Goldbach’s Conjecture Proof

Proving Goldbach’s Conjecture by Means of Prime Factorization

Guide

Proving Goldbach’s Conjecture by Means of Prime Factorization 
By Leon Botha   
021/08/05 

Introduction 

The unproved Goldbach conjecture states that every even number can be written as the sum of two prime numbers. To prove that every even number Q can be constructed out of a prime number A summed with prime number B, we will show that the sum of prime numbers is directly related to the product of prime numbers in such a way that y* (A+B)= x*(A*B). 

We will show that the proof can be summarized in the following statement: 

Because every odd natural number X must at least be composed out of two prime factors (A*B) and, can be found as the difference between two squares (Z ² -Y ² =A*B) means that every even number Q (2*Z) can be obtained by the sum of at least one set of two prime numbers (A+B) that differ by a square from Z ². 

Method 

To prove Goldbach’s conjecture we will first show how any odd number X, where X=A*B can be transformed to the set A*B=Z ² -Y ² that always has two integer solutions. We will show that A*B relates to A+B such that 4*A*B= (A+B) ² -Y ². Then we will show that every even number Q is implied in the set of Z such that Q=2*Z. Finally, to show that even number Q consists out of the sum of two prime numbers A+B because some numbers can only be prime factorized in one way, and we show how every even number will have a unique prime number set for A and B where A is the first prime number less than 2*Z and B= Z*2-A. 

For every statement of consideration, some examples are included for clarity purpose. 

1) Consider, every odd number can be factorized into at least one set of factors 
X=A*B 
1=1*1 
3=1*3 
5=1*5 
7=1*7 
9=1*9 
21=3*7 and 1*21 
33=3*11 and 1*33 

2)Consider, every odd number x is a perfect square different from another square. 

X= ((n+1)/2)²- ((n-1)/2) ² 
3= 2² -1²  
5= 3² -2² 
7= 4² -3² 
9= 5² -4²   
11=6² -5² 
13=7² -6² 
.... 

3) Consider every odd number composed of the product of two prime numbers always has two and only two factorization solutions  

x= ((n+1)/2) ² - ((n-1)/2) ²  
and  
if a*b= x where a>b then 
x= (a-(a-b)/2) ² -(a-b)/2) ² 

Ex: 
21=1*21                   and   21=3*7 
21= 11² -10²   and   21= (7-(7-3)/2) ² -((7-3)/2) ² 

Ex: 
33= 1*33 and 3*11 
33= 17² -16²     and  33= (11-(11-3)/2) ² -((11-3)/2) ² 

 Using another number as the example: 91

For any another number X, such as say 91, which is X=7*13, the formula X= ((A*B+1)/2) ² - ((A*B-1)/2) ² will result in the 1*X expression.
X= ((A*B+1)/2) ² - ((A*B-1)/2) ²
X= (92/2)² - (90/2)
X= 46² - 45²
X= (46+45)*(46-45)
X= 91*1

The second expression for the number X, such as 91, depends on the prime factors that exclude 1 and X itself, such as 7 and 13.
X=(A-(A-B)/2) ² -((A-B)/2) ²
X=(13-(13-7)/2)² - ((13-7)/2)²
X= (13-6/2)² - (6/2)²
X= (13-3)² - (3)²
X= 10² - 3²
X= (10+3)*(10-3)
X= 13*7

To generalize this formula, it is rewritten as  

x= Z ² -Y ² 

Thus 

Y ² =Z ² -X 

Where Z and Y is of high interest for the next point. 

4) 
We can define the Z-set as being all the values for X for where 0<X<Y<Z that satisfies Z² -Y² =X.  
Meaning, the Z-set for number N contains all the numbers that are exactly a perfect square away from N². 

Example of the first 5 Z-sets, 

Z(1) = {1,0} 
Z(2) = {4,3,0} 
Z(3) = {9,8,5,0} 
Z(4) = {16,15,12,7,0} 
Z(5) = {25,24,21,16,9} 

.. 

Z(n) = {n ²-0², n ² -1², n ² -2², ….. n ² -n ²} 

This “Z-Set” for a number N will be of primary concern. To be clear, the Z-set of a number N contains all the values numbers less than Z ² that are a square difference from Z ² For matter of convenience, the definition of “Z-Set” will be used extensively. The ‘Z-value' is the actual value for n in Z(n). 

For example, in the following set: 
Z(3) = {9,8,5,0} 
The ‘Z-value’ is  3 
The ‘Z-set’ are the numbers: 9,8,5,0 

5) Consider that the sum of the of the prime factors of a number is equal to 4 times the Z value of the number, and is as effectively as scaling/multiplying the resulting X value by 4 and the Z value by 2. 

If X=Z ² -Y ² 
and Q= a+b 
then Y= (Q*Q-4*X)/4 

 Ex: 21=7*3 
Then 21= 5² -2²  
Q= 7+3 = 10 
Y= (10² -4*21)/4 
Y=2*2 
Thus 21*4=4*(5² -2²) 
         21*4=     10² -4² 

 To generalize this formula, I rewrite it as: 
x= Z ² -Y ²  
Where X is any positive odd number. 

