Divisors of Product of Coprime Integers

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.

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 $$b \perp c$$.