Proof that gcd(a,b)*lcm(a,b) = ab (solve only by Bézout's identity) Hello all and good evening.
As part of the course's assignments, we received a task to prove the following sentence using only Bézout identity:

gcd$(a, b)$*lcm$(a, b)$  = ab

Although I have found a variety of solutions to this problem, I have not yet found a solution that uses only Bézout identity. Here's what I've had so far:
By Bézout's identity, we know that $$\text{gcd}(a,b) = ax + by$$ for some integers $x,y$.
Let $d$ be the common divisor of $a$ and $b$.
By definition, since $d$ divides $a$, we know that there exists $k_1$ so that $dk_1 = a$. The same is true for $b$, so $k_2$ exists so that $dk_2 = b$.
We also know that if lcm$(a, b)$ = $m$, it means that a divides m, and also b divides m: a|m, b|m.
Depending on the top of the proof, we can place the a and b in this area and get: $dk_1\mid m$, $dk_2\mid m$.
But in this part I get stuck, and do not understand how to continue the proof. I would be happy to receive advice for further solution.
Thanks :)
 A: Hint: Letting $d=\gcd(a,b)$ and $m=\operatorname{lcm}(a,b)$ you can show $dm\mid ab$ and $ab\mid dm.$
Showing $ab\mid dm$ will use the Bézout's identity. It only needs that $m$ is a common multiple of $a$ and $b$ and that $\gcd(a,b)\mid d.$
For showing $dm\mid ab,$ it is easier to show that $\frac{ab}{d}$ is a common multiple of $a$ and $b$ using that $d\mid a$ and $d\mid b.$
For this latter part, you will implicitly need the lemma, which you may already know, but here's a proof:

Lemma: Given integers $a,b$ and $N$ such that $a\mid N$ and $b\mid N,$ then $\operatorname{lcm}(a,b)\mid N.$
Proof: Let $m=\operatorname{lcm}(a,b).$
Apply the division algorithm to find pair $q,r$ so that $N=mq+r$, with $0\leq r<m.$
If $r\neq 0$, then  $r=N-mq>0$ is a common multiple of $a,b,$ and smaller than $m,$ contradicting that $m=\operatorname{lcm}(a,b).$ So $r=0$ and $m\mid N.$

A: Write $(a,b) := \gcd(a,b),\,$ and $\,[a,b] := {\rm lcm}(a,b).\,$  Then by GCD & LCM Universal Properties 
$\quad d\mid (a,b)\color{#0a0}\iff d\:|\:a,b \iff a,b\:|\:ab/d \color{#c00}\iff [a,b]\:|\:ab/d \iff d\:|\:ab/[a,b]$
Thus $\,(a,b)\,$ and $\,ab/[a,b]\,$ divide each other so they are equal naturals.
The third $\color{#c00}\iff$ is the LCM Universal Property and the first $\rm\color{#0a0}{\iff}$ is the GCD Universal Property, provable by Bezout (follow the links for various proofs). The other arrows arise from cofactor duality $\,x\mid y\iff y'\mid x'\,$ for $\,x' := ab/x.\,$ which yields a slick one-line proof - further explained here as
$$\gcd(a,b)\,  =  \,  {\rm lcm}(a',b')'=\, \dfrac{ab}{{\rm lcm}(b,a)}\qquad$$
which yields divisor-theoretic analogs of DeMorgan's Laws. See here for more, where I show how this viewpoint immediately yields the following natural generalization (above is case $\,m = ab)$
Theorem $\,\ \displaystyle \gcd\left(\frac{m}a,\frac{m}b\right) = \frac{m}{{\rm lcm}(a,b)}\,\ $ for $\,m\,$ any common multiple of $a,b\,$
