Prove that if $a \mid bc$ and $\gcd(a,b) = d$ then $a/d$ divides $c$.
Attempt at solution
\begin{align} &a \mid bc \implies \exists \ x \in \mathbb{Z} \ \text{s.t} \quad bc = ax \tag{1}\\ &\gcd(a,b) = d \implies a \mid b \implies \exists \ t \in \mathbb{Z} \ \text{s.t} \quad b = at\tag{2} \\ &\gcd(a,b) = d \implies a \mid d \implies \exists \ s \in \mathbb{Z} \ \text{s.t} \quad d = as\tag{3} \end{align} From $(2)$ , we get $a = b/t$. Substituting in $(3)$, we have $d = as = bs/t \implies b = dt/s$. Substituting this last result in $(1)$ ; $$ bc = ax \implies \frac{dt}{s} c = ax.$$ Since $t,s$ and $x$ are all arbitrary constants in $\mathbb{Z}$, define $sx/t := x^{\prime} \in \mathbb{Z}$. Finally, $$ dc = ax^{\prime}\implies c= \frac{a}{d}x' \implies \frac{a}{d} \mid c \qquad \square.$$
Can anyone confirm my proof ? I feel like there are some justifications missing. Particularly in the substitions part could there not be divisions by $0$?
