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The rational root theorem states that, given $P \in \mathbb{Z}[x], P = \displaystyle\sum_{j=0}^n a_j x^j$ with $a_0 \neq 0$ and $a_n \neq 0$, if $P(p/q)=0, p\perp q$ then $p|a_0$ and $q|a_n$.

As a follow-up to this question, I'd like to know how can I prove the following:

$$a_0 = a_n \displaystyle\prod_{j=1}^n (-r_j)$$, with $r_j$ being roots of P(x)

This is the same as saying:

$$a_0 = a_n \frac{-p}{q} \displaystyle\prod_{j=2}^n (-r_j)$$

Rearranging this, we get:

$$a_n = q\frac{a_0}{-p \displaystyle\prod_{j=2}^n (-r_j)}$$

I want to prove that $$\frac{a_0}{-p \displaystyle\prod_{j=2}^n (-r_j)} \in \mathbb{Z}$$ without using the rational root theorem (actually, I want to prove the rational root theorem using this identity).

So far I've tried to prove it by contradiction, assuming $a_0 = k(-p)\displaystyle\prod_{j=2}^n (-r_j) + \epsilon$ with $\epsilon \in (0, (-p)\displaystyle\prod_{j=2}^n (-r_j))$ but I don't seem to find anything weird rearranging that equation.

If you manage to find a proof please just post a small hint. Thanks in advance.

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    $\begingroup$ @fmartin: You are missing a minus sign; if $r_1 = \frac{p}{q}$, then you factor out $-\frac{p}{q}$, or put a minus sign somewhere... $\endgroup$ Commented Nov 5, 2010 at 19:20
  • $\begingroup$ @Arturo: You're right, I'll edit that. $\endgroup$ Commented Nov 5, 2010 at 19:27
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    $\begingroup$ Note that generally the roots $ r_j $ are not rationals but, rather, algebraic numbers. Why do you desire to pass to algebraic numbers to try prove a result that has a simple proof in integers via Bezout's theorem? $\endgroup$ Commented Nov 5, 2010 at 19:56
  • $\begingroup$ @Bill: On the other hand, the full product of the "other" roots will be a rational... $\endgroup$ Commented Nov 5, 2010 at 20:20
  • $\begingroup$ @Arturo. Of course but that's of no help for the matter at hand. $\endgroup$ Commented Nov 5, 2010 at 20:23

1 Answer 1

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Those are just Vieta's formulas for the coefficients in terms of the roots.

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