Section 1,a new mathematical idea to prove Goldbach's conjecture----3

Section 2, Insufficient approximation formula of the number of primes in the [n, 2n]-------------------34

Section 3, Goldbach's conjecture 2n=p_i+p_j,the basic properties of identity 2n=q+(2n-q) and Hou Shao-sheng's theorem--------43

Section 4, 4 inferences of Hou Shaosheng's theorem-----------------67

Section 5, The 4 genetic codes of the conjecture, the proof of Goldbach's conjecture---------------------83

Section 6, Summary of the proof process of Goldbach conjecture----176

References--------------------------------------------------------183

In 1977, the sufficient and necessary condition theorem of the conjecture was proved, and in 2016, it was proved that the conjecture has only four genetic codes, and as long as one genetic code exists, the conjecture continues to be true! Hou Shaosheng has theoretically proved that the genetic code Impossible to interrupt! This is the fundamental guarantee and mystery of the conjecture!

No scientist had theoretically proved that the genetic code of living things Impossible to interrupt.The sad history of the Chinese nation without mathematical theorems is over forever!! Hou Shaosheng invites mathematicians from all over the world to review and wait to answer any questions from the experts.

By Hou Shaosheng

Abstract In 1742, the German mathematician Goldbach, in a letter to Euler, the world's great mathematician, put forward two conjectures, in slightly modified language, which can be expressed as follows: A: Any even number that is not less than 6 is the sum of two odd prime numbers (that is“$1+1$”). B: Any odd number that is not less than 9 is the sum of three odd prime numbers.

This became known as Goldbach's conjecture. Euler hopes future mathematicians will prove the conjecture.Since conjecture B is A corollary to conjecture A, the conjecture mentioned in the following article refers to conjecture A, and the proof conjecture refers to proving conjecture A.Since there are an infinite number of n's, Goldbach's conjecture is an infinite conjecture. The difficulty in proving the conjecture is that you have to prove it for every even 2n. The title of this paper is: New Mathematical Ideas to Prove Goldbach's Conjecture. Now, this paper presents two new mathematical ideas to prove conjectures. They belong to two different mathematical idea systems,this paper discusses the topic in the end.

2n, 2 (n+1), Goldbach's conjecture is true, so Goldbach's conjecture has a genetic code just like humans and living things.

This paper proves that Goldbach conjecture has only 4 kinds of genetic code, and as long as there is 1 kind of genetic code, the conjecture can be maintained.Hou Shaosheng proved in theory that there are at least three kinds of genetic codes exist, which proved Goldbach's conjecture completely and comprehensively in theory.

The genetic code of Goldbach's conjecture, which is theoretically proved to be impossible to interrupt, is a major breakthrough in mathematical theory.No one has theoretically proved that the genetic code of living things to be impossible to interrupt.

In the follow-up paper, 8 mathematical examples are randomly given. With these 8 mathematical examples, all the theories above are tested. The examples are completely consistent with the theories.

1.1.1. Goldbach Conjecture is briefly Introduced.

Goldbach's conjecture is expressed mathematically as: 2n=p_i+p_j, p_i≤p_j,(3≤n∈N).

p represents the odd prime number, p_i and p_j respectively represent the i_th odd prime number and the j_th odd prime number, N represents the set of natural numbers.

Someone checks every even number up to 3 × 10⁶ one by one, the conjecture is valid.

Since there are infinitely many even numbers, Goldbach's conjecture is an infinite conjecture. Goldbach's conjecture can never be proved by computer verification alone.

In 1900, German mathematician Hilbert held in Paris the second International Congress of Mathematics speech, Goldbach conjecture as the past legacy of one of the most important problems, introduced to the 20th century mathematicians to solve. In 1912, at the Fifth International Mathematical Congress held in Cambridge, the German mathematician Landes, in his speech, the conjecture of A as one of the four unsolved problems in prime number theory was recommended. In 1921, the British mathematician Hardy said at the mathematical Congress held in Copenhagen that the difficulty of guessing A could be compared with any unsolved mathematical problem. Goldbach's conjecture is therefore not only number theory, but also one of the most famous and difficult problems in mathematics.

In March 2000, Faber and Bloomsbury jointly offered a $1 million reward for information, but no one has been able to prove the theory.

In 1963, Wang Yuan, Pan Seung-dong and Barbain all proved "1+4".

In 1965, A Vinogradoff, Buchtabb and the Italian mathematician Pombini proved "1+3".

In 1966, Chen Jingrun proved "1+2". He showed that any sufficiently large even number can be expressed as the sum of two numbers, one of which is prime, the other is either prime or the product of two prime numbers. This result is called Chen's theorem in the world, and is considered to be the "pinnacle of sieve development".

It is generally acknowledged in the world mathematical circle that it is impossible to prove the conjecture by the improvement of the past method, and a new mathematical thought is necessary to prove the conjecture. The Chinese Academy of Mathematics even believes that it is difficult to find new mathematical ideas to prove conjectures in several decades, even hundreds or thousands of years, let alone prove conjectures.

1.1.2. The study of the Goldbach Conjecture Is in a Dark Freezing Period

The history of proving Goldbach's conjecture can be divided into three stages.

The first period, from 1742 to 1920, involved many mathematicians in the verification of Goldbach's conjecture. But there is no theory to prove the conjecture.

The second period was from 1920 to 1977. In 1920, mathematicians proved 9+9, and many others followed until 1966, when Chen Jingrun proved 1+2. There has been no progress since.

The third period is from 1977 to the present. In 1977, Hou Shaosheng proved the sufficient and necessary condition theorem of Goldbach conjecture. And based on this theorem, after 29 years of preparation, the monograph "Proof of Goldbach Conjecture· Proof of Fermat's last conjecture " was published in 2006 in 1000 volumes. In 2007, two old professors at Sun Yat-sen University praised the proof of Goldbach's conjecture.

In 2016, Hou completed the second proof of Goldbach's conjecture. And this proof, and the proof in the book, belongs to two different systems of mathematical ideas. A few years ago, I have posted this Chinese certificate on the Internet. At least tens of thousands of people have seen it, but so far no one can deny it.

Wang Yuan was the president of the Chinese Academy of Mathematics and an academician of the Chinese Academy of Sciences. Wang Yuan has won the First Prize in China Natural Science, the Chen Jiageng Prize in Physical Science and the He Liang and He Li Prize in Mathematics. In 1963, he proved 1+4, and he signed and published Chen Jingrun's 1+2. He is the author of Wang Yuan on Goldbach conjecture. It is very appropriate to say that Wang Yuan is the representative figure of this stage.

In 1992, 4 academicians of the Chinese Academy of Sciences: Wang Yuan, Chen Jingrun, Pan Chengdong, Yang Le, held a press conference.For another 1,000 years, they said, it would be difficult to find a new mathematical idea to prove Goldbach's conjecture, let alone prove it!

In 2006, on CCTV4, Wang Yuan said, "I firmly oppose Chinese attempts to prove Goldbach conjecture! Because it's impossible!

Because of Wang Yuan's authoritative position, the proof of Goldbach's conjecture entered a dark frozen period: almost all magazines did not review the proof of conjecture! Almost all mathematicians consider conjectures unprovable! Even the reviewers hired by the magazine found no errors and dared not publish a proof of Goldbach's conjecture.

Hou once made a written agreement with West China Science and Technology Magazine: If the expert they hired could raise a problem and Hou could not solve it, Hou would send 10,000 yuan to the magazine; Otherwise, the proof of Hou Shao-sheng's Goldbach conjecture must be published in full. The magazine hired professors from two prestigious universities to review the paper and found no problems. The president of Western China Science and Technology magazine said that if it was published to you, it would be equivalent to the explosion of an atomic bomb in China, which would shake the whole of China and affect the whole world. We could not afford the responsibility. Not answering the phone after that.

There is no long night that will not end! There is never a day when the sun does not rise!

Now the proof of the Goldbach's conjecture: "The nature of the identity 2n=q + (2n-q) determines that the Goldbach's conjecture must be true", translated into English, about 180 pages of A4 paper. The vast majority of reviewers are not willing to review, afraid that after I admit, others find a mistake. To facilitate the magazine and reviewers, Hou had to publish a new mathematical idea to prove Goldbach's conjecture, with the support of some experts. After receiving some expert support, I will publish follow-up paper.

If any expert is willing to review the proof method 2, Hou will send you "the nature of the identity 2n=q+ (2n-q) determines that Goldbach's conjecture must be true." Any questions you may have are welcome. If there is no problem, please recommend and cooperate to publish.

1.1.3. The Proof of Goldbach's Conjecture Has Dawned

Hou Shaosheng's research, from the very beginning, focused on new discoveries, on linking theory with practice, on theoretical abstraction and mathematical proof of known facts, and always worked around sufficient and necessary conditions.

Sufficient and necessary conditions (n±Δ are all prime numbers) are not only the action guide for Hou Shaosheng to prove 1+1, but also the theoretical basis for judging whether Hou has proved 1+1.

In 1977, Hou Shaosheng proved the sufficient and necessary condition theorem of conjectures.However, the difficult task put forward, scared Hou a cold sweat. After 23 years of theoretical preparation, it was not until August 2000 that the first draft of the proof was completed.

Professors at more than 10 universities in China have reviewed Hou's manuscripts. Including Wang Tianze, China's first doctor of number theory trained by Chen Jingrun. Wang Tianze asked a question, Hou answered, he never asked a question again. But no university magazine dared to publish it!

Hou Shaosheng requested Anyang City mathematical Society to organize Anyang city mathematicians to review. From September 2004 to March 2005, after half a year of review, the mathematics society finally reported to the Anyang Municipal government in a red-headed document, admitting that no mistakes had been found. It is quite possible that Hou has proved the world-famous Goldbach conjecture, and the municipal government is requested to give high support and attention. We hope that the municipal government will hire famous experts at home and abroad to review it.

Henan Provincial Mathematical Society doubted the level of Anyang City mathematical Society, in 2005, Henan Provincial Mathematical Society reviewed again,Finally, the document admitted Hou Shaosheng's proof. And to recommend the document to the Chinese Mathematical journal and the Chinese science journal, I hope the journal published, let mathematicians around the world comment. The Chinese Mathematical Journal, however, requested that Wang Tianze's name be put first and Hou Shaosheng's name be put second, because Wang Tianze was an established mathematician. Goldbach Conjecture is the most famous mathematical problem in the world, Hou Shaosheng is not famous, the name in the first place, will not organize experts to review. Hou immediately accused them of using their power for personal gain and falsifying history.

On November 24, 2005, the editorial office of Today Science Yuan Magazine sent a book publication notice.

The notice said that the manuscript of "Proof of Goldbach's Conjecture" was submitted by our press to two academicians of the Institute of Mathematics of the Chinese Academy of Sciences for review. There was no objection, and it was decided to publish it officially by the national publishing house, Changzheng Press or Guangming Daily Press.

We think that the publication of the manuscript is of great significance for promoting the study of mathematics.

When you receive the notice of publication, please hurry up and raise the publication fund of thirty thousand yuan only so that the book can be published as soon as possible.

I hope you can continue your efforts to climb the scientific peaks and play an even bigger role in becoming a mathematical country.

China Association for Science and Technology

Editorial Department of Today Science Yuan Magazine (with official seal)

November 24, 2005.

Hou Shaosheng asked Anyang association for science and technology Association for support dozens, even hundreds of times, but there was no support. This time, the book publishing notice finally caused the attention of the city Association for Science and Technology, to the city leaders did a report.

After a debate, the Standing Committee of Anyang Municipal Committee finally decided to support Hou Shaosheng with 30,000 yuan and let Hou publish the book.

In 2006, Proof of Goldbach's Conjecture and Proof of Fermar's Last Conjecture was published by Guangming Daily Press in 1000 copies. One thousand Chinese books have been circulated to date, and no one has been able to deny the proof.

This proof of Goldbach's conjecture in the book is simply called proof Method 1.

Hou Shaosheng sent his monograph "The Proof of Goldbach's Conjecture and the Proof of Fermat's Last Conjecture" to Wang Yuan and others several times by registered mail. He also asked Wang Yuan's friends to bring books to Wang Yuan in the hope that Wang Yuan would ask questions. However, until his death, Wang Yuan did not respond.

In 2013, Hou Shaosheng1, Ma Linjun2 et al published "A New Mathematical idea to Prove Goldbach's Conjecture" and "Defects of Eratosthenes screen method ,and Ideal final screening method" in Beijing Today Science and Technology Journal.

Chen Jiazhong, vice president of the magazine and chief editor of the magazine, is very close to Wang Yuan. I asked Chen Jiazhong to give this issue to Wang Yuan. Wang Yuan promised, must give a statement. But until Wang died, he did not respond.

In April 2007, Li Baitian, a professor at Sun Yat-sen University, called Hou and said: 'Someone on the Internet said they had proved Goldbach's conjecture. We have reviewed almost every proof on the Internet and none of them is correct.' It only takes a few minutes, a few hours at most, to find their essential mathematical error. Now we need to review your testimony. Do you agree?

Hou Shaosheng and Li Baitian telephone agreement: if Li Baitian put forward a problem, Hou Shaosheng can not solve, Li Baitian can declare Hou Shaosheng's proof is wrong! If such a question cannot be raised, Li must confirm Hou's proof in writing.

In the afternoon of the same day, Hou Shaosheng gave Professor Li Baitian two books and asked Professor Li to invite another professor to review them. A month later, according to Li Baitian's request, another three books were given to Li Baitian.

Li Baitian and Ma Linjun, both born in 1932, began to teach at Sun Yat-sen University in 1957 and began to study Goldbach Conjecture. After more than six months of review, they have a question. Hou answered the question.Finally, the two old professors spoke highly of Hou Shaosheng's proof in writing.

(Note: The other three, not professors, did not sign. One of the three, Li Baitian's university classmate, also wrote a proof of Goldbach's conjecture and asked the two old professors to support it. The two old professors refused, pointing out that she had not solved the fundamental mathematical problem of conjecture. The researcher, who participated in the review of Hou's proof.

