.2 Proof of the Identities

There are other proofs, the one we presented here is intended to showcase different techniques.

**Lemma 26.** For $P > n$ the following holds:

$$\text{GoldbachNumber}_2(2n) = \frac{1}{P} \sum_{x \in T_{2P}} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi(p+q-2n)x).$$

**Proof.**

Note that $T_{2P} = R_2 \cup R_{2P}$, and note that $|R_{2P}| = \frac{P-1}{P}$.

$$\frac{1}{P} \sum_{x \in T_{2P}} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi(p+q-2n)x) = \frac{1}{P} \sum_{x \in R_2} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi(p+q-2n)x) + \\ \frac{1}{P} \sum_{x \in R_{2P}} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi(p+q-2n)x) \\ \text{Note that } \frac{1}{P} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{x \in R_2} \cos(2\pi(p+q-2n)x) = \frac{1}{P}(\pi_3(2n))^2.$$

Note that $\cos(2\pi(p+q-2n)x)$ is an even function on $x$, thus

$$\frac{1}{P} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{x \in R_{2P}} \cos(2\pi(p+q-2n)x) = \\ \frac{1}{P} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{x \in R_{2P}} \cos(2\pi|p+q-2n|x).$$

Note that by Ramanujan's sum, we have the following:

$$\frac{1}{P} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{x \in R_{2P}} e^{2\pi i |p+q-2n|x} = \\ \frac{1}{P} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{x \in R_{2P}} \cos(2\pi|p+q-2n|x) = \\ \frac{1}{P} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \mu\left(\frac{2P}{(2P, |p+q-2n|)}\right) \frac{\phi(2P)}{\phi\left(\frac{2P}{(2P, |p+q-2n|)}\right)}.$$
Thus we have the following:

$$
\begin{align*}
& \frac{1}{P} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{x \in \mathbb{R}_{2P}} \cos(2\pi|p+q-2n|x) \\
&= \frac{1}{P} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \mu\left(\frac{2P}{(2P, |p+q-2n|)}\right) \frac{\phi(2P)}{\phi\left(\frac{2P}{(2P, |p+q-2n|)}\right)} \\
&= \frac{1}{P} \left\{ \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \mu\left(\frac{2P}{(2P, |p+q-2n|)}\right) \frac{\phi(2P)}{\phi\left(\frac{2P}{(2P, |p+q-2n|)}\right)} \bigg|_{p+q-2n=0} + \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \mu\left(\frac{2P}{(2P, |p+q-2n|)}\right) \frac{\phi(2P)}{\phi\left(\frac{2P}{(2P, |p+q-2n|)}\right)} \bigg|_{p+q-2n \neq 0} \right\} \\
&= \frac{1}{P} \left\{ \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \mu\left(\frac{2P}{2P}\right) \frac{\phi(2P)}{\phi\left(\frac{2P}{2P}\right)} \bigg|_{p+q-2n=0} + \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \mu\left(\frac{2P}{2P}\right) \frac{\phi(2P)}{\phi\left(\frac{2P}{2P}\right)} \bigg|_{p+q-2n \neq 0} \right\} \\
&= \frac{1}{P} \left\{ \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \mu(1)\phi(2P) \bigg|_{p+q-2n=0} + \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \mu(P)\phi(2P) \bigg|_{p+q-2n \neq 0} \right\} \\
&= \frac{1}{P} \left\{ \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} (P-1) \bigg|_{p+q-2n=0} + \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} (-1) \bigg|_{p+q-2n \neq 0} \right\} \\
&= \frac{1}{P} \left\{ \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} (P-1) \bigg|_{p+q-2n=0} + \left( \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} (-1) \right) \bigg|_{p+q-2n=0} \right\} \\
&= \frac{1}{P} \left\{ \text{GoldbachNumber}_2(2n)(P-1) - (\pi_3(2n))^2 + \text{GoldbachNumber}_2(2n) \right\} \\
&= \frac{1}{P} \left\{ \text{GoldbachNumber}_2(2n)(P) - (\pi_3(2n))^2 \right\}.
\end{align*}
$$
Hence we have

$$
\begin{align*}
& \frac{1}{P} \sum_{x \in T_{2P}} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi(p+q-2n)x) \\
&= \frac{1}{P} \sum_{x \in R_2} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi(p+q-2n)x) + \frac{1}{P} \sum_{x \in R_{2P}} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi|p+q-2n|x) \\
&= \frac{1}{P}(\pi_3(2n))^2 + \frac{1}{P}\{\text{GoldbachNumber}_2(2n)(P) - (\pi_3(2n))^2\} \\
&= \text{GoldbachNumber}_2(2n).
\end{align*}
$$

This completes the proof of Lemma 26.

**Lemma 27.** The following holds:

$$
\begin{align*}
& \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi(p+q-2n)x) = \\
& [\sum_{3 \le p < 2n} \cos(2\pi px)]^2 \cos(2\pi(2n)x) + 2 \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\sin(2\pi qx)\sin(2\pi(2n)x) - \\
& \phantom{[\sum_{3 \le p < 2n} \cos(2\pi px)]^2 \cos(2\pi(2n)x) + 2} [\sum_{3 \le p < 2n} \sin(2\pi px)]^2 \cos(2\pi(2n)x).
\end{align*}
$$

Proof.
$$
\begin{align*}
& \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi(p+q-2n)x) \\
&= \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px + 2\pi qx - 4\pi nx) \\
&= \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos((2\pi px - 2\pi nx) + (2\pi qx - 2\pi nx)) \\
&= \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \{\cos(2\pi px - 2\pi nx)\cos(2\pi qx - 2\pi nx) - \\
& \phantom{= \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \{} \sin(2\pi px - 2\pi nx)\sin(2\pi qx - 2\pi nx)\} \\
&= \sum_{3 \le p < 2n} \cos(2\pi px - 2\pi nx) \sum_{3 \le q < 2n} \cos(2\pi qx - 2\pi nx) - \\
& \phantom{= \sum_{3 \le p < 2n} \cos(2\pi px - 2\pi nx) \sum_{3 \le q < 2n} \} \sum_{3 \le p < 2n} \sin(2\pi px - 2\pi nx) \sum_{3 \le q < 2n} \sin(2\pi qx - 2\pi nx) \\
&= [\sum_{3 \le p < 2n} \cos(2\pi px - 2\pi nx)]^2 - [\sum_{3 \le p < 2n} \sin(2\pi px - 2\pi nx)]^2 \\
&= [\sum_{3 \le p < 2n} \{\cos(2\pi px)\cos(2\pi nx) + \sin(2\pi px)\sin(2\pi nx)\}]^2 - \\
& \phantom{= [\sum_{3 \le p < 2n} \{} [\sum_{3 \le p < 2n} \{\sin(2\pi px)\cos(2\pi nx) - \cos(2\pi px)\sin(2\pi nx)\}]^2 \\
&= \{[\sum_{3 \le p < 2n} \cos(2\pi px)\cos(2\pi nx)]^2 + [\sum_{3 \le p < 2n} \sin(2\pi px)\sin(2\pi nx)]^2 + \\
& \phantom{= \{[\sum_{3 \le p < 2n} \cos(2\pi px)\cos(2\pi nx)]^2 + [} 2\sum_{3 \le p < 2n} \cos(2\pi px)\cos(2\pi nx) \sum_{3 \le p < 2n} \sin(2\pi px)\sin(2\pi nx)\} - \\
& \phantom{= \{[\sum_{3 \le p < 2n} \cos(2\pi px)\cos(2\pi nx)]^2 + [} \{[\sum_{3 \le p < 2n} \sin(2\pi px)\cos(2\pi nx)]^2 + [\sum_{3 \le p < 2n} \cos(2\pi px)\sin(2\pi nx)]^2 - \\
& \phantom{= \{[\sum_{3 \le p < 2n} \sin(2\pi px)\cos(2\pi nx)]^2 + [} 2\sum_{3 \le p < 2n} \sin(2\pi px)\cos(2\pi nx) \sum_{3 \le p < 2n} \cos(2\pi px)\sin(2\pi nx)\} \\
&= [\sum_{3 \le p < 2n} \cos(2\pi px)]^2 \{[\cos(2\pi nx)]^2 - [\sin(2\pi nx)]^2\} - \\
& \phantom{= [\sum_{3 \le p < 2n} \cos(2\pi px)]^2 \{} [\sum_{3 \le p < 2n} \sin(2\pi px)]^2 \{[\cos(2\pi nx)]^2 - [\sin(2\pi nx)]^2\} + \\
& \phantom{= [\sum_{3 \le p < 2n} \sin(2\pi px)]^2 \{} 2\{[\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\cos(2\pi nx)\sin(2\pi qx)\sin(2\pi nx)] + \\
& \phantom{= [\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\cos(2\pi nx)\sin(2\pi qx)\sin(2\pi nx)] +} [\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sin(2\pi px)\cos(2\pi nx)\cos(2\pi qx)\sin(2\pi nx)]\} \\
&= [\sum_{3 \le p < 2n} \cos(2\pi px)]^2 \cos(2\pi(2n)x) - [\sum_{3 \le p < 2n} \sin(2\pi px)]^2 \cos(2\pi(2n)x) + \\
& \phantom{= [\sum_{3 \le p < 2n} \cos(2\pi px)]^2 \cos(2\pi(2n)x) - [} 2\{[\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\sin(2\pi qx)\cos(2\pi nx)\sin(2\pi nx)] + \\
& \phantom{= [\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\sin(2\pi qx)\cos(2\pi nx)\sin(2\pi nx)] +} [\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sin(2\pi px)\cos(2\pi qx)\cos(2\pi nx)\sin(2\pi nx)]\} \\
&= [\sum_{3 \le p < 2n} \cos(2\pi px)]^2 \cos(2\pi(2n)x) - [\sum_{3 \le p < 2n} \sin(2\pi px)]^2 \cos(2\pi(2n)x) + \\
& \phantom{= [\sum_{3 \le p < 2n} \cos(2\pi px)]^2 \cos(2\pi(2n)x) - [} 2\{[\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\sin(2\pi qx)\cos(2\pi nx)\sin(2\pi qx)] + \\
& \phantom{= [\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\sin(2\pi qx)\cos(2\pi nx)\sin(2\pi qx)] +} [\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\sin(2\pi qx)\cos(2\pi nx)\sin(2\pi qx)]\} \\
&= [\sum_{3 \le p < 2n} \cos(2\pi px)]^2 \cos(2\pi(2n)x) - [\sum_{3 \le p < 2n} \sin(2\pi px)]^2 \cos(2\pi(2n)x) + \\
& \phantom{= [\sum_{3 \le p < 2n} \cos(2\pi px)]^2 \cos(2\pi(2n)x) - [} 2\{[\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\sin(2\pi qx)(2\cos(2\pi nx)\sin(2\pi nx))]\} \\
&= [\sum_{3 \le p < 2n} \cos(2\pi px)]^2 \cos(2\pi(2n)x) - [\sum_{3 \le p < 2n} \sin(2\pi px)]^2 \cos(2\pi(2n)x) + \\
& \phantom{= [\sum_{3 \le p < 2n} \cos(2\pi px)]^2 \cos(2\pi(2n)x) - [} 2\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px)\sin(2\pi qx)\sin(2\pi(2n)x)
\end{align*}
$$

This completes the proof of Lemma 27.

By Lemma 26, for $P > n$ we have

$$
GoldbachNumber_2(2n) = \frac{1}{P} \sum_{x \in T_{2P}} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi(p+q-2n)x).
$$

By Lemma 27, we have

$$
\sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi(p+q-2n)x) = \\
[ \sum_{3 \le p < 2n} \cos(2\pi px)]^2 \cos(2\pi(2n)x) + 2 \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px) \sin(2\pi qx) \sin(2\pi(2n)x) - \\
[ \sum_{3 \le p < 2n} \sin(2\pi px)]^2 \cos(2\pi(2n)x).
$$
Hence, for $P > n$ we have the following:

$$ \text{GoldbachNumber}_2(2n) = $$

$$ \frac{1}{P} \sum_{x \in T_{2P}} \{[ \sum_{3 \le p < 2n} \cos(2\pi px) ]^2 \cos(2\pi(2n)x) + 2 \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px) \sin(2\pi qx) \sin(2\pi(2n)x) - [ \sum_{3 \le p < 2n} \sin(2\pi px) ]^2 \cos(2\pi(2n)x)\}. $$

**Lemma 28.** By placing a stronger restriction on to $P$, i.e. $P > 3n$, the following holds.

$$ \frac{1}{P} \sum_{x \in T_{2P}} [ \sum_{3 \le p < 2n} \cos(2\pi px) ]^2 \cos(2\pi(2n)x) = \\ \frac{1}{P} \sum_{x \in T_{2P}} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px) \sin(2\pi qx) \sin(2\pi(2n)x), \\ \frac{1}{P} \sum_{x \in T_{2P}} [ \sum_{3 \le p < 2n} \cos(2\pi px) ]^2 \cos(2\pi(2n)x) = \\ - \frac{1}{P} \sum_{x \in T_{2P}} [ \sum_{3 \le p < 2n} \sin(2\pi px) ]^2 \cos(2\pi(2n)x) $$

and hence

$$ \frac{1}{P} \sum_{x \in T_{2P}} \{[ \sum_{3 \le p < 2n} \cos(2\pi px) ]^2 \cos(2\pi(2n)x) + 2 \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos(2\pi px) \sin(2\pi qx) \sin(2\pi(2n)x) - [ \sum_{3 \le p < 2n} \sin(2\pi px) ]^2 \cos(2\pi(2n)x)\} = $$

$$ \frac{4 \sum_{x \in T_{2P}} [ \sum_{3 \le p < 2n} \cos(2\pi px) ]^2 \cos(2\pi(2n)x) }{P} = \frac{4 \sum_{x \in T_{2P}} [ \text{Scos}(2n, x) ]^2 \cos(2\pi(2n)x) }{P}. $$

The proof of this lemma left to the reader.

