I'm trying to prove "the following generalization of Theorem 5 [ Th.5: if $a|bc$ and $(a,b)=1$ then $a | c$ ], which uses the same argument for its proof" (Sierpinski, The Theory of Numbers): if $a$, $b$, and $c$ are integers such that $b | ac$, then $b | (a,b)(b,c)$.
I haven't been able to prove it without any reference to prime numbers (which the author introduces way later), using only divisibility and facts like $(a,b)[a,b]=ab$. Here's what I've done so far:
Let $(a,b)=d_a$, $(b,c)=d_c$, $a=a'd_a$, $c=c'd_c$. Since $b | ac$ and $a | ac$ $\Rightarrow [b,a] | ac$ (this is the argument used in Th5's proof) $\Rightarrow ab/(a,b) | ac \Rightarrow a'b | ac \Rightarrow b | d_a c$. Doing the same for $c$, we get $b | ad_c$. From this we also have $b | (ad_c,d_a c)$.
Thank you for your help.