Written testimonies of the two old professors are as follows:

In the past six months, several other professors and I have carefully studied Hou Shaosheng's Proof of Goldbach Conjecture (hereinafter referred to as Proof). And raised a question, other than that, we haven't found anything else. Hou Shaosheng's answers to the questions raised were satisfactory and prompt beyond our expectation, which made us deeply impressed. His solution, together with part of the original text, satisfactorily proves that there are prime numbers with single digits of 1, 3, 7 and 9 in any interval [n,2n] (19≤n∈N). This is a very big problem in number theory, and one that is extremely difficult to prove. We are ready to recommend the relevant content for publication in the Journal of Sun Yat-sen University.

We believe that Hou Shaosheng proved that the mathematical thought of "1+1" not only conforms to the essence of the existing mathematical theory, but also has a series of innovations. The mathematical idea of how to prove "1+1" has been correctly answered. For more than 260 years, "1+1" can not be proved, and there is no correct mathematical thinking is the root cause. Famous mathematicians such as Wang Yuan and Chen Jingrun once expected to produce new mathematical ideas in the process of studying "1+1", Hou Shaosheng has already given answers to this question.

Hou Shaosheng and Wang Shunqing proved that all the composite numbers whose single digits are 1, 3, 7 and 9 are only the values of 10 functions. This is the theoretical basis for his proof of "1+1". Thus they easily answered the most fundamental distribution law of prime numbers: All integers whose values $\notin$ the 10 functions, and whose single digits are 1, 3, 7, and 9, are prime. They and 2 and 5 are all prime numbers, and others integers whose single digit are 1, 3, 7, and 9 are composite number.

Hou and Wang's theorems in the range of composite numbers, the theorems outside the range of composite numbers, and the theorems of sufficient and necessary conditions that positive integers are prime numbers have essentially answered the quantitative relation between any composite number and other composite numbers, and the quantitative relation between any composite number and other prime numbers, positive integers are necessary and sufficient conditions for prime numbers.

Since the conception of man and nature, man has been studying three relationships, that is, the relationship within man, the relationship within nature, and the relationship between man and nature.

Similarly, prime and composite numbers have been the basic objects of number theory since the conception of prime and composite numbers. The relationship between prime numbers, composite numbers, prime numbers and composite numbers is the basic question that number theory should answer, the first question.

It is now fair to say that Hou and Wang have answered these questions. This is a major breakthrough in the research of basic theories and a few big golden eggs laid in the process of "1+1" research. At the same time, it lays a reliable theoretical foundation for proving "1+1".

Hou Shaosheng divided n (≥3) into prime number and composite number at first, and when n is prime, "1+1" does not need to be proved. When n is composite, he divides it into 16 classes. Then for each class of composite numbers n, it is proved that there are values of the function △ (> 0), so that n±△ is a prime number, so 2n = (n+△) +(n-△) is the sum of two prime numbers.

Here he created an entirely new sieve: when he wanted to prove that a certain class of composite numbers n, such as n = (10a+3) (10b+11), must exist △ such that n±△ are all prime, he first screened out all the composite numbers $\notin$ {f = (10x+3) (10y+11)}. This makes the problem much simpler to solve (prove).

Then, in view of the particularity of n = (10a+3) (10b+11), the specific expression (function) of △ is given to prove that the value of △ must exist, so that n±△ is a prime number.

Hou shaosheng creatively used 8 equations (some 12) equations to form the equation system, making the equations both interrelated and restricting each other.

With its association (common solution), each desired value of △ is netted. Filter out all values that are not needed by its constraint. It is the culmination of the "1+1" screening method!

After he proved the case of n = (10a+3) (10b+11) in detail, he could prove the case of n = (10a+7) (10+9), n = (10a+7) (10+11), n = (10a+3) (10+9) in the same way. Should be detailed, he has been detailed, should be simple, he has been simple.The same is true for the other 12 types of n.

We should recommend Hou's proof to the Chinese Academy of Sciences. However, we are nearly 80 years old, and many diseases, every day to take several medicine, and the journey is long and bumpy, I am afraid that the heart is willing to help but the power is insufficient! The above words are left for the reference of future referees and readers. In order to show our contribution to science, for the revival of the Chinese nation's heart.

We have been studying "1+1" for decades, and Hou's research results are of the highest level to our knowledge. It has already established the basis for the research of expert meetings and deserves the support and help of the government.

Department of Mathematics, Sun Yat-sen University: Li Baitian, Ma Linjun

At CUHK on 15 November 2007.

Note: Professor Li Bai-tian has passed away; Professor Ma Lin-jun, 2023 Spring Festival, in good health, loud voice, clear mind.

Since 2015, Professor Ma Linjun has been repeatedly reviewing the proof of the second Goldbach conjecture, which is simply called proof Method 2. He affirmed that the proof was correct!

Proof Method 2, a few years ago, has been posted online, and no one has asked questions yet.

1.1.4. The Proof of Goldbach's Conjecture Cannot be Completed by Mathematical Induction

In the recent 12 years of research, Hou Shaosheng focused on the decomposition law of 2n=q+(2n-q). Because, when q and (2n-q) both take odd primes, 2n is decomposed into the sum of two odd primes, which is Goldbach's conjecture. So Goldbach's conjecture is only part of 2n=q+(2n-q). So we must find and prove the decomposition law of 2n=q+(2n-q). The decomposition law of 2n=q+(2n-q) is Hou Shaosheng's theorem.

On the other hand, Hou Shao-sheng always focused on: suppose that the conjecture is true when 2n, why is the conjecture also true when 2 (n+1)? If the conjecture is true for 2n, it can be proved that the conjecture is also true for 2 (n+1), which proves that Goldbach's conjecture must be true. This looks like mathematical induction, but it's a lot more complicated than mathematical induction.

Mathematical induction, we have induction 1 and induction 2. Mathematical induction has definite implementation steps. Since n in 2n is all the integers not less than 3, this presents an insurmountable difficulty in proving the conjecture by mathematical induction.

For example, if n is an odd prime, 2n=n+n, the conjecture is valid; To prove that k=n+1, the conjecture is also true, because 2(n+1)=n+2+n, but n+1 is not an odd prime, and we can't be sure that 2+n is an odd prime, so we can't continue to prove it.

Another example is that if 2n=p_i+p_j is true, and both p_i and p_j are odd prime numbers, To prove that the 2 (n+1) conjecture is also true , 2 (n+1) =2n+2=p_i+2+p_j, because there is no certainty that at least one of (p_i+2) and (2+p_j) is odd prime numbers, the proof cannot be completed.

The above practice tells us that it is impossible to prove Goldbach's conjecture by mathematical induction.

The previous checking calculation tells us that every even number which is no less than 6 and no more than 3×10⁶ satisfies the sufficient and necessary condition theorem of conjectures.

Previous calculations tell us that Goldbach conjecture is probably correct. However, since there are an infinite number of even numbers that are not less than 6, it is impossible to prove the conjecture completely with concrete checking calculations.

To prove the conjecture, one must make a new discovery, and abstract the discovery into a new mathematical proposition, and then prove the new mathematical proposition into a new mathematical theorem.

The necessary and sufficient condition theorem of the conjecture is proved below. And prove that the sufficient and necessary condition theorem can not be violated.

Therefore, as a proof of Goldbach's conjecture, it should answer all the questions raised by the sufficient and necessary condition theorem, or prove that the proof of Goldbach's conjecture satisfies the sufficient and necessary condition theorem.

We know that the mathematical expression of Goldbach's conjecture is 2n=p_i+p_j, and we know that 2n=p_i+p_j is only a part of 2n=q+(2n-q), so we should study the relationship between 2n=p_i+p_j and 2n=q+(2n-q).

The research on the necessary and sufficient conditions for the establishment of 2n=p_i+p_j, the relationship between 2n=p_i+p_j and 2n=q+(2n-q), and the relationship between the establishment of the conjecture at 2n and the establishment of the conjecture at 2 (n+1) are the main line of thinking throughout the proof method 2.

It is this main line that establishes connections between sections and forms a logical system where each step is based on theorems or axioms and leads to the final goal.

This paper emphasizes that a proof satisfying the theorems of sufficient and necessary conditions is a correct proof, and a proof not satisfying the theorems of sufficient and necessary conditions must be wrong.This is the core idea of Proof Method 1.

A professor asked: the proof that does not satisfy the sufficient and necessary condition theorem must be wrong, why? Please look at the following: The sufficient and necessary condition theorem cannot be violated.

If the conjecture is true for 2n, why is it true for 2 (n+1)? For 2n and 2 (n+1), both conjectures are true, which is a necessary and sufficient condition for Goldbach's conjecture to be valid continuously.

In order to uncover the relationship between the 2n conjecture and the 2 (n+1) conjecture, after 45 years of repeated research, we write: The nature of the identity 2n=q +(2n-q), q≤(2n-q), determines that Goldbach's conjecture must hold.

The nature of the identity 2n=q+(2n-q), q ≤ (2n-q), determines that Goldbach's conjecture must hold. This is the core idea of proof method 2.

Hou Shao-Sheng¹, Ma Lin-Jun²

(1 Anyang Mathematical Society, Henan Province, China; 2 Department of Mathematics, Sun Yat-sen University, Guangzhou City, Guangdong Province, China)

Hou Shao-Sheng Email: ayhss@163.com.

Abstract: There are prime numbers in the interval [n,2n], (3≤n∈N), which is a necessary condition for the establishment of Goldbach conjecture. In this section, we give the insufficient approximation formula of the number of primes in the interval [n,2n]. And prove that there must be prime numbers in the interval [n,2n], 3≤n.

Key words [n, 2n] interval; the number of prime；formula ；

Insufficient approximation value.

[Document ID] A [document number]

2.1.1. Definition of the Insufficient Approximation Value

There are prime numbers in the interval [n,2n], (3≤n∈N), which is a necessary condition for Goldbach's conjecture to be established. In this section, we first derive the formula for the number of primes in the interval [n,2n], and then prove that there must be primes in the interval [n, 2n], 3≤n.

The mathematical dictionary introduces the theorem of prime numbers. The prime number theorem is as follows: If $\pi$(x) represents the number of primes not exceeding x ( x is a positive integer), then there is $\underset{x\to \infty }{{\displaystyle \lim }}$=${\displaystyle \frac{\pi (x)}{\frac{x}{\ln x}}}=1.$

This theorem says that when x is very large, $\pi (x)$ is very close to ${\displaystyle \frac{x}{\ln x}}$. When x is large, $\pi (x)$≈${\displaystyle \frac{x}{\ln x}}$; It shows that the distribution of prime numbers has a certain law.

On the other hand, we see: the average density ${\displaystyle \frac{\pi (x)}{x}}$ of the distribution of prime numbers should be close to ${\displaystyle \frac{x}{\ln x}}$÷x=${\displaystyle \frac{1}{\ln x}}$. That means that as x gets bigger ,${\displaystyle \frac{1}{\ln x}}$ smaller and smaller , the average density ${\displaystyle \frac{\pi (x)}{x}}$ of the distribution of prime numbers gets smaller and smaller.

Also because n!+2, n!+3, n!+4,... ,n!+n, are composite numbers, which gives the illusion that the number of prime numbers in the interval [n,2n] is likely to decrease as n increases! However, this is not consistent with the facts as people know them.

When n≤5000, you can find through the experiment that in the interval [n,2n], with the increase of n, the number of prime numbers is increasing! So, in the interval [n, 2n], as n increases, does the number of prime numbers increase? Or fewer and fewer? Still no pattern?

We have proved that the necessary and sufficient conditions for Goldbach's conjecture are that for every integer n of no less than 3, there is at least a non-negative integer △, so that n-△ and n+△ are all odd prime! It is known that n-△, n+△, is two odd primes about n symmetry. And the n+△ is just within the [n, 2n] interval.

If the number of prime numbers in the interval [n,2n] increases with the increase of n, then the probability that the conjecture is true increases. On the contrary, the probability of the conjecture being true is decreasing!

For the reasons mentioned above, it is necessary to answer the regularity of the number of primes in the interval [n,2n].

2.1.2. Theorem 2. Let num[n≤p≤2n] Represent the Number of Prime p in the Interval [n, 2n]. Then

num[n≤p≤2n]≈${\displaystyle \frac{2\mathrm{n}}{\ln 2n}}$- ${\displaystyle \frac{\mathrm{n}}{\ln n}}$ =${\displaystyle \frac{0.8686\mathrm{n}}{\lg 2n}}$- ${\displaystyle \frac{0.4343\mathrm{n}}{\lg n}}$.

(Note: Theorem 1 is the necessary and sufficient condition theorem of Goldbach's conjecture.)

(Note: num is short for number.)

Proof: The prime number theorem states that if $\pi$(x) represents the number of primes that does not exceed x (x is a positive integer), then there is $\underset{x\to \infty }{{\displaystyle \lim }}$=${\displaystyle \frac{\pi (x)}{\frac{x}{\ln x}}}=1.$

The prime number theorem shows that when x is large, $\pi (x)$ with ${\displaystyle \frac{x}{\ln x}}$being very close to. Can be regarded as:$\pi (x)$≈${\displaystyle \frac{x}{\ln x}}$, $\pi (2x)$≈${\displaystyle \frac{2x}{\ln 2x}}$.

We use num[n≤p≤2n] to represent the number of prime p in the interval of [n, 2n], so has:

num[n≤p≤2n]=$\pi$(2n)-$\pi$(n-1)

≈$\pi (2n)$-$\pi (n)$ (2.1.0)

num[n≤p≤2n] ≈${\displaystyle \frac{2n}{\ln 2n}}$-${\displaystyle \frac{n}{\ln n}}$ (2.1.1)

=2(${\displaystyle \frac{n}{\ln 2n}}$-${\displaystyle \frac{n}{2\ln n}}$). (2.1.2)

Let x≥10 be a real number and the function(${\displaystyle \frac{\mathrm{x}}{\ln 2x}}$-${\displaystyle \frac{\mathrm{x}}{2\ln x}}$), when x is an integer not less than 10, its function value is (${\displaystyle \frac{n}{\ln 2n}}$-${\displaystyle \frac{n}{2\ln n}}$) . Now, by studying the derivative of (${\displaystyle \frac{\mathrm{x}}{\ln 2x}}$-${\displaystyle \frac{\mathrm{x}}{2\ln x}}$) , we will study the tendency of (${\displaystyle \frac{n}{\ln 2n}}$-${\displaystyle \frac{n}{2\ln n}}$) as n increases.