By Lemma 26, Lemma 27 and Lemma 28 for $P > 3n$ we have the following:
$$ \text{GoldbachNumber}_2(2n) = \frac{4 \sum_{x \in T_{2P}} [\text{Scos}(2n, x)]^2 \cos(2\pi(2n)x)}{P}. $$

That completes the roof of Theorem 24.

## 3.3 Propositions

**Proposition 29.** There exists at least one $P > 3(2n)$ such that

$$ \sum_{x \in T_{2P}} [\text{Scos}(2n, x)]^2 \cos(2\pi(2n)x) > 0. $$

# 4 The Fourth Set of Identities

## 4.1 Identities and Examples

**Theorem 30.** The following formulas hold:

Complex version:

$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{z \in V_{2n}} \text{Goldbach}z0(2n, z)}{2n} = $$

$$ \frac{\sum_{z \in V_{2n}} (\sum_{3 \le p < 2n} e^{2\pi ipz})^2}{2n} = $$

$$ \frac{\sum_{1 \le j \le 2n} (\sum_{3 \le p < 2n} e^{2\pi ip(j/2n)})^2}{2n}. $$

Real version:

$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{x \in V_{2n}} \text{Goldbach}x0(2n, x)}{2n} = $$

$$ \frac{\sum_{x \in V_{2n}} \{[\sum_{3 \le p < 2n} \cos(2\pi px)]^2 - [\sum_{3 \le p < 2n} \sin(2\pi px)]^2\}}{2n} = $$

$$ \frac{\sum_{1 \le j \le 2n} \{[\sum_{3 \le p < 2n} \cos(2\pi p j/2n)]^2 - [\sum_{3 \le p < 2n} \sin(2\pi p j/2n)]^2\}}{2n}. $$
Observe that $\text{Goldbachz}_0(2n, \frac{2n}{2n}) = \text{Goldbachz}_0(2n, 1) = (\pi_3(2n))^2$ and $\text{Goldbachx}_0(2n, \frac{2n}{2n}) = \text{Goldbachx}_0(2n, 1) = (\pi_3(2n))^2$. Hence, we have the following corollary.

**Corollary 31.** The following formulas hold:

$$ \text{GoldbachNumber}_2(2n) = \frac{(\pi_3(2n))^2}{2n} + \frac{\sum_{1 \le j \le 2n-1} \text{Goldbachz}_0(2n, \frac{j}{2n})}{2n} $$

$$ \text{GoldbachNumber}_2(2n) = \frac{(\pi_3(2n))^2}{2n} + \frac{\sum_{1 \le j \le 2n-1} \text{Goldbachx}_0(2n, \frac{j}{2n})}{2n}. $$

**Example 32.** Let $2n = 24$. Then $\text{GoldbachNumber}_2(24) = 6$, and $(\pi_3(24))^2 = 64$. By evaluating the functions we obtain the following:

$$ \frac{(\pi_3(2n))^2}{2n} + \frac{\sum_{1 \le j \le 2n-1} \text{Goldbachz}_0(2n, \frac{j}{2n})}{2n} = \frac{(\pi_3(24))^2}{24} + \frac{\sum_{1 \le j \le 23} \text{Goldbachz}_0(24, \frac{j}{24})}{24} = \\ \frac{64}{24} + \frac{80}{24} = 6, $$

and

$$ \frac{(\pi_3(2n))^2}{2n} + \frac{\sum_{1 \le j \le 2n-1} \text{Goldbachx}_0(2n, \frac{j}{2n})}{2n} = \frac{(\pi_3(24))^2}{24} + \frac{\sum_{1 \le j \le 23} \text{Goldbachx}_0(24, \frac{j}{24})}{24} = \\ \frac{64}{24} + \frac{80}{24} = 6. $$

Recall that $V_m = \bigcup_{d|m} R_d$, we have the following theorem, the proof is omitted.

**Theorem 33.** For large $2n$, with $d|2n$, a properly selected $k \ge 1$ (where $k$ may increase as $2n$ grows) and $d < (\log(2n))^k$, the following holds:

$$ \sum_{z \in R_d} \text{Goldbachz}_0(2n, z) \sim \frac{(\mu(d))^2}{\phi(d)} [\pi_3(2n)]^2. \quad (4.1) $$

We are unable to find approximate formulas for $\sum_{z \in R_d} \text{Goldbachz}_0(2n, z)$ when $d$ is not small compared to $2n$.
Next, we study Theorem 33 further. By the definition of Goldbach0(2n, x), we can see $\sum_{z \in R_1} \text{Goldbach}0(2n, x) = \sum_{z \in R_2} \text{Goldbach}0(2n, x) = [\pi_3(2n)]^2$. Hence, the Theorem 33 holds for $d=1$ and $d=2$.

The following is an example of approximation expression (4.1) for cases where $d|2n$, $d > 2$ and $d < (\log(2n))^k$.

**Example 34.** Randomly select $2n = 46886$ and take $k=2$, the following divisors satisfy $d|2n$ and $2 < d < (\log(2n))^2$:
$\{7, 14, 17, 34\}$.

By calculations we obtain the following:

$$ \sum_{z \in V_7} \text{Goldbach}0(46886, z) = 3897676, \frac{(\mu(7))^2}{\phi(7)} [\pi_3(46886)]^2 = 3909108.2, $$

$$ \sum_{z \in V_{14}} \text{Goldbach}0(46886, z) = 3897676, \frac{(\mu(14))^2}{\phi(14)} [\pi_3(46886)]^2 = 3909108.2, $$

$$ \sum_{z \in V_{17}} \text{Goldbach}0(46886, z) = 1452646, \frac{(\mu(17))^2}{\phi(17)} [\pi_3(46886)]^2 = 1465915.6, $$

$$ \sum_{z \in V_{34}} \text{Goldbach}0(46886, z) = 1452646, \frac{(\mu(34))^2}{\phi(34)} [\pi_3(46886)]^2 = 1465915.6 $$

## 4.2 Proof of the Identities

Proof of Theorem 30.

We first prove the following lemma for the complex version.

**Lemma 35.** The following holds for any $k$ and $m \ge 1$.

$$ \sum_{z \in V_m} e^{2\pi i k z} = \sum_{1 \le j \le m} e^{2\pi i k (j/m)} = \begin{cases} m & \text{if } m \mid k \\ 0 & \text{if } m \nmid k. \end{cases} $$

Proof.

Case 1: $k=0$, then $m \nmid k$, trivial.

Case 2: $k \ne 0$ and $m \nmid k$, trivial.

Case 3: $k \ne 0$ and $m \mid k$.

For the Case 3, $e^{2\pi i k(\frac{1}{m})} \ne 1$ and by summation of geometric series formula

$$ \sum_{z \in V_m} e^{2\pi i k z} = \sum_{1 \le j \le m} e^{2\pi i k (\frac{j}{m})} = \sum_{1 \le j \le m} (e^{2\pi i k (\frac{j}{m})})^j = e^{2\pi i k (\frac{1}{m})} \sum_{0 \le j \le m-1} (e^{2\pi i k (\frac{1}{m})})^j = $$

$$ e^{2\pi i k (\frac{1}{m})} \frac{1 - (e^{2\pi i k (\frac{1}{m})})^{(m-1)+1}}{1 - e^{2\pi i k (\frac{1}{m})}} = e^{2\pi i k (\frac{1}{m})} \frac{1 - e^{2\pi i k (\frac{m}{m})}}{1 - e^{2\pi i k (\frac{1}{m})}} = 0. $$

This complete the proof of the Lemma 35.
**Corollary 36.** The following holds for any $k$ and $m \ge 1$.

$$ \sum_{x \in V_m} \cos(2\pi k x) = \sum_{1 \le j \le m} \cos(2\pi k \left(\frac{j}{m}\right)) = \begin{cases} m & \text{if } m \mid k \\ 0 & \text{if } m \nmid k. \end{cases} $$

Proof is left to the reader.

Now, we continue the proof of Theorem 30.

Note that

$$ \begin{aligned} \sum_{z \in V_{2n}} \text{Goldbach}z0(2n, z) &= \sum_{z \in V_{2n}} \left[ \sum_{3 \le p < 2n} e^{2\pi i p z} \right]^2 = \\ & \sum_{z \in V_{2n}} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} e^{2\pi i (p+q)z} = \\ & \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{z \in V_{2n}} e^{2\pi i (p+q)z}. \end{aligned} $$

Furthermore, note that

$$ 6 \le (p+q) \le 2(2n-1), \text{ so } 2n|(p+q) \text{ if and only if } (p+q) = 2n. $$

So by the Lemma 35

$$ \sum_{z \in V_{2n}} e^{2\pi i (p+q)z} = 2n \text{ for every instance where } (p+q) = 2n, $$

$$ \text{and } \sum_{z \in V_{2n}} e^{2\pi i (p+q)z} = 0 \text{ for every instance where } (p+q) \neq 2n. $$

Thus,

$$ \sum_{z \in V_{2n}} \text{Goldbach}z0(2n, z) = \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{z \in V_{2n}} e^{2\pi i (p+q)z} = \text{GoldbachNumber}_2(2n) * (2n). $$

Hence,

$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{z \in V_{2n}} \text{Goldbach}z0(2n, z)}{2n}. $$

This completes the proof of the complex version of Theorem 30.

To prove the real version of Theorem 30, recall that $V_{2n} = \bigcup_{d|2n} R_d$.

By similar arguments used in proving the equation of (1.1), we can prove

$$ \sum_{d|2n, z \in R_d} \text{Goldbach}z0(2n, z) = \sum_{d|2n, x \in R_d} \text{Goldbach}x0(2n, x). $$

Hence

$$ \sum_{z \in V_{2n}} \text{Goldbach}z0(2n, z) = \sum_{x \in V_{2n}} \text{Goldbach}x0(2n, x), $$
and

$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{x \in V_{2n}} \text{Goldbachx0}(2n, x)}{2n}. $$

This completes the proof of the real version of Theorem 30.

The proof of Theorem 33 is omitted. Readers may follow similar techniques presented in [1, 2] to prove this theorem.

## 4.3 Propositions

**Proposition 37.** One of the following statements holds:

$$ \sum_{z \in V_{2n}} \text{Goldbachx0}(2n, z) = \sum_{1 \le j \le 2n} \left( \sum_{3 \le p < 2n} e^{2\pi i p \left( \frac{j}{2n} \right)} \right)^2 > 0 $$

or

$$ \sum_{z \in V_{2n}} \text{Goldbachx0}(2n, z) = \sum_{1 \le j \le 2n} \left[ \left( \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{2n}\right)\right) \right)^2 - \left( \sum_{3 \le p < 2n} \sin\left(2\pi p \left(\frac{j}{2n}\right)\right) \right)^2 \right] > 0. $$

**Proposition 38.** For $2n \ge 1720$, one of the following holds:

$$ \sum_{1 \le j \le 2n-1} \left( \sum_{3 \le p < 2n} e^{2\pi i p \left( \frac{j}{2n} \right)} \right)^2 > 0 $$

or

$$ \sum_{1 \le j \le 2n-1} \left[ \left( \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{2n}\right)\right) \right)^2 - \left( \sum_{3 \le p < 2n} \sin\left(2\pi p \left(\frac{j}{2n}\right)\right) \right)^2 \right] > 0, $$

furthermore,

$$ \text{GoldbachNumber}_2(2n) > \frac{(\pi_3(2n))^2}{2n}. $$

# 5 The Fifth Set of Identities

## 5.1 Identities and Examples

**Theorem 39.** The following formulas hold for $2n = 4k + 2$ (this identity is not true for $2n = 4k$).
Complex version:

$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{z \in V_n} \text{Goldbachz0}(2n, z)}{n}. $$

Real version:

$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{x \in V_n} \text{Goldbachx0}(2n, x)}{n}. $$

## 5.2 Proof of the Identities

*Proof.* By following arguments similar to those used in the proof of 30, we can prove Theorem 39. □

## 5.3 Propositions

**Proposition 40.** If one of the following holds, then the Goldbach Conjecture holds for $2n = 4k + 2$.

$$ \sum_{z \in V_n} \text{Goldbachz0}(2n, z) = \sum_{1 \le j \le 2n} \left( \sum_{3 \le p < 2n} e^{2\pi i p \left( \frac{j}{n} \right)} \right)^2 > 0 $$

or

$$ \sum_{z \in V_n} \text{Goldbachx0}(2n, z) > 0. $$

# 6 The Sixth Set of Identities

## 6.1 Identities and Examples

**Theorem 41.** The following formula holds for any $m > 2n$,

$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{z \in V_m} \text{Goldbachz}(2n, z)}{m}. $$

**Example 42.** Example for $2n = 12, m = 2n + 1 = 13$.

Note that $\text{GoldbachNumber}_2(12) = 2$ and

$$ \text{Goldbachz}(12, z) = (e^{2\pi i(3z)} + e^{2\pi i(5z)} + e^{2\pi i(7z)} + e^{2\pi i(11z)})^2 e^{2\pi i(-12z)}. $$

By evaluating the functions, we obtain the following:
$$ \frac{\sum_{z \in V_m} \text{Goldbachz}(2n, z)}{m} = \frac{\sum_{z \in V_{13}} \text{Goldbachz}(12, z)}{13} = \frac{\sum_{z \in V_{13}} (e^{2\pi i(3)z} + e^{2\pi i(5)z} + e^{2\pi i(7)z} + e^{2\pi i(11)z})^2 e^{2\pi i(-12)z}}{13} = \frac{26}{13} = 2. $$

Hence

$$ \text{GoldbachNumber}_2(12) = \frac{\sum_{z \in V_{13}} \text{Goldbachz}(12, z)}{13}. $$

## 6.2 Proof of the Identities

Proof.

