"]Lemma 1.$n\geq 6$Let $p_{1}<\dots<p_{k}$ be all primes not greater than $n$ then $\sum_{i=1}^{k}\frac{1}{p_{i}^{2}}<\frac{n-1}{2n}$ proof : \[\sum_{i=1}^{k}\frac{1}{p_{i}^{2}}=\frac{1}{4}+\frac{1}{9}+\sum_{i=3}^{k}\frac{1}{p_{i}^{2}}< \frac{1}{4}+\frac{1}{9}+\sum_{l=5}^{n}\frac{1}{l^{2}}=\frac{1}{4}+\frac{1}{9}+\frac12\sum_{l=5}^{n}{(\frac{1}{l-1}-\frac{1}{l+1})}=\frac14+\frac19+\frac18-\frac{1}{n+1}<\frac{1}{2}-\frac{1}{2n}=\frac{n-1}{2n}\] so there at most $\sum_{p\in P}{ \frac{n-1}{p^{2}}}<\frac{n-1}{2}$integer less than $n$ not being square-free integer.so there must exist $k$ such that $k$ and $n-k$ both square-free.(by a famous thoerem with the first letter "P",I forgot its name )

Probably the stupidest proof possible: We can check that it holds for $n \le 10^4$. Now, let $S$ be the set of all squarefree integers, except for the primes larger than $10^4$. Then by the fact that the Schnirelmann density of the set of squarefree integers equals $\dfrac{53}{88}$ $^{[1]}$ and some decent estimate on the prime counting function$^{[2]}$, we have that the Schnirelmann density of $S$ must be larger than $\dfrac{1}{2}$. By Mann's Theorem$^{[3]}$ we now have that every positive integer can be written as sum of at most $2$ elements of $S$. In particular, every prime number can be written as sum of $2$ elements of $S$, and every integer that is not squarefree can be written as sum of $2$ elements of $S$. All there is now left, is proving the theorem for composite squarefree numbers; $n = pq = (p_1 + p_2)q = p_1q + p_2q$, where $p$ is the smallest prime dividing $n$ and $p_1, p_2$ are squarefree integers. $^{[1]}$ http://www.jstor.org/pss/2034736 $^{[2]}$ http://en.wikipedia.org/wiki/Prime-counting_function#Inequalities $^{[3]}$ http://mathworld.wolfram.com/MannsTheorem.html

We can verify it manually for small values of $n\ge 2$, indeed: 2=1+1 3=2+1 4=3+1 5=2+3 6=5+1 7=5+2 8=5+3 9=7+2 10=7+3 11=10+1 12=10+2 13=10+3 Suppose then that $n\ge 14$. We claim that if $S:=\{n \in \mathbb{N}: \mu(n) \neq 0\}$ then $|S \cap [1,n]| > \frac{n}{2}$. Indeed, if it's true then there exists for pigeonhole a squarefree integer $x \le \frac{n}{2}$ such that $n-x$ is squarefree too. But this is true since: \[ |(\mathbb{N}\setminus S) \cap [1,n]| \le \sum_{i=1}^{\pi(n)}{\left\lfloor np_i^{-2} \right\rfloor} < n\sum_{i=1}^{\pi(n)}{p_i^{-2}} < \] \[ < n \left(2^{-2}+3^{-2}+5^{-2}+7^{-2}+11^{-2}+13^{-2}+\sum_{i=17}^{n}{i^{-2}}\right) < \] \[ < n \left[2^{-2}+3^{-2}+5^{-2}+7^{-2}+11^{-2}+13^{-2}+\sum_{i=17}^n\left(\frac{1}{i-1}-\frac{1}{i}\right)\right]\] \[ < n\left(2^{-2}+3^{-2}+5^{-2}+7^{-2}+11^{-2}+13^{-2}+16^{-1} \right) < \frac{n}{2}\].[]