Prove: If $a|c \wedge b | c \wedge (a, b) = d \Rightarrow ab | cd$

Question

I know that $(a,b)=d \Rightarrow ma+bn=d, (m,n\in Z)$.

$ma+bn=d/*c \Rightarrow cma+cnb=cd$

And I'm kinda stuck here. Any help or hint is appreciated.

Answer 1

You're very close. Note that $a\mid c$ and $b \mid c$, so you can do two substitutions: $c \mapsto as$ and $c \mapsto bt$ for some integers $s, t$. Now the last line you have reads $$ btma + asnb = cd $$ Can you finish?

Answer 2

Hint $$ mac+bnc=cd$$

Now show that $ab|mac$ and $ab|bnc$.