TITLE: sums of square free numbers , is this conjecture equivalent to goldbach's conjecture?
QUESTION [18 upvotes]: As one can notice every integer greater than $1$ is a sum of two squarefree numbers.(numbers that are not divided by some prime square  power). Can we prove that?
Edit: Can we have bounds for the length of these numbers? (meaning the number of the primes that divide it)
Chen's theorem asserts that for large enough even  numbers the length (2,1) is enough Goldbach's conjecture says that (1,1) would be enough too.
And one conjecture: every odd number can be written as a sum of two squarefree numbers of length at most (2,1) (meaning as a sum of a prime and a double of a prime or a sum of a prime plus 2 or as a sum of 1 plus a double of a prime) 
Questions

do i really need the prime plus 2 or the 1 plus the double of a prime in oredr to have all the odd numbers?I think i do not need them but can we prove that??
What is the relation of this conjecture to Goldbach's conjecture? does the one implies the other?

EDIT Searching wikipedia i realised that this is a well-known conjecture, for more details see http://en.wikipedia.org/wiki/Lemoine%27s_conjecture

REPLY [9 votes]: It is known that the Schnirelmann Density of square free numbers is $\displaystyle \frac{53}{88}$ for instance, as mentioned here: http://www.jstor.org/pss/2040089. Note this is different from the natural density which is known to be $\displaystyle \frac{6}{\pi^2}$.
This implies that for any $n$, the number of square free numbers $\displaystyle \le n$ is at least $\displaystyle \frac{53n}{88}$. Since $\displaystyle \frac{53}{88} \gt \frac{1}{2}$, this implies that $n+1$ can be written as the sum of exactly two squarefree numbers.
btw, your title does not match the question.
Additive Basis means that you also include $\displaystyle 0$ in the set which you seem to exclude by saying two instead of at most two.