Easiest and most complex proof of $\gcd (a,b) \times \operatorname{lcm} (a,b) =ab.$ I'm looking for an understandable proof of this theorem, and also a complex one involving beautiful math techniques such as analytic number theory, or something else. I hope you can help me on that. Thank you very much
Let $\gcd(a,b)=d$. Then for some $a_0,b_0$ such that $a_0$ and $b_0$ are relatively prime, we have $a=da_0$ and $b=d b_0$. If we can show that the lcm of $a$ and $b$ is $da_0b_0$, we will be finished.

Certainly $da_0b_0$ is a common multiple of $a$ and $b$. We must show that it is the least common multiple.

Let $m$ be a common multiple of $a$ and $b$. We will show that $da_0b_0$ divides $m$. 

Since $m$ is a multiple of $a$, we have $m=ka=ka_0d$ for some $k$. But $b$ divides $m$, so $db_0$ divides $ka_0d$, and therefore $b_0$ divides $ka_0$. Since $a_0$ and $b_0$ are relatively prime, it follows that $b_0$ divides $k$, and we are finished.
Easy proof by set theory: (I'm a beginner and thought of this, please correct if me if I'm wrong)
Let $a=xy, b=xz$.
So
$a = \{x,y\}$ and $b = \{x, y\}$,
where $x$ is the product of the common prime factors of $a, b$ which is $\gcd(a,b)$ and the product $xyz$ produces $\text{lcm}(a,b)$
By Set Theory,
$P(a \cup b) \cdot P(a \cap b) = P(a) \cdot P(b)$.
$xyz \cdot x = xy \cdot xz$
lcm(a,b) * gcd(a,b) = ab