Prove: If $a\mid m$ and $b\mid m$ and $\gcd(a,b)=1$ then $ab\mid m\,$ [LCM = product for coprimes] Prove: If $a\mid m$ and $b\mid m$ and $\gcd(a,b)=1$ then $ab\mid m$
I thought that $m=ab$ but I was given a counterexample in a comment below.
So all I really know is $m=ax$ and $m=by$ for some $x,y \in \mathbb Z$. Also, $a$ and $b$ are relatively prime since $\gcd(a,b)=1$. 
One of the comments suggests to use Bézout's identity, i.e., $aq+br=1$ for some $q,r\in\mathbb{Z}$. Any more hints?
New to this divisibility/gcd stuff. Thanks in advance! 
 A: Write $ax+by=1$, $m=aa'$, $m=bb'$. Let $t=b'x+a'y$.
Then $abt=abb'x+baa'y=m(ax+by)=m$ and so $ab \mid m$.
Edit: Perhaps this order is more natural and less magical:
$m = m(ax+by) = max+mby = bb'ax+aa'by = ab(b'x+a'y)$.
A: Hint $\rm\, \ a,b\mid m\iff ab\mid am,bm
 \!\!\overset{\rm\color{#0a0} U\!\!}\iff ab\mid \overbrace{(am,bm)}^{\large \color{#c00}{ (a,\,b)\,m}}\!\iff ab/(a,b)\mid m$
Remark $\ $ We used $\rm\color{#0a0} U$= gcd Universal Property and  the gcd distributive law $\rm\:\color{#c00}{(a,b)\,c} = (ac,bc).\ $ If above we use Bezout's Identity to replace the gcd $\rm\:(a,b)\:$ by  $\rm\:j\,a + k\,b\:$ (its linear representation) then we obtain the proof by Bezout in lhf's answer (here in divisibility language).
The above proof is more general than  Bezout-based proofs since there are rings with gcds not of linear (Bezout) form, e.g. $\,\rm \Bbb Z[x,y]\,$ the ring of polynomials in $\,\rm x,y\,$ with integer coefficients, where $\,\rm gcd(x,y) = 1\,$ but  $\rm\, x\, f + y\, g\neq 1\,$ (else evaluating at $\rm\,x,y = 0\,$ yields $\,0 = 1).\,$
The proof shows that $\rm\ a,b\mid m\iff ab/(a,b)\mid m,\ $  i.e. $\ \rm lcm(a,b) = ab/(a,b)\ $  using the universal definition of lcm. $ $ The OP is the special case $\rm\,(a,b)= 1.$
A: If $\gcd (a,b) =1$, then $a$ and $b$ have no prime factors in common. This means if we divide $m$ by $a$, the result is still divisible by $b$. So $b | \frac{m}{a}$, thus $ab|m$.
A: Ok first of all let me show you that m does not equal ab necessarily. suppose m=36. a=2 and $b=3$. $ab=6$ and a|m and b|m. Now lets go to the other part of the problem. lets put a in prime factorization. $a=2^{a_2}*3^{a_3}...p^{a_p}$ and $b=2^{b_2}*3^{b_3}...p^{b_p}$ where p is a prime number.So then $\gcd(a,b)= 1$ if and only if $(a_i+b_i)=max(a_i,b_i)$ for any prime i. Now lets do the same prime decomposition for m. $m=2^{m_2}*3^{m_3}...p^{m_p}$ so then a can only divide m if $a_i\leq m_i$ for any prime i. also $a*b=2^{a_2+b_2}*3^{a_3+b_3}...p^{a_p+bp}$ but since (a,b)=1 this is equal to $2^{max(a_i,b_i)}*3^{max(a_i,b_i)}...*p^{max(a_p*b_p)}$ and since $m_i>a_i $and $m_i>b_i $ then $m_i>max(a_i,b_i)$ as desired
A: HINT: You know that there is an integer $k$ such that $ak=m$. Now you have $b\mid ak$ and $(a,b)=1$; do you know a theorem that let’s you draw a conclusion about $b$ and $k$? (The theorem that I have in mind can be proved using Bézout’s lemma; the argument is the one that wj32 has in mind for your question, but it’s not necessary if you already know this theorem.)
