# Is this a valid proof of the Goldbach's theorem? [closed]

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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I'm afraid this has nothing to do with a proof of Goldbach's conjecture. –  Olivier Bégassat Oct 3 '12 at 2:29
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
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## closed as not a real question by Andres Caicedo, Thomas, Noah Snyder, Norbert, Douglas S. StonesOct 6 '12 at 9:21

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@DongMinSon: Julien's answer explains exactly what you're missing. You write "it follows that any pair of odd prime numbers gives an even number." (emphasis added) This is true, but Goldbach's conjecture states that we can obtain every even number $> 4$ in this way. –  Jonas Kibelbek Oct 3 '12 at 3:13
@DongMinSon: No, now you seem to be proving "any even integer can be written as the sum of two odd numbers" instead of "any even integer ($>2$) can be written as the sum of two primes". Every even integer can be written as $2m+2n+2=(2m+1)+(2n+1)$, but in many cases, $2m+1$ and $2n+1$ will not both be primes. For example, you write $4=2(1)+2(0)+2$, which we can rearrange to get $4=3+1$. But 1 is not a prime number! –  Jonas Kibelbek Oct 3 '12 at 3:32