Proof for goldbach conjecture
1. :
Goldbach conjecture states that every even number is the result of the sum of two prime numbers.The
formula which is proposed in this paper is a way to generate the two prime numbers for a given even
number n and it is not like the existing methods which check if goldbach conjecture is true for a given even
number using brute force.This formula have a potential to prove goldbach conjecture by checking for even
larger dataset.Like any other formula which is accepted in mathematical community by checking for a large
dataset,this formula can be checked like this which proves goldbach conjecture by presenting two prime
numbers for a given even number
2. :
The Goldbach Conjecture posits that every even number greater than 2 can be expressed as the sum of two
prime numbers. In this proof, we present a formula to support this conjecture.
3. :
- **Assumption 1:** The Goldbach Conjecture is assumed to be true.
- **Assumption 2:** Fundamental principles of number theory, such as the properties of even and odd
numbers, are considered valid.
4. :
- n = 2k, where n is a generic even number, and k is an integer.
- ± values are determined based on the range of n/2 as follows:
- If n/2 falls between 1 to 100 (inclusive), use ±1, ±2, or ±3.
- If n/2 falls between 101 to 200 (inclusive), use ±4, ±5, or ±6.
- Continue this pattern for subsequent ranges of 100 numbers where the lower and the higher number are
also considered in the range like 1-100(where 1 and 100 are also considered in the range).
5. :
No specific theorems or lemmas are required for this proof.
6. :
- Start with a generic even number, n = 2k, where k is a natural number.
- Define the range of ± values based on n/2 as described above.
- To find the two prime numbers that sum to n, follow these steps:
a. Calculate n/2 and add or subtract the appropriate values from the defined range to obtain candidates.
b.To ease the ± process do this:
▪If n/2 is even number then-The values of the range to be used are only odd numbers.
▪If n/2 is a odd number then-The values of the range to be used are only even numbers.
c. If any candidate falls within the new range due to the ± process, use the ± values of that range to obtain
a prime number accordingly.
d. Repeat this process until the final candidates fall inside a range and are stable(e.g.they should not fall
in a new range due to ± process or else the ± process should be repeated again and again).
e. If the final answer falls outside the defined range due to the ±process then perform the ± process again
based on the values of the new range and keep doing this until the final answer stays within a range.
- Calculate n minus the final answer to find the other prime number.
7. :
To generate ±values for a set of 100 numbers:
▪We consider here 1-100(1 and 100 are also considered in the range) as 1,101-200 as 2,etc.
▪Take 1,2,3,4,etc as per your requirements(the ±values you want to generate for a set you want)then multiply
it by three and then subtract two from it and you will get the starting ±value of the set you choosed
▪Example:If i want to generate ±values for the set 101-200(represented as 2 in the formula):(2*3)-2=the
starting value of the range 101-200(4)
8. :
For 78:
=78÷2=39±2(39 is odd so the values to be used from the range are only a even numbers,so from ±1,±2,±3
theonly even number is ±2)
=39+2=41(a prime number)
=39-2=37(a prime number)
=78-41=37(a prime number)
The two values for 78 are 41 and 37.
=78-37=41(a prime number)
▪The two values for 78 are 41 and 37.(You might get two prime numebers during the ±process but it does
not happens everytime,so don't rely on it and only choose one of the prime number to get the final answer)
▪In some case where you get the prime number by only doing the n÷2 step without doing the ±process like
14÷2=7,10÷2=5,etc.In these type of cases if the achived prime number is achived only and only by doing the
n÷2 step and not doing the ±process,the achieved prime number is considered the final answer.
9. :
This proof provides a formula that, based on the assumptions stated, yields two prime numbers that sum to
a given even number n. If this formula holds true for all even numbers, it supports the validity of the
Goldbach Conjecture.
10. :
- This proof contributes to the ongoing quest to validate the Goldbach Conjecture, which holds historical
and mathematical significance.
- Limitations include the need for further validation, potential refinements, and an expanded range of tested
numbers.
- The potential impact of this work on the field of number theory and mathematics should be considered.
- The formula has a potential to generate two prime numbers for even number,which when added gives the
even number.This formula thus can help prove goldbach conjecture if large quantity numbers are checked
for the correctness of the formula which also leaves us with a way to generate two prime numbers in
context of a given even number.
11. :
A new formula have been discovered for proving the goldbach conjecture(beacause it gives two prime
numbers for a even number):(n÷2)±1,2,3[n=a even number][if n÷2 results in a number between 1 to 100
including 1 and 100 then it is ±1,2,3 and if n÷2 is between 101 to 200 including 1 and 200 then it should be
±4,5,6 and so on the numbers will change according to the set of 100 numbers to which it belongs to](like 1
to 100 with 1 and 100:±1,2,3)(101 to 200 with 101 and 200:±4,5,6)and so on.
This a formula for proving goldbach conjecture for a number n(where n is even number).And if n÷2 is odd
number then you have to choose even number from the three numbers for every set of 100 numbers like
from ±1,2,3 you choose ±2 for further evaluation and the opposite if n÷2 is even number.This formula is
original.This formula proves golbach conjecture as it will always generate the two prime numbers which will
add up to the given even number.
Cases like 1212:1212÷2=606
=606±19,±21[result for ±21 are similiar like ±19]
=606+19=625(wrong),606-19=587(it is prime number but 1212-587=625[not a prime]
so you do ±16,±18[beacause 587 falled in the by the ± process and if it happens like this anytime then you
have to ±x number from which is assigned for the specific range in which it falls][basically after dividing and
after that subtracting or adding the values assigned for the range in which n÷2 falls and if the final answer
falls in other range due to the addition or subtraction of values from the answer of n÷2 and the values of the
range of 100 numbers is specified then the final answer should again go through the ± process but the
values are of the new range in which the final answer falls]Now do n-final answer=the other prime number.
:
Proof for goldbach conjecture by Chaitanya Bankar is licensed under Attribution-NonCommercial 4.0
International
Proof for goldbach conjecture © 2023 by Chaitanya Bankar is licensed under Attribution-NonCommercial
4.0 International. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc/4.0/