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September 13th, 2016, 02:32 AM   #1
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Goldbach's conjecture proof

Here I will proof Goldbach's conjecture by a simple way:
to prove the conjecture, we will prove that any odd natural number greater than (3) is a sum of a prime number and an even number, we take that an even number is a sum of two odd numbers, from the first proof we decompose the result to a sum of two prime numbers and two even numbers, then we substitute the sum of the two even numbers from the original even number we get an even number equal a sum of two prime numbers and that is the Goldbach's conjecture.

For the interval [10, +∞ [ we have:
Let ({prime, even, odd}, +, ×) be a field with three elements.
All odd numbers greater than 3 is a sum of a prime number and an even number:
(Odd= prime + even)……………………………………… ………………………….(1)
We have that every even number is a sum of two odd numbers:
Even = odd + odd
From (1) we get:
that every even number is a sum of two prime numbers and two even numbers:
Even = prime + even + prime + even
Sash as the sum of two even numbers is an even number we get:
=> Even = prime + prime + even
=> Even – even = prime + prime
Sash as the substitution of two even numbers is an even number we get:
Even = prime + prime……………………………………… …………………...……………….(2)
Sash as the set of even numbers and prime numbers are both an infinite sets the equation is correct for every even number belong to the interval [10, +∞] .

For the interval ]2, 10[:
We have only three even numbers : 4,6,8 :
And we know that:
4=2+2……………………………………… …………………………...…………(3)
6=3+3……………………………………… …………………………...…………(4)
8=5+3……………………………………… …………………………...…………(5)

Conclusion:
From (2) , (3) , (4) , (5) we have that every even number belong to the interval ]2, +∞ [ is a sum of two prime numbers and the Goldbach's conjecture is correct.
.
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Last edited by skipjack; September 13th, 2016 at 12:38 PM.
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September 13th, 2016, 04:32 AM   #2
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This is nice work. Good luck with publishing it.
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Last edited by 1ucid; September 13th, 2016 at 04:57 AM.
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September 13th, 2016, 05:46 AM   #3
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Quote:
Originally Posted by redos View Post
=> Even = prime + prime + even
In this line, you need to prove that the left-hand side is any even number. At the moment your derivation shows only that any even number less some particular even number (that depends on the choice of prime) is equal to the right-hand side.
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Last edited by skipjack; September 13th, 2016 at 12:37 PM.
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September 13th, 2016, 06:52 AM   #4
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Quote:
Originally Posted by v8archie View Post
you need to prove that the left hand side is any even number.
I think is already proven by the proofs of Olivier Ramaré and Harald Helfgott which is saying that any even number is the sum of 6 primes, that means:
even = prime + prime + (prime + prime + prime + prime)

we know that (prime + prime + prime + prime) is an even number.
But I still think that is proven by the derivation I wrote above.

Last edited by skipjack; September 13th, 2016 at 12:36 PM.
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September 13th, 2016, 11:06 AM   #5
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That idea suffers from the same problem as your proof, which explains why you think it is correct. Perhaps you should ask Olivier Ramaré and Harald Helfgott whether they think they have proved Goldbach. I can guarantee that if they thought they had, their paper giving the six prime proof would have had Goldbach as its main focus.

Last edited by skipjack; September 13th, 2016 at 12:35 PM.
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September 18th, 2016, 09:15 AM   #6
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Can you please give a numerical example, illustrating all steps of obtaining (even=prime+prime)?
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September 18th, 2016, 10:21 AM   #7
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The Goldbach Conjecture is a statement of the existence of such primes, not a claim that there exists any special method for finding them.
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September 11th, 2017, 07:14 AM   #8
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goldbach proof

The formula that generates whole prime numbers :


y=(2^(x-1)-1)/x ( formula-1)

If y is an integer , then x must be absolutely a prime number .
the set of x for any value of integer y ; x = { 3,5,7,11,13,....} and it generates all the prime numbers .

The question is that for the set of prime numbers ( x1 , x2) does the formula generates all the even numbers or not ?
y1 = (2^(x1-1) -1 ) /x1 + (2^(x2-1) -1 ) /x2
for ( x1 , x2) = ( 3,3) then y1 = 2 ;
for ( x1 , x2) = ( 3,5) then y2 = 4 ;
for ( x1 , x2) = ( 5,5) then y3 = 6 ;
for ( x1 , x2) = (5,7) then y4 = 8 ;
....................

The result for whole prime sets of ( x1 , x2) then you can generate all the even number's set .
P.S.: For the proof of formula-1 and to learn more about it please contact me . For example formula-1 must be always divided by 3 .

METE UZUN
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e-mail: meteuzun@hotmail.com
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September 11th, 2017, 08:44 AM   #9
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Quote:
Originally Posted by meteuzun View Post
If y is an integer, then x must be absolutely a prime number.
That's incorrect. For example, y is an integer if x = 341 = 11 × 31.
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September 11th, 2017, 10:12 AM   #10
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thank you skipjack .


Can you test this formula .please



(2^(x-1) + ( x-1)! ) /x
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