Hi I would like to ask whether this is a liable solution, $(a,b)$ meaning the greatest common divisor of $a$ and $b$:
Show that if $a>0$ then $(ab,ac)=a(b,c)$.
Assume that $(b,c)=n$ then $n$ is expressible as the smallest positive linear combination of $b$ and $c$. So we have $$ n=bk+cl $$ for some $k,l\in\mathbb{Z}$. Multiplying by $a$ on both sides we get $$ an=abk+acl $$ since $a>0$ this is still the smallest positive linear combination of $ab$ and $ac$ and therefore $an$ is the greatest common divisor of $ab$ and $ac$ which is the desired result.
Yes. Your argument is fine. However, I would put a little bit more effort in justifying
For instance, derive a contradiction from this: suppose we had $\tilde{k},\tilde{l}\in\mathbb{Z}$ s.t. the linear combination $ab\tilde{k}+ac\tilde{l}$ was positive and less than $abk+acl$. That is, $ab\tilde{k}+ac\tilde{l}<abk+acl$ or which amounts to the same thing, $a(b\tilde{k}+c\tilde{l})<a(bk+cl)$. Why can't this happen?