Note that

$$ \sum_{z \in V_m} \text{Goldbachz}(2n, z) = \sum_{z \in V_m} \left[ \sum_{3 \le p < 2n} e^{2\pi i p z} \right]^2 e^{-2\pi i (2n)z} = \\ \sum_{z \in V_m} \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} e^{2\pi i (p+q-2n)z} = \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{z \in V_m} e^{2\pi i (p+q-2n)z}. $$

Note that $6 - 2n \le (p + q - 2n) \le 2(2n - 1) - 2n = 2n - 2$, so $m|(p + q - 2n)$ if and only if $(p + q - 2n) = 0$. So by the Lemma 35,

$$ \sum_{z \in V_m} \text{Goldbachz}(2n, z) = \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{z \in V_m} e^{2\pi i (p+q-2n)z} = \\ \text{GoldbachNumber}_2(2n) * (m). $$

Hence,

$$ \text{GoldbachNumber}_2(2n) = \frac{\sum_{z \in V_m} \text{Goldbachz}(2n, z)}{m}. $$

This completes the proof of Theorem 41.

## 6.3 Propositions

**Proposition 43.** There exists an $m > 2n$ such that

$$ \sum_{z \in V_m} \text{Goldbachz}(2n, z) = \sum_{1 \le j \le m} \left( \sum_{3 \le p < 2n} e^{2\pi i p \left(\frac{j}{m}\right)} \right)^2 e^{-2\pi i (2n) \left(\frac{j}{m}\right)} > 0. $$
# 7 The Seventh Set of Identities

## 7.1 Identities and Examples

We define the set

$$ \mathrm{Seed}_{(2n)} = \{e^{2\pi i(\frac{p}{2n})}|3 \le p < 2n\}. $$

**Theorem 44.** The following formula holds for any $2n \ge 6$,

$$ \mathrm{GoldbachNumber}_2(2n) = \frac{\sum_{1 \le j \le 2n} (\sum_{z \in \mathrm{Seed}_{(2n)}} z^j)^2}{2n}. $$

**Example 45.** For $2n = 12$, we know that $\mathrm{GoldbachNumber}_2(2n) = \mathrm{GoldbachNumber}_2(12) = 2$.

$$ \mathrm{Seed}_{(2n)} = \mathrm{Seed}_{(12)} = \{e^{2\pi i(\frac{p}{12})}|3 \le p < 12\} = \{e^{2\pi i(\frac{3}{12})}, e^{2\pi i(\frac{5}{12})}, e^{2\pi i(\frac{7}{12})}, e^{2\pi i(\frac{11}{12})}\}. $$

We have

$$ \frac{\sum_{1 \le j \le 2n} (\sum_{z \in \mathrm{Seed}_{(2n)}} z^j)^2}{2n} = \frac{\sum_{1 \le j \le 12} (\sum_{z \in \mathrm{Seed}_{(12)}} z^j)^2}{12} = $$

$$ \frac{\sum_{1 \le j \le 12} [(e^{2\pi i(\frac{3}{12})})^j + (e^{2\pi i(\frac{5}{12})})^j + (e^{2\pi i(\frac{7}{12})})^j + (e^{2\pi i(\frac{11}{12})})^j]^2}{12}. $$

By direct calculation, we obtain

$$ \frac{\sum_{1 \le j \le 12} [(e^{2\pi i(\frac{3}{12})})^j + (e^{2\pi i(\frac{5}{12})})^j + (e^{2\pi i(\frac{7}{12})})^j + (e^{2\pi i(\frac{11}{12})})^j]^2}{12} = $$

$$ \frac{40 + 2ie^{-\frac{1}{6}(i\pi)} - 2ie^{\frac{i\pi}{6}} + 20e^{-\frac{1}{3}(2i\pi)} + 20e^{\frac{2i\pi}{3}} + 2ie^{-\frac{1}{6}(5i\pi)} - 2ie^{\frac{5i\pi}{6}}}{12} = 2. $$

So

$$ \mathrm{GoldbachNumber}_2(12) = \frac{\sum_{1 \le j \le 12} (\sum_{z \in \mathrm{Seed}_{(12)}} z^j)^2}{12}. $$

Note that for any $z = e^w = e^{2\pi i(\frac{p}{2n})} \in \mathrm{Seed}_{(2n)}$, $|2\pi i(\frac{p}{2n})| < 2\pi$. We can expand $z = e^w = 1 + w + \frac{1}{2}w^2 + \frac{1}{3}w^3 + \dots + \frac{1}{(2n)!}w^{2n} + \text{Remainder}_{2n+1}$, where
$$|\text{Remainder}_{2n+1}| = \left| \sum_{k=2n+1}^{\infty} \frac{1}{k!} w^k \right| \le \sum_{k=2n+1}^{\infty} \frac{1}{k!} |w^k| < \sum_{k=2n+1}^{\infty} \frac{1}{k!} (2\pi)^k.$$

We define the set

$$\text{EstimateSeed}_{(2n)} = \{1+(2\pi i(\frac{p}{2n}))+\frac{1}{2!}(2\pi i(\frac{p}{2n}))^2+\dots+\frac{1}{(2n)!}(2\pi i(\frac{p}{2n}))^{2n}|3 \le p < 2n\}$$

and

$$\text{EstimateGoldbachNumber}_2(2n) = \frac{\sum_{1 \le j \le 2n} \left( \sum_{z \in \text{EstimateSeed}_{(2n)}} z^j \right)^2}{2n}.$$

We want to examine

$\text{EstimateGoldbachNumber}_2(2n)$ and compare it with $\text{GoldbachNumber}_2(2n)$.

**Example 46.** For $2n = 12$, we know that $\text{GoldbachNumber}_2(2n) = \text{GoldbachNumber}_2(12) = 2$. We will examine $\text{EstimateGoldbachNumber}_2(12)$ and compare it with $\text{GoldbachNumber}_2(12)$.

$\text{EstimateSeed}_{(12)} =$

$$\{1 + (2\pi i(\frac{p}{12})) + \frac{1}{2!}(2\pi i(\frac{p}{12}))^2 + \dots + \frac{1}{12!}(2\pi i(\frac{p}{12}))^{12}|3 \le p < 12\} =$$

$$\{6.32147*10^{-9}+1.0i, -0.866017+0.499958i, -0.866017-0.499958i, 0.866025-0.5i\}.$$

By further calculation we obtain the following:

$$\text{EstimateGoldbachNumber}_2(12) = 1.99905 - 0.0000484838i.$$

We see even for the small $2n = 12$, $\text{EstimateGoldbachNumber}_2(12)$ is accurate.

**Example 47.** For $2n = 112$, we know that $\text{GoldbachNumber}_2(2n) = \text{GoldbachNumber}_2(112) = 14$. We will examine $\text{EstimateGoldbachNumber}_2(112)$.

$\text{EstimateSeed}_{(112)} =$

$$\{1 + (2\pi i(\frac{p}{112})) + \frac{1}{2!}(2\pi i(\frac{p}{112}))^2 + \dots + \frac{1}{112!}(2\pi i(\frac{p}{112}))^{112}|3 \le p < 112\} =$$

$$\{0.985871018518235874+0.167506223304736408i, 0.960917321945099543+0.276835511424849388i, \dots\}$$
$0.960917321945099543-0.276835511424849388i, 0.985871018518235874-0.167506223304736408i}.$

By further calculation we obtain the following:

$$EstimateGoldbachNumber_2(112) =$$

$$14 - 5.168951868456043445088963637330350966568975452437779 * 10^{-130} +$$

$3.076372359036790231486794978488018798909478371216591529484689477*10^{-131}i.$

We see even for the relatively small $2n = 112$, $EstimateGoldbachNumber_2(112)$ is extremely accurate.

## 7.2 Proof of the Identities

Proof.

Recall that

$$GoldbachNumber_2(2n) = \frac{\sum_{z \in V_{2n}} Goldbachz0(2n, z)}{2n}.$$

Note that

$$\sum_{z \in V_{2n}} Goldbachz0(2n, z) = \sum_{z \in V_{2n}} \left[ \sum_{3 \le p < 2n} e^{2\pi i p z} \right]^2 = \sum_{1 \le j \le 2n} \left[ \sum_{3 \le p < 2n} e^{2\pi i (p) \frac{j}{2n}} \right]^2 =$$

$$\sum_{1 \le j \le 2n} \left[ \sum_{3 \le p < 2n} e^{2\pi i (\frac{p}{2n}) j} \right]^2 = \sum_{1 \le j \le 2n} \left( \sum_{z \in Seed_{(2n)}} z^j \right)^2.$$

Hence,

$$GoldbachNumber_2(2n) = \frac{\sum_{z \in V_{2n}} Goldbachz0(2n, z)}{2n} = \frac{\sum_{1 \le j \le 2n} \left( \sum_{z \in Seed_{(2n)}} z^j \right)^2}{2n}.$$

This completes the proof of the complex version of Theorem 44.
## 7.3 Propositions

**Proposition 48.** The following holds

$$ \sum_{1 \le j \le 2n} \left( \sum_{z \in \text{Seed}_{(2n)}} z^j \right)^2 > 0. $$

**Proposition 49.** For sufficiently large $2n$:

$$ \text{Re}(\text{EstimateGoldbachNumber}_2(2n)) > 0 \text{ and} $$

$$ \text{GoldbachNumber}_2(2n) \sim \text{Re}(\text{EstimateGoldbachNumber}_2(2n)) \text{ and} $$

$$ \text{Im}(\text{EstimateGoldbachNumber}_2(2n)) = \varepsilon(\text{Re}(\text{EstimateGoldbachNumber}_2(2n))). $$

where $\text{Re}(z)$ is real part of $z$, $\text{Im}(z)$ is imaginary part of $z$.

# 8 The Eighth Set of Identities

## 8.1 Identities and Examples

**Theorem 50.** Let $P > 3n$, then the following holds.

$$ \text{GoldbachNumber}_2(2n) = $$

$$ \frac{4}{P} \sum_{x \in V_P} (\text{Scos}(2n, x))^2 \cos(2\pi(2n)x) = \frac{4}{P} \sum_{x \in V_P} \left[ \sum_{3 \le p < 2n} \cos(2\pi p x) \right]^2 \cos(2\pi(2n)x). $$

The following three identities surprisingly relate $\pi_3(2n)$, $\text{GoldbachNumber}_2(2n)$ and $\text{TiwnPrimeNumber}_2(2n)$ in such a simple and elegant form.

**Theorem 51.** The following holds.

$$ \pi_3(2n) = \frac{1}{2n} \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos(\pi 0), $$

$$ \text{GoldbachNumber}_2(2n) = \frac{1}{2n} \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos(\pi j), $$
$$ \text{TiwnPrimeNumber}_2(2n) = \frac{1}{2n} \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos\left(\pi \frac{j}{n}\right) - \left(\frac{1}{2}\right) \text{IsPrime}(2n-1), $$

where

$$ \text{TiwnPrimeNumber}_2(2n) = \sum_{p < 2n, p-2 \text{ is a prime}} 1 = $$

$$ \int_{0}^{1} \left( \sum_{3 \le p < 2n} e^{2\pi i p z} \right) \left( \sum_{3 \le p < 2n} e^{-2\pi i p z} \right) e^{2\pi i (-2)z} dz, $$

$p$ is an odd prime, $2n \ge 6$,

$$ \text{IsPrime}(l) = \begin{cases} 1 & \text{if } l \text{ is a prime} \\ 0 & \text{otherwise} \end{cases}. $$

**Example 52.** Take $2n = 8$, so $n = 4$, $4n = 16$ and $\text{IsPrime}(2n - 1) = \text{IsPrime}(7) = 1$, and we know that $\pi_3(2n) = \pi_3(8) = 3$, $\text{GoldbachNumber}_2(2n) = \text{GoldbachNumber}_2(8) = 2$, and $\text{TiwnPrimeNumber}_2(2n) = \text{TiwnPrimeNumber}_2(8) = 2$ (because the odd primes smaller than 8 are: $p = 3$, $p = 5$, $p = 7$; there are two cases where $p-2$ is also a prime: $5-3$ is a prime, and $7-5$ is a prime).

