Elementary Proof: If $a|m$, $b|m$ and $\gcd(a,b)=1$ then $ab|m$ So I am trying to make some way on this proof,but I cannot see how any of these relate. 
Here is the full question: Let $a,b,m$ be integers such that $a\mid m$, $b\mid m$ and $\gcd(a,b)=1$, then $ab\mid m$. 
Listing what I know I can see that by definition
$m = ap$ where $p$ is some integer,
$m= bq$ where $q$ is some integer,
$\gcd(a,b)=1=ax+by$ where $x,y$ are some integers and we are looking to prove
$ m = ab(r) $ where $r$ is some integer. 
What I thought about doing is either setting $ap$ = $bp$ and then doing something with that for noticing the fact that $m= m\cdot 1$ so by replacement you can do $m= m(ax+by)$ and then 
$$m(ax+by) = abr.$$ 
But I feel like this is leading nowhere. Am I on the right track? I can't seem to find any relevant propositions.
 A: I feel like the proof provided by vrugtehagel doesn't shed much light on what is actually going on, so I want to elucidate the theorem in a different way.
If $(a,b)=1$ then $a$ and $b$ have no common factors. IDK if you're allowed to use unique prime factorization in this proof, but even if you're not it gives an easy way to understand why this statement is true. If $gcd(a,b)=1$ then the list of their prime factors are disjoint. If $a|m,b|m$, then all of the prime factors of $a$ are prime factors of $m$ and the same is true for $b$. The reason that the GCD requirement is necessary is that if it doesn't hold, $a$ and $b$ could "double up" by both containing prime factors of $m$, which get double counted in the product. Consider $a=6, b=10, m=30$. It isn't true that $ab|m$ because the single occurrence of a $2$ in the factorization of $30$ occurs in both $6$ and in $10$.
This concept is encapsulated in the following theorem, which is a generalization of the one you have to prove:

Theorem: $lcm(a,b)=ab/gcd(a,b)$

The product of two numbers is the product of all of their prime factors. The LCM of two numbers is the smallest number that contains all the prime factors of both numbers. The gcd of two numbers is the product of all the shared prime factors. The ratio of the lcm and the gcd takes all the prime factors, and removes the extent to which they've been double-counted, yielding the smallest number containing all the prime factors of the two numbers.
A: $1=ax+by\Rightarrow m=max+mby\Rightarrow m=bqax+apby\Rightarrow m=ab(qx+py)$
A: Let $ap=m$ and $bq=m$. Then let $x,y$ be integers such that $ax+by=1$. Now note $apqx+bpqy=AB$, or $mqx+mpy=pq$. This means $pq=m(qx+py)$, which we can rewrite to $$apbq=abm(qx+py)$$
or
$$m^2=abm(qx+py)$$
which simplifies to $m=ab(qx+py)$, resulting in $ab\mid m$.
A: This solution doesn't use Bézout's identity and only uses properties of $\gcd$ and thus it's generalizable to GCD domains. 
Let's set $\text{lcm}(a,b)=c$. It's enough to prove that $|ab|=|c|$ because as $a\mid m$ and $b\mid m$, then by the universal definition of $\text{lcm}$ it follows that $c\mid m$ and hence $ab\mid m$.
To prove that $|ab|=|c|$ we first note that since $a\mid ab$ and $b\mid ab$, then applying again the universal definition of $\text{lcm}$ we have $c\mid ab$. On the other hand, $$a=\frac{ab}{c}\cdot \frac{c}{b}\implies \frac{ab}{c}\mid a$$ $$b=\frac{ab}{c}\cdot \frac{c}{a}\implies \frac{ab}{c}\mid b$$
Therefore by the universal definition of $\gcd$ we deduce that $ab/c\mid \gcd(a,b)=1$, then $|ab/c|=|1|$ and thus $|ab|=|c|$.
A: maybe worth noting that the result follows quickly from the rule describing how the distributive lattice structure on the positive integers interacts with ordinary multiplication:
$$
(a\land b)\cdot (a\lor b) = a \cdot b
$$
these properties are straightforward once you have the unique prime decomposition theorem for the positive integers. 
A: Let $a\alpha=m$ and $b\beta=m$.
Since $\gcd(a, b)=1$, there exist integers $x,y$ such that $ax+by=1$. Thus $\alpha\beta(ax+by)=\alpha\beta$, i.e., $m\beta x+m\alpha y=\alpha\beta$.
This means that $\alpha\beta=m(\beta x+\alpha y)$, so that $m^2= a\alpha b\beta=abm(\beta x+\alpha y)$, which gives $m=ab(\beta x+\alpha y)$.
Hence $ab\mid m$.
