Let $A=1+\sum\_{n=1}^\infty \alpha\_nx^n\in\mathbb Z[[x]]$ and $B=\frac{1}{A}=1+\sum\_{n=1}^\infty\beta\_n x^n$ two mutually inverse power series
having bounded integral coefficients (ie. $\vert \alpha\_n\vert,\vert \beta\_n\vert<C$ for some constant $C$ and for all $n$). 


Examples are given by $A=\frac{P}{Q}$ where $P$ and $Q$ are both finite products of cyclotomic polynomials having only simple roots.


Are there other, exotic, examples?


More generally, consider again $A=1+\sum\_{n=1}^\infty \alpha\_nx^n\in\mathbb Z[[x]]$ with inverse $B=\frac{1}{A}=1+\sum\_{n=1}^\infty\beta\_n x^n$ and require that the integral coefficients $\alpha\_n,\beta\_n$ have at most polynomial growth (ie. there exists a constant $C$ such that $\vert\alpha\_n\vert,\vert\beta\_n\vert<Cn^C+C$ for all $n$). 


Examples are now arbitrary rational fractions $A=\frac{P}{Q}$ involving only cyclotomic polynomials. Again, I know of no other examples. Do they exist?


Added: This question is closely linked to [The sum of integers being a bijection](https://mathoverflow.net/questions/50798/the-sum-of-integers-being-a-bijection/50803#50803), see Zaimi's comment after Venkataramana's anwser.



Which power series have bounded integral coefficients and have an inverse given by a series having bounded integral coefficients
Consider the set $S$ of nonnegative integers whose $2$-adic expansion involves only square powers of two (e.g. $n=1+16$, and $n=2^{25}+2^{49}+2^{64}$ belong to $S$). Let $T$ be the set of integers whose $2$-adic expansion never involves any square power of two. Then, $(S\cup\{0\})+(T\cup\{0\})$ represents each positive integer only once. Write $f(x)=\sum \_{k\in S\cup\{0\}}x^k$ and $g(x)=\sum \_{k\in T\cup\{0\}}x^k$. Clearly,
$f(x)g(x)=\frac{1}{1-x}$. This shows that $f(x)$ and $g(x)(1-x)$ have bounded integer coefficients, but are not ratios of products of cyclotomic polynomials. 