First, we begin by evaluating the right-hand side of the following expression:

$$ \text{GoldbachNumber}_2(2n) = \frac{1}{2n} \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos(\pi j) = $$

$$ \frac{1}{2n} \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 (-1)^j = $$

$$ \frac{1}{8} \sum_{1 \le j \le 16} \left[ \sum_{3 \le p < 8} \cos\left(2\pi p \left(\frac{j}{16}\right)\right) \right]^2 (-1)^j = $$

$$ \frac{1}{8} \sum_{1 \le j \le 16} \left[ \cos\left(2\pi(3)\left(\frac{j}{16}\right)\right) + \cos\left(2\pi(5)\left(\frac{j}{16}\right)\right) + \cos\left(2\pi(7)\left(\frac{j}{16}\right)\right) \right]^2 (-1)^j. $$

That is, we evaluate the following:

$$ \frac{1}{8} \sum_{1 \le j \le 16} \left[ \cos\left(2\pi(3)\left(\frac{j}{16}\right)\right) + \cos\left(2\pi(5)\left(\frac{j}{16}\right)\right) + \cos\left(2\pi(7)\left(\frac{j}{16}\right)\right) \right]^2 (-1)^j. $$

We now evaluate the following expression for each $j$.
$$ \sum_{1 \le j \le 16} \left[ \cos\left(2\pi\left(3\left(\frac{j}{16}\right)\right)\right) + \cos\left(2\pi\left(5\left(\frac{j}{16}\right)\right)\right) + \cos\left(2\pi\left(7\left(\frac{j}{16}\right)\right)\right) \right]^2 (-1)^j . $$

$$ j = 1 : \left[ \cos\left(2\pi\left(3\left(\frac{1}{16}\right)\right)\right) + \cos\left(2\pi\left(5\left(\frac{1}{16}\right)\right)\right) + \cos\left(2\pi\left(7\left(\frac{1}{16}\right)\right)\right) \right]^2 (-1)^1 = -(\cos(\frac{\pi}{8}))^2 $$

...

$$ j = 16 : \left[ \cos\left(2\pi\left(3\left(\frac{16}{16}\right)\right)\right) + \cos\left(2\pi\left(5\left(\frac{16}{16}\right)\right)\right) + \cos\left(2\pi\left(7\left(\frac{16}{16}\right)\right)\right) \right]^2 (-1)^{16} = 9. $$

Hence,

$$ \frac{1}{8} \sum_{1 \le j \le 16} \left[ \cos\left(2\pi\left(3\left(\frac{j}{16}\right)\right)\right) + \cos\left(2\pi\left(5\left(\frac{j}{16}\right)\right)\right) + \cos\left(2\pi\left(7\left(\frac{j}{16}\right)\right)\right) \right]^2 (-1)^j = $$

$$ \frac{1}{8} \left[ -\cos^2\left(\frac{\pi}{8}\right) + \frac{1}{2} - \sin^2\left(\frac{\pi}{8}\right) + 0 - \sin^2\left(\frac{\pi}{8}\right) + \frac{1}{2} - \cos^2\left(\frac{\pi}{8}\right) + 9 \right. $$

$$ - \cos^2\left(\frac{\pi}{8}\right) + \frac{1}{2} - \sin^2\left(\frac{\pi}{8}\right) + 0 - \sin^2\left(\frac{\pi}{8}\right) + \frac{1}{2} - \cos^2\left(\frac{\pi}{8}\right) + 9 = \frac{1}{8}(16) = 2, $$

and

$$ \text{GoldbachNumber}_2(8) = 2 = \frac{1}{8} \sum_{1 \le j \le 16} \left[ \sum_{3 \le p < 8} \cos\left(2\pi p \left(\frac{j}{16}\right) \right) \right]^2 \cos(\pi j). $$

Next, we have the following:

$$ \pi_3(8) = 3 = \frac{1}{8} \sum_{1 \le j \le 16} \left[ \sum_{3 \le p < 8} \cos\left(2\pi p \left(\frac{j}{16}\right) \right) \right]^2 = $$

$$ \frac{1}{8} \left[ \cos^2\left(\frac{\pi}{8}\right) + \frac{1}{2} + \sin^2\left(\frac{\pi}{8}\right) + 0 + \sin^2\left(\frac{\pi}{8}\right) + \frac{1}{2} + \cos^2\left(\frac{\pi}{8}\right) + 9 \right. $$

$$ + \cos^2\left(\frac{\pi}{8}\right) + \frac{1}{2} + \sin^2\left(\frac{\pi}{8}\right) + 0 + \sin^2\left(\frac{\pi}{8}\right) + \frac{1}{2} + \cos^2\left(\frac{\pi}{8}\right) + 9 = \frac{1}{8}(24) = 3. $$

Finally, by function evaluation (details omitted) we obtain:
$$ TiwnPrimeNumber_2(8) = \frac{1}{8} \sum_{1 \le j \le 16} \left[ \sum_{3 \le p < 8} \cos\left(2\pi p \left(\frac{j}{16}\right)\right) \right]^2 \cos\left(\pi \frac{j}{4}\right) - \frac{1}{2} = \frac{1}{8}(20) - \frac{1}{2} = 2. $$

Note that when $j = 4n$,

$$ \frac{1}{2n} \sum_{j=4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos(\pi j) = \frac{(\pi_3(2n))^2}{2n}, $$

$$ \frac{1}{2n} \sum_{j=4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos\left(\pi \frac{j}{n}\right) = \frac{(\pi_3(2n))^2}{2n}, $$

thus we obtain the following Corollary.

**Corollary 53.** *The following hold*

$$ GoldbachNumber_2(2n) = \frac{(\pi_3(2n))^2}{2n} + \frac{1}{2n} \sum_{1 \le j \le 4n-1} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos(\pi j), $$

and

$$ TiwnPrimeNumber_2(2n) = \frac{(\pi_3(2n))^2}{2n} + \frac{1}{2n} \sum_{1 \le j \le 4n-1} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos\left(\pi \frac{j}{n}\right). $$

## 8.2 Proof of the Identities

The proof of Theorem 50 is similar to that of Theorem 51.

We now prove the following identity from Theorem 51:

$$ GoldbachNumber_2(2n) = \frac{1}{2n} \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos(\pi j). $$

Proof.

Note that

$$ \begin{aligned} \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos(\pi j) &= \\ & \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos\left(2\pi\left(2n\left(\frac{j}{4n}\right)\right)\right) = \end{aligned} $$

$$ \begin{aligned} & \sum_{x \in V_{4n}} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p x\right) \right]^2 \cos\left(2\pi\left(2n\right)x\right) = \\ & \sum_{x \in V_{4n}} \left[ \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \cos\left(2\pi p x\right) \cos\left(2\pi q x\right) \cos\left(2\pi\left(2n\right)x\right) \right] = \end{aligned} $$
$$ \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{x \in V_{4n}} \cos(2\pi px)\cos(2\pi qx)\cos(2\pi(2n)x). $$

Observe that

$$ \cos(2\pi px)\cos(2\pi qx)\cos(2\pi(2n)x) = $$

$$ \frac{1}{4}\cos(2\pi(2n-p-q)x) + \frac{1}{4}\cos(2\pi(2n+p-q)x) + $$

$$ \frac{1}{4}\cos(2\pi(2n-p+q)x) + \frac{1}{4}\cos(2\pi(2n+p+q)x). $$

Thus,

$$ \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{x \in V_{4n}} \cos(2\pi px)\cos(2\pi qx)\cos(2\pi(2n)x) = $$

$$ \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{x \in V_{4n}} \frac{1}{4} \cos(2\pi(2n-p-q)x) + \frac{1}{4} \cos(2\pi(2n+p-q)x) + $$

$$ \frac{1}{4}\cos(2\pi(2n-p+q)x) + \frac{1}{4}\cos(2\pi(2n+p+q)x). $$

By the Corollary 36 we have the following:

$$ \sum_{x \in V_{4n}} \frac{1}{4} \cos(2\pi(2n-p-q)x) = \frac{1}{4}(4n) \text{ if and only if } (2n-p-q) = 0, $$

$$ \sum_{x \in V_{4n}} \frac{1}{4} \cos(2\pi(2n+p-q)x) = 0, $$

$$ \sum_{x \in V_{4n}} \frac{1}{4} \cos(2\pi(2n-p+q)x) = 0, $$

and

$$ \sum_{x \in V_{4n}} \frac{1}{4} \cos(2\pi(2n+p+q)) = \frac{1}{4}(4n) \text{ if and only if } (2n+p+q) = 4n. $$

Each instance of $(2n-p-q) = 0$ and each instance of $(2n+p+q) = 4n$ contributes a count of 1 to $GoldbachNumber_2(2n)$, thus
$$ \sum_{3 \le p < 2n} \sum_{3 \le q < 2n} \sum_{x \in V_{4n}} \frac{1}{4} \cos(2\pi(2n-p-q)x) + \frac{1}{4} \cos(2\pi(2n+p-q)x) + $$

$$ \frac{1}{4} \cos(2\pi(2n - p + q)x) + \frac{1}{4} \cos(2\pi(2n + p + q)x) = $$

$$ \text{GoldbachNumber}_2(2n) \frac{1}{4}(4n) + 0 + 0 + \text{GoldbachNumber}_2(2n) \frac{1}{4}(4n) = $$

$$ \text{GoldbachNumber}_2(2n)(2n), $$

in other words,

$$ \text{GoldbachNumber}_2(2n)(2n) = \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos(\pi j), $$

or

$$ \text{GoldbachNumber}_2(2n) = \frac{1}{2n} \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos(\pi j). $$

This completes the proof of the identity.

The other two identities in Theorem 51 can be proved in a similar fashion.

## 8.3 Propositions

**Proposition 54.** There exists a $P > 3n$ such that

$$ \sum_{x \in V_P} (\text{Scos}(2n, x))^2 \cos(2\pi(2n)x) > 0. $$

**Proposition 55.** The following holds for $2n \ge 6$.

$$ \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos(\pi j) = \sum_{1 \le j \le 4n} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 (-1)^j > 0. $$

**Proposition 56.** The following holds for $2n \ge 1720$.
$$ \sum_{1 \le j \le 4n-1} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 \cos(\pi j) = \sum_{1 \le j \le 4n-1} \left[ \sum_{3 \le p < 2n} \cos\left(2\pi p \left(\frac{j}{4n}\right)\right) \right]^2 (-1)^j > 0, $$

furthermore,

$$ \mathrm{GoldbachNumber}_2(2n) > \frac{(\pi_3(2n))^2}{2n}. $$

As a related and intriguing result, we present the following statement concerning the number of odd primes less than n. This statement may offer a new avenue for exploring prime number theory and Riemann's Hypothesis. Further details are provided in the paper by Yue [7].

*Claim 57.* The following holds for sufficiently large n:

$$ \pi_3(n) \sim \frac{n}{4} \left(1 - \sqrt{1 - 8w(n)}\right), $$

where

$$ w(n) = \frac{1}{\log(n)} - \frac{1}{(\log(n))^2} - \frac{2}{(\log(n))^3} - \frac{4}{(\log(n))^4} - \frac{8}{(\log(n))^5} - \frac{8}{(\log(n))^6} + \frac{48}{(\log(n))^7} + \dots $$

**Example 58.** Let $n = 68,678,345,546,768$, and consider the first seven terms of $w(n)$

$$ w(n) = \frac{1}{\log(n)} - \frac{1}{(\log(n))^2} - \frac{2}{(\log(n))^3} - \frac{4}{(\log(n))^4} - \frac{8}{(\log(n))^5} - \frac{8}{(\log(n))^6} + \frac{48}{(\log(n))^7} $$

we know that

$$ \pi_3(n) = \pi_3(68678345546768) = 2227962662424, $$

and by calculation, we obtain the following:

$$ \frac{n}{4} \left(1 - \sqrt{1 - 8w(n)}\right) = 2227962638535. $$

# 9 The Ninth Set of Identities

## 9.1 Identities and Examples

**Theorem 59.** If one wishes to apply Vinogradov's method to real function identities, the following three real function identities hold:

$$ \mathrm{GoldbachNumber}_2(2n) = 4 \int_0^1 \left[ \sum_{3 \le p < 2n} \cos(2\pi p x) \right]^2 \cos(2\pi 2nx) dx $$
$$ GoldbachNumber_2(2n) = 4 \int_0^1 -[ \sum_{3 \le p < 2n} \sin(2\pi px) ]^2 \cos(2\pi 2nx) dx $$

and

$$ GoldbachNumber_2(2n) = 4 \int_0^1 [ \sum_{3 \le p < 2n} \cos(2\pi px) \sin(2\pi px) ] \sin(2\pi 2nx) dx. $$

## 9.2 Proof of the Identities

Proof is left to the reader.

