Euclid's Lemma/Proof 2

Proof
Let $a, b, c \in \Z$.

We have that $a \perp b$.

That is:
 * $\gcd \set {a, b} = 1$

where $\gcd$ denotes greatest common divisor.

From Bézout's Identity, we may write:
 * $a x + b y = 1$

for some $x, y \in \Z$.

Upon multiplication by $c$, we see that:
 * $c = c \paren {a x + b y} = c a x + c b y$

Now note that $c a x + c b y$ is an integer combination of $a c$ and $b c$.

So, since:
 * $a \divides a c$

and:
 * $a \divides b c$

it follows from Common Divisor Divides Integer Combination that:
 * $a \divides \paren {c a x + c b y}$

However:
 * $c a x + c b y = c \paren {a x + b y} = c \cdot 1 = c$

Therefore:
 * $a \divides c$

Also see

 * Euclid's Lemma for Prime Divisors