Sufficient and Necessary Conditions for A(X)Q(X)=P(X) I am given two formal power series.
$$ P(X) = \sum_{n \ge r} p_nx^n $$
$$ Q(X) = \sum_{n \ge s} q_nx^n $$
With the conditions that $p_r \ne 0$ and $q_s \ne 0$
I'm looking for a necessary and sufficient condition for the equation $Q(X)A(X)=P(X)$ to have a solution, as well as to find out that when a solution exists whether or not it is unique.
So far I understand that $s=0$ is a sufficient condition, otherwise there would be a zero-constant term of $Q(X)$. I am having trouble finding a necessary one though. 
My initial guess is that its $ r \ne s $ but I'm having trouble working with power series to show this.
Any help is appreciated.
 A: First, we may pull out powers of $X$ in both power series to get series starting at $0$:
$$P(X) = \sum_{n\geq r} p_n X^n = X^r \cdot \sum_{n \geq 0} p_{r+n} X^n \\ Q(X) = \sum_{n\geq s} q_n X^n = X^s \cdot \sum_{n \geq 0} q_{s+n} X^n.$$
Now, to get a solution $A(X)$ to $Q(X) A(X) = P(X)$ we need to be able to find a solution $A(X) = P(X) / Q(X)$. In terms of the above series, this means
$$A(X) = \frac{P(X)}{Q(X)} = \frac{X^r \cdot \sum_{n \geq 0} p_{r+n} X^n}{X^s \cdot \sum_{n \geq 0} q_{s+n} X^n} = X^{r-s} \cdot \frac{\sum_{n \geq 0} p_{r+n} X^n}{\sum_{n \geq 0} q_{s+n} X^n}.$$
Since $q_s \neq 0$ the series $\sum_{n \geq 0} q_{s+n} X^n$ is invertible. So all we need is that the powers of $X$ are all non-negative. From $p_r \neq 0$ it follows that we need $r - s \geq 0$ or $\boxed{r \geq s}$. This condition is also sufficient: if $r \geq s$, then the unique solution $A(X)$ to the equation is given by
$$A(X) = X^{r-s} \cdot \left(\sum_{n \geq 0} p_{r+n} X^n\right) \cdot \left(\sum_{n \geq 0} q_{s+n} X^n\right)^{-1}.$$
If you want, you can even find somewhat explicit formulas for the coefficients $a_i$ of $A(X)$, i.e.,
$$\begin{align}
a_{n} &= 0 \qquad \qquad (n = 0, \ldots, r-s-1) \\
a_{r-s} &= \frac{p_r}{q_s} \\ 
a_{r-s+1} &= \frac{p_r}{q_s}\left(1 - \frac{q_{s+1}}{q_s}\right) \\ 
a_{r-s+2} &= \frac{p_r}{q_s}\left(1 - \frac{q_{s+1}}{q_s} - \frac{q_{s+2}}{q_s} + \frac{q_{s+1}^2}{q_s^2}\right) \\
\vdots \quad &= \quad \vdots
\end{align}$$
Note that these terms only contain divisions by $q_s$, which is ok, since we know $q_s \neq 0$.
