Here is the theorem they prove:
$$|ab| = \operatorname{lcm}(a,b) \operatorname{gcd}(a,b)$$
where
- $\operatorname{lcm}(a,b)$ denotes the lowest common multiple of $a$ and $b$.
- $\operatorname{gcd}(a,b)$ denotes the greatest common divisor of $a$ and $b$.
Proof 1 is below (in case the original site changes). I have highlighted the end part that I don't understand in red.
It is sufficient to prove that $\operatorname{lcm}(a,b) \operatorname{gcd}(a,b) = ab$ where $a,b \in \mathbb{Z}_{>0}$.
$d = \operatorname{gcd}(a,b)$
$\implies d | ab$
$\exists n \in \mathbb{Z}_{>0}: ab = dn$
$d | a \land d | b$
$\implies \exists u,v \in \mathbb{Z}:a =du \land b = dv$
$\implies dub = dn \land adv = dn$
$\implies n=bu \land n = av$
$\implies a | n \land b | n$
Now we have $a|m \land b | m \implies m = ar=bs$.
Also, by Bézout's Lemma we have $d = ax + by$.
So: \begin{align}md &= axm + bym \\ &= bsax + arby \\ &= ab(sx + ry) \\ &= dn (sx + ry) \end{align}
So:
$m = n(sx + ry)$
Thus:
$n|m \implies n \leq |m|$
while:
$\color{red}{ab = dn = \operatorname{gcd}(a,b) \operatorname{lcm}(a,b)}$
as required.
$\blacksquare$
I get that we are meant to conclude that $n = \operatorname{lcm}$, but I don't yet see why the workup to this point entails that $n$ must be the least common multiple.