## 9.3 Propositions

**Proposition 60.** One of the following statements holds:

$$ \int_0^1 \left[ \sum_{3 \le p < 2n} \cos^2(2\pi px) \right] \cos(2\pi 2nx) dx > 0, $$

or

$$ \int_0^1 - \left[ \sum_{3 \le p < 2n} \sin^2(2\pi px) \right] \cos(2\pi 2nx) dx > 0, $$

or

$$ \int_0^1 \left[ \sum_{3 \le p < 2n} \cos(2\pi px) \sin(2\pi px) \right] \sin(2\pi 2nx) dx > 0. $$

# 10 Asymptotic Formula

Please note that $Goldbachz(2n, \frac{2n}{2n}) = Goldbachz(2n, 1) = (\pi_3(2n))^2$.

**Proposition 61.** All the work leads to the following conjecture:

$$ GoldbachNumber_2(2n) = \frac{\sum_{z \in V_{2n}} Goldbachz(0, 2n, z)}{2n} = \frac{\sum_{z \in V_{2n}} \left( \sum_{3 \le p < 2n} e^{2\pi ipz} \right)^2}{2n} = $$

$$ = \frac{\sum_{1 \le j \le 2n} \left( \sum_{3 \le p < 2n} e^{2\pi ip(j/2n)} \right)^2}{2n} \sim 2C_2 \frac{(\pi_3(2n))^2}{2n} \prod_{p|2n} \frac{p-1}{p-2}. \quad (10.1) $$
It is worth noting that the following conjectured asymptotic formula is widely accepted in the mathematical community:

$$ \text{GoldbachNumber}_2(2n) \sim 2C_2 \frac{2n}{(\log(2n))^2} \prod_{p|2n} \frac{p-1}{p-2}. \quad (10.2) $$

<table><caption>Table 1. Comparison of the Goldbach Number, its evaluation from the complex version identity, and its approximation by the asymptotic formulas (randomly selected values of 2n)</caption><thead><tr><th>2n</th><th>GdNo</th><th>GdNoV<sub>2n</sub></th><th>Est</th><th>OurEst</th><th>Est/GdNo</th><th>OurEst/GdNo</th></tr></thead><tbody><tr><td>22</td><td>5</td><td>5</td><td>3.38</td><td>3.27</td><td>0.68</td><td>0.653</td></tr><tr><td>64</td><td>10</td><td>10</td><td>4.89</td><td>0.489</td><td>0.49</td><td>0.596</td></tr><tr><td>158</td><td>9</td><td>9</td><td>8.25</td><td>10.97</td><td>0.92</td><td>1.219</td></tr><tr><td>728</td><td>30</td><td>30</td><td>28.97</td><td>38.90</td><td>0.97</td><td>1.297</td></tr><tr><td>1462</td><td>58</td><td>58</td><td>39.72</td><td>52.66</td><td>0.68</td><td>0.908</td></tr><tr><td>4240</td><td>144</td><td>144</td><td>109.10</td><td>141.92</td><td>0.76</td><td>0.986</td></tr><tr><td>6530</td><td>194</td><td>194</td><td>149.21</td><td>191.88</td><td>0.77</td><td>0.989</td></tr><tr><td>25904</td><td>398</td><td>398</td><td>331.39</td><td>413.68</td><td>0.83</td><td>1.039</td></tr><tr><td>86814</td><td>2854</td><td>2854</td><td>2366.28</td><td>2887.96</td><td>0.83</td><td>1.012</td></tr><tr><td>140036</td><td>1734</td><td>1734</td><td>1437.01</td><td>1741.59</td><td>0.83</td><td>1.004</td></tr></tbody></table>

$$ \sum_{z \in V_{2n}} \text{Goldbach} z 0(2n, z) \\ \text{where } \text{GoldbachNumber}_2(2n), \text{GdNoV}_{2n} = \frac{z \in V_{2n}}{2n}, \\ \text{Est} = 2C_2 \frac{2n}{(\log(2n))^2} \prod_{p|2n} \frac{p-1}{p-2}, \text{OurEst} = 2C_2 \frac{(\pi_3(2n))^2}{2n} \prod_{p|2n} \frac{p-1}{p-2}. $$

The ultimate goal of the work on the Goldbach Conjecture is to prove this asymptotic formula.

For $2n > 1400$, our conjecture given in Equation 10.1 provides significantly higher accuracy than the conjecture given in Equation 10.2; see Table 1.

# Part II

## Weak Goldbach Conjecture

We study identities related to the Weak Goldbach Conjecture because they may provide a pathway toward proving the Goldbach Conjecture through contraction techniques. These identities may also suggest a new approach to proving the Weak Goldbach Conjecture itself. Interestingly, many of their simpler forms cease to hold once we move beyond the domain of the Goldbach and Weak Goldbach Conjectures.

We define Weak Goldbach Conjecture ($2n \ge 8$) number as

$$ \text{GoldbachNumber}_3(2n+1) = \sum_{p+q+r=2n+1} 1 = \int_0^1 \left( \sum_{3 \le p < 2n+1} e^{2\pi ipz} \right)^3 e^{2\pi i(-2n-1)z} dz. $$
# 11 Identities

## 11.1 Identities and Examples

**Theorem 62.** The following identity holds for $n \ge 4$.

$$ \text{GoldbachNumber}_3(2n + 1) $$

$$ = - \frac{\sum_{z \in D_{2(2n+1)}} (\sum_{3 \le p < 2n+1} e^{2\pi i p z})^3}{2n + 1}. $$

**Example 63.** For $2n + 1 = 15$, we have GoldbachNumber$_3(15) = 7$, $D_{2(2n+1)} = D_{30} = \{1/2, 1/6, 5/6, 1/10, 3/10, 7/10, 9/10, 1/30, 7/30, 11/30, 13/30, 17/30, 19/30, 23/30, 29/30\}$

$$ - \frac{\sum_{z \in D_{2(2n+1)}} (\sum_{3 \le p < 2n+1} e^{2\pi i p z})^3}{2n + 1} = - \frac{\sum_{z \in D_{30}} (\sum_{3 \le p < 15} e^{2\pi i p z})^3}{15} = $$

$$ - \frac{\sum_{z \in D_{30}} (e^{2\pi i (3z)} + e^{2\pi i (5z)} + e^{2\pi i (7z)} + e^{2\pi i (11z)} + e^{2\pi i (13z)})^3}{15} = - \frac{-105}{15} = 7. $$

**Theorem 64.** The following identity holds for $n \ge 4$.

$$ \text{GoldbachNumber}_3(2n + 1) $$

$$ = \frac{\sum_{z \in V_{2n+1}} (\sum_{3 \le p < 2n+1} e^{2\pi i p z})^3}{2n + 1}. $$

**Example 65.** For $2n + 1 = 15$, we have GoldbachNumber$_3(15) = 7$, $V_{2n+1} = V_{15} = \{1/15, 2/15, 1/5, 4/15, 1/3, 2/5, 7/15, 8/15, 3/5, 2/3, 11/15, 4/5, 13/15, 14/15, 1\}$

$$ \frac{\sum_{z \in V_{2n+1}} (\sum_{3 \le p < 2n+1} e^{2\pi i p z})^3}{2n + 1} = \frac{\sum_{z \in V_{15}} (\sum_{3 \le p < 15} e^{2\pi i p z})^3}{15} = $$

$$ \frac{\sum_{z \in V_{15}} (e^{2\pi i (3z)} + e^{2\pi i (5z)} + e^{2\pi i (7z)} + e^{2\pi i (11z)} + e^{2\pi i (13z)})^3}{15} = \frac{105}{15} = 7. $$

**Theorem 66.** The following more general identity holds for any $n \ge 4$, $2m+1 \ge 2n+1$
GoldbachNumber$_3(2n + 1)$

$$
\begin{align*}
&= \frac{\sum_{z \in V_{2m+1}} \left( \sum_{3 \le p < 2n+1} e^{2\pi i p z} \right)^3 e^{2\pi i \left(-\left(\frac{2n+1}{2}\right)z\right)}}{2m+1} \\
&= \frac{\sum_{1 \le j \le 2m+1} \left( \sum_{3 \le p < 2n+1} e^{2\pi i p z} \right)^3 e^{2\pi i \left(-\left(\frac{2n+1}{2}\right)\frac{j}{2m+1}\right)}}{2m+1}.
\end{align*}
$$

**Example 67.** For $2n + 1 = 15$, $2m + 1 = (2n + 1) + 2 = 17$, we have
$\textit{GoldbachNumber}_3(15) = 7, V_{2m+1} = V_{17} = \{1/17, 2/17, 3/17, 4/17,$
$5/17, 6/17, 7/17, 8/17, 9/17, 10/17, 11/17, 12/17, 13/17, 14/17, 15/17, 16/17, 1\}$

$$
\frac{\sum_{z \in V_{2m+1}} \left( \sum_{3 \le p < 2n+1} e^{2\pi i p z} \right)^3 e^{2\pi i \left(-\left(\frac{2n+1}{2}\right)z\right)}}{2m+1} = \frac{\sum_{z \in V_{17}} \left( \sum_{3 \le p < 15} e^{2\pi i p z} \right)^3 e^{2\pi i \left(-15\right)z}}{17} = \\
\frac{\sum_{z \in V_{17}} \left( e^{2\pi i \left(3\right)z} + e^{2\pi i \left(5\right)z} + e^{2\pi i \left(7\right)z} + e^{2\pi i \left(11\right)z} + e^{2\pi i \left(13\right)z} \right)^3 e^{2\pi i \left(-15\right)z}}{17} = \frac{119}{17} = 7.
$$

**Theorem 68.** *The following identity holds for $n \ge 4$, $m \ge 4n$ (requirement of $m$ must be odd is no longer necessary when $m \ge 4n$)*

GoldbachNumber$_3(2n + 1)$

$$
= \frac{\sum_{z \in V_m} \left( \sum_{3 \le p < 2n+1} e^{2\pi i p z} \right)^3 e^{2\pi i \left(-\left(\frac{2n+1}{2}\right)z\right)}}{m}.
$$

**Example 69.** For $2n + 1 = 15$, $m = 4n = 28$, we have GoldbachNumber$_3(15) =$
$7, V_m = V_{28} = \{1/28, 1/14, 3/28, 1/7, 5/28, 3/14, 1/4, 2/7, 9/28, 5/14, 11/28, 3/7,$
$13/28, 1/2, 15/28, 4/7, 17/28, 9/14, 19/28, 5/7, 3/4, 11/14, 23/28, 6/7, 25/28, 13/14, 27/28, 1\}$

$$
\frac{\sum_{z \in V_m} \left( \sum_{3 \le p < 2n+1} e^{2\pi i p z} \right)^3 e^{2\pi i \left(-\left(\frac{2n+1}{2}\right)z\right)}}{m} = \frac{\sum_{z \in V_{28}} \left( \sum_{3 \le p < 15} e^{2\pi i p z} \right)^3 e^{2\pi i \left(-15\right)z}}{28} = \\
\frac{\sum_{z \in V_{28}} \left( e^{2\pi i \left(3\right)z} + e^{2\pi i \left(5\right)z} + e^{2\pi i \left(7\right)z} + e^{2\pi i \left(11\right)z} + e^{2\pi i \left(13\right)z} \right)^3 e^{2\pi i \left(-15\right)z}}{28} = \frac{196}{28} = 7.
$$
**Theorem 70.** The following identity holds for any $2n + 1$,

$$ \text{GoldbachNumber}_3(2n + 1) = \frac{\sum_{1 \le j \le 2n+1} \left( \sum_{z \in \text{Seed}_{(2n+1)}} z^j \right)^3}{2n + 1}, $$

where

$$ \text{Seed}_{(2n+1)} = \{e^{2\pi i(\frac{p}{2n})} | 3 \le p < 2n + 1\}. $$

**Example 71.** For $2n + 1 = 15$, we have $\text{GoldbachNumber}_3(2n+1) = \text{GoldbachNumber}_3(15) = 7$.

$$ \text{Seed}_{(2n+1)} = \text{Seed}_{(15)} = \{e^{2\pi i(\frac{p}{15})} | 3 \le p < 15\} = \{e^{2\pi i(\frac{3}{15})}z, e^{2\pi i(\frac{5}{15})}z, e^{2\pi i(\frac{7}{15})}z, e^{2\pi i(\frac{11}{15})}z, e^{2\pi i(\frac{13}{15})}z\}. $$

We have

$$ \frac{\sum_{1 \le j \le 2n+1} \left( \sum_{z \in \text{Seed}_{(2n+1)}} z^j \right)^3}{2n + 1} = \frac{\sum_{1 \le j \le 15} \left( \sum_{z \in \text{Seed}_{(15)}} z^j \right)^3}{15} = $$

$$ \frac{\sum_{1 \le j \le 15} \left[ \left( e^{2\pi i(\frac{3}{15})} \right)^j + \left( e^{2\pi i(\frac{5}{15})} \right)^j + \left( e^{2\pi i(\frac{7}{15})} \right)^j + \left( e^{2\pi i(\frac{11}{15})} \right)^j + \left( e^{2\pi i(\frac{13}{15})} \right)^j \right]^3}{15} $$

By direct calculation, we obtain

$$ \frac{\sum_{1 \le j \le 15} \left[ \left( e^{2\pi i(\frac{3}{15})} \right)^j + \left( e^{2\pi i(\frac{5}{15})} \right)^j + \left( e^{2\pi i(\frac{7}{15})} \right)^j + \left( e^{2\pi i(\frac{11}{15})} \right)^j + \left( e^{2\pi i(\frac{13}{15})} \right)^j \right]^3}{15} = $$

$$ \frac{1}{15}(351+69e^{-\frac{1}{15}(2i\pi)}+69e^{\frac{2i\pi}{15}}+69e^{-\frac{1}{15}(4i\pi)}+69e^{\frac{4i\pi}{15}}+171e^{-\frac{1}{5}(2i\pi)}+171e^{\frac{2i\pi}{5}}+69e^{-\frac{1}{15}(8i\pi)}+69e^{\frac{8i\pi}{15}}+144e^{-\frac{1}{3}(2i\pi)}+144e^{\frac{2i\pi}{3}}+171e^{-\frac{1}{5}(4i\pi)}+171e^{\frac{4i\pi}{5}}+69e^{-\frac{1}{15}(14i\pi)}+69e^{\frac{14i\pi}{15}}) = 7. $$

So

$$ \text{GoldbachNumber}_3(15) = \frac{\sum_{1 \le j \le 15} \left( \sum_{z \in \text{Seed}_{(15)}} z^j \right)^3}{15}. $$
## 11.2 Proof of the Identities

The proofs can be obtained using the same techniques outlined in Part I.

