sums of square free numbers , is this conjecture equivalent to goldbach's conjecture? 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
 A: A majority of numbers are squarefree.  More precisely, you can show that there is a constant $c<\frac{1}{2}$ such that for each positive integer $n$, less than $cn$ of the numbers $1$ through $n$ are nonsquarefree.  
One way to get a rough bound for $c$ is to observe that at most $n/4$ of the numbers from $1$ to $n$ are divisible by $4$, at most $n/9$ divisible by $9$, at most $n/25$ divisible by $25$, etc.  If we add up all these inequalities, there are less than $(1/4+1/9+1/25+\cdots)n\lt .46n$ nonsquarefree numbers from $1$ to $n$.  The sum $\displaystyle{\sum_{p\text{ prime}}\frac{1}{p^2}=.4522474200410654985065...}$ is the prime zeta function evaluated at $2$.  
So you can take $c=.46$.  For $n\geq 13$, $\lfloor n/2\rfloor \gt .46n$, so not all of the $\lfloor n/2\rfloor$ pairwise disjoint sets $\{1,n-1\},\{2,n-2\},\ldots,\{\lfloor n/2\rfloor,n-\lfloor n/2\rfloor\}$ can contain a nonsquarefree number.
A: 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.
A: According to this paper, the number of ways to represent a positive integer $N$ as a sum of a prime and a $k$-free integer is asymptotic to
$$
{N\over\log N}\prod_{p\nmid N}\left(1-{1\over p^{k-1}(p-1)}\right)
$$
as $N\to+\infty$.
By looking into the details of Chen's argument (see last chapter of Halberstam & Richert's Sieve Methods published in 1974), one can observe what Chen actually proved is that for a given large even integer $N$, the number of primes $p\le N$ such that
$$
N-p=\begin{cases}
p_1p_2 & N^{1/10}<p_1<p_2<N & p_1,p_2\text{ prime} \\
p_1 & N^{1/10}<p_1<N & p_1\text{ prime}
\end{cases}
$$
is greater than
$$
{0.67N\over\log^2N}\prod_{\substack{p|N\\p>2}}{p-1\over p-2}\prod_{p>2}\left(1-{1\over(p-1)^2}\right).
$$
I hope these two results can answer your question.
