prove that $(k,mn)=(k,m)(k,n)$ $\forall k\in \mathbb Z$ and $(m,n)=1$:
My attempt:
let $b=(k,m)$, $c=(k,n)$ and $a=(k,mn)$then there exist $x_{1},x_{2},x_{3},y_{1},y_{2},y_{3}\in \mathbb Z$ so that $b=kx_{1}+my_{1}$, $c=kx_{2}+ny_{2}$ and $a=kx_{3}+mny_{3}$ hence:
$$bc=(kx_{1}+my_{1})(kx_{2}+ny_{2})=k(kx_{1}x_{2}+nx_{1}y_{2}+my_{1}x_{2})+mny_{1}y_{2}$$
then by hypothesis we have that $(m,n)=1$ then $\forall k\in \mathbb Z$, $(k,m,n)=1$ hence:
$$kx_{1}x_{2}+nx_{1}y_{2}+my_{1}x_{2}=1$$
and: $$bc=k+mny_{1}y_{2}$$
then $x_{3}=1$ and $y_{1}y_{2}=y_{3}$ we have that $$bc=kx_{3}+mny_{3}=a$$
Is this proof correct? I would appreciate your help.
Hint $\,\ (k,m)(k,n) = (kk,km,kn,mn)= \overbrace{(k\color{#c00}{(k,m,n)},mn) = (k,mn)}^{\textstyle \color{#c00}{(k,m,n)=1}\ {\rm by}\ (m,n)=1}$
Remark $\ $ You can replace the above gcds by their Bezout rep's if your prefer to use integer arithmetic laws vs. gcd laws (distributive, associative, etc). However, this will be more messy due to the obfuscatory Bezout coefficients.