Let x≥10 be a real number because:

(${\displaystyle \frac{\mathrm{x}}{\ln 2x}}$-${\displaystyle \frac{\mathrm{x}}{2\ln x}}$)′= (${\displaystyle \frac{\mathrm{x}}{\ln 2x}}$)′-(${\displaystyle \frac{\mathrm{x}}{2\ln x}}$)′

=${\displaystyle \frac{\ln 2x-\mathrm{x}\times \frac{2}{2\mathrm{x}}}{{{\displaystyle (\ln 2x)}}^2}}$-${\displaystyle \frac{2\ln x-\mathrm{x}\times 2\frac{1}{\mathrm{x}}}{{{\displaystyle (2\ln x)}}^2}}$=${\displaystyle \frac{\ln 2x-1}{{{\displaystyle (\ln 2x)}}^2}}$-${\displaystyle \frac{2\ln x-2}{{{\displaystyle (2\ln x)}}^2}}$

=${\displaystyle \frac{1}{\ln 2x}}$-${\displaystyle \frac{1}{2\ln x}}$-${\displaystyle \frac{1}{{{\displaystyle (\ln 2x)}}^2}}$+${\displaystyle \frac{2}{{{\displaystyle (2\ln x)}}^2}}$

=${\displaystyle \frac{1}{\ln 2x}}$-${\displaystyle \frac{1}{2\ln x}}$+${\displaystyle \frac{1}{{{\displaystyle (2\ln x)}}^2}}$-${\displaystyle \frac{1}{{{\displaystyle (\ln 2x)}}^2}}$+${\displaystyle \frac{1}{{{\displaystyle (2\ln x)}}^2}}$

=${\displaystyle \frac{1}{\ln 2x}}$-${\displaystyle \frac{1}{2\ln x}}$+(${\displaystyle \frac{1}{2\ln x}}$+${\displaystyle \frac{1}{\ln 2x}}$)(${\displaystyle \frac{1}{2\ln x}}$-${\displaystyle \frac{1}{\ln 2x}}$)+${\displaystyle \frac{1}{{{\displaystyle (2\ln x)}}^2}}$

=(${\displaystyle \frac{1}{\ln 2x}}$-${\displaystyle \frac{1}{2\ln x}}$)[1-(${\displaystyle \frac{1}{\ln 2x}}$+${\displaystyle \frac{1}{2\ln x}}$)]+${\displaystyle \frac{1}{{{\displaystyle (2\ln x)}}^2}}$

＞(${\displaystyle \frac{1}{\ln 2x}}$-${\displaystyle \frac{1}{2\ln x}}$)[1-(${\displaystyle \frac{1}{\ln 2x}}$+${\displaystyle \frac{1}{2\ln x}}$)]

＞0. (2.1.3)

Note: Because when x≥10, ${\displaystyle \frac{1}{\ln 2x}}$-${\displaystyle \frac{1}{2\ln x}}$＞0; And 1-(${\displaystyle \frac{1}{\ln 2x}}$+${\displaystyle \frac{1}{2\ln x}}$)＞0.

We know from (2.1.3) that 2(${\displaystyle \frac{n}{\ln 2n}}$-${\displaystyle \frac{n}{2\ln n}}$) is an increasing function of n, so num[n≤p≤2n] is, on the whole, an increasing function of n.

Here it is not written as: so num[n≤p≤2n] is an increasing function of n because: $\pi (x)$≈${\displaystyle \frac{x}{\ln x}}$, Instead of: $\pi (x)=$${\displaystyle \frac{x}{\ln x}}$.

(2.1.1), (2.1.4) is the calculation formula of num[n≤p≤2n].

Hou Shao-Sheng¹, Ma Lin-Jun²

(1 Anyang Mathematical Society, Henan Province, China; 2 Department of Mathematics, Sun Yat-sen University, Guangzhou City, Guangdong Province, China)

Hou Shao-Sheng Email: ayhss@163.com.

Abstract 2n=q+(2n-q), we call it Hou Shaosheng's identity, that is, when the variable q takes any value, this equation is always true.

Let 3≤q≤n, and q be odd. When q takes all the odd numbers in the interval [3,n], (2n-q) takes all the odd numbers in the interval [n,2n-3].

In order to fully grasp the law of 2n=p_i+p_j, we must study the basic law of 2n=q+(2n-q). The basic law of 2n=q+(2n-q) is Hou Shaosheng's theorem.

The basic task of this section is to introduce Hou Shaosheng's identity and prove Hou Shaosheng's theorem. Then the mathematical meaning of the mathematical formula of Hou Shaosheng's theorem is explained.

Key words Identity; Odd number; Odd prime numbers; Odd composite number.

[Document ID] A [Document number]

Hou Shao-Sheng¹, Ma Lin-Jun²

(1 Anyang Mathematical Society, Henan Province, China; 2 Department of Mathematics, Sun Yat-sen University, Guangzhou City, Guangdong Province, China)

Hou Shao-Sheng Email: ayhss@163.com.

Abstract The mathematical formula of Hou Shaosheng's theorem is as follows:

num(2n=q+(2n-q))=num(2n=p_i+p_j)+ num(2n=h_t+h_u)

+ num(2n=p_s+h_r)+ num(2n=h_k+p_d). (3.3.0)

This formula includes:

num(2n=q+(2n-q)),num(2n=p_i+p_j),num(2n=h_t+h_u),num(2n=p_s+h_r),num(2n=h_k+p_d).Formulas contain too much mathematical content. It is very inconvenient to prove Goldbach's conjecture directly by formula. So it is necessary to derive the 4 inferences of Hou Shaosheng's theorem.

The 4 inferences of Hou Shaosheng's theorem are as follows:

num(3≤p≤n)=num(2n=p_i+p_j)+num(2n=p_s+h_r); (4.0.1)

num(3≤h≤n)=num(2n=h_t+h_u)+num(2n=h_k+p_d); (4.0.2)

num(n≤p≤2n-3)=num(2n=p_i+p_j)+num(2n=h_k+p_d); (4.0.3)

num(n≤h≤2n-3)=num(2n=h_t+h_u)+num(2n=p_s+h_r). (4.0.4)

Among them, the convention is: 3≤q≤n,q is odd.

p_i, p_j, p_s, p_d are all odd prime numbers; h_t, h_u, h_r, h_k are all odd composite numbers.

And p_i≤p_j, h_t≤h_u, p_s≤h_r, h_k≤p_d.

p_i, h_t, p_s, h_k, ∈[3,n]; p_j, h_u, h_r, p_d∈[n,2n-3].

The basic task of this section is to prove the 4 inferential formulas of Hou Shaosheng's theorem. The second is to explain the mathematical meaning of the 4 inferential formulas. And the 4 inferential formulas are verified by mathematical examples.

The 4 inferential formulas are the main theoretical tools for proving Goldbach's conjecture.

Therefore, this section is the central link between the preceding and the following.

[Document code] A [Document number]

About the formula:

num(2n=q+(2n-q))=num(2n=p_i+p_j)+ num(2n=h_t+h_u)

+ num(2n=p_s+h_r)+ num(2n=h_k+p_d). (3.3.0)

always have:

num(3≤p≤n)= num(2n=p_i+p_j)+ num(2n=p_s+h_r); (4.0.1)

num(3≤h≤n)= num(2n=h_t+h_u)+ num(2n=h_k+p_d); (4.0.2)

num(n≤p≤2n-3)= num(2n=p_i+p_j)+ num(2n=h_k+p_d); (4.0.3)

num(n≤h≤2n-3)= num(2n=h_t+h_u)+ num(2n=p_s+h_r). (4.0.4)

Among them, the convention is: 3≤q≤n,q is odd.

p_i, p_j, p_s, p_d are all odd prime numbers; h_t, h_u, h_r, h_k are all odd composite numbers.

And p_i≤p_j, h_t≤h_u, p_s≤h_r, h_k≤p_d.

p_i, h_t, p_s, h_k, ∈[3,n]; p_j, h_u, h_r, p_d∈[n,2n-3].

(4.0.1) is stated as follows: The number of odd primes p in the interval [3,n] is equal to the number of 2n decomposed into the sum of two odd primes, plus the number of 2n decomposed into the sum of odd primes and odd composite numbers.

(4.0.2) is stated as follows: the number of odd composite number h in the interval [3, n] is equal to the number of 2n decomposed into the sum of two odd composite number, plus the number of 2n decomposed into the sum of odd composite number and odd prime.

(4.0.3) is stated as follows: The number of odd primes p in the interval [n,2n-3] is equal to the number of 2n decomposed into the sum of two odd prime number, plus the number of 2n decomposed into the sum of odd composite number and odd prime.

(4.0.4) is stated as follows: The number of odd composite number h in the interval [n,2n-3] is equal to the number of 2n decomposed into the sum of two odd composite number,plus the number of 2n decomposed into the sum of odd prime number and odd composite number.

Proof: Because we agreed: p_i≤p_j, h_t≤h_u, p_s＜h_r, h_k＜p_d.

So there must be a causal relationship:

If 2n=p_i+p_j, because 3≤p_i≤p_j, therefore p_i≤n; n≤p_j≤2n-3.

If 2n=p_s+h_r, because 3≤p_s＜h_r, therefore p_s＜n; n＜h_r≤2n-3.

If 2n=h_t+h_u, because 3≤h_t≤h_u, therefore h_t≤n; n≤h_u≤2n-3.

If 2n=h_k+p_d, because 3≤h_k＜p_d, therefore h_k＜n; n＜p_d≤2n-3.

From the above we can see:

The only odd prime numbers in the interval [3,n] are: p_i and p_s. So (4.0.1) is true.

The only odd composite numbers in the interval [3,n] are: h_t and h_k. So (4.0.2) is true.

The only odd prime numbers in the interval [n,2n-3] are: p_j and p_d. So (4.0.3) is true.

The only odd composite numbers in the interval [n,2n-3] are: h_u and h_r. So (4.0.4) is true.

End of proof.

There are 4 mathematical formulas for the inference of Hou Shaosheng's theorem. These 4 mathematical formulas are the theorem in

num(3≤p≤n), num(3≤h≤n), num(n≤p≤2n-2), num(n≤h≤2n-2),

4 quantitative deepening and refinement.

In these 4 quantities, more specific is given:

num(2n=p_i+p_j), num(2n=p_s+h_r), num(2n=h_t+h_u), num(2n=h_k+p_d),

the relationship between. It plays an important role in proving Goldbach's conjecture.

Let's look at the mathematical implications of these 4 mathematical formulas.

(4.0.1), (4.0.2), (4.0.3), (4.0.4), not only means that the quantity on both sides of the equal sign is equal, but also has a deeper mathematical meaning, as follows.

4.1.1 (4.0.1) Mathematical meaning of the formula

num(3≤p≤n)= num(2n=p_i+p_j)+ num(2n=p_s+h_r). (4.0.1)

The mathematical meaning of (4.0.1) is: The number of all odd primes p in the interval [3,n] is equal to the number of p_i in num (2n=p_i+p_j) + the number of p_s in num (2n=p_s+h_r). p_i,p_s are all the odd prime numbers p in the interval [3,n]. This is the most basic mathematical meaning of (4.0.1).

(4.0.1) True, not only is the quantity on both sides of the equal sign equal. There are deeper mathematical implications, as follows.

(4.0.1) is true, that is, the number of odd primes p in the interval [3,n] on both sides of the equal sign is equal. On the left side of the equal sign, num (3≤p≤n) represents the number of all odd prime numbers p in the interval [3,n]; As an equation, the right side of the equal sign must also be the number of all the odd primes p in the interval [3,n]. p_i,p_s, is all the odd prime numbers p in the interval [3,n]. This is the most basic mathematical meaning of (4.0.1).

(4.0.1) indicates that all the odd prime numbers p in the interval [3,n] are divided into p_i and p_s,2 parts.

p_i is combined with the odd prime number p_j in the interval [n,2n-3] to form 2n=p_i+p_i;

p_s is combined with the odd composite number h_r in the interval [n,2n-3] to form 2n=p_s+h_r.

There must be no other possibility than 2n=p_i+p_j and 2n=p_s+h_r.

p_i,p_s, are all part of the odd prime number p in the interval [3,n]. One is written as p_i and the other as p_s to distinguish between two different combinations: 2n=p_i+p_j, 2n=p_s+h_r. If we write them all as p, it will be difficult when we need to specify p_i, or when we need to specify p_s.

(4.0.1) True, contains two meanings:

First, the numbers on both sides of the equal sign are equal;

Second, things on both sides of the equal sign have the same name, and are the odd prime p in the interval [3,n].

If one of the two is missing, it cannot be an equation. This mathematical meaning is completely consistent with that in primary school arithmetic, only the numbers with the same name can be added or subtracted, and only the numbers with the same name can be equal.

The above properties of (4.0.1) play an important role in the process of proving the conjecture.

The above properties of (4.0.1) have been tested with 8 examples in the paper.

4.1.2 (4.0.2) Mathematical meaning of the formula

num (3≤h≤n)= num(2n=h_t+h_u)+ num(2n=h_k+p_d). (4.0.2)

The mathematical meaning of (4.0.2) is: The number of all odd composite number h in the interval [3,n] is equal to the number of h_t in num (2n=h_t+h_u) + the number of h_k in num (2n=h_k+p_d). h_t, h_k are all the odd composite numbers h in the interval [3,n]. This is the most basic mathematical meaning of (4.0.2).

(4.0.2) True, not only is the quantity on both sides of the equal sign equal. There are deeper mathematical implications, as follows.

(4.0.2) is true, that is, the number of odd composite number h in the interval [3,n] on both sides of the equal sign is equal. On the left side of the equal sign, num (3≤h≤n) represents the number of all odd composite h in the interval [3,n]; As an equation, the right side of the equal sign must also be the number of all the odd composite h in the interval [3,n]. h_t,h_k, is all the odd composite numbers h in the interval [3,n].

