TITLE: The sum of integers being a bijection
QUESTION [19 upvotes]: What are the pairs $(P,Q)$ of subsets of $\mathbb N$ for which the map
\begin{eqnarray*}
P\times Q & \rightarrow & {\mathbb N} \\\\
(p,q) & \mapsto & p+q
\end{eqnarray*}
is a bijection ?
Obvious examples are $P=\mathbb N$ with $Q=\{0\}$, or $P=2\mathbb N$  with $Q=\{0,1\}$. Are there others ?
This question is related to a puzzle given in EMISSARY (fall 2010), asking to find infinite series $f(x)$ and $g(x)$ with coefficients $0$ and $1$, whose product equals $\frac{1}{1-x}$. I suspect that the word infinite was written on purpose, and therefore $P$ and $Q$ must be infinite.
Later. After the answers, I understand that one can find a sequence $(P_j)_{j\ge0}$ of subsets of $\mathbb N$ with $0\in P_j$, such that every $n\in\mathbb N$ writes $\sum_{j\ge0}p_j$ with $p_j\in P_j$, in a unique way. Of course, all but finitely many $p_j$'s are zeros. Now, I feel dumb, because this follows for instance from the writing of integers in some basis.

REPLY [4 votes]: If you accept that 0 is not a natural number, then there is a very simple answer to your question: take $P$ to be all numbers whose expansions base 4 contain only digits 0 and 1 and $Q$ to contain only digits 0 and 2. Then $P\cap Q=\{0\}$, which we have boldly excluded. 
Also, both sets have the lowest possible asymptotic density of order $1/\sqrt n$, which is kinda nice.