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 $\ell= \text{lcm}(a,b), g=\text{gcd}(a,b)$ for some $a,b$ positive integers.


Division algorithm: exists $q,r$ integers with $0\leq r < \ell$ such $ab = q\ell + r$. Observing that both $a$ and $b$ divide both $ab$ and $q\ell$ we conclude they both divide $r$. As $r$ is a common multiple, we must have $\ell \leq r$ or $r\leq 0$ so $r=0$.


Therefore $s=\frac{ab}{\ell}$ is an integer. Observe that $\frac{a}{s}=\frac{\ell}{b}$ and $\frac{b}{s}=\frac{\ell}{a}$ are both integers, so $s$ is a common divisor. Thus $g\geq s = \frac{ab}{\ell}$.


On the other hand, $\frac{ab}{g}=a\frac{b}{g}=b\frac{a}{g}$ where $\frac{b}{g}$ and $\frac{a}{g}$ are both integers, so $\frac{ab}{g}$ is a common multiple of $a$ and $b$. As $\frac{ab}{g}>0$ we conclude $\frac{ab}{g}\geq \ell$, therefoire $\frac{ab}{\ell}\geq g$ and we are done.
