# Conditions on Rational Solution to Polynomial Equation

## Theorem

Let $P$ be the polynomial equation:

$a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0 = 0$

where $a_0, \ldots, a_n$ are integers.

Let $\dfrac p q$ be a root of $P$ expressed in canonical form.

Then $p$ is a divisor of $a_0$ and $q$ is a divisor of $a_n$.

## Proof

By definition of the canonical form of a rational number, $p$ and $q$ are coprime.

Substitute $\dfrac p q$ for $z$ in $P$ and multiply by $q^n$:

$(1): \quad a_n p^n + a_{n - 1} p^{n - 1} q + \cdots + a_1 p q^{n - 1} + a_0 q^n = 0$

Dividing $(1)$ by $p$ gives:

$(2): \quad a_n p^{n - 1} + a_{n - 1} p^{n - 2} q + \cdots + a_1 q^{n - 1} = -\dfrac {a_0 q^n} p$

The left hand side of $(2)$ is an integer and therefore so is the right hand side.

We have that $p$ and $q$ are coprime.

By Euclid's Lemma it follows that $p$ divides $a_0$.

Similarly, dividing $(1)$ by $q$ gives:

$(3): \quad -\dfrac {a_n p^n} q = a_{n - 1} p^{n - 1} + \cdots + a_1 p q^{n - 2} + a_0 q^{n - 1}$

By Euclid's Lemma it follows that $q$ divides $a_n$.

$\blacksquare$