supposition: $A_1=\{k \in \mathbb{Z}\ \colon\ k | (bc +a) \text{ and }k |b\}$, $A_2=\{k \in \mathbb{Z}\ \colon\ k|a \text{ and }k|b\}$
claim: $A_1=A_2 $
(my) proof: Let's show that $A_1 \implies A_2 $. Let $k \in A_1$, so $k \mid (bc +a)$ and $k \mid b$. Now if $k \mid (bc+a)$, then $k \mid 1 \cdot (bc+a) -bc$ $\implies$ $ k \mid a $. So $A_1 \implies A_2 $.
I'm not sure how to show $k \mid 1 \cdot (bc+a) -bc$?