If we have that $[a,b]=k$ meaning $a|k$ and $b|k$. Is there an equation that we can get from this relation? My problem says "find and prove the sufficient condition in which $(a,b)=[a,b]$". I think it is when $a=b$ but I am trying to prove a biconditional statement that $(a,b)=[a,b]$ if and only if $a=b$. I am stuck in the forward direction.
Any input will help.
On
The main relation between greatest common divisor and lowest common multiple is $$ (a,b)[a,b]=ab $$ (when $a$ and $b$ are positive integers). Suppose $(a,b)=[a,b]$; then $$ (a,b)^2=ab $$ On the other hand, $(a,b)\le a$, so $(a,b)^2\le a^2$. In particular $$ ab\le a^2 $$ so $b\le a$. Similarly, $a\le b$.
See, if $a=b$ then it is obvious that $(a,b)= [a,b]$.
The other way, if $(a,b) = [a,b]$, then because $(a,b) \leq a \leq [a,b]$, it follows that $(a,b) = a$ (because above,equality holds in our case).
Similarly, if $(a,b) = [a,b]$, then because $(a,b) \leq b \leq [a,b]$, it follows that $(a,b) = b$.
Finally, $a = (a,b) = b = [a,b]$ is true.
Therefore, $a = b \iff [a,b] = (a,b)$.
EDIT: If $a=b$, then the "least common multiple" of $a$ and $b$ is obviously $a$ (or $b$). (Because both $a$ and $b$ divide $a$, and $a$ is the smallest number with this property).
Similarly, if $a=b$, the "greatest common divisor" of $a$ and $b$ is obviously $a$ (or $b$).(Because $a$ divides both $a$ and $b$, and $a$ is the largest number with this property).
Hence, $(a,b) = a = [a,b]$, nearly by definition of $\operatorname{lcm}$ and $\gcd$.