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?



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 $\,{\rm lcm}(a,b) = ab/\gcd(a,b).$


Write a = k(a,b), and b = s(a,b) where (k,s) = 1. So a divides c ==> c = ma =mk(a,b),

and b divides c ==> c = nb = ns(a,b) ==> c/(a,b) = mk = ns. This means k divides c/(a,b), and

s also divides c/(a,b). Since (k,s) = 1, ks divides c/(a,b) . This means:

[a/(a,b)]*[b/(a,b)] divides c/(a,b) , and cancel out (a,b) we get the answer.


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