How can I prove that $a|bc$ if and only if $\frac{a}{(a,b)}|c$? I tried to start by showing that $ \frac{a}{\gcd(a,b)} $ is always an integer, let's call it $d$, because $a$ is always a multiple of $(a,b)$ based on the definition of a g.c.d. I then tried to show that if $a | b$ then $a|bc$ from the theorem that states that $a|b$ implies that $a|bc$ for any integer $c$, but I cant seem to prove that $a | b$ or that $a|bc$ only if $d \mid c$ (where $d = \frac{a}{\gcd(a,b)}$ ). Any help would be much appreciated.
Note that $x|y$ if and only if $(x,y) = x$; then use the properties of the gcd.

\begin{align*}
a|bc &amp;\Longleftrightarrow (a,bc) = a\\\
&amp;\Longleftrightarrow \left(\frac{a}{(a,b)}, \frac{bc}{(a,b)}\right) = \frac{a}{(a,b)}\\\
&amp;\Longleftrightarrow \left(\frac{a}{(a,b)}, \frac{b}{(a,b)}c\right) = \frac{a}{(a,b)}\\\
&amp;\Longleftrightarrow \left(\frac{a}{(a,b)}, c\right) = \frac{a}{(a,b)}&amp;&amp;\mbox{(by Euclid's Lemma)}\\\
&amp;\Longleftrightarrow \frac{a}{(a,b)} \Bigm| c,
\end{align*}
using Euclid's Lemma ("If $(x,y)=1$, then $x|yz\Leftrightarrow x|z$") and 
$$1 = \frac{1}{(a,b)}(a,b) = \left(\frac{a}{(a,b)},\frac{b}{(a,b)}\right).$$

The manipulations are consequences of $x(y,z) = (xy,xz)$; see Bill Dubuque's proof here.
$$a\mid bc \Longleftrightarrow \frac{a}{(a,b)}\mid c$$


  Show:
  
  $\Longrightarrow$
  
  $a\mid bc\Rightarrow\exists k\in\mathbb{N}$ such that $bc=ak$$$bc=ak\Rightarrow\frac{bc}{(a,b)}=\frac{ak}{(a,b)}\Rightarrow\frac{b}{(a,b)}c=\frac{a}{(a,b)}k\Longrightarrow$$$$\frac{a}{(a,b)}\mid\frac{b}{(a,b)}c$$
  
  Two theorems that we will use $$\text{If}\;\;x\mid yz\;\;\text{and}\;\;(x,y)=1\Longrightarrow x\mid z\;\;\;\;(\text{Theorem}\;*)\\\text{and}\\(\frac{x}{(x,y)},\frac{y}{(x,y)})=1\;\;\;\;(\text{Theorem}\;**)$$
  
  Returning to the question, we have $$\frac{a}{(a,b)}\mid\frac{b}{(a,b)}c$$ by theorems $*$ and $**$ we have $$\frac{a}{(a,b)}\mid c\;\;\;\;\;\Box$$
  
  $\Longleftarrow$
  
  $$\frac{a}{(a,b)}\mid c\Longrightarrow a\mid bc$$
  
  $$\frac{a}{(a,b)}\mid c\Longrightarrow\exists k'\in\mathbb{N}\;\;\text{such that}\;\; c=\frac{a}{(a,b)}\cdot k'\\bc=\frac{ab}{(a,b)}\cdot k'\Longrightarrow bc=a(\frac{b}{(a,b)}\cdot k')\Longrightarrow a\mid bc\;\;\;\;\;\Box$$