In the post dated October 3, 2008 we showed the validity of Goldbach's conjecture in cases where there were only two primes less of it first with it.
In the case of the first three we can reach an even more convenient, however, having to consider the restriction of n even.
Then, let n be an even integer and p, q re the first three children of n (in ascending order) and the first with it.
Let P, Q and R in the corresponding full symmetry with respect to n, or is
P = 2n - p = 2n
Q - R
q = 2n - r
Suppose by contradiction that P, Q and R are composed and not before.
to comments made in the post of October 3 r> n / 2 and thus r can not belong to the factorization of P. It follows that P can only be powers of q.
So, P = pot (q)
Q for its part can be produced by some power of p for r (which implies very special case p = 3 ... but possible).
If Q then n = p * r - q = Q - n is a number less than n that may not have the factors nor the factors of n (otherwise they would also q), nor the factors of Q (otherwise they would even n), nor does it have the same q (which in this case should also be a factor of n ).
born on the absurd as n - q is a number less than n which has no factor for any prime less than n.
If Q = pot (p) then R = pot (p) * pot (q) and the speech made shortly before for n - q it would do the same for n - r, which is a number less than n is not divisible by any No less than the first
The proof of the case of the first three is complete.
The restriction to the case of n even is necessary because, without it, n - q (on - r as the case may be considered) could easily be a power of 2, and then collapse on the absurd.
Obviously, this procedure can be adapted to the case of two primes (shown in the post of 3 October with another procedure), only that it should add the restriction to n even.
In general, therefore, the proof of Goldbach for even n could be based on the idea that, whatever the number of primes less than it first with it, for at least one of them necessarily symmetrical with respect to n should consider all of these early (except for its clearly symmetrical) in its factorization. And
'That's what we try to work with.
