Square Divides Product of Multiples

Theorem

Let $a, b, c, \in \Z$ be integers.

Let:

$a \divides b, a \divides c$

where $\divides$ denotes divisibility.

Then:

$a^2 \divides b c$

We have that:

$\ds a$

$\divides$

$\ds b$

$\ds \leadsto \ \ $

$\ds \exists k_1 \in \Z: \, $

$\ds k_1 a$

$=$

$\ds b$

$\ds a$

$\divides$

$\ds c$

$\ds \leadsto \ \ $

$\ds \exists k_2 \in \Z: \, $

$\ds k_2 a$

$=$

$\ds c$

$\ds \leadsto \ \ $

$\ds b c$

$=$

$\ds \paren {k_1 a} \paren {k_2 a}$

$\ds $

$=$

$\ds \paren {k_1 k_2} a^2$

$\ds \leadsto \ \ $

$\ds a^2$

$\divides$

$\ds b c$

$\blacksquare$

1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $2 \ \text{(b)}$