This can be rewritten as: 

Y ² =Z ² -X. 

6) Therefore, it can be stated that for every odd number x there must exist a positive integer Z and Y value such that x=Z ² -Y ². 

Z ² -Y ² =x 
1² -0² = 1  
2² -1² = 3 
3² -2² = 5 
4² -3² =7 
5² -4² = 9 
6² -5² = 11 
7² -6² = 13 

... 

7) Therefore, every odd integer number has a solution for the formula 4*X=Z*Z-Y*Y well. 
If we multiply an odd number by 4, the result is an even number. 
4*(Z ² -Y ²)=4*x 
4*(1² -0²)=   1*4=4 
4*(2² -1²)=   3*4=12 
4*(3² -2²)=   5*4=20 
4*(4² -3²)=  7*4=28 
4*(5² -4²)=  9*4=36 
4*(6² -5²)= 11*4=44 
4*(7² -6²)= 13*4=52 

Also, If we multiply an odd number by 4, the Z-value doubles, so Z will always be an even number. 
Z*Z-Y*Y=x 
(2*1² -2*0²)=   1*4=4 
(2*2² -2*1²)=   3*4=12 
(2*3² -2*2²)=   5*4=20 
(2*4² -2*3²)=  7*4=28 
(2*5² -2*4²)=  9*4=36 
(2*6² -2*5²)= 11*4=44 
(2*7² -2*6²)= 13*4=52 

8) The sum of A+B relates to A*B directly by scaling the value of Z so that the result is 4*A*B, and hence Z is doubled but also equal be twice the sum of the products A+B.  
If Z ² -Y ² =A*B then (A+B) ² -Y ² = =4*A*B 

Example: 
X=21 
Z =A+B       X=A*B 
10=7+3      21=7*3 
Z ² -Y ² =A*B 
10² -Y ² =4*(7*3) 
10² -4² = 4*(5² -2²) 
 

Example: 
X=91 
Z =A+B       X=A*B 
20=7+13      91=7*13 
Z ² -Y ² =A*B 
20² -Y ² =4*(7*13) 
20² -6² = 4*(10² -3²) 

Example: 
X= 493 
Z =A+B       X=A*B 
46=17+29      493=7*13 
Z ² -Y ² =A*B 
46² -Y ² =4*(17*29) 
46² -12² = 4*(23² -12²) 

Example: 
X= 6 
Z =A+B       X=A*B 
5=2+3      6=2*3 
Z ² -Y ² =A*B 
5² -Y*Y =4*(2*3) 
5² -1² = 4*(6) 

Point 9) If it is known that a number has only two prime factors such that X=A*B then, multiplying that number by a perfect square C ², would mean the only factors for A*B*C is A, B and C. Thus, multiplying and semi-prime number by 4 (which is 2²) cannot add any additional prime factors that would satisfy the equation X=A*B*C other than A, B,or 2. 

Point 10) Every even number is divisible by two. Some even numbers if divided by two results in an odd number. Some odd numbers are composed out of prime factors that cannot be factorized further. Therefore, there are some even numbers that can only be obtained by doubling two multiplied prime numbers for x=A*B*2. 

Example: 

1*1*2=2 
1*3*2=6       
1*5*2=10 
1*7*2= 14 
.. 
3*5*2=30 
3*7*2=42 
 

Point 11) Because the difference in two prime numbers results in being 1 less than another prime number, it means that we can find the unique pair of prime numbers in a Z-set directly that satisfies the solution of Z ² -Y ² =A*B where A and B are prime numbers. 
To find A and B such that A and B are prime numbers and add up to equal 2*Z, the prime number less than 2*Z is deducted by A and the difference is added to B. 

Example  
Z(7) 
First prime less than 2*7 is 13.  
Thus 2*7=13+1 and 7² -1*13=6² 

Thus, the first prime set of A and B in set Z(7) is 1*13. 

Example 
Z(14) 
First prime less than 2*14 is 23.  
Thus 2*7=23+5 and 14² -5*23=8² 

Thus, the first prime set of A and B in set Z(14) is 5*23. 

Example 
Z(6) 
First prime less than 2*6 is 11.  
Thus 2*6=11+1 and 6² -1*11=5² 

Thus, the first prime set of A and B in set Z(6) is 1*11. 

Example: 
Z(39) 
First prime less than 2*39 is 67.  
Thus 2*39=67+11 and 39² -11*67=28² 

Thus, the first prime set of A and B in set Z(39) is 11*67. 
 

12) Therefore, every Z-value has to have a solution for which A and B are prime numbers in all of these equations Z ² -Y ² =4*A*B and (A+B) ² -Y ² =4*A*B and 2*Z=A+B. 
 

13) Since every Z-function has to have a solution where the products A*B are prime numbers in the set, and any even number can be represented by 2*Z, which can be obtained also be obtained from the sum A+B, then it means we have shown that 2*Z=A+B has a solution for every even number such that A and B are prime numbers in the solution. 

Conclusion  
Every even number can be written as the sum of two prime numbers and the Goldbach conjecture is true.