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Let $p(x) \in \mathbb{Z}[x]$ Let $n$ be a positive integer with $\operatorname{deg} p(x)=n$ Let $c_0, \ldots , c_n \in \mathbb{Z}$ such that $p(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots +c_1x+c_0$
Let $a,b \in \mathbb{Z}$ with $\operatorname{gcd}(a,b)=1$ and $b>0$ such that $\displaystyle \frac{a}{b}$ is a root of $p(x)$ Then
$\begin{array}{ll} 0 & \displaystyle =p\left( \frac{a}{b} \right) \\ \\ & \displaystyle =c_n \left( \frac{a}{b} \right)^n+c_{n-1} \left( \frac{a}{b} \right)^{n-1} +\cdots +c_1 \cdot \frac{a}{b} +c_0 \\ \\ & \displaystyle =c_n \cdot \frac{a^n}{b^n}+c_{n-1} \cdot \frac{a^{n-1}}{b^{n-1}} +\cdots +c_1 \cdot \frac{a}{b} +c_0. \end{array}$
Multiplying through by $b^n$ and rearranging yields:
$\begin{array}{rl} c_na^n+c_{n-1}a^{n-1}b+\cdots +c_1ab^{n-1} +c_0b^n & =0 \\ \\ c_0b^n & =-c_na^n-c_{n-1}a^{n-1}b -\cdots -c_1ab^{n-1} \\ \\ c_0b^n & =a \left( -c_na^{n-1}-c_{n-1}a^{n-2}b -\cdots -c_1b^{n-1} \right) \\ \end{array}$
Thus, $a|c_0b^n$ and, by hypothesis, $\operatorname{gcd}(a,b)=1$ This implies that $a|c_0$
Similarly:
$\begin{array}{rl} c_na^n+c_{n-1}a^{n-1} b+\cdots +c_1ab^{n-1} +c_0b^n & =0 \\ \\ c_na^n & =-c_{n-1}a^{n-1} b-\cdots -c_1ab^{n-1} -c_0b^n \\ \\ c_na^n & =b \left( -c_{n-1}a^{n-1} -\cdots -c_1ab^{n-1} -c_0b^{n-1} \right) \\ \end{array}$
Therefore, $b|c_na^n$ and $b|c_n$
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