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Re: questions about a "proof" of the Goldbach Conjecture.
Posted:
Nov 28, 2012 11:24 PM


On Wednesday, November 28, 2012 3:36:51 PM UTC8, locorodri...@gmail.com wrote: > In www.primepuzzles.net/puzzles/puzz_088.htm you can see the minimum "Goldbach sets" than can produce the first 50,100,250 even numbers. > > It interesting to know that there are 24 odd primes available for produce the 50 even numbers < 100 but 13 are sufficient. Same, there are 45 odd primes available for the 100 even numbers < 200 but 19 are sufficient. > > Ludovicus
Thanks. That was great.
I'm just beginning to explore this stuff. The following is quite vague. I've been thinking about an approach that goes something like this:
Suppose 3 is met by a composite number. The absolute maximum factor is achieved when one of the factors is 3. That is n is a multiple of 3 so 2n3 is divisible by 3. The other factor is 2n/3  1. Now that is just the set of odd numbers greater than 3. Suppose that number is a prime number. Its equadistant number is 4n/3 + 1. That number won't be divisible by 3 because 2n/3  1 is a prime. Since the idea is to attempt to prove Goldbach's conjecture false so that number has to be a composite. The smallest factor is 5....
Now suppose 2n/3  1 isn't a prime. The next smallest facotr for 2n3 is 5. This only happens for n=(5x+3)/2 where x>1 (n has to be greater than 5 for 2n5 to be > 5) and the sequence is again the set of odd number > 2 (3 is allowed this time because it hasn't been handled)...
What I'm trying to do is set up a race condition where I can't avoid equadistant primes no matter how big I make n. Maybe that's how Chen proved all even numbers > 3 can be expresses as the sum of two primes or a prime and semiprime.



