As one can easily prove http://math.stackexchange.com/questions/20564/sums-of-square-free-numbers-is-this-conjecture-equivalent-to-goldbachs-conjec every integer greater than $1$ is a sum of two squarefree numbers. Can we have bounds for the length of these numbers? I write $(n,m)$ to denote the sum of a squarefree number of n prime factors to one of m prime factors, when $n$ or $m$ $=0$ then i mean that the summand is $1$. Chenn'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. >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 >1 Is there any easy counterexample? >2 do i really need the prime plus 2 or the 1 plus the double of a prime in order to have all the odd numbers? It seems too difficult to me to prove that i do not need them. >3 What is the relation of this conjecture to Goldbach's conjecture? does the one implies the other? I apologise for the elementary style of my question , i think that this conjecture is well known but i haven't met it. If it is well known maybe it is known the relation to the Goldbach Conjecture. Maybe i miss something obvious... NOTE: From the second question we have one new conjecture >CONJECTURE:Every prime $p$ is $p=p1+2(p2-1)$ and $p=p3+1/2(p4-1)$ for some primes $p1,p2,p3,p4$ . Of course someone could ask many questions about these form for the consecutive primes ,etc. EDIT:after asking this question i found this related article en.wikipedia.org/wiki/Lemoine%27s_conjecture at wikipedia.