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By Gauss Lemma, the roots of a polynomial $P(x) = a_nx^n + \cdots + a_1x + a_0 \in \Bbb Z[x]$ are either integer, irrational or complex. Vietà's formulas imply that the product of all roots equals $(-1)^na_0/a_n$.

However, it's not true that each individual root must divide $a_0/a_n$ as some group of Galois conjugate roots may multiply to a non-integer rational number. For example, consider the polynomial $$-84^2/28 + (28-b)(27-b)(2b-27) = 2 b^3-137 b^2+2997 b-20664$$ which has $b = 24$ as a root, but $24 \nmid 20664/2 = 2^2 \cdot 3^2 \cdot 7 \cdot 41$.

I suspect that such a problem can only occur with non-monic polynomials, because these may have a non-monic irreducible factor. But can we estimate what will be the leading coefficients $\alpha_{n_1}, \beta_{n_2}, \dots$ of the irreducible factors from $a_n$? And from that can we say that each integer root of $P(x)$ will divide $a_0$?

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  • $\begingroup$ Your statement of the Gauss Lemma is not correct; consider the polynomial $P(x) = 2 x + 1 \in \Bbb Z[x]$. $\endgroup$ – Travis Willse Apr 13 '16 at 17:10
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    $\begingroup$ Yes: if $f(x) \in \Bbb{Z}[x]$ is a monic polynomial, and $a$ is an integer root of $f(x)$, then $a$ must divide $f(0)$ (the constant term of $f(x)$). $\endgroup$ – Crostul Apr 13 '16 at 17:10
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If $p/q$ is a rational root of $P(x)\in \mathbf Z[x]$, $p$ and $q$ coprime, this means $$q^nP(p/q)=(a_np^n+a_{n-1}p^{n-1}q+\dots+a_1pq^{n-1}+a_0q^n)=0,$$ whence \begin{align*} (a_np^{n-1}+a_{n-1}p^{n-2}q+\dots+a_1q^{n-1})p=-a_0q^n\\ (a_{n-1}p^{n-1}+\dots+a_1pq^{n-2}+a_0q^{n-1})q=-a_np^n \end{align*} which means $p$ divides $a_0q^n$ and $q$ divides $a_np^n$. Since $p$ and $q$ are coprime, each power of one of them is coprime with each power of the other. Now Gauß's lemma ensures $p$ divides $a_0$ and $q$ divides $a_n$.

The same result is valid in any GCD domain.

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  • $\begingroup$ Why did you need Gauss Lemma to ensure $p$ divides $a_0$ and $q$ divides $a_n$? $\endgroup$ – Rodrigo Apr 14 '16 at 18:33
  • $\begingroup$ that's because $p$, being coprime to $q$, is coprime to $q^n$. Hence we can apply Gauß's lemma to $a_0q^n$. Similarly for $q$ and $a_np^n$. $\endgroup$ – Bernard Apr 14 '16 at 18:49
  • $\begingroup$ If I know that $p$ divides $a_0q^n$ and $p$ is coprime to $q^n$, then I know that $p$ divides $a_0$ by the definition of primeness, right? I don't need Gauss lemma... $\endgroup$ – Rodrigo Apr 15 '16 at 21:36

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