Prove that if $\gcd(a,c)=1$ and $\gcd(b,c)=1$ then $\gcd(ab,c)=1$ Prove that if $\gcd(a,c)=1$ and $\gcd(b,c)=1$ then $\gcd(ab,c)=1$. My attempt is let $\gcd(ab,c)=d$. Since $d \mid ab$ and $d \mid c$ , $d \mid (abt+cs)$ for some integers $s$ and $t$. Then by definition of divisibility, we have $$abt+cs=dk$$ for some integers $k$. Then I got stuck here.
 A: Hint 1: First approach. Assume by contradiction that $d \neq 1$. Pick some prime $p$ which divides $d$. Then $p\mid ab$ and $p\mid c$. Do you see the contradiction?
Hint 2: Second approach. You know that 
$$ax+cy=1$$
$$bz+ct=1$$
For some positive integers.The second equation yields
$$abz+act=a$$
Plugging in the first equation you get
$$(abz+act)x+cy=1$$
But this implies that $\gcd(ab,c)=1$ (Why?)...
A: Using the fundamental theorem of arithmetic, if $a$ and $c$ are coprime then there is no common prime in their prime factorisations. The same goes for $b$ and $c$, and hence the same goes for $ab$ and $c$.
A: Theorem $\rm\qquad gcd(ab,c)\,|\,\gcd(a,c)\,gcd(b,c)$
$\begin{eqnarray}\rm{\bf Proof}\qquad\quad\ \ gcd(a,c) &=&\:\rm j\:a+m\:c \quad for\ some\,\ j,m\in \Bbb Z\ \ by\ Bezout\\
\rm gcd(b,c) &=&\:\rm k\:b+n\:c \quad for\ some\,\ k,n\in \Bbb Z\ \ by\ Bezout\\
\rm\ \  \Rightarrow\ \  gcd(a,c)\,gcd(b,c) &=&\:\rm jk\,\color{#C00}{ ab} + \color{#C00}c\,(\cdots)\ \ \ for\, \ \ (\cdots)\in \Bbb Z
\end{eqnarray}$
Since $\rm\:gcd(ab,c)\,|\,\color{#C00}{ab,c},\:$ it divides the right-side of prior, so also the left. $\ $ QED
A: By the fundamental theorem of arithmetic, every integer greater than $1$ has a unique prime factorization, except for the order of the prime factors. 
Now as gcd($a$, $c$) and gcd($b$, $c$) are both $1$, it is obvious that $a$ and $c$ have no common prime factors and $b$ and $c$ have no common prime factors. So $ab$ and $c$ can have no common prime factors. Hope this helps. 
A: (I’m assuming that you don’t have the fundamental theorem of arithmetic available; if you do, the result is trivial, since $c$ has no prime factors in common with $a$ and $b$. I’m assuming that you do have the fundamental result that if $m\mid kn$ and $\gcd(m,k)=1$, then $m\mid n$.) 
HINT: Suppose that $d\mid c$. Since $\gcd(c,a)=1$, we know that $\gcd(d,a)=1$. If $d\mid ab$, therefore, we can conclude that $d\mid b$, and hence ... ?
A: Since $$\gcd(a,c)=1=\gcd(b,c),$$
then there exists $m,m',n,n' \in \mathbb{Z}$ such that $$ma+nc=1=m'b+n'c.$$
Thus because 
\begin{align} 
1&=1\times1 \\&=(ma+nc)\times(m'b+n'c)\\&=(m)(m')ab+(man'+nm'b+ncn')c,
\end{align} 
with $mm',man'+nm'b+ncn'\in \mathbb{Z}$
hence we must have $\gcd(ab,c)=1$.
