
Re: Vindication of Goldbach's conjecture
Posted:
Jun 21, 2012 1:50 PM


El jueves, 14 de junio de 2012 10:24:12 UTC4:30, Count Dracula escribió: > On Jun 14, 5:28 am, mluttgens <lutt...@gmail.com> wrote: > > In P, the number 26 = 3+23, 5+21, 7+19, 9+17, 11+15, 13+13, 15+11, > > 17+9, 19+7, 21+5 and 23+3, all the terms coming from the successive > > uneven numbers of S and S' from 3 to 23, where 23 = 26  3. > > As S and S' contain all the prime numbers inferior or equal to 263 , > > the even number 26 is necessarily the sum of two primes. > > Let?s note that it is not necessary to indentify those primes. Knowing > > that they exist in S and S? is enough.. > > The argument you give here is fallacious. The sets S and S' also > contain odd > numbers that are not prime. This makes it necessary to identify the > particular > primes and it is not enough to know that primes exist in S and S'. In > your > example for 26, it is not possible to use the sums 5+21, 9+17, 11+15, > 15+11, > 17+9, 21+5 since each contains a composite integer. My earlier > question to you was: what if there is an integer for which _all_ the > sums > are inadmissable, not just some as in the case of 26? Nothing in your > argument rules out this possibility. > > If a counterexample exists it will have to be larger that 10^17 > because > computer verifications have been performed up to about that value. A > correct > proof is highly likely to require some highly technical innovations. > > Count Dracula
The true difficult with Goldbach's Conjecture is this: Given a even number 2k to warrant that the pairing of the two arithmetic progressions 2n + 1 and 2k  (2n+1) will produce nor less than one coincidence of two primes. Example. Given the even number 98. The two progressions are: 3, 5, 7, 9, 11, 13, 15, 17,.........95 95, 93, 91, 89, 87, 85, 83, 81,......... 3
Who guarantee that there will be a column with two primes?
Ludovicus

