Many attempts to prove than any even number (>3) is the sum of two primes try to find a contradiction, but I believe a direct proof is more likely to be the right approach. I have attached such an approach. It uses mod(6) to show for all 2n, (n>1) there are only three patterns of eligible pairs. And, these pairs can be examined to determine a prime-limit above which a prime pair must exist and below which a 2n could have no prime pairs.
No real surprise - but no 2n (n>1) can have a number of primes below its prime-limit. In fact, this proof shows that all 2n (n>6) must have TWO prime pairs.
Note: the "discussion" at the end is optional reading. It is not necessary for the proof, but it was useful to me in getting a count of composites for any (large 2n). I developed this approach, as a substitute for p-series arithmetic progressions, by identifying composites by their lowest common prime.
(See attached file.)
No real surprise - but no 2n (n>1) can have a number of primes below its prime-limit. In fact, this proof shows that all 2n (n>6) must have TWO prime pairs.
Note: the "discussion" at the end is optional reading. It is not necessary for the proof, but it was useful to me in getting a count of composites for any (large 2n). I developed this approach, as a substitute for p-series arithmetic progressions, by identifying composites by their lowest common prime.
(See attached file.)