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
Best Answer
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.