how do i prove $ab|n$ if $\gcd(a, b) = 1$ and $a|n $ and $b|n$? Suppose that, for integers $a, b,$ and $n,$
$$\gcd(a, b) = 1\text{ and }a|n\text{ and }b|n.$$ 
How do I prove that $ab|n$ using linear Diophantine equations?
Can I extend the above result to the case where $\gcd(a, b) \ne 1$, but $\gcd(a, b) < a$ and
$\gcd(a, b) < b$? If I can't, is there a counter-example for it?
Thanks!
 A: Hint $\,\ b\mid a\,(n/a)\,\overset{\large\rm\color{#c00}{(E)}}\Rightarrow\, b\mid n/a\,\Rightarrow\, ab\mid n\ $ by $\rm\color{#c00}{(E)} = $ Euclid's Lemma and $\,(a,b)= 1.$
A: Let  $$a\cdot A=n=b\cdot B$$ where $A,B$ are integers
Now $\displaystyle \frac{b\cdot B}a=A$ which is an integer
As $(a,b)=1,a$ must divide $B\implies B=C\cdot a$ for some integer $C$
$\displaystyle\implies n=b\cdot B=b\cdot C\cdot a $

HINT:
Let the highest power of prime $p$ that divides $a$ is $A$
and the highest power of prime $p$ that divides $b$ is $B$
$\implies $ the highest power of prime $p$ that divides $(a,b)=$min$(A,B)$
As $(a,b)=1,$ min$(A,B)=0$
If $A>0,B$ must be $0$ and vice versa
$\implies $ the highest power of prime $p$ that divides $b\cdot a$ must be $A+0$
If the highest power of prime $p$ that divides $n$ is $N$
As $a|n, N\ge A$
Clearly, this will hold true for any prime $q$ that divides $a$ or $b$
A: Regarding your second question, the result is not true if $(a,b)\neq 1$. For example, take $a=6$ and $b=14$. Then $(a,b)=2$ and we have $a|42$ and $b|42$ but $ab=84$ and certainly $84\not|42$.
A: Let $$ap = n$$ $$bq = n$$ $$ap = bq$$
$gcd(a , b) = 1$ so a must divide q , q is a multiple of a
Let $$q = ak$$ $$ap = b(ak)$$ $$n = (ab)k$$
This shows n is a multiple of ab therefore $(ab) | n$
