Prove that if $a\mid(b-1)$ and $a^2\mid(b^2-2b+4)$, then $a\mid12$.

Question

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!

Answer 1

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 $3\mid12$!

Answer 2

Well:

$$a|(b-1)\implies a^2|(b^2-2b+1)$$

So for a particular $t$ we have $a^2|t$ and $a^2|(t+3)$

Can you continue this?

Answer 3

As has been stated in other answers,

$$a \mid b - 1 \Rightarrow a^2 \mid b^2 - 2b + 1$$

as $\left(b - 1\right)^2 = b^2 - 2b + 1$, so

$$a^2 \mid \left(b^2 - 2b + 4\right) - \left(b^2 - 2b + 1\right) \Rightarrow a^2 \mid 3$$

Since $3$ is a prime, this can only be true if $a = 1$, which of course means that $a \mid 12$.