<s>[INST] For the polynomial $P(x) = x^{2004}$, finding $Q(0)Q(1)$ Let's assume that $P(x)\in \mathbb{R}[x]$ is a polynomial such that $P(x) = x^{2004}$, and that $Q(x)$ is the quoitent of the division of $P$ by $x^2-1$. How could we find $Q(0)Q(1)$?
$$x^2\equiv 1\pmod{x^2-1}$$
$$P(x)  = x^{2004}\equiv 1\pmod{x^2-1}$$
Then, $P(x) = (x^2-1)Q(x)+1$ and
$$\begin{align}\frac{P(x)-1}{x^2-1} = \frac{x^{2004}-1}{x^2-1} = \frac{\biggr(x^{1002}-1\biggr)\biggr(x^{1002}+1\biggr)}{x^{2}-1} &= \frac{\biggr(x^{501}-1\biggr)\biggr(x^{501}+1\biggr)\biggr(x^{1002}-1\biggr)}{x^{2}-1} \\ &= \cdots\end{align}$$
Which will get progressively worse.
 [/INST] As you already showed, $Q(x)=\frac{x^{2004}-1}{x^2-1}.$ Therefore:

*

*$Q(0)=1$ and

*$Q(1)=\lim_{y\to1}\frac{y^{1002}-1}{y-1}=f'(1)$ where $f(y)=y^{1002},$ hence
$Q(1)=1002.$ "Alternatively", $Q(\sqrt{1+h})=\frac{(1+h)^{1002}-1}h=\sum_{k=1}^{1002}\binom{1002}kh^{k-1}$ hence $Q(1)=\binom{1002}1=1002.$
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