# 12 Relationship Between the Goldbach Conjecture and the Weak Goldbach Conjecture

## 12.1 Identities Relating the Goldbach Conjecture and the Weak Goldbach Conjecture

**Theorem 72.** For any $(2n \ge 8)$ and any $2m + 1 \ge 2n + 1 \ge 9$, the following identity holds:

$$ \text{GoldbachNumber}_3(2n+1) = \text{GoldbachNumber}_2(2n) * \pi_3(2n-3) - \frac{\sum_{1 \le j \le 2m} A_j(b_{j+1} - b_j)}{2m+1} = $$

$$ \text{GoldbachNumber}_2(2n) * \pi_3(2n-3) + \text{Gd3Rest}(2n+1, 2m+1), $$

where

$$ \text{Gd3Rest}(2n+1, 2m+1) = - \frac{\sum_{1 \le j \le 2m} A_j(b_{j+1} - b_j)}{2m+1}. $$

Some clarifications are needed for the identity above.

First, we note that

$$ \text{GoldbachNumber}_2(2n) = \int_0^1 \left( \sum_{3 \le p < 2n} e^{2\pi ipz} \right)^2 e^{2\pi i(-2n)z} dz = \\ \int_0^1 \left( \sum_{3 \le p \le 2n-3} e^{2\pi ipz} \right)^2 e^{2\pi i(-2n)z} dz \text{ where } 2n \ge 8 \quad (12.1) $$

The above change for the summation range is valid because if $p = (2n - 1)$ is a prime, then $(2n - 1) + q \ne 2n$, where $q$ is an odd prime.

Second, we note that

$$ \text{GoldbachNumber}_3(2n+1) = \int_0^1 \left( \sum_{3 \le p < 2n+1} e^{2\pi ipz} \right)^3 e^{2\pi i(-2n-1)z} dz = \\ \int_0^1 \left( \sum_{3 \le p \le 2n-3} e^{2\pi ipz} \right)^3 e^{2\pi i(-2n-1)z} dz \text{ where } 2n \ge 8 \quad (12.2) $$
The above change for the summation range is valid because if $p = (2n - 1)$ is a prime, then $(2n - 1) + q + r \neq 2n + 1$, where $q$ and $r$ are odd primes.

The purpose of making these two changes in (12.1) and (12.2) is to ensure that the expression of

$$ (\sum_{3 \le p \le 2n-3} e^{2\pi ipz})^2 $$

is in both $GoldbachNumber_2(2n)$ and $GoldbachNumber_3(2n + 1)$.

Third, we define

$$ a_j = \left( \sum_{3 \le p \le 2n-3} e^{2\pi ip \left( \frac{j}{2m+1} \right)} \right)^2 e^{2\pi i(-2n) \left( \frac{j}{2m+1} \right)} $$

hence by Theorem 41 we have

$$ GoldbachNumber_2(2n) = \frac{\sum_{1 \le j \le 2m+1} \left( \sum_{3 \le p \le 2n-3} e^{2\pi ip \left( \frac{j}{2m+1} \right)} \right)^2 e^{2\pi i(-2n) \left( \frac{j}{2m+1} \right)}}{2m+1} = \frac{\sum_{1 \le j \le 2m+1} a_j}{2m+1}. \quad (12.3) $$

Fourth, we define for $1 \le k \le 2m + 1$, $A_k = \sum_{1 \le j \le k} a_j$, and hence

$$ GoldbachNumber_2(2n) = \frac{\sum a_j}{\frac{1 \le j \le 2m+1}{2m+1}} = \frac{A_{2m+1}}{2m+1}. \quad (12.4) $$

Fifth, we define

$$ b_j = \sum_{3 \le p \le 2n-3} e^{2\pi i(p-1)\left(\frac{j}{2m+1}\right)}. $$

**Example 73.** (For Theorem 72.) For $2n + 1 = 15$, $2m + 1 = 2n + 1 = 15$, we know that $GoldbachNumber_3(2n + 1) = GoldbachNumber_3(15) = 7$, $GoldbachNumber_2(2n) = GoldbachNumber_2(14) = 3$, $\pi_3(2n - 3) = \pi_3(11) = 4$, so

$$ GoldbachNumber_2(2n) * \pi_3(2n - 3) = GoldbachNumber_2(14) * \pi_3(11) = 3 * 4 = 12. $$

By calculation, we obtain the following:

$$ Gd3Rest(2n+1, 2m+1) = -\frac{\sum_{1 \le j \le 2m} A_j(b_{j+1} - b_j)}{2m+1} = -\frac{\sum_{1 \le j \le 14} A_j(b_{j+1} - b_j)}{15} = -\frac{75}{15} = -5. $$

Hence,

$$ GoldbachNumber_3(2n+1) = GoldbachNumber_2(2n) * \pi_3(2n-3) - \frac{\sum_{1 \le j \le 14} A_j(b_{j+1} - b_j)}{15} = 12 - 5 = 7. $$
**Example 74.** (For Theorem 72.) For $2n + 1 = 15$, $2m + 1 = 21$, we know that
$\text{GoldbachNumber}_3(2n+1) = \text{GoldbachNumber}_3(15) = 7$, $\text{GoldbachNumber}_2(2n) = \text{GoldbachNumber}_2(14) = 3$, $\pi_3(2n-3) = \pi_3(11) = 4$, so
$\text{GoldbachNumber}_2(2n) * \pi_3(2n-3) = \text{GoldbachNumber}_2(14) * \pi_3(11) = 3 * 4 = 12$.

By calculation, we obtain the following:

$$ \text{Gd3Rest}(2n+1, 2m+1) = -\frac{\sum_{1 \le j \le 2m} A_j(b_{j+1} - b_j)}{2m+1} = -\frac{\sum_{1 \le j \le 20} A_j(b_{j+1} - b_j)}{21} = -\frac{105}{21} = -5. $$

Hence,

$$ \text{GoldbachNumber}_3(2n+1) = \text{GoldbachNumber}_2(2n) * \pi_3(2n-3) - \frac{\sum_{1 \le j \le 20} A_j(b_{j+1} - b_j)}{21} = 12 - 5 = 7. $$

## 12.2 Proof of the Identity

**Lemma 75.** *Abel Summation*

For two given complex sequences ${a_j}_{j=1}^{j=m}$ and ${b_j}_{j=1}^{j=m}$

$$ \sum_{1 \le j \le m} a_j b_j = A_m b_m - \sum_{1 \le j \le m-1} A_j (b_{j+1} - b_{j+1}). $$

*Proof.* See G. Tenenbaum[3].

**Example 76.** For ${a_j}_{j=1}^{j=4}$ and ${b_j}_{j=1}^{j=4}$

$$ a_1 b_1 + a_2 b_2 + a_3 b_3 + a_4 b_4 = $$

$$ (a_1+a_2+a_3+a_4)*b_4 - (a_1)*(b_2-b_1) - (a_1+a_2)*(b_3-b_2) - (a_1+a_2+a_3)*(b_4-b_3). $$

We now continue the proof of Theorem 72. By Theorem 66 for $2m+1 \ge 2n+1$ we have the following:

$$ \text{GoldbachNumber}_3(2n+1) = \frac{\sum_{z \in V_{2m+1}} \left( \sum_{3 \le p \le 2n-3} e^{2\pi ipz} \right)^3 e^{2\pi i(-2n-1)z}}{2m+1} = $$

$$ \frac{\sum_{z \in V_{2m+1}} \left( \sum_{3 \le p \le 2n-3} e^{2\pi ipz} \right)^2 e^{2\pi ipz} e^{2\pi i(-2n-1)z}}{2m+1} = $$
$$ \frac{\sum_{z \in V_{2m+1}} \left( \sum_{3 \le p \le 2n-3} e^{2\pi i p z} \right)^2 e^{2\pi i (-2n)z} e^{2\pi i p z} e^{2\pi i (-1)z}}{2m+1} = $$

$$ \frac{\sum_{z \in V_{2m+1}} \left( \sum_{3 \le p \le 2n-3} e^{2\pi i p z} \right)^2 e^{2\pi i (-2n)z} e^{2\pi i (p-1)z}}{2m+1} = $$

$$ \frac{\sum_{1 \le j \le 2m+1} \left( \sum_{3 \le p \le 2n-3} e^{2\pi i p \left(\frac{j}{2m+1}\right)} \right)^2 e^{2\pi i (-2n)\left(\frac{j}{2m+1}\right)} e^{2\pi i (p-1)\left(\frac{j}{2m+1}\right)}}{2m+1} = \frac{\sum_{1 \le j \le 2m+1} a_j b_j}{2m+1} $$

where

$$ a_j = \left( \sum_{3 \le p \le 2n-3} e^{2\pi i p \left( \frac{j}{2m+1} \right)} \right)^2 e^{2\pi i (-2n) \left( \frac{j}{2m+1} \right)} $$

and

$$ b_j = \sum_{3 \le p \le 2n-3} e^{2\pi i (p-1) \left(\frac{j}{2m+1}\right)}. $$

By the Lemma 75

$$ \frac{\sum_{1 \le j \le 2m+1} a_j b_j}{2m+1} = \frac{A_{2m+1} b_{2m+1} - \sum_{1 \le j \le 2m+1-1} A_j (b_{j+1} - b_j)}{2m+1} = $$

$$ \frac{A_{2m+1} b_{2m+1} - \sum_{1 \le j \le 2m+1-1} A_j (b_{j+1} - b_j)}{2m+1} $$

where for $1 \le k \le 2m+1$, $A_k = \sum_{1 \le j \le k} a_j$.

Note that

$$ \frac{A_{2m+1}}{2m+1} = \frac{\sum_{1 \le j \le 2m+1} a_j}{2m+1} = \text{GoldbachNumber}_2(2n), $$

and

$$ b_{2m+1} = \sum_{3 \le p \le 2n-3} 1 = \pi_3(2n-3), $$

thus we have the following:

$$ \text{GoldbachNumber}_3(2n+1) = \text{GoldbachNumber}_2(2n) * \pi_3(2n-3) - \frac{\sum_{1 \le j \le 2m} A_j(b_{j+1} - b_j)}{2m+1}. $$

This completes the proof of Theorem 72.
## 12.3 Propositions

Suppose to the contrary, for some large $n$, we have $GoldbachNumber_2(2n) = 0$.

