Shangerbiao is the 13th generation grandson of Nanjing Prince of Qingping. His ancestral home is Hongdong, Shanxi Province. Liaoning Liaoyang lighthouse people. Born on October 9, 1960. Han nationality, master's degree, major, senior engineer. He used to be a guidance technician, staff officer, instructor, team leader, director, director, TV station director, general manager, etc. He was clever and kind-hearted. When I was a monitor in middle school, I was a good student for three years. Shortly after the resumption of the college entrance examination, he was admitted to the air force ground to Air Missile Institute (Air Force Engineering University) and studied new media in the school of information management of Peking University.
Shang's experience in Mathematics: in the early examination of junior high school, he got full marks in arithmetic; after joining the army, he taught trigonometric function and plural number to the soldiers in the 13th Chengkong regiment; when he was in military school, he got the best results in mathematics among the 107 students in the class; he took professional courses and mathematics courses in the 11th Chengkong mixed brigade teaching team; he got the highest score in Mathematics in the postgraduate entrance examination; he got 70 in Mathematics in the doctoral examination; he was able to teach junior high school mathematics and plural number Senior high school mathematics; can speak "advanced mathematics", "Engineering Mathematics" and "postgraduate entrance examination mathematics"; has talked about Grade 1-9 Olympic mathematics. Unfortunately, he didn't graduate from mathematics, but he still liked mathematics best. His actual major is second artillery, which requires a lot of mathematics. He began to use his spare time to contact the three conjectures of world mathematics in 2012.
Goldbach conjecture was put forward by German mathematician Goldbach in 1742. The content of Goldbach's conjecture is divided into two parts: the first part is that any even number greater than or equal to 6 is the sum of two odd primes; the second part is that any odd number greater than or equal to 9 is the sum of three odd primes. Up to now, conjecture has been put forward for 278 years, but it has not been proved whether the theory is correct or not. As long as the conjecture "1 + 1" is proved, the second part will be solved.
How to establish mathematical model to prove the difficulty of Goldbach's conjecture "1 + 1"? He has explored in series, exponential function, logarithmic function, linear equation and plane equation, but has not got satisfactory conclusion. Try to prove it with mathematical induction and counter proof (only one condition is different). By using definite integral and multiple integral, series and Fourier expansion, differential equation (2k = P1 + P2 can be used as the solution of the equation), linear algebra can build even matrix A, prime matrix B and C, and perform matrix operation. Most efforts have been made in the complex variable function. The integral, series and residue of the complex variable function have also tried to find a breakthrough through number theory and integral transformation, but they are useless.
Do you remember Weida theorem for quadratic equation of one variable? Let the quadratic equation of one variable be:
Two of them have the following relations:
(Note: the formula for finding the root of a quadratic equation of one variable is applicable in the field where the coefficient of the equation is rational, real, complex or any number.)
Y = A x^2 + BX + C (A ≠ 0) is a quadratic function whose image is symmetric with respect to x = -
B/ 2A.
It can be seen from the increase and decrease of quadratic function that the roots ofAx^ 2 + BX + C = 0 (a ≠ 0) are the only pair, sometimes odd prime roots. For example:
x²-6x+9=0 x²-（3+3）x+3*3=0 x1=3 x2=3
x²-8x+15=0 x²-（3+5）x+3*5=0 x1=3 x2=5
x²-10x+21=0 x²-（3+7）x+3*7=0 x1=3 x2=7
x²-12x+35=0 x²-（5+7）x+5*7=0 x1=5 x2=7
x²-14x+49=0 x²-（7+7）x+7*7=0 x1=7 x2=7
x²-16x+39=0 x²-（3+13）x+3*13=0 x1=3 x2=13
x²-18x+45=0 x²-（5+13）x+5*13=0 x1=5 x2=13
x²-20x+51=0 x²-（3+17）x+3*17=0 x1=3 x2=17
x²-22x+57=0 x²-（3+19）x+3*19=0 x1=3 x2=19
x²-24x+95=0 x²-（5+19）x+5*19=0 x1=5 x2=19
x²-26x+133=0 x²-（7+19）x+7*19=0 x1=7 x2=19
.
.
X^ 2+2kx + C = 0 x^2 - (x^ 1 + X^ 2) x + X1 * x 2 = 0 (K ≥ 3 integer)
Proof 1: equation A x^ 2 + BX + C = 0 (a ≠ 0), take a = 1, B = - 2K (K ≥ 3 integer), when two X 1, x2 are odd prime numbers, that is, x 1 = P 1, x 2 = P 2, the Veda theorem is as follows:
x1+ x2=-B/A=-B/1=-(-2k)=2k=p1+p2
x1 x2=C/A=C/1=C=p1 p2
Since K is an integer greater than or equal to 3, 2K ≥ 6 even
That is: 2K = P 1 + P 2 = 1 + 1
Therefore, any even number greater than or equal to 6 is the sum of two odd prime numbers. A proof of Goldbach's conjecture I
Add 3 to 2K = P 1 + P 2 at the same time A proof of Goldbach's conjecture I
2k+3=p1+p2+3
Since 2K + 3 is an odd number greater than or equal to 9, and p1, p2and 3 are all odd prime numbers, Goldbach conjecture II is derived from conjecture I: any odd number greater than or equal to 9 is the sum of three odd prime numbers.
Proof 2: it is known that x ∈ R, y ∈ R, z = f (x, y), Z is continuous and ∂ Z / ∂ x = z∂ x ≠ 0, ∂ Z / ∂ y = z∂ y ≠ 0; Z = f (x, y) is determined by the following integral equations.
try to find z = f (x, y)
The partial derivation of X on both sides of equation (1) is obtained
.....
Z=(x+y)/2 ， ∂z/∂x=∂z/∂y=1/2
....
So z = (x + y) / 2 is the common solution of equations (1) and (2).
Next, it is explained by z = (x + y) / 2 that 2K = P1 + P2=1+1
Because x ∈ R, y ∈ R, Z ∈ R. When x ∈ n +, y ∈ n +, 2Z ∈ n +, and 2Z is positive
Let's say 2k is equal to 2z, x is equal to p1, and y is equal to p2, so obviously 2k is equal to p1 plus p2.
Let the set of even Numbers greater than or equal to 6 be A, and the set of odd primes be B and C, respectively
A = {6,8,10,12,14,16,18,20,22,24,... . 2 k}
B={3,5,7,11,13,17... P1}, C={3,5,7,11,13,17... P2}
Because: 6 = 3 + 3
8 = 3 + 5 (5 + 3)
10 = 5 + 5
12 = 5 + 7 (7 + 5)
14 = 3 + 11 (11 + 3)
16 + 11 = 5 + 5 (11)
.
2k=p1+p2 (2k∈A, p1∈B, p2∈C, p1+p2 ≥6)
That any even number greater than or equal to 6 is the sum of two odd primes.
2k is greater than or equal to 6,2 k plus 3 is greater than or equal to 9
2k+3=p1+p2+3 (p1, p2 and 3 are odd primes)
That any odd number greater than or equal to 9 is the sum of three odd prime Numbers.
The solution to many mathematical problems is not unique, and so is the solution to the goldbach conjecture. Fortunately, the conclusion of the above two arguments is not myopic, and there are not many restrictions on the two prime Numbers, when k goes to +∽ infinity,2k=p1+p2 is still true.
Head of shangerbiao is in Shilin
The end
Author: shang erbiao
Sunday, March 24, 2020
