greatest common divisor proves I have two exercises for my mathematic study, and I really can't prove them:
Let $a, b$ be in $\mathbb{Z}$. Prove:
(a) If $\gcd(a, b) = \gcd(a, c) = 1$ , then $\gcd(a, bc) = 1$ 
(b) If $\gcd(a, b) = 1$, and $a\mid c$ and $b\mid c$ then $ab\mid c$ 
I'm trying for two days to prove these exercises, but I'm not able to do it.
Thanks in advance!
 A: Hint:
(a), suppose $\gcd(a,b)=\gcd(a,c)=1$. Suppose $\gcd(a,bc)=d\ne 1$. So there is a prime $p$ such that $p|d$. How can you use this to get a contradiction?
(b) Have you heard of the theorem which says, if $a|c$ and $b|c$ then $lcm(a,b)|c$? You could use that very well. Or just go with the fundamental theorem of arithmetic.
Edit
You can also solve (b) using this
Suppose $a|c$ and $b|c$ and $\gcd(a,b)=1$
Then, $aq=c$ and $b|aq$. And Since $\gcd(a,b)=1\ldots\ldots$?
A: Use Bezout's identity, $ gcd(a,b)=1 \iff \exists k,s \in Z : ka +sb =1 $. 
a. $\alpha a+\beta b = 1$ ,$ \gamma a + \delta c =1 $. Writing $b=\frac{1-\alpha a}{\beta}$ and $c=\frac {1- \gamma a}{\gamma} $. Taking the product $ bc = \frac{(1-\alpha a) (1-\gamma a)}{\beta \delta} $. Multiplying and taking the terms containing a to the other side.
$\therefore \beta \delta bc +(\alpha +\gamma -\alpha \gamma a)a =1 $  Thus, gcd(a,bc)=1. 
b. $a|c\implies c=ka$ and $b|c \implies c=lb$. Also, $\alpha a +\beta b =1$ . Multiplying by kl gives $ kla\alpha +klb\beta =kl$. As $ kl = \frac{c^2}{ab}, \frac{kl}{c}=\frac{c}{ab}$.
$\therefore kl = \alpha cl + \beta ck \implies \frac{c}{ab}=\alpha l +\beta k \in Z $. Thus, $ab|c$
