Let $a,b,m,n \in \mathbb Z$. If $(m,n)=1$ ( $m,n$ are coprime integers) prove that $(ma+ nb, mn)=(a,n)(b,m)$
I started the proof like this:
Let $c,d,e$ be the greatest common divisors of $(ma+nb,mn), (a,n), (b,m)$ respectively that is :
$c=x_{1}(ma+nb)+ y_{1}(mn)$, $d=x_{2}a+ y_{2}n$, $e=x_{3}b + y_{3}m$ for some $ x_{1}, y_{1}, x_{2}, y_{2}, x_{3}, y_{3}\in \mathbb Z$.
Then $$de=(x_{2}a+ y_{2}n)(x_{3}b + y_{3}m)=x_{2}x_{3}ab + x_{2}y_{3}am + y_{2}x_{3}nb + y_{2}y_{3}nm$$ but then I dont know what else to do. I can't use prime factorization in this proof.
I really would appreciate your help.
Hint $\ $ By $\,(m,n)=1\,$ we have $\,(ma\!+\!nb,mn) = (ma\!+\!nb,m)\,(ma\!+\!nb,n),\ $ and
$\qquad\quad (ma\!+\!nb,m) = (nb,m) = (b,m),\ $ and $\ \ \ (ma\!+\!nb,n) = (ma,n) = (a,n)$
Remark $\ $ The inference in the first line is the special case $\,(m,n) = 1\ $ of this
Lemma $\ \ (k,m)(k,n) = (k,mn)\ \ $ if $\ \ \color{#c00}{(k,m,n) = 1}$
Proof $\quad\ (k,m)(k,n) = (k^2,km,kn,mn) = (k\color{#c00}{(k,m,n)},mn) = (k,mn)$