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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.

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    Hint: $Q(x)=\sum_{n=0}^{1001}x^{2n}$. – J.G. Jan 22 '23 at 23:18
  • I do not see how to arrive at that $Q$, though. –  Jan 22 '23 at 23:19
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    You will if you follow that link. – J.G. Jan 22 '23 at 23:20
  • Then we should have something like: $$P(x) = (x^2+1)Q(x) + 1 = (x^2+1)(1+ x + x^2+x^3+x^4+\cdots x^{1001}) + 1 = (x^2+1)(\sum_{0\leq i\leq 1001}x^{i}) + 1$$ Why are we excluding the odd terms in the quoitent? –  Jan 22 '23 at 23:26
  • Use $y=x^2$ in $y^m-1=(y-1)\sum_{n=0}^{m-1}y^n$ to understand why. – J.G. Jan 22 '23 at 23:29
  • Oh, I think that follows from the quoitent being a symmetric polynomial, right? If $P(x), Q(x)\in \mathbb{R}[x]$ are a symmetric polynomial, then so is their quoitent $\frac{P(x)}{Q(x)}\in \mathbb{R}[x]$ –  Jan 22 '23 at 23:29
  • If by "symmetric" you mean "even", yes. Obviously, similar problems could exist with e.g. $\frac{x^{3t}-1}{x^3-1}$. – J.G. Jan 22 '23 at 23:31
  • Yes, by "symmetric", I had meant even polynomial. –  Jan 22 '23 at 23:32

1 Answers1

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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 $Q(1)=\binom{1002}1=1002.$
Anne Bauval
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