Approaches to a Goldbach’s conjecture

I forgot how long this question has been lingering in my mind, it is the famous Goldbach’s conjecture that every even number greater than six is the sum of 2 prime number. My suggested direction of thinking is: A. Since it only require the existence of at least one pair of prime number adding up to an even number greater than 6. i.e. We only need to concern ourselves with the distribution of prime number at n (where 2n is the targeted number). I remember that from

THE PRIME NUMBER THEOREM

           lim   ( pi(x).log(x)/x )  =  1

saying that the number of prime number is exponentially related to the total number of numbers. Then this question would become a question of probability. Then we can break this question for 2 parts: First the smaller number, then for the even numbers greater than an amount.

B. Since it only require the existence of at least one pair of prime number adding up to an even number greater than 6. i.e. We need to find one pair of prime number that is co-prime to each other adding up to 2k. i.e.

(A) k+n, k-n is co-prime;
(B) k-n is prime;
(C) k+n is indivisible by any number between k+n and k-n.

Part C could be quicken by the facts that the greatest divisor of any number must be equal to or lesser than its square root.

1. k-n, k+n is co-prime iff 2n and k-n are co-prime. Now since k-n, k+n is odd and 2n is an even number, therefore, n either must be an odd number that is a factor of both k-n and k+n, or n=2(n1) which n1 is an odd number that is a factor of both k-n and k+n. To summarize, let the m=2n, we have

M=2^S*factor of both k-n, k+n

S could be ranging from 1 to log (k-n), which (k-n), 2n and k+n share a common factor, it follows that
[(k+n)-2n and (k-n)-2n] or [(k+n)-2n and 2n-(k-n)] must also share a common factor, the later case is unlikely since we want √(k+n)≈√(k-n), since we assume here it is impossible for √(k+n)≈√(k-n) if 2k>k-n.

So we have k-n and k-3n share a common factor, i.e. (k-n)/(k-3n)=A/B, where A, B are both integer.
i.e. we have Bk-Bn=Ak-3An
or (B-A)k=(B-3A)n
or k/n=(B-3A)/(B-A)
Given n and k are real number, we could thus solve for B and A that satisfy this condition, then exclude this case in the comparison.

Okay, now assume that we have √(k+n)≈√(k-n) and 2k>(k-n),
we use similar reasoning as above, thus we have
k/n=(B-3A)/(A-B)
Given n and k are real number, we could thus solve for B and A that satisfy this condition, then exclude this case in the comparison.

C. Somehow related to the way number expressed in binary system, all odd and prime number must have 1 in its last digit. Given in order for a odd number to be prime, it must be indivisible by another odd number other than 1(i.e. those ending with 1). What is the characteristic of odd binary number indivisible by all odd number (i.e. ending with 1)?

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