Prove if $(a,b) = 1$ implies $a|n$ and $b|n \implies ab|n$.
I'm pretty sure this has been asked before but I cannot find anything online.... I also have no idea how to solve it, I get stuck with al = bk for some l,k in the integers.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
As I've been commenting that this is simpler than some other answers suggest, let me be fair and say what minimum knowledge I think is required for this conclusion. I'll assume (though the question does not say so) $a,b$ are supposed to be nonzero; anyway either being $0$ would force $n=0$, a trivial case.
You must (as is often the case when reasoning about integers) at some point use the possibility of integer division: for integer $k,l$ with $l>0$ there exist integer $q,r$ with $k=ql+r$ and $0\leq r<l$. Here it can be used to show that the smallest positive common multiple $m$ of $a,b$ (which certainly exists because $|ab|$ is a positive common multiple) divides any other common multiple $c$ of $a,b$: writing $c=qm+r$ with $0\leq r<m$ one easily sees that $r=c-qm$ is a common multiple of $a,b$, so $r=0$ by minimality of$~m$.
Now things are easy. If the least common multiple$~m$ of $a,b$ were smaller than $|ab|$, then by the above it would divide $|ab|$, and $|ab|/m>1$ would be a common divisor of $a,b$, contradicting the hypothesis $\gcd(a,b)=1$. So $m=|ab|$, and as we saw it divides any common divisor of $a,b$, such as$~n$.