Product of Coprime Factors

Theorem

Let $a, b, c \in \Z$ such that $a$ and $b$ are coprime.

Let both $a$ and $b$ be divisors of $c$.

Then $a b$ is also a divisor of $c$.

That is:

$a \perp b \land a \divides c \land b \divides c \implies a b \divides c$

By definition of divisor:

$a \divides c \implies \exists r \in \Z: c = a r$

$b \divides c \implies \exists s \in \Z: c = b s$

So:

$\ds a$

$\perp$

$\ds b$

$\ds \leadsto \ \ $

$\ds \exists m, n \in \Z: \, $

$\ds m a + n b$

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds c m a + c n b$

$=$

$\ds c$

$\ds \leadsto \ \ $

$\ds b s m a + a r n b$

$=$

$\ds c$

$\ds \leadsto \ \ $

$\ds a b \paren {s m + r n}$

$=$

$\ds c$

$\ds \leadsto \ \ $

$\ds a b$

$\divides$

$\ds c$

$\blacksquare$

1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Some Properties of $\Z$: Exercise $2.6$
1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 4$: Corollary $2$