Let $a,b,c,r,s\in\mathbb{Z}$ such that $(a,b)=r$, $(a,c)=s$ and $(b,c)=1$. Prove that $(a,bc)=rs$. Let $a,b,c,r,s\in\mathbb{Z}$ such that $(a,b)=r$, $(a,c)=s$ and $(b,c)=1$. Prove that $(a,bc)=rs$.
Proof: $ax_1+by_1=r$ and $ax_2+cy_2=s$ for some $x_i\in\mathbb{Z}$. This gives
\begin{align*}
by_1=r-ax_1\\
cy_2=s-ax_2\\
\end{align*}
Multiplying the two equations, we obtain
\begin{align*}
bcy_1 y_2&=(r-ax_{1})(s-ax_{2})\\
&=rs-a(rx_{2}+sx_{1}-ax_{1}x_{2})\\
a(rx_{2}+sx_{1}-ax_1 x_2)+bcy_1 y_2 &=rs\\
\end{align*}
Therefore, $(a,bc)=rs$.
Is this proof enough?
 A: Restriction $(b,c)=1$ is critical, so it's impossible to get the answer without using it! But you can add this to your statement to get the complete proof:
$$(a,b)=r \Longrightarrow r|a \ \& \ r|b$$
$$(a,c)=s \Longrightarrow s|a \ \& \ s|c$$
Since $r,s$ are divisors of coprime integers $b,c$ they are coprime too, $(r,s)=1$.
Now $r,s$ are coprime divisors of $a$, so $rs$ is a divisor of $a$ too, $rs|a$.
So we have:
$$r|b \ \& \ s|c \ \Longrightarrow rs|bc$$
$$rs|a \ \& \ rs|bc \ \Longrightarrow rs|(a,bc)$$
Now the inverse of what you proved (that @mathcounterexamples said in comment) is shown, and this complete your proof.
A: $ rs = (a,b)(a,c) = ((a,b)a,(a,b)c) = (aa,ab,ac,bc) = (a(a,\color{#c00}{b,c}),bc) = (a,bc)\  $ by $\,\color{#c00}{(b,c)\! =\! 1}$
Remark $ $ We used only basic gcd laws (associative, commutative, distributive). Note that the proof needs only $\,(a,b,c) = 1,\,$ which is weaker than $\,(b,c) = 1$
Your argument only shows one direction $\,(a,bc)\mid rs.\,$ You also  need to show the reverse $\,rs\mid (a,bc)\,$ (which follows from $\,r,s\mid (a,bc)\,$ and $\,(r,s) = (a,\color{#c00}{b,c}) = 1\,$ and Euclid's Lemma or gcd laws or similar results closely related to uniquness of prime factorizations.
Note how much simpler the proof is using gcd arithmetc (vs, Bezout equations), whose arithmetical laws are intuitive being very close to normal arithmetic. In fact we could choose notation to make this similarity clearer, e.g. see here.
