Goldbach's Conjecture - proposed proof

I'd really would appreciate it if someone would take a look at this and tell me where I am wrong or have over-simplified things. It's short and a straight-forward approach. (I hope the 2 figures transfer okay to this post. I'll upload a file with them just in case.)

Goldbach's Conjecture: Every 2n (n>1) is a sum of two primes.

Introduction: Modular arithmetic enables a useful way to display all 2n’s with their integer pairs (p1,p2) totaling 2n. If we assume the existence of a 2n that has no prime pairs a contradiction results.

Proof: If we pick any n > 1 then some primes p1 and p2 are easy to identify, such that p1 + p2 = 2n. For example, if n=2 then 2n=4; p1=2=p2; and. 2+2=4 is the solution. Also, any pair of twin primes (e.g., 3,5, 11,13, 17,19, or 41,43) is a trivial case where primes p1 = n + 1 and p2 = n – 1. So, p1+p2= n+1+n-1 = 2n.

Similarly for n=3 then 3n=6; we have: 3 mod 3 = 0; 4 mod 3 = 1; 5 mod 3 = 2; and

6 mod 3 = 3. This is the complete series for mod 3 and the only pair that adds to 2n is 3,3. So, p1 = p2 = 3 is the solution to p1+p2=2n.

For 4 = 2n, (and all n<1), The pairs of each modulus n can be viewed as a circle (and all higher integers are concurrent with the elements of these sets).

Four Six Eight Ten Twelve

0 0 0 0 0

1 3 1 5 1 7 1 9 1 11

2 2 4 2 6 2 8 2 10

3 3 5 3 7 3 9

4 4 6 4 8

5 5 7

6
Figure 1 – 2n prime-pair solutions (highlighted)

Within each “circle”, the sum of every pair on a row = 2n (and 2 times the middle element (at the bottom) or 2 x (2n/2) = 2n). So, for all n if 2n/2 is prime then (n,n) is a prime pair solution (i.e., 2n = n + n = p1 + p2.)

So, while 2n/2 is a solution for n=2 and 3, Eight has a solution (3,5); Ten has two solutions (5,5) and (3,7); and Twelve has a solution (5,7). The question is: Does every 2n have at least one prime pair solution (i.e., p1+p2=2n)? (Note: Figure 1 also suggests why ‘1’ was historically considered a prime.)

Since 1 is not a prime, a (1,x) pair will never be a solution. Also, since 2 (a prime) is always paired with an even number, a (2,x) pair will never be a solution (except 4=2,2 ). In fact, 2, 4, 6, 8, … all even numbers will be paired to even numbers. So, no (2n,2m) pair will ever be a solution (for m, n >1). Again 15, 18, 21, … 3x), paired with any other number will never be a solution. Okay. You see the pattern, any composite paired with another number (prime or not) will never be a solution, because by definition p1 and p2 must both be prime numbers.

If there exists an n such that 2n has no prime pair p1+ p2 =2n, then every pair for that modulus n has at least one composite number (i.e., no pair has 2 primes) and each pair summed = 2n. Therefore, any of the pairs with a prime for the first element must not have a prime for the second element of the pair. (Otherwise, that pair would be a solution.)

For any n …consider just the pairs which have a prime p1 for the first element:

p1 + p2 = 2n.

1 + (2n – 1) = 2n

2 + (2n – 2) = 2n

3 + (2n – 3) = 2n

… up to (n + n) = 2n (However, n may or may not be prime.)

So, assume for the above pairs for some x=2n when the first element is prime that none of the second elements are a prime. So, … 2n-2 is not prime, and 2n-3 is not prime, and 2n-5 is not prime, and 2n-7 is not prime, and
… 2n-n is not prime (but only when n is not prime)

Then if for at least one x, 1 < x <= n, all of the following are not prime:
2x-2, 2x-3, 2x-5, 2x-7, 2x-11, …, (2x-n) = n.

For example, for each row ‘n’ in Figure 2 the non-primes include all values in that row and all rows above it.…

n 2x-2 2x-3 2x-5 2x-7 2x-11 2x-13 2x-17 2x-19 2x-13 2x-29 2x-31

2 2 1

3 4 3 1

4 6 5 3 1

5 8 7 5 3

6 10 9 7 5 1

7 12 11 9 7 3 1

8 14 13 11 9 5 3

9 16 15 13 11 7 5 1

10 18 17 15 13 9 7 3 1

11 20 19 17 15 11 9 5 3

12 22 21 19 17 13 11 7 5 1

13 24 23 21 19 15 13 9 7 3

14 26 25 23 21 17 15 11 9 5

15 28 27 25 23 19 17 13 11 7 1

16 30 29 27 25 21 19 15 13 9 3 1

17 32 31 29 27 23 21 17 15 11 5 3

Figure 2 – Table of “non-primes” if a 2n exists.

In fact, just the first two columns cover all integers >1. But, if these second elements of all the first-prime pairs are not prime then all n, from 2 to n are not prime. So, the first element of each pair isn’t prime. This is a contradiction. So, no ‘x’ exists for any n > 1.

Therefore, for any n > 1 there exist a 2n with a prime pair p1, p2 such that p1+p2 = 2n. Q.E.D.

 

I found my error (as I suspected). When I conclude from Figure 2 that, "the non-primes include all values in that row and all rows above it.…." I have not backed up that statement with any proof.

I will have to dig a little deeper.