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Let us in front of the conjecture Goldbach and try to understand, from a purely algorithmic point of view, what is the question. Suppose we have an even number, denoted by 2n such that p and q that do not exist the first two added together give just 2n. Imagine that the Goldbach conjecture is verified for a given 2n.

say this is like saying that n is not in the middle of the first two, namely that the symmetry with respect to n, for every prime less than n is a number made the other side.

The point is that, in this case, each of these numbers can only be made by the same compound that the first child of n. Indeed, a prime greater than n, multiplied by any other first (even doubled) makes a greater number of 2n and clearly the numbers in question are between them 2n. In addition, these compounds can not be among their prime factors which are also factors of n, for obvious reasons.

So everything is summarized in the following way:

If for any even number p = 2n Goldbach not true, then the primes less than n (and first with n!) Must compose, combine only with each other, all the numbers the results in the symmetry with respect to n.

To be more precise, ik first under its first coat should combine it with a producer ik places equidistant from them n the other hand, those who give them added up to 2n.

The conjecture then says that those k may not hold each of the first k places that are symmetrical with respect to at least one of the jump ..... No!

speech is a very difficult course to be pursued in an attempt to extract a proof from this account. Just think how many possible combinations of k prime numbers, how difficult it is to define the restrictions on those combinations in order to bring these products in the range of its 2n ...

A vision that makes us understand how other results (to Chen, for example) do not fall completely on that interpretation.

This is not to assess the likelihood of achieving results apparently near you .... Goldbach question is whether you can show that given any number n, ik first child early with it and it will not be able to cover all ik places that are symmetrical with respect to n.