(4.0.2) indicates that all the odd composite number h in the interval [3,n] are divided into h_t and h_k,2 parts.

h_t is combined with the odd composite number h_u in the interval [n,2n-3] to form 2n=h_t+h_u;

h_k is combined with the odd prime number p_d in the interval [n,2n-3] to form 2n=h_k+p_d.

There must be no other possibility than 2n=h_t+h_u and 2n=h_k+p_d.

In (4.0.2),h_t, h_k, is all the odd composite numbers h in the interval [3,n].

h_t,h_k, are all odd composite numbers h in the interval [3,n]. One is written as h_t and the other as h_k to distinguish between two different combinations: 2n=h_t+h_u, 2n=h_k+p_d. If we write both h, it will be difficult when we need to specify h_t, or when we need to specify h_k.

(4.0.2) True, contains two meanings:

First, the numbers on both sides of the equal sign are equal;

Second, things on both sides of the equal sign have the same name, and are the odd composite number h in the interval [3,n].

If one of the two is missing, it cannot be an equation. This mathematical meaning is completely consistent with that in primary school arithmetic, only the numbers with the same name can be added or subtracted, and only the numbers with the same name can be equal.

The above properties of (4.0.2) have been tested with 8 examples in the paper.

4.1.3 (4.0.3) The mathematical meaning

num (n≤p≤2n-3)= num(2n=p_i+p_j)+ num(2n=h_k+p_d). (4.0.3)

The mathematical meaning of (4.0.3) is: The number of all odd prime p in the interval [n,2n-3] is equal to the number of p_j in num (2n=p_i+p_j) + the number of p_d in num (2n=h_k+p_d). p_j , p_d are all the odd prime numbers p in the interval [n,2n-3]. This is the most basic mathematical meaning of (4.0.3).

(4.0.3) indicates that all the odd prime number p in the interval [n,2n-3] are divided into p_j and p_d,2 parts.

P_j is combined with the odd prime number p_i in the interval [3,n] to form 2n=p_i+p_j;

p_d is combined with the odd composite number h_k in the interval [3,n] to form 2n=h_k+p_d.

There must be no other possibility than 2n=p_i+p_j and 2n=h_k+p_d.

In (4.0.3), p_j, p_d, is all the odd prime numbers p in the interval [n,2n-3].

p_j, p_d, are all odd prime numbers p in the interval [n,2n-3]. One is written p_j and the other p_d to distinguish between two different combinations: 2n=p_i+p_j, 2n=h_k+p_d. If we write them all as p, it will be difficult when we need to specify p_j, or when we need to specify p_d.

(4.0.3) is true, that is, the number of odd primes p in the interval [n,2n-3] on both sides of the equal sign is equal. On the left side of the equal sign, num (n≤p≤2n-2) represents the number of all odd prime numbers p in the interval [n,2n-3]; As an equation, the right side of the equal sign must also be the number of all the odd primes p in the interval [n,2n-3].

(4.0.3) True, contains two meanings:

First, the numbers on both sides of the equal sign are equal;

Second, things on both sides of the equal sign have the same name, and are the odd prime p in the interval [n,2n-3].

If one of the two is missing, it cannot be an equation. This mathematical meaning is completely consistent with the mathematical idea that only numbers with the same name can add or subtract, and only numbers with the same name can be equal in elementary school arithmetic.

The above properties of (4.0.3) have been tested with 8 examples in the paper.

num (n≤h≤2n-3)= num(2n=h_t+h_u)+ num(2n=p_s+h_r). (4.0.4)

The mathematical meaning of (4.0.4) is: The number of all odd composite number h in the interval [n,2n-3] is equal to the number of h_u in num(2n=h_t+h_u) + the number of h_r in num (2n=p_s+h_r). h_u, h_r are all the odd composite numbers h in the interval [n,2n-3]. This is the most basic mathematical meaning of (4.0.4).

(4.0.4) indicates that all the odd composite numbers h in the interval [n,2n-3] are divided into h_u and h_r,2 parts. h_u combines with the odd composite number h_t in the interval [3,n] to form 2n=h_t+h_u; h_r is combined with the odd prime p_s in the interval [3,n] to form 2n=p_s+h_r.

There must be no other possibility than the above two: 2n=h_t+h_u, 2n=p_s+h_r.

In (4.0.4),h_u, h_r, are all the odd composite numbers h in the interval [n,2n-3].

The odd composite number h in the interval [n,2n-3], one written h_u and the other written h_r, is to distinguish between two different combinations: 2n=h_t+h_u, 2n=p_s+h_r. If we write both h, it will be difficult when we need to specify h_u, or when we need to specify h_r.

(4.0.4) is true, that is, the number of odd composite number h in the interval [n,2n-3] on both sides of the equal sign is equal. Before the equal sign, num (n≤h≤2n-3) indicates the number of all odd composite number h in the interval [n,2n-3]; As an equation, after the equal sign, it must also be the number of all the odd composite number h in the interval [n,2n-3].

(4.0.4) True, contains two meanings:

First, the numbers on both sides of the equal sign are equal;

Second, things on both sides of the equal sign have the same name, and are the odd composite number h in the interval [n,2n-3].

If one of the two is missing, it cannot be an equation. This mathematical meaning is completely consistent with the mathematical idea that only numbers with the same name can add or subtract, and only numbers with the same name can be equal in elementary school arithmetic.

The above properties of (4.0.4) have been tested with 8 mathematical examples in the paper.

The above meanings of the 4 formulas play a very important role in the process of proving the conjecture.

The following example, 3, is the third of the eight mathematical examples in 3.2: Instance 3. Now test (verify) formula (3.3.0), (4.0.1), (4.0.2), (4.0.3), (4.0.4) with the relevant data in Example 3.

Example 3. 2n=2×51=102

=1+101=3+99=5+97=7+95=9+93=11+91=13+89=15+87=17+85=

=19+83=21+81=23+79=25+77=27+75=29+73=31+71=33+69=35+67

=37+65=39+63=41+61=43+59=45+57=47+55=49+53=51+51.

Note: Numbers underlined, or red numbers, odd prime numbers. Numbers without underscores, or blue numbers, odd composite numbers.

In the example above, the data is as follows:

num(2n=q+(2n-q))=25.

These 25 number of 2n=q+(2n-q) are as follows:

2n=3+99； 2n=5+97； 2n=7+95； 2n=9+93； 2n=11+91；

2n=13+89； 2n=15+87； 2n=17+85； 2n=19+83； 2n=21+81；

2n=23+79； 2n=25+77； 2n=27+75； 2n=29+73； 2n=31+71；

2n=33+69； 2n=35+67； 2n=37+65； 2n=39+63； 2n=41+61；

2n=43+59； 2n=45+57； 2n=47+55； 2n=49+53； 2n=51+51.

num(2×51=p_i+p_j)=8.

These 8 number of 2×51=p_i+p_j, as follows:

2n=5+97； 2n=13+89； 2n=19+83； 2n=23+79； 2n=29+73；

2n=31+71； 2n=41+61； 2n=43+59；

num(2×51=h_t+h_u)=9.

These 9 number of 2n=2×51=h_t+h_u are as follows:

2n=9+93； 2n=15+87； 2n=21+81； 2n=25+77； 2n=27+75；

2n=33+69； 2n=39+63； 2n=45+57； 2n=51+51.

num(2n=p_s+h_r)=6.

These 6 number of 2n=p_s+h_r are as follows:

2n=3+99； 2n=7+95； 2n=11+91； 2n=17+85； 2n=37+65；

2n=47+55；

num(2n=h_k+p_d)=2.

These 2 number of 2n=h_k+p_d are as follows:

2n=35+67； 2n=49+53；

num(3≤p≤51)=14.

The 14 odd prime numbers in the interval [3,51] are as follows:

3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 .

num(n≤p≤2n-3)=10.

The 10 odd prime numbers in the interval [51, 99] are as follows:

97, 89, 83, 79, 73, 71, 67, 61, 59, 53.

num(3≤h≤n)= num(3≤h≤51)=11.

The 11 odd composite numbers h in the interval [3,51] are as follows:

num (n≤h≤2n-3)=15.

The 15 odd composite numbers h in the interval [51, 99] are as follows:

After testing one by one, the above data fully satisfy each of the following formulas:

(3.3.0), (4.0.1), (4.0.2), (4.0.3), (4.0.4).

num (2n=q+(2n-q))= num (2n=p_i+p_j)+ num (2n=h_t+h_u)

+ num (2n=p_s+h_r)+ num (2n=h_k+p_d). (3.3.0)

num (3≤p≤n)= num(2n=p_i+p_j)+ num(2n=p_s+h_r). (4.0.1)

num (3≤h≤n)= num(2n=h_t+h_u)+ num(2n=h_k+p_d); (4.0.2)

num (n≤p≤2n-3)= num(2n=p_i+p_j)+ num(2n=h_k+p_d); (4.0.3)

num (n≤h≤2n-3)= num(2n=h_t+h_u)+ num(2n=p_s+h_r). (4.0.4)

After testing (verification), the above data fully satisfy each of the above formulas. In the paper, there are other examples, the reader is also free to use examples to test (verify) the above formula.

From example 3, statistics show that:

num(3≤p≤n)= num(3≤p≤51)=14.

The 14 odd prime numbers p are:

num(2n=p_i+p_j)=8.

These 8 number of 2×51=p_i+p_j are:

2×51=5+97, 2×51=13+89, 2×51=19+83, 2×51=23+79, 2×51= 29+73, 2×51=31+71, 2×51=41+61, 2×51=43+59.

8 of these p_i are: 5,13,19,23, 29,31,41,43.

num(2n=p_s+h_r)=6.

The 6 number of 2×51=p_s+h_r are:

2×51=3+99, 2×51=7+95, 2×51=11+91,

2×51=17+85, 2×51=37+65, 2×51=47+55.

6 of these p_s are: 3, 7, 11, 17, 37, 47.

The above 8 of p_i and 6 of p_s are all the odd prime numbers in the interval [3,51].

Arrange the above 8 of p_i and 6 of p_s in order from small to large to get:

These 14 odd prime numbers are the 14 odd prime numbers obtained by statistics from example 3.

This example, once again, illustrates the mathematical meaning of (4.0.1) :

The number of all odd primes p in the interval [3,n] is equal to the number of p_i in num (2n=p_i+p_j) + the number of p_s in num (2n=p_s+h_r). p_i, p_s are all the odd prime numbers p in the interval [3,n]. This is the most basic mathematical meaning of (4.0.1).

(4.0.1) True, not only is the quantity on both sides of the equal sign equal. There are deeper mathematical implications, as follows.

(4.0.1) is true, that is, the number of odd primes p in the interval [3,n] on both sides of the equal sign is equal. On the left side of the equal sign, num (3≤p≤n) represents the number of all odd prime numbers p in the interval [3,n]; As an equation, the right side of the equal sign must also be the number of all the odd primes p in the interval [3,n]. p_i,p_s, is all the odd prime numbers p in the interval [3,n]. This is the most basic mathematical meaning of (4.0.1).

(4.0.1) indicates that all the odd prime numbers p in the interval [3,n] are divided into p_i and p_s,2 parts.

p_i is combined with the odd prime number p_j in the interval [n,2n-3] to form 2n=p_i+p_j;

p_s is combined with the odd composite number h_r in the interval [n,2n-3] to form 2n=p_s+h_r.

There must be no other possibility than 2n=p_i+p_j and 2n=p_s+h_r.

p_i,p_s, are all part of the odd prime number p in the interval [3,n]. One is written as p_i and the other as p_s to distinguish between two different combinations: 2n=p_i+p_j, 2n=p_s+h_r. If we write them all as p, it will be difficult when we need to specify p_i, or when we need to specify p_s.

(4.0.1) True, contains two meanings:

First, the numbers on both sides of the equal sign are equal;

Second, things on both sides of the equal sign have the same name, and are the odd prime p in the interval [3,n].

If one of the two is missing, it cannot be an equation. This mathematical meaning, and primary school arithmetic, only the number with the same name can be added or subtracted , only the number with the same name can be equal, completely consistent.

The above properties of (4.0.1) play an important role in the process of proving the conjecture.

Please use Example 3 to test (verify) the mathematical meaning of (4.0.2), (4.0.3), (4.0.4) for yourself.

The mathematical meanings of (4.0.1),(4.0.2),(4.0.3), and (4.0.4) are tested and repeated here because their mathematical meanings play a very important role in the proof of conjectures. I want the reader to remember their mathematical implications.

You may ask: num (3≤p≤51)=14, are both sides equal?

Answer: Equal. num(3≤p≤51), 14, followed by the name p, is already implied after num(3≤p≤51) and 14.

Above, we give inferences of Hou Shaosheng's theorem at 2n. Now we give the inference of Hou Shaosheng's theorem at 2k, k= n+1.

Definition: 2k=q+(2k-q), we call it 2k time Hou Shao Sheng identity.

Where: 3≤q≤k; q is odd.

The inference of Hou Shaosheng's theorem at 2k is as follows.

Theorem 5: 2k The inference of Hou Shaosheng's theorem is as follows:

num (3≤p≤k)= num(2k=p_i++p_j+)+ num(2k=p_s++h_r+). (4.3.1)

num (3≤h≤k)= num(2k=h_t++h_u+)+ num(2k=h_k++p_d+). (4.3.2)

num (k≤p≤2k-3)= num(2k=p_i++p_j+)+ num(2k=h_k++p_d+). (4.3.3)

num (k≤h≤2k-3)= num(2k=h_t++h_u+)+ num(2k=p_s++h_r+). (4.3.4)

Among them, the agreement:

p_i+, p_j+, p_s+, p_d+, are all odd prime numbers; h_t+, h_u+, h_r+, h_k+, are all odd composite numbers.

And p_i+≤p_j+, h_t+≤h_u+, p_s+≤h_r+, h_k+≤p_d+.

p_i+, p_s+, h_t+, h_k+, ∈[3,k]; p_j+, h_r+, h_u+, p_d+, ∈[n,2k-3].

Proof By replacing n with k in the Hou Shaosheng identity, we can

get the Hou Shaosheng identity at 2k.

