Divisors of Product of Coprime Integers
Contents |
Theorem
Let $a \backslash b c$, where $b \perp c$.
Then $a = r s$, where $r \backslash b$ and $s \backslash c$.
Corollary
Let $p$ be a prime.
Let $p \backslash b c$, where $b \perp c$.
Then $p \backslash b$ or $p \backslash c$, but not both.
Proof
Let $r = \gcd \left\{{a, b}\right\}$.
By Divide by GCD for Coprime Integers, $\exists s, t \in \Z: a = r s \land b = r t \land \gcd \left\{{s, t}\right\} = 1$.
So we have written $a = r s$ where $r$ divides $b$.
We now show that $s$ divides $c$.
Since $a$ divides $b c$ there exists $k$ such that $b c = k a$.
Substituting for $a$ and $b$:
- $r t c = k r s$
which gives:
- $t c = k s$
So $s$ divides $t c$.
But $s \perp t$ so by Euclid's Lemma $s \backslash c$ as required.
$\blacksquare$
Proof of Corollary
From the main result, $p = r s$, where $r \backslash b$ and $s \backslash c$.
But as $p$ is prime, either:
- $r = 1$ and $s = p$, or:
- $r = p$ and $s = 1$.
So $p \backslash b$ or $p \backslash c$.
But $p$ can not divide both as $b \perp c$.
$\blacksquare$