If $a$|$c$ and $b$|$c$, then $\frac {ab}{(a,b)}$|$c$ How can I prove that for $a,b,c \in ℕ^*$, if $a$|$c$ and $b$|$c$, then $\frac {ab}{(a,b)}$|$c$? This is what I've tried: $a$|$c$ and $b$|$c$ implies that $ba$|$bc$ and $ab$|$ac$, so $ab$|$bcx + acy$ and $ab$|$c(bx+ay)$. We know that for some $x$ and $y$, $bx+ay$$=(a,b)$, so $\frac {ab}{(a,b)}$|$c$. Is this right? Thanks.

Yes, $\ a,b\mid c\,\Rightarrow\, ab\mid ac,bc\,\Rightarrow\, ab\mid(ac,bc)=(a,b)c\,\Rightarrow\,ab/(a,b)\mid c,\ $ where, for variety, I have replaced your use of the Bezout Identity by the GCD Distributive Law.

**Remark** $\ $ The reverse implications are also true, which shows $\,{\rm lcm}(a,b) = ab/\gcd(a,b).$