Replace n in Hou Shao-sheng's theorem (theorem 4) with k, and at the same time

Replace 2n in 2n=p_i+p_j with 2k, p_i, and p_j with p_i+, p_j+;

Replace 2n in 2n=p_s+h_r with 2k,p_s, h_r with p_s+, h_r+;

Replace 2n in 2n=h_t+h_u with 2k, h_t, h_u with h_t+,h_u+;

Replace 2n in 2n=h_k+p_d with 2k, h_k, and p_d with h_k+, p_d+.

The formula (4.3.1), (4.3.2), (4.3.3), (4.3.4) can be obtained.

Please note that: (4.3.1), (4.3.2), (4.3.3), (4.3.4), are directly derived from Hou Shaosheng's theorem at 2k, so the expression of these 4 equations is always the same, and it has no relationship with whether num (2n=p_i+p_j) =0; It does not matter whether num(2k=p_i++p_j+) =0.

It has nothing to do with whether the conjecture is true. In other words, even if num (2n=p_i+p_j) =0, num (2k=p_i++p_j+) =0, (4.3.1), (4.3.2), (4.3.3), (4.3.4), this is still true.

Special emphasis is placed on:

2k=h_t++h_u+, where h_t+ and h_u+ are odd composite numbers.

2k=p_s++h_r+, where p_s+ is an odd prime and h_r+ is an odd composite number.

num (k≤h≤2k-3) equals the number of h_u+ + the number of h_r+.

We call (4.3.4) the standard representation of num (k≤h≤2k-3). In this formula, h_t+ must be an odd composite number in the interval [3,k]; p_s+ must be an odd prime in the interval [3,k]. h_u+, h_r+ must be odd composite numbers in the interval [k,2k-3].

End of proof

Proof We want to prove that (4.3.4) must be true when h_u+, h_r+, are both odd composite numbers in the interval [k,2k-3].

To prove that (4.3.4) holds, it is necessary to prove that the quantities on both sides of the equality sign are equal and that the names of things on both sides of the equality sign are the same.

On the left side of the equal sign, num (k≤h≤2k-3) represents the number of all odd composite numbers in the interval [k,2k-3].

Let h_x be any odd composite number in the interval [k,2k-3]. w is an odd number in the interval [3,k].

Set again:

2k-h_x=w (4.4.1)

2k is a definite constant. For any h_x, there is always w to make (4.4.1) true.

Order h_x to take each odd composite number in the interval [k,2k-3] in order from smallest to largest.

2k-h_x , either an odd composite number, or an odd prime number, there are only two possibilities.

If 2k-h_x is odd composite number, then 2k-h_x=h_t+ , where h_t+ is odd composite number. And write h_x as h_u+ , that is, h_x=h_u+, h_u+ odd composite number.

So we get 2k=h_t++h_u+.

The number of (2k=h_t++h_u+) is denoted by num(2k=h_t++h_u+).

If 2k-h_x is an odd prime, then 2k-h_x=p_s+, and p_s+ is an odd prime. And write h_x as h_r+, that is, h_x=h_r+, h_r+ odd composite number.

So we get 2k=p_s++h_r+.

The number of (2k=p_s++h_r+) is denoted by num (2k=p_s++h_r+).

Since h_x is any odd composite number in the interval [k,2k-3], now h_x is written either h_u+ or h_r+, so h_u+, h_r+ is all odd composite numbers in the interval [k,2k-3].

Because h_u+, h_r+ , are all odd composite numbers in the interval [k,2k-3].

So there must be:

num (k≤h≤2k-3)= num(2k=h_t++h_u+)+ num(2k=p_s++h_r+). (4.3.4)

Now num (k≤h≤2k-3) is the total number of odd composite numbers in the interval [k, 2k-3].

num(2k=h_t++h_u+)+ num(2k=p_s++h_r+)is also the number of all odd composite numbers in the interval [k,2k-3].

On both sides of the equal sign, the quantity is the same, the name of the thing is the same. So (4.3.4) must be true.

(4.3.4) Must be established, end of proof.

(4.0.1), (4.0.2), (4.0.3), (4.0.4); (4.3.1), (4.3.2), and (4.3.3) can be proved by referring to the proof method of (4.3.4).

brief summary: The title of this section is: 4 Inferences of Hou Shaosheng's Theorem.

Therefore, proving the 4 inferences of Hou Shaosheng's theorem is the main task of this section. We have proved 4 mathematical formulas in the inference of the theorem. Then the mathematical meanings of the 4 mathematical formulas are explained. Finally, 4 mathematical formulas in the theorem are tested by mathematical examples.

The test shows that the mathematical example is completely consistent with the mathematical formula of Hou Shaosheng's theorem and the 4 mathematical formulas in the inference.

The 4 mathematical formulas in inference are the main mathematical tools we use to prove conjectures. The mathematical meanings of the 4 mathematical formulas presented in this section will play an important role in proving the conjecture. The 4 formulas have made the necessary theoretical preparation for the proof of the conjecture.

Finally, we prove 4 inferences of Hou Shaosheng's theorem in 2k time. And give an independent proof of (4.3.4); We also pointed out: (4.0.1),(4.0.2),(4.0.3),(4.0.4); (4.3.1), (4.3.2), (4.3.3) can be proved by referring to the method of proving (4.3.4).

The theorems of sufficient and necessary conditions of conjectures make clear the mathematical thought of proving conjectures: a proof that satisfies the theorems of sufficient and necessary conditions must be a correct proof, and a proof that does not satisfy the theorems of sufficient and necessary conditions must be a wrong proof.

The formula for calculating the insufficient approximation of the number of odd primes in the interval [n,2n] proves that there must be odd primes in the interval [n,2n]. One of the necessary conditions for the hypothesis to be valid is satisfied.

Hou Shao-sheng's identity 2n=q+(2n-q) contains all the changes of Goldbach's conjecture 2n=p_i+p_j. We have proved Hou Shaosheng's theorem, which has laid a reliable theoretical foundation for proving the conjecture.

This section proves 4 inferences of Hou Shao-sheng's theorem. Concrete theories are prepared for the proof of the conjecture. Now we should prove our conjecture.

Hou Shao-Sheng¹, Ma Lin-Jun²

(1 Anyang Mathematical Society, Henan Province, China; 2 Department of Mathematics, Sun Yat-sen University, Guangzhou City, Guangdong Province, China)

Hou Shao-Sheng Email: ayhss@163.com.

Abstract The mathematical formula of Hou Shaosheng's theorem is as follows:

num(2n=q+(2n-q))=num(2n=p_i+p_j)+ num(2n=h_t+h_u)

+ num(2n=p_s+h_r)+ num(2n=h_k+p_d). (3.3.0)

The 4 inferences of Hou Shaosheng's theorem are as follows:

num(3≤p≤n)=num(2n=p_i+p_j)+num(2n=p_s+h_r); (4.0.1)

num(3≤h≤n)=num(2n=h_t+h_u)+num(2n=h_k+p_d); (4.0.2)

num(n≤p≤2n-3)=num(2n=p_i+p_j)+num(2n=h_k+p_d); (4.0.3)

num(n≤h≤2n-3)=num(2n=h_t+h_u)+num(2n=p_s+h_r). (4.0.4)

Among them, the convention is: 3≤q≤n,q is odd.

p_i, p_j, p_s, p_d are all odd prime numbers; h_t, h_u, h_r, h_k are all odd composite numbers.

And p_i≤p_j, h_t≤h_u, p_s≤h_r, h_k≤p_d.

p_i, h_t, p_s, h_k, ∈[3,n]; p_j, h_u, h_r, p_d∈[n,2n-3].

4 inferential formulas provide a reliable theoretical basis for proving Goldbach's conjecture.

The basic task of this section is to uncover the 4 genetic codes that make Goldbach's conjecture valid, based on 4 inferential formulas. Moreover, as long as one genetic code exists, Goldbach's conjecture will continue to hold true. The paper proved that there were at least three of the four genetic codes, which proved Goldbach's conjecture once and for all.

Keyword Goldbach conjecture; Continue to be established;

genetic code.

Proof Since 2n=n+n, if n is an odd prime, then n+n is already the sum of two odd prime numbers, so the conjecture works.

As for Goldbach's conjecture 2n=p_i+p_j, some sources say that the computer has verified that 6≤2n≤10¹⁰, and Goldbach's conjecture is valid. As we have already said, if n is an odd prime, then the conjecture naturally holds. Therefore, in the following study of the conjecture, n is generally assumed to be composite number and relatively large, for example, 1000≤2n.

5.1 Let num(3≤p≤2k-3)=num(3≤p≤2n-3)+1, k=n+1,

(After that, if not specified, always set k=n+1)

Theorem 7.If num(3≤p≤2k-3)=num(3≤p≤2n-3)+1, where k=n+1, then 2k must can be expressed as the sum of two odd prime numbers, that is, if 2k=2 (n+1), Goldbach's conjecture must be true.

Proof because

num(3≤p≤2k-3)=num(3≤p≤2n-3)+1.

And k=n+1, so we have:

num(3≤p≤2k-3)

=num(3≤p≤2n-1)

=num(3≤p≤2n-3)+1 (5.1.1)

(5.1.1) indicates that the number of odd prime p in the interval

[3,2n-1] is 1 more than that in the interval [3,2n-3].

There are only two more integers in the interval [3,2n-1] than in the interval [3,2n-3] : 2n-2,2n-1. Obviously, 2n-2 is even and 2n-1 is odd number. So, one extra odd prime can only be 2n-1.

Because k=n+1, 2k-3=2n-1. So in the figure below, 2k-3 and 2n-1 are the same point.

as shown in the figure: In the interval [3,2n-1], there are only two more integers than in the interval [3,2n-3] : 2n-2,2n-1.

So there must be:

2k=2(n+1)=3+(2n-1). (5.1.2)

Since 3 and (2n-1) are both odd prime numbers, so for 2k=2(n+1), the conjecture is true.

Theorem 7 has been proved.

The above proof reveals an important secret: let k=n+1, if (2n-1) is an odd prime number, regardless of whether n is odd or even, then 2k must can be expressed as the sum of two odd prime numbers, that is, 2k conjecture must be true.

If (2n-1) is an odd prime, then 2k=2 (n+1) must can be expressed as the sum of two odd prime numbers, So the conjecture must be true at 2k.

This situation, in practice, is not uncommon.

See 1:

When n=3, 2n-1=2×3-1= 5, 5 is an odd prime number.

So we have 2k=2 (n+1)

=2×4= (2×3-1)+3.

See 2:

When n=4, 2n-1=2×4-1= 7, 7 is an odd prime.

So we have 2k=2 (n+1)

=2×5= (2×4-1)+3.

See 3:

when n=6, 2n-1=2×6-1=11,11 is an odd prime.

So we have 2k=2(n+1)

=2×7=(2×6-1)+3.

See 4:

When n=7, 2n-1=2×7-1= 13,13 is an odd prime number.

So we have 2k=2(n+1)

=2×8=(2×7-1)+3.

Above, we give 4 examples. These 4 examples show that the situation expressed by theorem 7 exists in practice.

However, not every n applies to theorem 7. For example, n=5, 2n-1=9. 9 is not an odd prime and does not satisfy that (2n-1) is an odd prime, so it does not apply to theorem 7.

If n is an odd prime, 2n=n+n, then obviously the conjecture works.

And since (2n-1) is an odd prime,2k=3+(2n-1),2k is the sum of two odd prime numbers.

That is, as long as (2n-1) is an odd prime, for such n we already have: 2(n+1)=2k=3+ (2n-1),2(n+1) is already the sum of two odd prime numbers.

So, if we assume that the conjecture is true for 2n and to prove that the conjecture must be true for 2k, we can assume that (2n-1) is odd composite. As for (2n-1) being an odd composite number, whether this condition must be used depends on the specific situation. Put this condition here in case you need it.

And we know that for 6≤2n≤10⁶, our conjecture is true. So we can assume that 2n is true.

Let 3≤n, k=n+1, and if k is an odd prime, then the conjecture works at 2k. So we'll work on the conjecture Always set k=n+1, and k is composite number.

Please read this section carefully, and do not ignore it because it has no proof of theorems and no derivation of formulas. Actually, this section is very important. Because this section is the link between the previous and the next, is the most important logical process in the proof of conjecture.

5.4 Minimum and maximum values of num(2k=p_i++p_j+).

num(3≤p≤n)=num(2n=p_i+p_j)+ num(2n=p_s+h_r); (4.0.1)

num(n≤p≤2n-3)=num(2n=p_i+p_j)+ num(2n=h_k+p_d); (4.0.3)

num(2n=p_i+p_j)≥1.

A: If one of (p_i+2), (p_j+2), (h_r+2), (h_k+2) is an odd prime, then the conjecture holds for 2k, k=n+1.

B: And (p_i+2), (p_j+2), (h_r+2), (h_k+2) are the sum of the number of odd primes, which is the maximum value of num(2k=p_i++p_j+). To wit:

num(2k=p_i++p_j+)≤num((p_i+2)∈P)+num((p_j+2)∈P)

+num((h_r+2)∈P)+num((h_k+2)∈P)

(p_i+2)∈P means that p_i+2 is an odd prime, num((p_i+2)∈P) means that p_i+2 is the number of odd prime numbers, and the same applies to the rest. (P is the set of odd prime numbers).

(p_i, p_j, p_s, p_d, p_i+, p_j+, are all odd prime numbers).

C: And num((pj+2)∈P)+num((hr+2)∈P), is the minimum value of

num(2k=p_i++p_j+). To wit:

num((p_j+2)∈P)+num((h_r+2)∈P) ≤ num(2k=p_i++p_j+).

