If $(a,b) = 1$ then $a,b|n \implies ab|n$

Solution 1:

If $\gcd(a,b) = 1$ then the Bezout identity gives some $x,y$ such that $ax+by = 1$. Hence $axn+byn = n$, and since $ab \mid axn$ and $ab \mid byn$ we have $ab \mid n$.

Solution 2:

Modular arithmetic to the rescue!

If

$$ n \equiv 0 \bmod a $$ $$ n \equiv 0 \bmod b $$

then by the Chinese Remainder Theorem,

$$ n \equiv 0 \bmod \operatorname{lcm}(a,b) $$

Solution 3:

The proof comes out of intuition as follows:

Since $a \mid n$, Let $q$ the quotient.

$b \mid n$ $ \implies b \mid q$. Since $a$ doesn't remove any of the factors of $b$ from $n$ as there are no common factors between $a$ and $b$.

$ \implies ab \mid n$

Solution 4:

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$~n$.