Prove that if $a\mid(b-1)$ and $a^2\mid(b^2-2b+4)$, then $a\mid12$. Prove that if $a\mid(b-1)$ and $a^2\mid(b^2-2b+4)$, then $a\mid12$.
I am not sure where to start for this question, any help would be greatly appreciated, thanks!

Suppose $b-1=ak$, $k\in\mathbb Z$. Then $b^2-2b+4=(b-1)^2+3=a^2k^2+3$, hence $a^2$ divides both $a^2k^2$ and $a^2k^2+3$. Therefore it divides $3$, but $3\mid12$!