Proof Firstly, part of the inference of Hou Shao-sheng's theorem (Theorem 4) is quoted as follows:

About the formula:

num(2n=q+(2n-q))=num(2n=p_i+p_j)+ num(2n=h_t+h_u)

+ num(2n=p_s+h_r)+ num(2n=h_k+p_d). (3.3.0)

always have:

num (3≤p≤n)= num(2n=p_i+p_j)+ num(2n=p_s+h_r); (4.0.1)

num (n≤p≤2n-3)= num(2n=p_i+p_j)+ num(2n=h_k+p_d). (4.0.3)

(1) The following proves theorem content A, as follows:

because

2k=2n+2

=(p_i+2)+p_j =p_i+(p_j+2) . (5.4.1)

So, as long as one of (p_i+2) is an odd prime, or one of (p_j+2) is an odd prime, 2k is already the sum of two odd prime numbers, so at 2k, the conjecture is true.

also because

2k=2n+2

=p_s+h_r+2=p_s+(h_r+2). (5.4.2)

That is to say, as long as one of (h_r+2) is an odd prime, 2k is already the sum of two odd prime numbers, so at 2k, the conjecture is true.

(h_r+2) is an odd prime number, which exists in all 8 examples. And the larger num (2n=p_s+h_r), the more odd prime numbers (h_r+2) are.

also because

2k=2n+2

=h_k+p_d+2=(h_k+2)+p_d. (5.4.3)

That is to say, as long as one of (h_k+2) is an odd prime, 2k is already the sum of two odd prime numbers, So at 2k, the conjecture is true.

At lest one of (h_k+2) is an odd prime number, which exists in all 8 examples. And the larger num(2n=h_k+p_d)is, the more odd primes (h_k+2) are.

Note that if num (2n=p_i+p_j) =m, and m is a positive integer, then there are m number of 2n=p_i+p_j, then there are m number of p_i, then there are m of p_j, and so on.

From (4.0.1), (4.0.3) we can see that if there is a p_i+2=p_i+1 that is an odd prime, or a p_j+2=p_j+1 that is an odd prime, or a h_r+2=p that is an odd prime, or a h_k+2=p that is an odd prime, then the conjecture must be true at 2k.

We have noticed that when (p_i+2), (p_j+2), (h_r+2), (h_k+2) are odd prime numbers,then

2k=(p_i+2)+p_j, 2k=p_i+(p_j+2), 2k=p_s+(h_r+2), 2k=(h_k+2)+p_d,

Every 2k is the sum of two odd prime numbers. As long as one of them exists, it proves that the 2k time conjecture is true.

Theorem content A proves the end.

From the above, if (p_i+2), (p_j+2), (h_r+2), (h_k+2) are odd prime numbers, then 2k=(p_i+2)+p_j, 2k=p_i+(p_j+2), 2k=p_s+(h_r+2), 2k=(h_k+2)+p_d,

Each 2k is the sum of two odd prime numbers. As long as one of them exists,

it proves that the 2k time conjecture is true.

And (p_i+2), (p_j+2), (h_r+2), (h_k+2) are the sum of the number of odd primes, which is the maximum value of num(2k=p_i++p_j+). To wit:

num(2k=p_i++p_j+)≤num((p_i+2)∈P)+num((p_j+2)∈P)

+num((h_r+2)∈P)+num((h_k+2)∈P)

This provedtheoremcontentB.

(3) The following proof theorem content C, as follows:

Let num((p_i+2) ∈P) indicate that p_i+2 is the number of odd prime numbers.

Let num((p_j+2)∈P)=m₁, (m₁ is a non-negative integer), that is, there are m₁ odd primes of p_j+2.

If 1 number of p_j+2 is an odd prime, then 1 number of p_i+(p_j+2) is the sum of 2 odd prime numbers, and there is 1 of 2k=p_i++p_j+.

Let num((p_j+2) ∈P)=m₁, that is, there are m₁ number of p_j+2 which is odd prime numbers,then there are m₁ number of p_i+(p_j+2) is the sum of two odd prime numbers, there are m₁ number of 2k=p_i++p_j+.

Replacing m₁ with num((p_j+2)∈P) in the above sentence gives the following statement:

If num((p_j+2)∈P). there are num((p_j+2)∈P) of p_j+2 which is odd prime, there are num((p_j+2)∈P) of p_i+(p_j+2) which is the sum of two odd primes, there are num((p_j+2)∈P) of 2k=p_i++p_j+ . The same goes for others.

Let num((h_r+2)∈P)=m₂, (m₂ is a non-negative integer),that is, there are m₂ number of h_r+2 which is odd prime numbers,then there are m₂ number of h_r+2 that is odd prime numbers.

If one number of h_r+2 is an odd prime, then there is one number of p_s+(h_r+2) is the sum of two odd prime numbers, and there is one 2k=p_i++p_j+ .

If m₂ number of h_r+2 is odd primes,then there are m₂ number of p_s+(h_r+2) is the sum of two odd primes, and there are m₂ number of 2k=p_i++p_j+ .

If num((h_r+2)∈P), there are num((h_r+2)∈P) of h_r+2 which is odd prime, there are num((h_r+2)∈P) of p_s+(h_r+2) which is the sum of two odd primes, there are num((h_r+2)∈P) of 2k=p_i++p_j+ . The same goes for others.

We notice that since there is no repetition between p_i,p_s , there is no repetition between p_i+(p_j+2),p_s+(h_r+2).

So num((p_j+2)∈P)+num((h_r+2)∈P)is the minimum value of

num(2k=p_i++p_j+).

num((p_j+2)∈P)+num((h_r+2)∈P) ≤ num(2k=p_i++p_j+).

This proved theorem content C.

Theorem 8 has been proved.

Proof Because the 2n conjecture is true, there must be:

num(3≤p≤n)= num(2n=p_i+p_j)+ num(2n=p_s+h_r). (4.0.1)

Among num(2n=p_i+p_j)≥1.

However, it is possible that num(2n=p_s+h_r)=0.

If num(2n=p_s+h_r)=0,

there must be: num(3≤p≤n)=num(2n=p_i+p_j).

This shows that p_i in (2n=p_i+p_j) is all the odd prime numbers p in the interval [3,n].

Let the smallest odd prime in p_i be p₁, then p₁=3, that is, p₁ is the first odd prime 3 in the interval [3,n]. So at least one of p_i+2 is an odd prime.

Because at least one of p_i+2 is an odd prime, and because:

num(2n=p_i+p_j)= num(2n+2=p_i+2+p_j)

= num(2k=(p_i+2)+p_j).

We know that there are sister primes in the interval [3,n].

Let the i_th odd prime be p_i, and p_i+2=p_i+1 is the odd prime, and p_i+1 is the (i+1)_th odd prime.

Here, p_i is the i_th odd prime number, and p_i+1 is the (i+1)_th odd prime number. p_i, p_i+1 is a sister prime number.

Then:

2k=(p_i+2)+p_j

=p_i+1+p_j.

So 2k time is a good guess.

From the above proof, we can see that when num(2n=p_s+h_r)=0, Goldbach's conjecture must be true.

Therefore, when proving the conjecture below, it can be assumed that num (2n=p_s+h_r) ≥1, And just consider that k=(n+1) is composite. Since k is an odd prime, the conjecture is obviously true at 2k.

Theorem 9, complete proof.

5.6 Theorem 10. If p_i+2is an odd prime, or h_r+2is an odd prime, Goldbach's conjecture must be true.

Proof Since k=n+1 is composite number, there must be:

num(3≤p≤k)=num(3≤p≤n)     (5.6.1)

Quote (4.0.1) as follows:

num(3≤p≤n)=num(2n=p_i+p_j)+num(2n=p_s+h_r). (4.0.1)

From (5.6.1) and (4.0.1), there must be:

num(3≤p≤k)= num(3≤p≤n)

=num(2n=p_i+p_j)+num(2n=p_s+h_r)

=num(2n+2=p_i+2+p_j)+num(2n+2=p_s+h_r+2) (5.6.2)

=num(2k=p_i+2+p_j)+num(2k=p_s+h_r+2), k=n+1. (5.6.3)

From (5.6.3) :

2k=p_i+2+p_j

=(p_i+2)+p_j

=p_i+(2+p_j)

If one of(p_i+2)is an odd prime, then(p_i+2)+p_j is already the sum of 2 odd prime numbers. So 2k=(p_i+2)+p_j, 2k is already the sum of 2 odd prime numbers. So the 2k time conjecture works.

If one of(2+p_j)is an odd prime, then p_i+(2+p_j)is already the sum of 2 odd prime numbers. So 2k=p_i+(2+p_j), 2k is already the sum of 2 odd prime numbers. So the 2k time conjecture works.

From (5.6.3) :

2k=p_s+h_r+2

=p_s+(h_r+2)

If one of(h_r+2)is an odd prime, then p_s +(h_r+2)is already the sum of 2 odd prime numbers. So 2k=p_s+(h_r+2), 2k is already the sum of 2 odd prime numbers. So the 2k time conjecture works.

This proves that if there is a p_i+2 that is an odd prime, or a 2+p_j that is an odd prime, or a h_r+2 that is an odd prime, then the conjecture must be true at 2k. That Goldbach's conjecture must be true.

Theorem 10 is proved.

Theorem 10 proves that if p_i+2 is an odd prime, or h_r+2 is an odd prime, Goldbach's conjecture must be true. However, this theorem is not answered: there must be p_i+2 which is an odd prime numbers, or there must be h_r+2 which is an odd prime numbers, to ensure that Goldbach's conjecture is true.

The task that p_i+2 must be an odd prime, or that h_r+2 must be an odd prime, requires theorem 17.

If someone checks every even number up to 3×$10{}_6$ one by one, Goldbach's conjecture is true. This fact tells us that Goldbach's conjecture is true for every even number in the interval [6, 3×$10{}_6$].

Let's say 2n,2(n+1), no less than 6, no greater than 3×$10{}_6$, why is the conjecture true for 2n, and true for 2(n+1)? It works for 2n, it works for 2(n+1), so what's the secret? The result of our research is the following 4 major theoretical questions raised by the authors.

Goldbach's conjecture at 2n is expressed as 2n=p_i+p_j; Goldbach's conjecture for 2 (n+1), we use 2(n+1)=p_i++p_j+, p_i+, p_j+, are odd prime numbers.

In the study of 4 major theoretical problems, the term 2(n+1)=p_i++p_j+ is involved. To do this, we need to give it a definition.

Definition: Let q_i and q_j both be odd prime numbers, and 2(n+1)=q_i+q_j, then we say (define) 2(n+1)=q_i+q_j, which is the source of 2(n+1)=p_i++p_j+.

We're working on Goldbach's conjecture. We have to ask: what is the nature of Goldbach's conjecture? Some people would say that the essence is: as long as 3≤n, then 2n=p_i+p_j can be true. We say this answer is correct, but it misses the point.

So, what's the point? The key is: why is the conjecture true for 2n, and true for 2(n+1)? Or, if 2n=p_i+p_j is true, 2k=p_i++p_j+ where does come from? 2k=p_i++p_j+ How is composed?

This part will reveal the most mysterious and essential recurrence relation of Goldbach's conjecture. The recursion relation here means that if the conjecture is true at 2n, it must lead to the conjecture being true at 2 (n+1).

5.7.1 Question 1: Assuming 2n is true, where does 2k=p_i++p_j+ come from? Or, why do assumptions persist?

We have theorem 11 to answer as follows:

Theorem 11: If the 2n conjecture is true, then2k=p_i++p_j+ has the following 4 sources, as long as 1 source exists, the 2k conjecture is guaranteed to continue to be true.

The 4 sources of2k=p_i++p_j+ are as follows:

If one of (p_i+2) is an odd prime, then 2k=(p_i+2)+p_j becomes one of 2k=p_i++p_j+.

If one of (p_j+2) is an odd prime, then 2k=p_i+(p_j+2) becomes one of 2k=p_i++p_j+.

If one of (h_r+2) is an odd prime, then 2k=p_s+(h_r+2) becomes one of 2k=p_i++p_j+.

If one of (h_k+2) is an odd prime, then 2k=(h_k+2)+p_d becomes one of 2k=p_i++p_j+.

Each of these 4 sources is a theorem. Because they have the same form, they're put together as a theorem.

4 sources of expression, very simple, are what if, then what. If what is the condition of the theorem, then what is the conclusion of the theorem. From the condition of the theorem, to the conclusion of the theorem, there is no other intermediate link, causality, very direct, so it is no longer proved.

At this point, the only question the reader should ask is, is it possible that none of the 4 ifs exist? This is the most central issue, and this is the issue that we are most concerned about. As a proof of the conjecture, this question must be answered.

As an answer to the core question, theorem 17 will prove:

If the Goldbach conjecture holds for 2n and k (=n+1) is composite number, then at least one of (p_i+2), (h_r+2) is an odd prime, which guarantees that the conjecture holds for 2k.

Theorem 11, theoretically gives 4 sources of 2k=p_i++p_j+.

Note that 2k=h_t+2+h_u, either 2k=(h_t+2)+h_u, or 2k=h_t+(2+h_u), cannot be the sum of 2 odd prime numbers. So it can't be one of 2k=p_i++p_j+, so 2k=p_i++p_j+, there are only 4 sources above.

The 4 sources of 2k=p_i++p_j+ essentially answer the close relationship between the 2n conjecture and the 2k conjecture.

Note that the first 2k=(p_i+2)+p_j of the 4 sources. when (p_i+2) is an odd prime, because p_i, (p_i+2), are both odd primes, so p_i, (p_i+2), is a sister prime, and p_i is the smaller of the sister primes.

The second 2k=p_i+(p_j+2) of the 4 sources , when (p_j+2) is an odd prime, since p_j, (p_j+2), are both odd primes, so p_j, (p_j+2), is a sister prime, and p_j is the smaller of the sister primes.

The third 2k=p_s+(h_r+2) of the 4 sources, originally h_r is an odd composite number, when (h_r+2) is an odd prime, p_s+(h_r+2) becomes the sum of 2 odd primes. The significance of this event is that it connects 2n=p_s+h_r with 2k=p_s+(h_r+2), completing the transition from the non-Goldbach conjecture (2n=p_s+h_r) to Goldbach's conjecture(2k=p_i++p_j+)!

The fourth 2k=(h_k+2)+p_d of the 4 sources, originally h_k is an odd composite number, and when (h_k+2) is an odd prime number, (h_k+2)+p_d becomes the sum of 2 odd primes. The significance of this event is that it connects 2n=h_k+p_d with 2k=(h_k+2)+p_d, completing the transition from the non-Goldbach conjecture (2n=h_k+p_d) to Goldbach's conjecture(2k=p_i++p_j+)!

