Proof involving lcm and biconditional statement.

Question

Suppose $a,b\in\mathbb{Z}$. Then $a = \operatorname{lcm}(a,b)$ if and only if $b\mid a$

Unsure of how to approach this problem.

Answer 1

$\newcommand{\lcm}{\operatorname{lcm}}$ By definition $\lcm(a,b)=bk$ for some $k\in\Bbb Z$. So if $bk=a$ then $b\mid a$ by definition.

Conversely, $b\mid a$ implies $\lcm(a,b)\le a$ and $a\mid a$ implies $\lcm(a,b)\ge a$.

Answer 2

Hint $\,\ b\mid a \iff a,b\mid a \iff {\rm lcm}(a,b)\mid a\iff {\rm lcm}(a,b) = a$

Answer 3

As shown in this answer, $$ \operatorname{lcm}(a,b)\gcd(a,b)=ab $$ Thus, we have that $$ \gcd(a,b)=b $$ This means that $b$ is a divisor of $a$; i.e. $b\mid a$.