Question

If the rational number $\frac{p}{q}$, $q\ne 0$ (where $p$ and $q$ are relatively prime) is a root of the equation, $a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_{1}x+a_0=0$,where $a_0$, $a_1$, $a_2,...,a_n$ are integers and $a_n\ne 0$, then show that $p$ is a divisor of $a_0$ and $q$ that of $a_n$.

Answer 1

Given $\frac{p}{q}$ is a root.
$\Rightarrow$ $a_n{\left(\frac{p}{q}\right)}^n+a_{n-1}{\left(\frac{p}{q}\right)}^{n-1}+...+a_1\left(\frac{p}{q}\right)+a_0=0$
$\Rightarrow$ $a_n{p}^n+a_{n-1}q{p}^{n-1}+...+a_1{q}^{n-1}p+a_0{q}^n=0$ (i)
$\Rightarrow$ $a_{n-1}{p}^{n-1}+a_{n-2}{p}^{n-2}q+...+a_1{q}^{n-2}p+a_0{q}^{n-1}=\frac{-a_n{p}^n}{q}$ (ii)
$\Rightarrow$ Here, $a_0$, $a_1$, ..., $a_{n-2}$, $a_{n-1}$, $p,q\in$Integers.
$\Rightarrow$ LHS is an integer, so RHS is also an integer.
ie, $-\frac{a_n{p}^n}{q}$ is an integer, where $p$ and $q$ are relatively prime to each other.
Thus, $q$ must divide $a_n$.
Again, $a_n{p}^n+a_{n-1}{p}^{n-1}q+...+a_1{q}^{n-1}p=a_0{q}^n$
$\Rightarrow$ $a_n{p}^{n-1}+a_{n-1}q{p}^{n-2}+...+a_1{q}^{n-1}=\frac{a_0{q}^n}{p}$ (iii)
As from above; $\frac{a_0{q}^n}{p}\in$Integer
$\Rightarrow p$ is divisor of $a_0$ (as $p$ and $q$ are relatively prime)
thus if the rational number $\frac{p}{q}$ is root of
$a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_{1}x+a_0=0$,
then $p$ divides $a_0$ and $q$ divides $a_n$.