Show that $\rm lcm(a,b)=ab \iff gcd(a,b)=1$ Show that 
$$\rm lcm(a,b)=ab \iff \gcd(a,b)=1.$$
My attempt: If $\gcd(a,b)=1$ then there exist two integers $r$ and $s$ such that $$ar+bs=1.$$
and then I'm stuck...
any advice?
 A: Theorem: For positive integers $a$ and $b$ we have:
$$\gcd(a,b)\text{lcm}(a,b)=ab.$$
Using this, suppose $\text{lcm}(a,b)=ab$ it follows that $\gcd(a,b)=1$.
Also on the other way suppose $\gcd(a,b)=1$, using the above theorem we conclude that $\text{lcm}(a,b)=ab$. 
A: If we have the prime decompositions of $a$ and $b$:
$$
a = 2^{a_2} \cdot 3^{a_3}\cdot 5^{a_5}\cdots \qquad b = 2^{b_2} \cdot 3^{b_3}\cdot 5^{b_5}\cdots
$$
where only a finite number of $a_p, b_p$ are non-zero, then we get
$$
\gcd(a, b) = 2^{\min(a_2, b_2)}\cdot 3^{\min(a_3, b_3)}\cdot 5^{\min(a_5, b_5)}\cdots \\
\operatorname{lcm}(a, b) = 2^{\max(a_2, b_2)}\cdot 3^{\max(a_3, b_3)}\cdot 5^{\max(a_5, b_5)}\cdots
$$
From there it's not difficult to conclude that $\operatorname{lcm}(a, b)\cdot \gcd(a, b) = ab$ (they must be equal, since they have the same prime decomposition), and then what you want to show follows easily.
A: If you want to do this without prime factorization and without the gcd-lcm theorem, here is another approach.
By the way, I am assuming here that $a$ and $b$ are positive integers.
Let $d=\gcd(a,b)$ and let $m=\mbox{lcm}(a,b)$.
Note that $\frac{ab}{d}=\frac{a}{d}\cdot b=\frac{b}{d}\cdot a$ is a common multiple of $a$ and $b$.  So if $d>1$, $\frac{ab}{d}$ is a common multiple of $a$ and $b$ that is less than $ab$.  So in this case, we would have $m<ab$.  Then by contrapositive, $m=ab\Rightarrow d=1$.
For the other direction, if $m<ab$, then $m=ax=by$ for some positive integers $x,y$ with $x<b$ and $y<a$.  Then $\frac{a}{y}=\frac{b}{x}$, and this common value is a common factor of $a$ and $b$.  However, since $y<a$, this common factor, $\frac{a}{y}$, is greater than $1$.  Thus $d>1$.  So again by contrapositive, $d=1\Rightarrow m=ab$.
A: Hint $\, $ Show $\,n\mapsto ab/n\,$ bijects the common divisors of $\,a,b\,$ with the common multiples $\le ab.$ Being order-$\rm\color{#c00}{reversing}$, it maps the $\rm\color{#c00}{Greatest}$ common divisor to the $\rm\color{#c00}{Least}$ common multiple, i.e.
$$\begin{align}
{\rm\color{#c00}{G}CD}(a,b)\,\mapsto\ &\bbox[5px,border:2px solid #c00]{ab/\color{#0a0}{{\rm GCD}(a,b)} = {\rm \color{#c00}{L }CM}(a,b)}\\[.4em]
&{\rm so}\ \  \color{#0a0}{{\rm GCD}(a,b)\!=\!1}\!\iff\! {\rm LCM}(a,b)\!=\!ab \end{align}\quad$$
Remark $\ $ For more on this (involution) duality between gcd and lcm see here and here.
A: $\dfrac{a\cdot b}{gcd(a,b)}=a\cdot\left(\dfrac b{gcd(a,b)}\right)=\left(\dfrac a{gcd(a,b)}\right)\cdot b$ is a multiple of both $a$ and $b$.
As $a\cdot b$ is known to be the smallest multiple, then $gcd(a,b)=1$.
