Let $a, b, x \in Z$, let $n ∈ N$ and let $h = hcf(a, n)$. Suppose that $h | b$, let $a'=\frac{a}{h}, b'=\frac{b}{h}$ and $n'=\frac{n}{h}$. Prove that $a'$ is comprime to $n'$.
By Bezout's lemma, there exist integers $u$ and $v$ such that $ax+nv=h$. Dividing by $h$ and using the properties given we get $a'x+n'y=1$. Let $h'=hcf(a',n')$. $h' | a$ and $h'|n'$, so by a lemma $h'|a'x+n'y=1$. Hence, if $h'|LHS$, $h'|RHS$, so $h' = 1$ or $h'=-1$. Since highest common factor must be positive, $h'=1$, so $a'$ and $n'$ are coprime.
I'm unsure of my method after dividing by $h$. Could someone please check this ?