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I have been stuck on this question for quite some time, I have tried several methods but to no avail. I attempted to use prime factorization but I couldn't really see where to go with it.

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    $\begingroup$ Try showing $(a^2 - 1) \mid (a^4 - 1)$. The $m$ just distracts. $\endgroup$ Oct 21, 2015 at 10:17

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$a^4-1=(a^2)^2-1^2=(a^2-1)(a^2+1) \Rightarrow (a^2-1)|(a^4-1).$

If $m|(a^2-1)\Rightarrow \exists n$ sunch that $a^2-1=mn\Rightarrow a^4-1=mn(a^2+1)\Rightarrow m|(a^4-1)$

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  • $\begingroup$ Thank you! all makes perfect sense now! $\endgroup$
    – Geoff
    Oct 21, 2015 at 10:54

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