So, two of the 4 sources of 2k=p_i++p_j+ are realized with the help of the smaller of the sister primes; Two are realized by means of the non-Goldbach conjecture.

Question 1.1: As n increases, the number of (2n=p_i+p_j) increases in a wave.

2k=2(n+1)=p_i++p_j+ has 4 sources, which is the fundamental reason why num(2n=p_i+p_j) gradually waves up as n increases.

In some materials, especially at

https://en.wikipedia.org/wiki/Goldbach’s_conjecture the odd primes slash the intersection of graphics, further enriches our perceptual knowledge. Although these data know that num (2n=p_i+p_j) increases with the increase of n, they do not indicate the reason for the gradual increase.

Numerous researchers of Goldbach's conjecture, through a limited number of mathematical examples, have found that as n increases, the number of (2n=p_i+p_j) gradually increases in a wave pattern. This discovery is a contribution and deserves recognition. However, no one knows that 2k=2(n+1)=p_i++p_j+ has 4 sources, so no one has mathematically explained the gradual increase in the number of (2n=p_i+p_j).

To explain the gradual increase in num(2n=p_i+p_j), we use the following 2 mathematical formulas.

num(3≤p≤n)=num(2n=p_i+p_j)+num(2n=p_s+h_r); (4.0.1)

num(n≤p≤2n-3)=num(2n=p_i+p_j)+num(2n=h_k+p_d); (4.0.3)

Let's say k=n+1, and let's compare num(3≤p≤n), num(3≤p≤k).

If k is composite number, whether k is odd composite number or even composite number, there is:

num(3≤p≤n)=num(3≤p≤k).

That is to say, if k is composite number, num(3≤p≤n) does not change as n increases by l from n to k. That is, num(3≤p≤n) does not change.

When n changes to k,num(3≤p≤n)=num(3≤p≤k), it is known from (4.0.1) that the total number of num(2n=p_i+p_j) +num (2n=p_s+h_r) does not change. However,num(2n=p_i+p_j), num(2n=p_s+h_r), can be changed. This change will cause a wave change in num(2n=p_i+p_j), num(2n=p_s+h_r).

If k is an odd prime, there is always:

num(3≤p≤n)+1=num(3≤p≤k).

That is to say, if k is an odd prime, num(3≤p≤n), because n increases by 1, and as n changes from n to k, num(3≤p≤k) increases by 1 compared to num(3≤p≤n), the change occurs.

When n changes to k,num(3≤p≤k) increases by 1 compared with

num(3≤p≤n), and it can be seen from (4.0.1) that the total number of num(2n=p_i+p_j)+ num(2n=p_s+h_r)increases by 1. num (2n=p_i+p_j), num (2n=p_s+h_r), must change. It is likely to increase num (2n=p_i+p_j).

⑴ If the conjecture is true for 2n, it is accepted that

1≤num (2n=p_i+p_j). There is no doubt that the greater num (2n=p_i+p_j), the greater the number of (2n=p_i+p_j). The more 2n=p_i+p_j, the more p_i+2, the more p_j+2. Not every p_i+2, not every p_j+2, is an odd prime, but the more chances p_i+2, p_j+2, is an odd prime, the more p_i+2, p_j+2, is an odd prime.

If p_i+2, p_j+2, are odd prime numbers, each 2k=(p_i+2)+p_j, each 2k=p_i+ (2+p_j), becomes one of(2k=p_i++p_j+). Causes an increase in the number of (2k=p_i++p_j+).

⑵ Assuming that the conjecture is true for 2n, in general, 1≤

num(2n=p_s+h_r). The greater num (2n=p_s+h_r), the greater the number of (2n=p_s+h_r). The more we have (2n=p_s+h_r), the more we have (h_r+2). Although not every (h_r+2) is an odd prime, the more chances (h_r+2) is an odd prime, the more (h_r+2) is an odd prime.

If (h_r+2) is an odd prime, every 2k=p_s+(h_r+2) becomes one of(2k=p_i++p_j+). Causes an increase in the number of (2k=p_i++p_j+).

2n=p_s+h_r, not one of (2n=p_i+p_j). If 2k=p_s+(h_r+2) becomes one of (2k=p_i++p_j+), this is a process from 2n=p_s+h_r to(2k=p_i++p_j+). This is an important factor that the num(2k=p_i++p_j+)of increase .

⑶ Assuming that the conjecture is true for 2n, in general, 1≤num (2n=h_k+p_d). There is no doubt that the greater num (2n=h_k+p_d), the greater the number of (2n=h_k+p_d). The more (2n=h_k+p_d) there is, the more (h_k+2) there is. Although not every (h_k+2) is an odd prime, the more chances (h_k+2) is an odd prime, the more (h_k+2) is an odd prime.

If (h_k+2) is an odd prime, every 2k=(h_k+2)+p_d becomes one of(2k=p_i++p_j+). Causes an increase in the number of (2k=p_i++p_j+).

2n=h_k+p_d, not one of (2n=p_i+p_j). If 2k=(h_k+2)+p_d becomes one of (2k=p_i++p_j+), this is a process from 2n=h_k+p_d to(2k=p_i++p_j+). This is an important factor that the num(2k=p_i++p_j+)of increase .

Theorem 11 theoretically proves that 2k=2(n+1)=p_i++p_j+ has 4 sources. Now, let's test theorem 11 with a mathematical example.

Verification question 1: 2k=2(n+1)=p_i++p_j+, there are four sources.

The following example, 3, is the third of the 8 mathematical examples in 3.2: Instance 3. Now verify the 4 sources of 2k=p_i++p_j+ with example 3.

Example 3. 2n=2×51=102

=1+101=3+99=5+97=7+95=9+93=11+91=13+89=15+87=17+85=

=19+83=21+81=23+79=25+77=27+75=29+73=31+71=33+69=35+67

=37+65=39+63=41+61=43+59=45+57=47+55=49+53=51+51.

Note: Numbers underlined are prime numbers. If the number is not underlined, it is a composite number.

Theorem 11 proves that 2k=2(n+1)=p_i++p_j+ has 4 sources. These 4 sources are fully confirmed in Example 3. See below for specific verification.

Verify the first source of 2k=2(n+1)=p_i++p_j+as follows:

num(2×51=p_i+p_j)=8.

These 8 number of 2×51=p_i+p_j are as follows:

2×51 =5+97 =13+89 =19+83 =23+79 =29+73 =31+71 =41+61 =43+59.

2n=p_i+p_j, (p_i +2)are odd prime numbers:

(5+2)； (29+2)；(41+2).

2k=2(n+1)=(p_i+2)+p_j. 2k is the sum of 2 odd prime numbers, such as:

2k=(5+2)+97 ；2k=(29+2)+73 ； 2k=(41+2)+61.

Verify the second source of2k=2(n+1)=p_i++p_j+, as follows:

2n=p_i+p_j,(p_j+2)are odd prime numbers:

(71+2)；(59+2).

2k=2(n+1)=p_i+(p_j+2), 2k is the sum of 2 odd prime numbers, which is:

2k=31+(71+2)； 2k=43+(59+2).

Note: (29+2)+73, repeated with 31+(71+2). (41+2)+61, repeat with 43+(59+2). We're shading the duplicators.

Verify the third source of 2k=2(n+1)=p_i++p_j+, as follows:

num(2n=p_s+h_r)=6.

The 6 number of 2×51=p_s+h_r are: 2×51= 3+99= 7+95= 11++91= 17+85= 37+65= 47+55.

2n=p_s+h_r,(h_r+2) are odd prime numbers: (99+2); (95+2); (65+2).

2k=2(n+1)=p_s+(h_r+2), 2k is the sum of 2 odd prime numbers, which is:

2k=3+(99+2); 2k=7+(95+2); 2k=37+(65+2).

Verify the fourth source of 2k=2(n+1)=p_i++p_j+, as follows:

num(2n=h_k+p_d)=2.

These 2 number of 2n=h_k+p_d are as follows:

2n=35+67； 2n=49+53；

2n=h_k+p_d, (h_k+2)are odd prime numbers: (35+2).

2k=2(n+1)=(h_k+2)+p_d, 2k is the sum of 2 odd prime numbers, which is:

2k=(35+2)+67.

(The character shading is repeated and can be ignored.)

For example, the above facts are taken as examples:

(5+2)+97, like 7+(95+2), is the sum of the same 2 numbers, and 7+(95+2) is the repeat.

37+(65+2)and (35+2)+67 are the sum of the same 2 numbers, and (35+2)+67 is the repeat.

The above facts show that the 4 sources of 2k=2(n+1)=p_i++p_j+ all appear in Example 3, and the theory and practice are completely consistent.

Above we get 9 number of 2k=2(n+1)=p_i++p_j+, of which there are 4 repetitions. So there are no repetitions(2k=p_i++p_j+),only 5.

After comparison, these 5 are exactly the same as the actual situation in example 2 (2(51+1)).

Us from(2n=2×51=102=1+101=3+99=······=49+53=51+51.), got 2k=

2(51+1)=p_i++p_j+ is all about.

The reader should appreciate that the 4 sources would rather repeat the same data (with character shading), and never give up the opportunity to ensure that the conjecture continues to hold.

The appearance of the duplicate data explains the 4 sources and provides multiple guarantees for the establishment of Goldbach's conjecture. If only one(2k=p_i++p_j+)occurs once, the conjecture is guaranteed. If(2k=p_i++p_j+)occurs m times in total, the conjecture is guaranteed m times. This is one of the mysteries of the persistence of conjecture.

Since(2k=p_i++p_j+)has only 4 sources,(2k=p_i++p_j+)can occur at most 4 times for the same (2k=p_i++p_j+). Because in the same source, there can be no duplication. For example, 2k=2(n+1)=p_i++p_j+ the fourth source: 2n=h_k+p_d, when one (h_k+2) is an odd prime, 2k= (h_k+2) +p_d, forming a 2k=p_i++p_j+, it is impossible to appear 2 identical 2k=(h_k+2)+p_d.

5.7.2 Question 2: Where does 2k=p_s++h_r+ come from? Will the 4 sources of 2k=2(n+1)=p_i++p_j+ use up 2n resources and hinder the source of 2k=p_s++h_r+?

We have theorem 12 answers as follows:

Theorem 12: If the conjecture is true for 2n, then 2k=p_s++h_r+has4sources:

If one of(p_j+2)is an odd composite number, then 2k=p_i+(p_j+2) becomes one of 2k=p_s++h_r+.

If one of (h_r+2) is an odd composite number, then 2k=p_s+(h_r+2) becomes one of 2k=p_s++h_r+.

If one of (p_s+2) is an odd prime, then 2k=(p_s+2)+h_r becomes one of 2k=p_s++h_r+ .

If one of (h_t+2) is an odd prime, then 2k=(h_t+2)+h_u becomes one of 2k=p_s++h_r+.

Each of the 4 sources of 2k=p_s++h_r+ above is a theorem. Because the causal relationship between the condition and the conclusion is very direct, it can no longer be proved.

2k=p_s++h_r+ There are only 4 sources mentioned above. Because 2k=h_k+2+p_d, neither 2k=(h_k+2)+p_d, nor 2k=h_k+(2+p_d), can be the sum of 1 odd prime + 1 odd composite number. So it can't be one of 2k=p_s++h_r+, it can't be a source of it.

2k=2(n+1)=p_s++h_r+ has 4 sources, which is why num(2n=p_s+h_r) is gradually increasing in waves.

The reason why num(2n=p_s+h_r) gradually increases in waves can still be explained by mathematical formulas. In this regard, we see above (Problem 1.1: As n increases, the number of (2n=p_i+p_j) increases in a wave.) Examples have been given, which are omitted here for the sake of shortening space.

Theorem 12, in theory, proves that 2k=2(n+1)=p_s++h_r+ has 4 sources. Let's test theorem 12 with a mathematical example.

Verification question 2: 2k=2(n+1)=p_s++h_r+ 4 sources.

Example 3 below, the third of the 8 mathematical examples in 3.2, now uses example 3 to verify the 4 sources of 2k=p_s++h_r+.

Example 3. 2n=2×51=102

=1+101=3+99=5+97=7+95=······=41+61=43+59=45+57=47+55=49+53=51+51.

The number is underlined. The number is an odd prime. None digits with underscores are odd composite numbers.

Theorem 12 proves that 2k=2(n+1)=p_s++h_r+ has 4 sources. These 4 sources are fully confirmed in Example 3. See below for specific verification.

Verify the first source of 2k=2(n+1)=p_s++h_r+, as follows.

num(2×51=p_i+p_j)=8.

These 8 number of 2×51=p_i+p_j as follows:

2×51=5+97=13+89=19+83=23+79=29+73=31+71=41+61=43+59.

2n=p_i+p_j, (p_j+2)are odd composite numbers:

5+(97+2)=13+(89+2)=19+(83+2)=23+(79+2)=29+(73+2)=41+(61+2).

Verify the second source of 2k=2(n+1)=p_s++h_r+, as follows.

num (2n=p_s+h_r)=6.

These 6 number of 2n=p_s+h_r are as follows:

2×51=3+99=7+95=11+91=17+85=37+65=47+55.

2n=p_s+h_r,(h_r+2) are odd composite numbers:

11+(91+2)=17+(85+2)=47+(55+2).

Verify the third source of 2k=2(n+1)=p_s++h_r+, as follows.

num(2n=p_s+h_r)=6.

These 6 number of 2n=p_s+h_r are as follows:

2×51=3+99=7+95=11+91=17+85=37+65=47+55.

2n=p_s+h_r,(p_s+2) are odd prime numbers:

(3+2)+99=(11+2)+91=(17+2)+85.

(3+2)+99, repeat with 5+(97+2), (11+2)+91 repeat with 13+(89+2),

(17+2)+85 is repeated with 19+(83+2).

Verify the fourth source of 2k=2 (n+1) =p_s++h_r+, as follows.

num(2×51=h_t+h_u)=9.

