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

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*}