By Theorem 72,

$$ GoldbachNumber_3(2n+1) = 0 + Gd3Rest(2n+1, 2m+1) = -\frac{\sum_{1 \le j \le 2m} A_j(b_{j+1} - b_j)}{2m+1}. $$

**Proposition 77.** For sufficiently large $n$ and for some $2m+1 \ge 2n+1$, the following holds:

$$ Gd3Rest(2n + 1, 2m + 1) = -\frac{\sum_{1 \le j \le 2m} A_j(b_{j+1} - b_j)}{2m + 1} < 0. $$

**Proposition 78.** For sufficiently large $n$ and for some $2m+1 \ge 2n+1$, the following holds:

$$ |Gd3Rest(2n + 1, 2m + 1)| = \left| \frac{\sum_{1 \le j \le 2m} A_j(b_{j+1} - b_j)}{2m + 1} \right| \sim 1.150481 \prod_{p|(2n+1)} \left(1 - \frac{1}{p(p-3)+3}\right) \frac{(2n+1)^2}{(\log(2n+1))^3} \text{ see (16.1).} $$

# Part III

## Beyond Goldbach Conjecture

For any integer $k$ with $k \ge 2$, we define the following functions:

$$ F_{2k}(2n, 2k, z) = \left( \sum_{3 \le p < 2n} e^{2\pi i p z} \right)^{2k} e^{2\pi i (-2n)z}, $$

and

$$ F_{2k+1}(2n+1, 2k+1, z) = \left( \sum_{3 \le p < 2n+1} e^{2\pi i p z} \right)^{2k+1} e^{2\pi i (-(2n+1))z}. $$

We define the Goldbach Number as follows

for $2n \ge (2k) * 3$

$$ GoldbachNumber_{2k}(2n) = \int_{0}^{1} \left( \sum_{3 \le p < 2n} e^{2\pi i p z} \right)^{2k} e^{2\pi i (-2n)z} dz = \int_{0}^{1} F_{2k}(2n, 2k, z) dz $$

and for $2n+1 \ge (2k+1) * 3$
$$
\begin{gather*}
\text{GoldbachNumber}_{2k+1}(2n+1) = \int_0^1 \left( \sum_{3 \le p < 2n+1} e^{2\pi ipz} \right)^{2k+1} e^{2\pi i(-(2n+1))z} dz = \\
\int_0^1 F_{2k+1}(2n+1, 2k+1, z) dz.
\end{gather*}
$$

13 The First Set of Identities

13.1 Identities and Examples

Theorem 79. The following identities hold:

$$
\text{GoldbachNumber}_{2k}(2n) = \frac{\sum_{z \in D_{2(1-k-n+2kn)}} F_{2k}(2n, 2k, z) - \frac{1}{2} \sum_{z \in E_{2(1-k-n+2kn)}} F_{2k}(2n, 2k, z)}{1 - k - n + 2kn}
$$

and

$$
\text{GoldbachNumber}_{2k+1}(2n+1) =
$$

$$
\frac{\sum_{z \in D_{2(k+2kn)}} F_{2k+1}(2n+1, 2k+1, z) - \frac{1}{2} \sum_{z \in E_{2(k+2kn)}} F_{2k+1}(2n+1, 2k+1, z)}{k + 2kn}.
$$

To illustrate the above identities, we present two examples: the first for
GoldbachNumber<sub>2k</sub>(2n), and the second for GoldbachNumber<sub>2k+1</sub>(2n + 1).

**Example 80.** For $2n = 14$ and $k = 2$, we explicitly evaluate both the left-hand side (LHS) and the right-hand side (RHS) of the following identity.

$$
\text{GoldbachNumber}_{2k}(2n) = \frac{\sum_{z \in D_{2(1-k-n+2kn)}} F_{2k}(2n, 2k, z) - \frac{1}{2} \sum_{z \in E_{2(1-k-n+2kn)}} F_{2k}(2n, 2k, z)}{1 - k - n + 2kn}.
$$

Note that $2k = 4$ and $1 - k - n + 2kn = 20$.

LHS = GoldbachNumber<sub>2k</sub>(2n) = GoldbachNumber<sub>4</sub>(14), which counts the
number of solutions of p<sub>1</sub> + p<sub>2</sub> + p<sub>3</sub> + p<sub>4</sub> = 14.

Note that $5+3+3+3=3+5+3+3=3+3+5+3=3+3+3+5=14$.

Hence, LHS = 4.
$$RHS = \frac{\sum_{z \in D_{2(1-k-n+2kn)}} F_{2k}(2n, 2k, z) - \frac{1}{2} \sum_{z \in E_{2(1-k-n+2kn)}} F_{2k}(2n, 2k, z)}{1 - k - n + 2kn}$$

$$= \frac{\sum_{z \in D_{40}} F_{4}(14, 4, z) - \frac{1}{2} \sum_{z \in E_{40}} F_{4}(14, 4, z)}{20}$$

$$= \frac{\sum_{z \in D_{40}} (\sum_{3 \le p < 14} e^{2\pi i p z})^4 e^{2\pi i (-14)z} - \frac{1}{2} \sum_{z \in E_{40}} (\sum_{3 \le p < 14} e^{2\pi i p z})^4 e^{2\pi i (-14)z}}{20}.$$

Note that

$$( \sum_{3 \le p \le 14} e^{2\pi i p z})^4 e^{2\pi i (-14)z} = (e^{2\pi i 3z} + e^{2\pi i 5z} + e^{2\pi i 7z} + e^{2\pi i 11z} + e^{2\pi i 13z})^4 e^{2\pi i (-14)z},$$

$$D_{40} = \{1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, 1/10, 3/10, 7/10, 9/10, 1/20, 3/20, 7/20, 9/20, 11/20, 13/20, 17/20, 19/20, 1/40, 3/40, 7/40, 9/40, 11/40, 13/40, 17/40, 19/40, 21/40, 23/40, 27/40, 29/40, 31/40, 33/40, 37/40, 39/40\}$$

and

$$E_{40} = \{1/4, 3/4, 1/8, 3/8, 5/8, 7/8, 1/20, 3/20, 7/20, 9/20, 11/20, 13/20, 17/20, 19/20, 1/40, 3/40, 7/40, 9/40, 11/40, 13/40, 17/40, 19/40, 21/40, 23/40, 27/40, 29/40, 31/40, 33/40, 37/40, 39/40\}.$$

By direct evaluation, we have

$$\sum_{z \in D_{40}} (e^{2\pi i 3z} + e^{2\pi i 5z} + e^{2\pi i 7z} + e^{2\pi i 11z} + e^{2\pi i 13z})^4 e^{2\pi i (-14)z} = -495,$$

and

$$\sum_{z \in E_{40}} (e^{2\pi i 3z} + e^{2\pi i 5z} + e^{2\pi i 7z} + e^{2\pi i 11z} + e^{2\pi i 13z})^4 e^{2\pi i (-14)z} = -1150.$$

Hence,

$$RHS = \frac{(-495) - \frac{1}{2}(-1150)}{20} = 4.$$

Thus,

$$LHS = RHS.$$
**Example 81.** For $2n + 1 = 93$ and $k = 4$, we evaluate both sides (LHS and RHS) of the identity below.

$$ \text{GoldbachNumber}_{2k+1}(2n+1) = $$

$$ \frac{\sum_{z \in D_{2(k+2kn)}} F_{2k+1}(2n+1, 2k+1, z) - \frac{1}{2} \sum_{z \in E_{2(k+2kn)}} F_{2k+1}(2n+1, 2k+1, z)}{k+2kn} $$

Note that $2k + 1 = 9$, $k + 2kn = 372$.

By calculations, we obtain LHS = GoldbachNumber$_{2k+1}$(2n+1) = GoldbachNumber$_{9}$(93) = 7342581.

$$ \text{RHS} = \frac{\sum_{z \in D_{2(k+2kn)}} F_{2k+1}(2n+1, 2k+1, z) - \frac{1}{2} \sum_{z \in E_{2(k+2kn)}} F_{2k+1}(2n+1, 2k+1, z)}{k+2kn} $$

$$ = \frac{\sum_{z \in D_{744}} F_{9}(93, 9, z) - \frac{1}{2} \sum_{z \in E_{744}} F_{9}(93, 9, z)}{372} $$

$$ = \frac{\sum_{z \in D_{744}} \left( \sum_{3 \le p < 93} e^{2\pi ipz} \right)^9 e^{2\pi i(-93)z} - \frac{1}{2} \sum_{z \in E_{744}} \left( \sum_{3 \le p < 93} e^{2\pi ipz} \right)^9 e^{2\pi i(-93)z}}{372} = $$

$$ \frac{-1728891947442 - \frac{1}{2}(-3463246775148)}{372} = 7342581. $$

Thus,

$$ \text{LHS} = \text{RHS}. $$

## 13.2 Proof of the Identities

The proofs of these identities follow arguments similar to those used in the proof of Theorem 7.

# 14 The Second Set of Identities

## 14.1 Identities and Examples

**Theorem 82.** The following hold for $m \ge 2k(2n)$, where $k > 1$.
$$ \text{GoldbachNumber}_{2k}(2n) = \frac{\sum_{z \in V_m} F_{2k}(2n, 2k, z)}{m} = \frac{\sum_{z \in V_m} \left( \sum_{3 \le p < 2n} e^{2\pi ipz} \right)^{2k} e^{2\pi i(-2n)z}}{m} $$

and

$$ \text{GoldbachNumber}_{2k+1}(2n+1) = \frac{\sum_{z \in V_m} F_{2k}(2n+1, 2k+1, z)}{m} = $$

$$ \frac{\sum_{z \in V_m} \left( \sum_{3 \le p < 2n+1} e^{2\pi ipz} \right)^{2k+1} e^{2\pi i(-2n-1)z}}{m} $$

**Example 83.** For $2n = 14$ and $k = 2$, thus $2k = 4$, let $m = 5(2n) = 70$. We evaluate both the left-hand side (LHS) and the right-hand side (RHS) of the following identity.

$$ \text{GoldbachNumber}_4(14) = \frac{\sum_{z \in V_{70}} \left( \sum_{3 \le p < 14} e^{2\pi ipz} \right)^4 e^{2\pi i(-14)z}}{70} $$

We know from example 80. LHS = 4.

Note that

$$ \text{RHS} = \frac{\sum_{z \in V_{70}} \left( \sum_{3 \le p < 14} e^{2\pi ipz} \right)^4 e^{2\pi i(-14)z}}{70} = $$

$$ \frac{\sum_{z \in V_{70}} (e^{2\pi i 3z} + e^{2\pi i 5z} + e^{2\pi i 7z} + e^{2\pi i 11z} + e^{2\pi i 13z})^4 e^{2\pi i(-14)z}}{70} = $$

$$ \frac{\sum_{1 \le j \le 70} (e^{2\pi i 3(j/70)} + e^{2\pi i 5(j/70)} + e^{2\pi i 7(j/70)} + e^{2\pi i 11(j/70)} + e^{2\pi i 13(j/70)})^4 e^{2\pi i(-14)(j/70)}}{70} = $$

$$ \frac{280}{70} = 4. $$

Thus,

$$ \text{LHS} = \text{RHS}. $$

**Example 84.** For $2n + 1 = 93$ and $k = 4$, $2k + 1 = 9$, let $m = 2k(2n) = 736$. We evaluate both the left-hand side (LHS) and the right-hand side (RHS) of the following identity.
$$ \text{GoldbachNumber}_9(93) = \frac{\sum_{z \in V_{736}} (\sum_{3 \le p < 93} e^{2\pi ipz})^9 e^{2\pi i(-93)z}}{736} $$

We know from example 81. LHS = 7342581.

Note that by calculation,

$$ \text{RHS} = \frac{\sum_{z \in V_{736}} (\sum_{3 \le p < 93} e^{2\pi ipz})^9 e^{2\pi i(-93)z}}{736} = \frac{5404139616}{736} = 7342581. $$

Thus,

$$ \text{LHS} = \text{RHS}. $$

## 14.2 Proof of the Identities

These identities can be proved using reasoning similar to that presented in Part I.

# 15 General Goldbach Number

In this section, we briefly discuss several generalized Goldbach Conjectures that we have investigated. It is clear that the range of potential research problems in this area is virtually unbounded.

In the most general form, let the set $N_n = \{1, 2, 3, ..., n - 1\}$, let $A\text{SubSet}N_n \subseteq N_n$, $l \ge 2$ and $m \ge 1$. We then define a General Goldbach Number based on the subset $A\text{SubSet}N_n$, and the parameters $l$ and $m$ as follows:

$$ \text{GeneralGoldbachNumber}_l(\text{ASubSet}N_n, l, m) = \int_0^1 \left( \sum_{s \in \text{ASubSet}N_n} e^{2\pi i sz} \right)^l e^{2\pi i (-m)z} dz $$

which counts number of ways to express $m$ as sum of $l$ elements from $A\text{SubSet}N_n$.

**Remark:** The reason we define $N_n = \{1, 2, 3, ..., n - 1\}$ rather than $N_n = \{1, 2, 3, ..., n - 1, n\}$ is because, with this definition, certain expressions become significantly simpler.

The key questions to ask are: for predefined $A\text{SubSet}N_n$ and predetermined values of $l$ and $m$, whether or not $\text{GeneralGoldbachNumber}_l(\text{ASubSet}N_n, l, m) > 0$ holds for all sufficiently large $n$, and if so, what the corresponding asymptotic formulas are. Some of these questions are easy to answer, while others can be very challenging. The classical Goldbach Conjecture is one such question.

The following examples illustrate this general framework.
For ASubSetN<sub>n</sub> = N<sub>n</sub>, l = 2, m = n, we have that the number of ways to express n as sum of two numbers from N<sub>n</sub> = {1, 2, 3, ..., n - 1} is

$$
GerenalGoldbachNumber_2(N_n, 2, n) = \int_0^1 \left( \sum_{s \in N_n} e^{2\pi i sz} \right)^2 e^{2\pi i (-n)z} dz = n - 1.
$$

Note that both $1+(n-1)=n$ and $(n-1)+1=n$ are counted as one way.

