Partition $\mathbb{N}_0$ into two sets where each set must have the same number of ways to add to m For any set that is a subset of the natural numbers $S \subseteq \mathbb{N}_0$ we may consider the number of possible ways to form $m$ by adding two distinct elements in $S$. Call this $c_S(m)$. For example, if $S = \{1, 2, 3, 4, 5\}$ there are 2 ways to form $6$ : $1 + 5$ and $2 + 4$. So $c_S(6) = 2$.
Is there a way to partition $\mathbb{N}_0$ into two sets $A$, $B$ such that $c_A(m) = c_B(m)$ for all $m$?
I played around with this and tried splitting the sets into the evens and the odds and it seems to work for any number that is not of the form $4k+2$, so I'm beginning to think that it's not possible but I don't know how to show it.

Lambek and Moser in "On some two way classifications of integers" proved that there is exactly one such partition: into evil and odious numbers (numbers that have even and odd 1s in binary respectively). It uses generating function, as Thomas suggested in comments.
(the way I found this paper is generating few elements of each set - it's easy to do by hand, as if we put $0$ to $A$, we have to put $1$ and $2$ in $B$, then $3$ in $A$ and so on, and checked if OEIS contains corresponding sequence (it does, and also contains statement from the question and a link to the paper above))
