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Goldbach's conjecture: every even integer greater than 2 can be expressed as the sum of two primes.

Since every uneven number is x+1, where x is an even number, and every such number can be expressed as the sum of 3 even integers, namely x + y + 2, where x and y are even numbers, or 2n + 2m + 2, which can be expressed as a summation of twos (2+2+...) or simply 2p, which by definition is divisible by two and therefore is an even number, we can conclude that since every prime number is uneven, except 2, this proves that every even number, except 4, can be obtained by 2 prime numbers.

Every summation of two prime numbers can be expressed as 2n + 2m + 2, since it is always divisible by 2. Thus any prime number is the summation of an even number n and 1. This can be easily seen in the simple fact that 2 is and can be the only even prime number. Think of an even number as a summation of twos (2+2+...). It is always divisible by 2, 1 and itself. Thus we have proven that any prime number cannot be an even number, except 2.

By exhaustion, we can prove the conjecture. We only need to prove it for 4, since we've already proven it for all other even integers.

We simply remove 2m from the equation, which gives 2n + 2, which comes from 2n + 1 +1, where n must be 1. 3 and 1 are primes as required, thus we have proven the conjecture.

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closed as not a real question by Andres Caicedo, Thomas, Noah Snyder, Norbert, Douglas S. Stones Oct 6 '12 at 9:21

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I'm afraid this has nothing to do with a proof of Goldbach's conjecture. –  Olivier Bégassat Oct 3 '12 at 2:29
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Can you explain a little bit more on your second paragraph? –  Patrick Li Oct 3 '12 at 2:30
    
I don't see where you have proved anything. Your statement at the end of the 3rd paragraph (any prime number cannot be an even number) follows directly from the definition of a prime number. You have not shown anything relating to Goldbach's conjecture. –  Barry Oct 3 '12 at 2:36
    
Every summation of two prime numbers can be expressed as 2n + 2m + 2, which comes from the prime numbers 2n + 1 and 2m + 1. We can actually use that to prove it for 4 and 2. We just need to set m or n to 0. –  Dong Min Son Oct 3 '12 at 2:37
    
Basically, I am saying that prime numbers must be odd, so if we prove that the summation of two odd numbers gives an even integer, we also prove the conjecture, since all prime numbers are odd, except 2 and 4, which we can prove by exhaustion. –  Dong Min Son Oct 3 '12 at 2:40

1 Answer 1

If you think you've proven a famous 270 year-old conjecture in just a few lines of arithmetic, it's time to go back and seriously check what you have written.

The conjecture isn't that any sum of two odd primes is an even integer (which is what you seem to be showing - incidentally, this can be done in a single sentence), but that any even integer can be written as the sum of two primes. These are two very different statements.

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I'm aware of that thanks. –  user39572 Oct 3 '12 at 2:44
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@Dong, you're really missing the point big time here. You were correctly and fairly adviced to study way deeper and better around this matter and, of course, about number theory so that you'll realize how far you are of having proved such a huge and important conjecture in mathematics. –  DonAntonio Oct 3 '12 at 3:05
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I am just a freshman and am taking discrete mathematics this semester, so perhaps I don't know what I am talking about. But if mathematics cannot prove something as obvious as 2 + 2 = 4, then there's something wrong with math. Perhaps, the concepts we use right now are TOO LIMITED in scope. –  Dong Min Son Oct 3 '12 at 3:07
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Ah, ok now I get it. We need the formula to obtain all the prime numbers, and that's the thing that makes the theorem so difficult to prove. Ok, thanks. –  Dong Min Son Oct 3 '12 at 3:38
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@DongMinSon: No -- "We need the formula to obtain all the prime numbers" is still wrong. Godbach's conjecture does not require that all prime numbers are used. What you need to obtain is all the even numbers. –  Henning Makholm Oct 3 '12 at 16:25

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