For ASubSetN<sub>2n</sub> = {k|k ∈ N<sub>2n</sub>, k is an odd number}, l = 2, m = 2n ≥ 2, we have that the number of ways to express 2n as the sum of two odd numbers from N<sub>2n</sub> = {1, 2, 3, ..., 2n - 1} is

$$
GerenalGoldbachNumber_2(\{k|k \in N_{2n}, k \text{ is an odd number}\}, 2, 2n) =
$$

$$
\int_0^1 \left( \sum_{s \in \{k | k \in N_{2n}, k \text{ is an odd number}\}} e^{2\pi i sz} \right)^2 e^{2\pi i (-2n)z} dz = n.
$$

For ASubSetN<sub>2n</sub> = {p|p ∈ N<sub>2n</sub>, p is an odd prime}, l = 2, m = 2n ≥ 6, we
have that the number of ways to express 2n as sum of two old primes from
N<sub>2n</sub> = {1, 2, 3, ..., 2n - 1} is

$$
GerenalGoldbachNumber_2(\{p|p \in N_{2n}, p \text{ is an odd prime}\}, 2, 2n) =
$$

$$
\int_0^1 \left( \sum_{p \in \{p | p \in N_{2n}, p \text{ is an odd prime}\}} e^{2\pi i p z} \right)^2 e^{2\pi i (-2n)z} dz =
$$

GoldbachNumber₂(2n).

For ASubSetN<sub>2n</sub> = {p|p ∈ N<sub>2n</sub>, p is a twin prime}, l = 2, m = 2n ≥ 6, we
have that the number of ways to express 2n as the sum of two twin primes from
N<sub>2n</sub> = {1, 2, 3, ..., 2n - 1} is

$$
GerenalGoldbachNumber_2(\{p|p \in N_{2n}, p \text{ is a twin prime}\}, 2, 2n) =
$$

$$
\int_0^1 \left( \sum_{p \in \{p | p \in N_{2n}, p \text{ is a twin prime}\}} e^{2\pi i p z} \right)^2 e^{2\pi i (-2n)z} dz.
$$

For ASubSetN_{2n+1} = \{p|p \in N_{2n+1}, p \text{ is an odd prime}\},

l = 3, m = 2n + 1 ≥ 9, we have that the number of ways to express 2n + 1
as the sum of three old primes from N_{2n+1} = {1, 2, 3, ..., (2n + 1) - 1} is

$$
GerenalGoldbachNumber_3(\{p|p \in N_{2n+1}, p \text{ is an odd prime}\}, 3, 2n+1) =
$$
$$
\int_0^1 \left( \sum_{p \in \{p | p \in N_{2n+3}, p \text{ is an odd prime}\}} e^{2\pi ipz} \right)^3 e^{2\pi i(-2n-1)z} dz \\
= \text{GoldbachNumber}_3(2n+1).
$$

For

$$
A\text{SubSet}N_{2n+1} = \{p|p \in N_{2n+1}, p \text{ is a twin prime}\}, l = 3, m = 2n+1 \geq 9,
$$

we have that the number of ways to express $2n + 1$ as sum of three twin
primes from $N_{2n+1} = \{1, 2, 3, ..., (2n + 1) - 1\}$ is

$$
\text{GeneralGoldbachNumber}_3(\{p|p \in N_{2n+1}, p \text{ is a twin prime}\}, 3, 2n+1)
$$

$$
= \int_0^1 \left( \sum_{p \in \{p | p \in N_{2n+1}, p \text{ is a twin prime}\}} e^{2\pi i p z} \right)^3 e^{2\pi i (-2n-1)z} dz.
$$

It is evident that the list of such questions is unbounded.

16 Explorations for the General Goldbach Num-
ber

The number of ways to express $2n$ as the sum of two odd numbers from the
set $N_{2n} = \{1, 2, 3, ..., 2n - 1\}$ is

$$
\text{GeneralGoldbachNumber}_2(\{k | k \in N_{2n}, k \text{ is an odd number}\}, 2, 2n) =
$$

$$
\int_0^1 \left( \sum_{z \in \{k | k \in N_{2n}, k \text{ is an odd number}\}} e^{2\pi i p z} \right)^2 e^{2\pi i (-2n)z} dz = n.
$$

The number of ways to express $2n$ as the sum of two old primes from $N_{2n} = \{1, 2, 3, ..., 2n - 1\}$ is

$$
\text{GeneralGoldbachNumber}_2(\{p|p \in N_{2n}, p \text{ is an odd prime}\}, 2, 2n) =
$$

$$
\int_0^1 \left( \sum_{z \in \{p | p \in N_{2n}, p \text{ is an odd prime}\}} e^{2\pi i p z} \right)^2 e^{2\pi i (-2n)z} dz = \\
\text{GoldbachNumber}_2(2n).
$$

**Conjecture 85.** We conjecture the following (see Table 1 in Proposition 61):
$$
\text{GoldbachNumber}_2(2n) \sim 2C_2 \frac{(\pi_3(2n))^2}{2n} \prod_{p|2n} \frac{p-1}{p-2}.
$$

**Conjecture 86.** For the number of ways to express $2n$ as the sum of two twin primes from $N_{2n} = \{1, 2, 3, ..., 2n-1\}$, we propose the following conjecture:

$$
\text{GerenalGoldbachNumber}_2(\{p|p \in N_{2n}, p \text{ is a twin prime}\}, 2, 2n) =
$$

$$
= \int_{0}^{1} \left( \sum_{z \in \{p|p \in N_{2n}, p \text{ is a twin prime}\}} e^{2\pi i p z} \right)^{2} e^{2\pi i (-2n)z} dz = f(n) * \frac{n}{\log(n)} + \varepsilon \frac{n}{\log(n)^4},
$$

where $f(n)$ is a function on $n$ such that $f(n) \gg 1$.

The number of ways to express $2n + 1$ $(2n + 1 \ge 9)$ as the sum of three old primes from $N_{2n+1} = \{1, 2, 3, ..., (2n + 1) - 1\}$ is given by:

$$
\text{GerenalGoldbachNumber}_3(\{p|p \in N_{2n+1}, p \text{ is an odd prime}\}, 3, 2n+1) =
$$

$$
\int_0^1 \left( \sum_{z \in \{p|p \in N_{2n+1}, p \text{ is an odd prime}\}} e^{2\pi i p z} \right)^3 e^{2\pi i (-2n-1)z} dz
$$

$$
= \text{GoldbachNumber}_3(2n + 1).
$$

The following can be proved see [1]:

$$
\text{GoldbachNumber}_3(2n + 1) \sim
$$

$$
1.15048077235 \prod_{p|(2n+1)} \left(1 - \frac{1}{p(3-3)+3}\right) \frac{(2n+1)^2}{(\log(2n+1))^3} \quad (16.1)
$$

**Conjecture 87.** For the number of ways to express $2n + 1$ as the sum of three twin primes from $N_{2n+1} = \{1, 2, 3, ..., (2n+1)-1\}$, we conjecture the following:

$$
\text{GerenalGoldbachNumber}_3(\{p|p \in N_{2n+1}, p \text{ is a twin prime}\}, 3, 2n+1) =
$$

$$
= \int_{0}^{1} \left( \sum_{z \in \{p|p \in N_{2n+1}, p \text{ is a twin prime}\}} e^{2\pi i p z} \right)^{3} e^{2\pi i (-2n-1)z} dz =
$$

$$
f(n) * n^2 / (\log(n))^6 + \varepsilon(n^2 / (\log(n))^6),
$$
where $f(n)$ is a function on $n$ such that $f(n) \gg 1$.

In particular, we conjecture that for sufficiently large $n$, the number of ways to express $2n+1$ as the sum of three twin primes from

$$N_{2n+1} = \{1, 2, 3, \dots, (2n+1)-1\}$$

exceeds the number of ways to express $2n$ as the sum of two odd primes from

$$N_{2n} = \{1, 2, 3, \dots, 2n-1\},$$

which can be formally expressed as follows:

$$\text{GeneralGoldbachNumber}_3(\{p | p \in N_{2n+1}, p \text{ is a twin prime}\}, 3, 2n+1) >$$

$$\text{GoldbachNumber}_2(2n).$$

Furthermore, we can ask: what is the smallest subset

$A\text{SubSet}N_{2n} \subseteq \{p | p \in N_{2n}, p \text{ is an odd prime}\}$ such that, for sufficiently large $n$, the number of ways to express $2n$ as the sum of two odd primes from $A\text{SubSet}N_{2n}$ remains positive?

**Conjecture 88.** We conjecture that for any $k \ge 2$, there exists a subset

$A\text{SubSet}N_{2n} \subseteq \{p | p \in N_{2n}, p \text{ is an odd prime}\}$ with

$$|A\text{SubSet}N_{2n}| = O(n/(\log(n))^k),$$

(That is $A\text{SubSet}N_{2n}$ can be made “arbitrarily small”),

such that the Goldbach Conjecture still holds. Moreover, the number of representations of $2n$ as a sum of two odd primes from this subset satisfies the asymptotic formula:

$$\text{GeneralGoldbachNumber}_2(\{p | p \in A\text{SubSet}N_{2n}\}, 2, 2n) =$$

$$= \int_0^1 \left( \sum_{z \in A\text{SubSet}N_{2n}} e^{2\pi i p z} \right)^2 e^{2\pi i (-2n)z} dz =$$

$$= f(n) * \frac{n}{(\log(n))^{2k}} + \varepsilon(\frac{n}{(\log(n))^{2k}}),$$

where $f(n)$ is a function on $n$ such that $f(n) \gg 1$.

Recall that the number of ways to express $2n$ as the sum of two odd numbers from the set $N_{2n} = \{1, 2, 3, \dots, 2n-1\}$ is given by

$$\text{GeneralGoldbachNumber}_2(\{k | k \in N_{2n}, k \text{ is an odd number}\}, 2, 2n) =$$

$$\int_0^1 \left( \sum_{z \in \{k | k \in N_{2n}, k \text{ is an odd number}\}} e^{2\pi i p z} \right)^2 e^{2\pi i (-2n)z} dz = n.$$
We can show that the number of ways to express $2m$, where $6 \le 2m \le 4n - 2$, as the sum of two odd numbers $n_1 \ge 3$ and $n_2 \ge 3$, chosen from the set $N_{2n} = \{1, 2, 3, ..., 2n - 1\}$ is greater than zero.

Another direction for generalizing the Goldbach Conjecture is to ask the following question:

What can be said about the number of ways to express $2m$, where $6 \le 2m \le 4n - 2$, as the sum of two odd prime numbers $p_1 \ge 3$ and $p_2 \ge 3$, chosen from the set $N_{2n} = \{1, 2, 3, ..., 2n - 1\}$?

The Goldbach Conjecture asserts that the number of ways to express $2m$, where $6 \le 2m \le 2n$, as the sum of two odd primes from the set $N_{2n} = \{1, 2, 3, ..., 2n - 1\}$ is greater than zero. But what about values of $2m$ in the range $2n + 2 \le 2m \le 4n - 2$?

**Conjecture 89.** We conjectured that for sufficiently large $n$, the number of ways to express $2m$ as the sum of two odd primes from the set $N_{2n} = \{1, 2, 3, ..., 2n - 1\}$ is greater than zero for all values of $2m$ in the range $2n + 2 \le 2m \le (4n - 2) - c(\log(n))^2$, where $c$ is a real positive number.

In other words,

$$ \int_{0}^{1} \left( \sum_{3 \le p < 2n} e^{2\pi ipz} \right)^2 e^{2\pi i(-2m)z} dz > 0 $$

for all $2m$ in the range $2n + 2 \le 2m \le (4n - 2) - c(\log(n))^2$.

**Conjecture 90.** We conjectured that for sufficiently large $n$, the number of ways to express $2m$, where $2n + 2 \le 2m \le (4n - 2)$, as the sum of two odd primes from the set $N_{2n} = \{1, 2, 3, ..., 2n - 1\}$ is greater than zero with $c*\log(n)$ exceptions, where $c$ is a positive real number.

In other words, with $c * \log(n)$ exceptions of $2m$,

$$ \int_{0}^{1} \left( \sum_{3 \le p < 2n} e^{2\pi ipz} \right)^2 e^{2\pi i(-2m)z} dz > 0 $$

for all $2m$ satisfying $2n + 2 \le 2m \le (4n - 2)$.

## References

[1] Pan Chengdong & Pan Chengbiao, Goldbach Conjecture.

[2] I., M. Vinogradov, The Method of Trigonometrical Sums in the Theory of Numbers.

[3] G. Tenenbaum, Introduction to Analytic and Probabilistic Number Theory

[4] Noga Alon and Joel Spencer, 3rd ed., The Probabilistic Method.

[5] Helfgott, Harald A. (2013). The ternary Goldbach Conjecture is true.
[6] Yue, Bill Quan (2025). On a Possible Approach to the Twin Prime Conjecture. Preprint. Available online: ResearchGate.

[7] Yue, Bill Quan (2025). On the Number of Odd Primes. Preprint.