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.



The sum of integers being a bijection
To comment on Qiaochu's answer, one can show that all such factorizations come from [mixed radix](http://en.wikipedia.org/wiki/Mixed_radix) representations (different bases, factorial base etc.). That is if $$\frac{1}{1-x}=P(x)Q(x)$$ then there must be a sequence $1=a\_0\le a\_1 \le a\_2\le\cdots$ so that $a\_i$ divides $a\_{i+1}$ and disjoint subsets $A,B$ with $\mathbb N=A\cup B$, so that $$P(x)=\prod\_{i\in A}\frac{1-x^{a\_{i+1}}}{1-x^{a\_i}},Q(x)=\prod\_{i\in B}\frac{1-x^{a\_{i+1}}}{1-x^{a\_i}}.$$
The proof is simple, suppose $P(x)=1+x+\cdots +x^{a\_1-1}+\cdots$ then $Q(x)=Q\_1(x^{a\_1})$ and $P(x)=\frac{x^{a\_1}-1}{x-1}P\_1(x)$. Then we proceed by induction.


