Is the Goldbach Conjecture Finally Proven?

[fibonacci235]

I have a question concerning Goldbach's Conjecture. First, it is easy to demonstrate that the sum of any two odd integers will always be even.
For example,
let an odd integer q=2k+1 and an odd integer p=2m+1
It then follows that q+p= 2(k+m+1) = 2n which is even.

Now, it is true that any prime number >2 is odd. So, couldn't you simply use this fact to prove a substantial amount of Goldbach's conjecture? I don't understand people who sit down and add two arbitrary primes >2 to find an even number as the sum thinking that they may find an exception to the conjecture, but they won't because of what I just proved.

I know that prime numbers don't have a neat little general form like 2k+1 or 2n, but they are a subgroup of the odd numbers. All prime numbers >2 are odd but not all odd numbers are prime. Primes >2 are odd simply by definition of a prime number.

Then after you prove that the only time you will get an even number as a sum by using the prime number 2 is when you add 2 to itself. I'm probably missing some subtle logical step, and if so, please enlighten me.

In conclusion, wouldn't it be simpler to focus on parity instead of the primality of numbers to prove the conjecture?

 

[dodo]

Hi, fibonacci235,
there are many ways of adding two primes and obtaining an even number. The point of the conjecture is, can *all* even numbers (> 2) be written as the sum of two primes? In other words, given *any* arbitrary even number greater than 2, say, 123456, why is it mandatory that there exists a prime "p" such that 123456-p is also prime? That's the problem.

 

[fibonacci235]

Hi, fibonacci235,
there are many ways of adding two primes and obtaining an even number. The point of the conjecture is, can *all* even numbers (> 2) be written as the sum of two primes? In other words, given *any* arbitrary even number greater than 2, say, 123456, why is it mandatory that there exists a prime "p" such that 123456-p is also prime? That's the problem.

Ok, that does make more sense. Now, I understand why this is so hard to prove.

 

[Sievert]

Goldbach Conjecture is 2n = Prime (a+n)+ Prime (a-n), 1 is here prime


2 = (0+1)+(1-0)

4 = (1+2)+(2-1)

6 = (2+3)+(3-2)

8 = (3+4)+(4-3)

10= (2+5)+(5-2)

12= (1+6)+(6-1)

14 =(4+7)+(7-4)

16 =(3+8)+(8-3)

18 =(4+9)+(9-4)

20=(3+10)+(10-3)

22=(6+11)+(11-6)
.
.
.
2n=(a+n)+(n-a)

Proof:

(a+n) = 2n+(a-n)=2n-(n-a)

q.e.d.

 

[PAllen]

Forget proof, you don't even posit any argument at all that an 'a' with right properties exists. Listing that it does for N cases is irrelevant, unless you list all infinity cases. Certainly, there is no induction in your (total non) argument.

 

[HallsofIvy]

Goldbach Conjecture is 2n = Prime (a+n)+ Prime (a-n), 1 is here prime


2 = (0+1)+(1-0)

4 = (1+2)+(2-1)

6 = (2+3)+(3-2)

8 = (3+4)+(4-3)

10= (2+5)+(5-2)

12= (1+6)+(6-1)

14 =(4+7)+(7-4)

16 =(3+8)+(8-3)

18 =(4+9)+(9-4)

20=(3+10)+(10-3)

22=(6+11)+(11-6)
.
.
.
2n=(a+n)+(n-a)

Yes, there exist many numbers "a" that will fit here. What about "prime"?

Proof:

(a+n) = 2n+(a-n)=2n-(n-a)

q.e.d.[/QUOTE]
So, essentially, you are telling us that you do not know what a "proof" is.

 

[micromass]

After this silliness, I'm locking the thread.