Prove or disprove $d\mid (a^2-1)\Longrightarrow d\mid (a^4-1)$

Question

I need to prove or to give a counter-example:

$$d\mid (a^2-1)\Longrightarrow d\mid (a^4-1)$$

My attempt:

Yes, this is correct,

First: $(a^2-1)=(a-1)(a+1)\\ (a^4-1)=(a-1)(a+1)(a^2+1)$

If $d\mid (a-1)(a+1)$ so $\exists k_1 : d\cdot k_1=(a-1)(a+1)$

Therfore $d$ must divide also $(a-1)(a+1)(a^2+1)$

Is my attempt correct?

Answer 1

The factorization $a^2-1=(a-1)(a+1)$ is not helpful in proving or disproving this claim. Essential is the question if $a^2-1$ divides $a^4-1$. So, the factorization of \begin{align*} a^4-1 \end{align*} is relevant. In case the following Lemma is already known to you, you could state:

Since the following Lemma holds \begin{align*} d|m\Rightarrow d|mn\quad\qquad \text{for all } d,m,n\in\mathbb{Z} \end{align*} and \begin{align*} a^4-1=(a^2-1)(a^2+1) \end{align*} the claim is valid by taking \begin{align*} &m=a^2-1\\ &n=a^2+1\\ &\qquad\qquad\qquad\qquad\Box \end{align*}

Answer 2

Hint:

See that $a^4-1=(a^2+1)(a^2-1)$

Answer 3

Here is another take: $$ a^2=1+kd \implies a^4=(1+kd)^2=1+2kd+k^2d^2 \implies a^4-1=d(2k+k^2d) $$

The fact that we're dealing with $a^2$ is not relevant. The same proof works in general: $$ d \mid b-1 \implies d \mid b^n-1 $$ for all $n \in \mathbb N$. You just need the binomal theorem to get $(1+kd)^n = 1+td$.