Show that if a|c, b|c and gcd(a,b)=1, then ab|c Let $a,b,c \in \mathbb{Z}$ such that $a|c$, $b|c$, gcd$(a,b)=1$. Prove that $ab|c$.
My thoughts so far: By the Unique Factorization Theorem, we can rewrite $a$ as $p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$ where $p_i$'s are primes that make up $a$. Similarly, we can write $b=q_1^{\beta_1}q_2^{\beta_2}...q_k^{\beta_k}$, but with the restriction that $p_i\neq q_j$, $\forall i,j$. Then, by definition of "$|$," we can say that $a|c$ implies $\exists k \in \mathbb{Z}$ such that $k(p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k})=c$. (Similarly for $b$).
From here I am unsure where to go, but it feels like I am only a lemma away from my conclusion.
 A: Any time you see $gcd(a,b)=1$, it can often be helpful to think of Bezout's identity. 
$$
gcd(a,b)=1\Rightarrow \exists \; x,y\in\mathbb{Z}\;s.t\; 1=ax+by
$$
Then we can scale to get
$$
c=cax+cby
$$
And now using that $a\vert c,\;b\vert c \Rightarrow c=ak_1=bk_2$ our identity becomes
$$
c=cax+cby=axbk_2+aybk_1\Rightarrow c=ab(k_2x+k_1y)
$$
A: The idea is right. We can first add that it's not restrictive to assume $a,b,c>0$ (verify it).
Since $a\mid c$, we have $c=ak$; every prime power that divides $b$ must so divide $k$.

Alternatively, write $c=abq+r$, with $0\le r<ab$. Since $a\mid c$, we deduce that $a\mid r$ and, similarly, that $b\mid r$. Suppose $r>0$; then $r$ is a multiple of the lowest common multiple of $a$ and $b$, which is $ab$. Hence $r\ge ab$, a contradiction. Therefore $r=0$.
A: A quick way of going at it is using Gauss Lemma:
If $a$ and $b$ are coprime ($\gcd(a,b)=1$) then  $a|bc\implies a|c$.
Reminder of the proof: By Bézout you have $u,v\in\mathbb{Z}$ such that
$ua+vb=1$ and you just multiply by $c$ to see it through.
Now your exercice: $b|c$ so $c=bk$; $a|c$ implies by Gauss Lemma that $a|k$, so $ab|c$.
A: Since $a\mid c$ and $b\mid c$, we must have $\text{lcm}(a,b)\mid c$.
We also have $ab=\gcd(a,b)\text{lcm}(a,b)$ and $\gcd(a,b)=1$, therefore $ab=\text{lcm}(a,b)$ and $ab\mid c$.
A: Since
$a|c$ and
$b | c$,
there are
$j, k$
such that
$aj = c$
and
$bk = c$.
Subce $gcd(a, b) = 1$,
there are
$u, v$
such that
$au-bv = 1$.
$aju-bjv = j$
and
$aku-bkv = k$
or
$cu-bjv = j$
and
$aku-cv = k$.
Once more,
$acu-abjv = aj = c$
and
$baku-bcv = bk = c$.
Adding these
(because subtracting
didn't do it),
$2c
=c(au-bv)-ab(ku-jv)
=c-ab(ku-jv)
$
or
$c
=ab(jv-ku)
$
so
$ab | c$.
Wadda ya know?
I played around for a while
and it worked!
