# Goldbach conjecture proof

by praharmitra
Tags: conjecture, goldbach, proof
 P: 3 I can prove the goldbach conjecture with a simple method but first I prove square of any prime -1 can be divided by 12 primexprime -1 = 12k k is an integer .
P: 177
 Quote by metheor I can prove the goldbach conjecture with a simple method but first I prove square of any prime -1 can be divided by 12 primexprime -1 = 12k k is an integer .
This is not really hard (and works only for primes greater than 3):

Let p be a prime number.
p$^{2}$-1 = (p+1)(p-1)

Since p is prime, it is not divisible by 2 (unless p is 2 but 2$^{2}$-1=3 is not 12k).
So p = 2n+1 for some positive integer n.

Any integer number can be written in the form 3k, 3k+1 or 3k+2.

Try n = 3k. Then p = 6k+1

Try n = 3k+1. Then p = 6k+3 = 3(2k+1) = 3m is impossible since p is prime. (unless k = 0 then p = 3 is prime, but 3$^{2}$-1 = 8 is not 12k)

Try n = 3k+2. Then p = 6k+5.

We see that p must be of the form 6k+1 or 6k+5.
Then (p+1)(p-1) = (6k+2)(6k) = 12(3k+1)k = 12m
or (p+1)(p-1) = (6k+6)(6k+4) = 12(k+1)(3k+2) = 12m

Therefore, for any prime number p greater than 3, p$^{2}$-1 is divisible by 12.
 P: 94 The complication of all of this: is proving there is no space between primes through infinity: such that for the binary part, the last prime before the spacing is such that it added to itself ( or times 2): is greater than or equal to the spacing been it and the next prime. Of course this comes down to prime numbers distribution, and the maximum spacing between primes.
 P: 69 FYI, someone submitted a proof of Goldbach Conjecture and Riemann's Hypothesis in Arxiv. It's under General Mathematics section.
 P: 94 ode_to_joy got a link? . .
 P: 124 http://arxiv.org/ftp/math/papers/0005/0005185.pdf Is this the paper you were talking about ode_to_joy?
 P: 69 http://arxiv.org/abs/1110.3465 goldbach conjecture http://arxiv.org/abs/1110.2952 riemann hypothesis
Mentor
P: 18,244
 Quote by ode_to_joy http://arxiv.org/abs/1110.3465 goldbach conjecture
On the first look, there seems to be some kind of mistake. He proves the Goldbach conjencture for big n. Thus if $n\rightarrow \infty$, then it is true. But he never states what "big" exactly is. He reasons that it has been numerically checked for n going until $\sim 10^{18}$ and that this large so it holds for larger integers. This is not enough.
PF Gold
P: 5,059
 Quote by CRGreathouse The binary implies the ternary, that's why they're both considered versions of the same problem. But the binary version is much harder.
I'm confused. Suppose the ternary is true. Then, for an even integer, you have that it is the sum of 3 primes. For n>6, one of these must be 2, the other two being odd primes. So then, for every even n>6, the ternary implies n-2 satisfies the binary. Isn't that enough to say that ternary implies binary?
 Sci Advisor PF Gold P: 5,059 According to: http://mathworld.wolfram.com/GoldbachConjecture.html what Vinogradov proved for sufficiently large odd n was the ternary form of odd numbers. Restricted to odd numbers only, the ternary does not imply the binary. However, it is correct, as I argued above, that the ternary for all numbers implies the binary; and the binary trivially implies the ternary.
 P: 48 the proof of the riemann hypothesis is pretty interesting, although at first glance I'm not sure about one of his derivatives, and there seems to be ALOT of assumptions...
 Mentor P: 18,244 From the first glance, the proof of the Riemann hypothesis suffers from the same problem. He checks it for the first 23 trillion values and he proves it for sufficiently large numbers. That is not enough...
 P: 15 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?
 P: 688 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.
P: 15
 Quote by 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.
Ok, that does make more sense. Now, I understand why this is so hard to prove.
 P: 1 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.
 Sci Advisor PF Gold P: 5,059 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.
Math
Emeritus
Thanks
PF Gold
P: 39,502
 Quote by 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)
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.

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