The Goldbach Conjecture (an unsolved problem)

The Goldbach conjecture (GBC) has been around in one form or another since the mid-1700s. 
It is easily stated, but has yet to be proven...

The statement:
"Every even number is the sum of two prime numbers"

Is what is known as the "strong Goldbach conjecture".  This was noted in correspondence between two very well known mathematicians of the time, Leonhard Euler and Christian Goldbach.

I am always drawn to study unsolved problems in mathematics because in the past, solutions to such problems have given insight to many other fields, and even require invention of new techniques and models to solve, which advances the science of mathematics even further. 

So here is an introduction to my research on the unsolved Goldbach Conjecture.

From the beginning,
"Every even number is the sum of two prime numbers"

So letting each even number be represented by 2n, we can rewrite this statement as:

2n = p + q    (where p and q are prime numbers less than 2n)

now, it turns out that you can do a little algebraic manipulation to this equation to obtain:

n + n = p + q

which then yields:

n - p = q - n

Now, closely consider this equation...  What this means is that if GBC is true, there is an underlying "symmetry" in the primes.  How?

This arrangement of the equation lets one see that p and q are equidistant from N (which is half of 2n, obviously).  This picture makes it easy to imagine:

So now, what's the big deal?  Why should somebody try rearranging 2n = p + q at all?

IF the GBC is true, this means that for every N, there exist 2 primes, p and q, such that:

1. p < n
2. n < q < 2n    ------>  (see below how this can be n < q < 2n-2 by Bertrand's postulate)
3. neither p or q divide 2n
4. d(p,n) = d(q,n)  (distance function d(x,y)...)

So my goal is to try to force a contradiction in the following manner:

We know that p exists, (any arbitrary prime less than N).
According to Bertrand's postulate,
(if n > 3 is an integer, then there always exists at least one prime p with n < p < 2n − 2)
We know that there must exist some prime q such that n < q < 2n-2
So we can close our bounds on condition 2 above just a little further as indicated above.

It is only left to prove that there exists one prime Q greater than n, less than 2n-2, such that Q-N = N-P, and P does not divide 2n.

IF it can be shown that no such Q can exist, we have contradicted Bertrand's postulate, and therefore GBC must be correct.

But that's the hard part.  I still have yet to find a successful route of attack to force this contradiction. 

Some things to consider if you're going to try to attack it:

1.  Non-prime numbers can be equidistant from prime numbers too...  i.e. 20 = 11+9   but 9 is not prime... How can one guarantee that at least one prime Q (where n < Q < 2n-2) is the same distance from N as a prime less than N?

2.  For each prime P < N, there exists a number N + (N - p) = 2N - P = Q' 
Is there an effective way to check primality of this Q' in each case where P is prime and P < N?  We know that at least one prime number Q must exist in (n, 2n-2), but how do we know which P' < N it is associated with?

From this line of attack I can see two possible ways of proving GBC:
Firstly, by proving the existence of Q for every N and 2N (induction? pigeon-hole principle?).
Or secondly, by assuming Q does not exist and forcing a contradiction to Bertrand's postulate saying that there must be a prime in (n, 2n-2). 

So that's a good introduction so far.  Still no solution as of yet, but I'll keep working on it and see where it goes!