Revisted: GCD - $(a,c)=1=(b,c) \overset{?}{\implies} (ab,c)$ How should I show that if $(a,c)=1=(b,c)$ then $(ab,c)$?
How should I show that if $a|bc$ and $(a,b)|c$, then $a|c^2$. I think I have the answer, but I'm not sure.
 A: If $(a,c)=(b,c)=1$ then $(ab,c)=1$. Here is the proof: 
By Bézout's identity, $(a,c)=1$ if and only if there exists $x$, $y$ such that $ax+cy=1$, and $(b,c)=1$ so there exists integers $u$ $v$ such that $bu+cv=1$. Therefore
$$(ax+cy)(bu+cv)=(ab)(xu) + c(axv+byu+cyv)$$
and by Bézout's identity we get $(ab,c)=1$.
A: It is true that $(a,b)=1\iff xa+yb=1$ for some $x,y\in\mathbb Z$.
From $(a,c)=(b,c)=1$ it follows that $ax_1+cy_1=1, \ bx_2+cy_2=1$ for some $x_1,x_2,y_1,y_2\in\mathbb Z$. Multiply the last two equations to show that $(ab,c)=1$.
A: Suppose there exists a prime factor $p$, dividing both $ab$ and $c$. Then $p$ divides the product $ab$, so it must divide either $a$ (and $c$, contradicting coprimality of $a$ and $c$) or $b$ (and $c$, contradicting coprimality of $b$ and $c$). Hence $(ab,c)=1$
A: In the second problem, you wanted to show that is $a|bc$ and $(a,b)|c$, then $a|c^2$.
The natural thing, albeit somewhat clumsy thing  to do is to let the $\gcd$ $(a,b)$ of $a$ and $b$ be $d$. Let $a=da_1$, $b=db_1$, and $c=dc_1$. Then $(a_1,b_1)=1$.
From $a|bc$ we conclude that $a_1$ divides $b_1c$. Since $a_1$ and $b_1$ are relatively prime, it follows that $a_1|c$. Let $c=a_1c_2$. 
So $c^2=(dc_1)(a_1c_2)=a(c_1c_2)$, and the result follows.
A: I hope to provide intuition. This is how I think about prime factors, anyway.
Let the $\color{red}{b}$ be the set of all distinct prime factors that make up $b$ (e.g $\{2,5\}$ represents $10$ or $20$...). Now, the initial condition of $\gcd(a,c)=\gcd(a,b)=1$ is equivalent to 

(Note that the intersection of $a$ and $b$ may be empty if $a$ and $b$ are coprime). $a,c$ and $b,c$ share no prime factors, so the intersection between $\color{blue}{a},\color{gold}{c}$ and $\color{red}{b},\color{gold}{c}$ is empty: they don't overlap.
Now, the product of prime factors in $ab$ is the product of the prime factors of $a$ multiplied by those of $b$.  Euclid says this is the only product of primes that makes $ab$. So an element in $ab$ is in $\color{blue}{a}$ or $\color{red}{b}$ or both. Incorporating this into the diagram:

Approaching the second question, and switching our diagrams to sets of non-distinct prime factors (using capitals), $a|bc $ is equivalent to

and $(a,b)|c$ is equivalent to saying that the intersection $A \cap B$ should be within $C$. Therefore the yellow blob will move rightwards to encompass $A$, and $a$ will divide $c$ (meaning there exists some $k$ so that $ak=c$, let $k=qc$ to see that $a|c^2$).