These 9number of 2n=2×51=h_t+h_uare as follows:

2n=9+93； 2n=15+87； 2n=21+81； 2n=25+77； 2n=27+75；

2n=33+69； 2n=39+63； 2n=45+57； 2n=51+51.

2n=h_t+h_u, (h_t+2) are odd prime numbers:

(9+2)+93=(15+2)+87=(21+2)+81=(27+2)+75=(39+2)+63=(45+2)+57.

(The character shading is repeated and can be ignored.)

The above facts show that the 4 sources of 2k=2(n+1)=p_s++h_r+ are exactly consistent with mathematical examples.

Us from(2n=2×51=102=1+101=3+99=······=49+53=51+51.), got it

num(2(51+1)=p_s++h_r+)=9. The data is exactly the same as in example 2

(2(51+1)).

Above, we have completed the verification task of 2k=2 (n+1) =p_s++h_r+ 4 sources.

In order to facilitate later references if necessary, we make the following summary.

Us from(2n=2×51=102=1+101=3+99=······=49+53=51+51.), got it

num(2(51+1)=p_s++h_r+)=9.

These 9 number of 2(51+1)=2(n+1)=p_s++h_r+ as follows.

From the first source of 2k=2(n+1)=p_s++h_r+, as follows:

2n=p_i+p_j, and p_j+2 are odd composite numbers:

5+(97+2)=13+(89+2)=19+(83+2)=23+(79+2)=29+(73+2)=41+(61+2).

From the second sourceof2k=2(n+1)=p_s++h_r+, as follows:

2n=p_s+h_r,(h_r+2) are odd composite numbers:

11+(91+2)=17+(85+2)=47+(55+2).

From the third source of 2k=2(n+1)=p_s++h_r+,as follows:

2n=p_s+h_r, (p_s+2) are odd prime numbers:

(3+2)+99=(11+2)+91=(17+2)+85.

From the fourth source of 2k=2(n+1)=p_s++h_r+, as follows:

2n=h_t+h_u, (h_t+2) are odd prime numbers:

(9+2)+93=(15+2)+87=(21+2)+81=(27+2)+75=(39+2)+63=(45+2)+57.

(The character shading is repeated and can be ignored.)

Us from(2n=2×51=102=1+101=3+99=······=49+53=51+51.), got it

num(2(51+1)=p_s++h_r+)=9.The data is exactly the same as in example 2

(2(51+1)).

From the above 4 sources, the obtained 2k=2(n+1)=p_s++h_r+ is summarized as follows:

5+(97+2)=13+(89+2)=19+(83+2)=23+(79+2)=29+(73+2)=41+(61+2).

11+(91+2)=17+(85+2)=47+(55+2).

(3+2)+99=(11+2)+91=(17+2)+85.

(9+2)+93=(15+2)+87=(21+2)+81=(27+2)+75=(39+2)+63=(45+2)+57.

The shadow is the duplicator.

5+(97+2)=13+(89+2)=19+(83+2)=23+(79+2)=29+(73+2)=41+(61+2).

2k=p_i +(p_j+2) is the sum of an odd prime number and an odd composite number.

Please pay attention to 2, the following:

11+(91+2)=17+(85+2)=47+(55+2).

2k=p_s+(h_r+2) is the sum of an odd prime number and an odd composite number.

It should also be noted that there is no repetition between 2k=p_i+(p_j+2)and 2k=p_s+(h_r+2)! And they're all part of 2k=2(n+1)=p_s++h_r+ . This example shows that in 2k=2(n+1)=p_s++h_r+ , if you want to reject 2k=p_i+(p_j+2), or reject 2k=p_s+(h_r+2), there is no reason.

5.7.3 Question 3: 2k=h_t++h_u+ Where does come from? We have theorem 13 answers as follows.

Theorem 13: If the conjecture holds for 2n, then2k=h_t++h_u+ has the following 4 sources:

If one of (p_s+2) is odd composite number, then 2k=(p_s+2)+h_r becomes one of 2k=h_t++h_u+.

If one of (h_t+2) is odd composite number, then 2k=(h_t+2)+h_u becomes one of 2k=h_t++h_u+.

If one of (h_u+2) is odd composite number, then 2k=h_t+(h_u+2) becomes one of 2k=h_t++h_u+.

If one of (p_d+2) is odd composite number, then 2k=h_k+(p_d+2) becomes one of 2k=h_t++h_u+.

Above, 2k=h_t++h_u+ 4 sources, each proposition is a theorem. Because the causal relationship between the condition and the conclusion is very direct, it can no longer be proved.

Above, we give 4 sources for 2k=h_t++h_u+. This is why num(2n=h_t+h_u), the wave form, increases rapidly as n increases! It is still possible to explain the reason for the rapid increase of num(2n=h_t+h_u) in a mathematical formula, but for the sake of compression, it is omitted here.

2k=h_t++h_u+, there are only 4 sources mentioned above. Because 2k=p_i+2+p_j, neither 2k= (p_i+2) +p_j, nor 2k=p_i+ (2+p_j), can be the sum of one odd composite number plus one odd composite number. So it can't be a source of 2k=h_t++h_u+.

Theorem 13 theoretically proves that 2k=2(n+1)=h_t++h_u+ has 4 sources. Let's test theorem 13 with a mathematical example.

Verification question 3: 2k=2(n+1)=h_t++h_u+ 4 sources.

Example 3, which is the third of the 8 mathematical examples in 3.2, is now used to verify 2k=2(n+1)=h_t++h_u+ 4 sources.

Example 3. 2n=2×51=102

=1+101=3+99=5+97=7+95=······=41+61=43+59=45+57=47+55

=49+53=51+51.

Theorem 13 proves that 2k=2(n+1)=h_t++h_u+ has 4 sources. These 4 sources are fully confirmed in Example 3. See below for specific verification.

Verify the first source of 2k=2(n+1)=h_t++h_u+, as follows.

num(2×51=h_t+h_u)=9.

These 9number of2n=2×51=h_t+h_uare as follows:

2n=9+93； 2n=15+87； 2n=21+81； 2n=25+77； 2n=27+75；

2n=33+69； 2n=39+63； 2n=45+57； 2n=51+51.

2n=h_t+h_u, and (h_t+2) are odd composite numbers:

(25+2)+77=(33+2)+69.

Verify the second source of 2k=2(n+1)=h_t++h_u+, as follows.

2n=h_t+h_u, (h_u+2) are odd composite numbers:

9+(93+2)=27+(75+2)=39+(63+2).

Verify the third source of 2k=2(n+1)=h_t++h_u+, as follows.

num(2n=p_s+h_r)=6.

These 6 number of 2n=p_s+h_r are as follows:

2n=3+99； 2n=7+95； 2n=11+91； 2n=17+85； 2n=37+65；

2n=47+55；

2n=p_s+h_r, (p_s+2) are odd composite numbers:

(7+2)+95=(37+2)+65=(47+2)+55.

Verify the fourth source of 2k=2(n+1)=h_t++h_u+, as follows.

num(2n=h_k+p_d)=2.

These two number of 2n=h_k+p_d are as follows:

2n=35+67； 2n=49+53；

2k=h_k+p_d, and (p_d+2) are odd composite numbers:

35+(67+2)=49+(53+2).

(The character shading is repeated and can be ignored.)

The above facts show that the 4 sources of 2k=2(n+1)=h_t++h_u+ are exactly consistent with mathematical examples.

Us from (2n=2×51=102=1+101=3+99=······=49+53=51+51.), obtained the num(2(51+1)=h_t++h_u+)=5. The data is exactly the same as in example 2 (2 (51+1)).

We should note that 2k=2(n+1)=h_t++h_u+ 4 sources, in mathematical practice, it is preferable to repeat the data, but never give up the data that should be.

5.7.4 Question 4: Where does2k=h_k++p_d+ come from? We have theorem 14 answers as follows.

Theorem 14: If the conjecture is true for 2n,2k=h_k++p_d+ has the following 4 sources:

If one of (p_i+2) is odd composite number , then 2k=(p_i+2)+p_j becomes one of 2k=h_k++p_d+;

If one of (h_u+2) is odd prime, then 2k=h_t+(h_u+2) becomes one of 2k=h_k++p_d+;

If one of (h_k+2) is odd composite number, then 2k=(h_k+2)+p_d becomes one of 2k=h_k++p_d+;

If one of (p_d+2) is odd prime, then 2k=h_k+(2+p_d) becomes one of 2k=h_k++p_d+;

Above, there are 4 sources of 2k=h_k++p_d+, each of which is a theorem. Because the causal relationship between the condition and the conclusion is very direct, it can no longer be proved.

Above we give 4 sources of 2k=h_k++p_d+. This is why num (2n=h_k+p_d) waves gradually increase as n increases! The reason for the gradual increase of num (2n=h_k+p_d) wave can still be explained by mathematical formula, but for the sake of space, it is omitted here!

2k=h_k++p_d+, there are only 4 sources mentioned above. Because 2k=p_s+2+h_r, neither 2k=(p_s+2)+h_r, nor 2k=p_s+ (2+h_r), can be the sum of one odd composite number plus one odd prime number. So it can't be a source of 2k=h_k++p_d+.

Theorem 14 theoretically proves that 2k=2(n+1)=h_k++p_d+ has 4 sources. Let's verify theorem 14 with a mathematical example.

Verification question 4: 2k=2(n+1)=h_k++p_d+ 4 sources.

Example 3 below, which is the third of the 8 mathematical examples in 3.2, is now used to verify 2k=2(n+1)=h_k++p_d+ 4 sources.

Example 3. 2n=2×51=102

=1+101=3+99=5+97=7+95=······=41+61=43+59=45+57=47+55

=49+53=51+51.

Theorem 14 theoretically proves that 2k=2(n+1)=h_k++p_d+ has 4 sources. These 4 sources are fully confirmed in Example 3. See below for specific verification.

Verify the first source of 2k=2(n+1)=h_k++p_d+, as follows.

num(2×51=p_i+p_j)=8.

These 8 number of 2×51=p_i+p_j, as follows:

5+97=13+89=19+83=23+79=29+73=31+71=41+61=43+59.

2n=p_i+p_j, (p_i+2) are odd composite numbers:

(13+2)+89=(19+2)+83=(23+2)+79=(31+2)+71=(43+2)+59.

Verify the second source of 2k=2(n+1)=h_k++p_d+, as follows.

num(2×51=h_t+h_u)=9.

These 9 number of 2n=2×51=h_t+h_u are as follows:

2n=9+93； 2n=15+87； 2n=21+81； 2n=25+77； 2n=27+75；

2n=33+69； 2n=39+63； 2n=45+57； 2n=51+51.

2n=h_t+h_u, (h_u+2) are odd prime numbers:

15+(87+2)=21+(81+2)=25+(77+2)=33+(69+2)=45+(57+2)=51+(51+2).

Verify the third source of 2k=2(n+1)=h_k++p_d+, as follows.

num(2n=h_k+p_d)=2.

These two number of 2n=h_k+p_d are as follows:

2n=35+67； 2n=49+53；

2n=h_k+p_d, and (h_k+2) are odd composite numbers: (49+2)+53.

Verify the fourth source of2k=2(n+1)=h_k++p_d+, as follows.

num(2n=h_k+p_d)=2.

These two number of 2n=h_k+p_d are as follows:

2n=35+67； 2n=49+53；

2n=h_k+p_d, (p_d+2) is an odd prime number, not one in this case. However, this cannot be used to negate the existence of odd prime numbers (p_d+2), much less to negate the theorem 14. It is now confirmed by example 2 that there are indeed (p_d+2) odd prime numbers.

Example 2. 2×52=104

=1+103=3+101=5+99=7+97=9+95=11+93=13+91=15+89=17+87=

=19+85=21+83=23+81=25+79=27+77=29+75=31+73=33+71=35+69

=37+67=39+65=41+63=43+61=45+59=47+57=49+55=51+53.

In this case, num(2n=h_k+p_d)=6.

These 6 number of 2n=h_k+p_d are as follows:

2n=15+89, 2n=21+83, 2n=25+79, 2n=33+71, 2n=45+59, 2n=51+53.

2n=h_k+p_d, (p_d+2) are odd prime numbers: 33+(71+2), 45+(59+2).

The above facts show that the four sources of 2k=2(n+1)=h_k++p_d+ are exactly consistent with mathematical examples.

Us from(2n=2×51=102=1+101=3+99=······=49+53=51+51.), obtained the num(2(n+1)=2(51+1)=h_k++p_d+)=6. The data is exactly the same as in example 2 (2(51+1)).

We should note that 2k=2(n+1)=h_k++p_d+ 4 sources, in mathematical examples, it is preferable to repeat the data, but never give up the data that should be.

Four big theoretical questions, and that concludes.

The 4 major theoretical problems are the core theoretical problems in the proof of Goldbach conjecture, which is a major breakthrough and great harvest in our study of Goldbach conjecture.

Goldbach's conjecture holds, essentially, because it holds for 2n, and it still holds for 2(n+1).

The proof of Goldbach's conjecture must answer: If the conjecture is true for 2n, why is it still true for 2(n+1)? What is the close relationship between the establishment of 2n and the establishment of 2k?

4 major theoretical questions are theoretically answered: the close relationship between 2n holds and 2k(k=n+1) holds.

Theorem 11 answers: If the conjecture is true at 2n, then 2k=p_i++p_j+ has 4 sources, and as long as one source exists, the conjecture will continue to be true at 2k. This is the most important discovery and central theory in the proof of the conjecture.

Theorem 12 answers: If the conjecture is true for 2n, then 2k=p_s++h_r+ has 4 sources.

Theorem 13 answers: If the conjecture is true for 2n, then 2k=h_t++h_u+ has 4 sources.

Theorem 14 answers: If the conjecture is true for 2n, then 2k=h_k++p_d+ has four sources.

The 4 major theoretical questions not only theoretically answer the close relationship between the establishment of 2n and the establishment of 2k(k=n+1), but also accept the strict test of case 3 and case 2.

2k=p_i++p_j+ has 4 sources; fully unified with complex mathematical examples.

2k=p_s++h_r+ has 4 sources; fully unified with complex mathematical examples.

2k=h_t++h_u+ has 4 sources; fully unified with complex mathematical examples.