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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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    There are other generalizations of Gauss's Lemma, e.g. Kronec$k$er's Lemma (a.$k$.a Dedeki$n$d's Prague Theorem) that can be proved by way of algebraic numbers. But that would be extreme over$k$ill here and, furthermore, probably beyond the OP's knowledge of algebra and number theory.2010-11-05

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