Prove by induction that $a^{4n+1}-a$ is divisible by 30 for any a and $n\ge1$ It is valid for n=1, and if I assume that $a^{4n+1}-a=30k$ for some n and continue from there with $a^{4n+5}-a=30k=>a^4a^{4n+1}-a$ then I try to write this in the form of $a^4(a^{4n+1}-a)-X$ so I could use my assumption but I can't find any $X$ that would set the two expressions equal.
Then I tried factoring $a^{4n+1}-a=30k$ as $a(a^n-1)(a^n+1)(a^{2n}+1)=30k$ and using that as my assumption. Then I also factor as $a^{4n+5}-a=a(a^{n+1}-1)(a^{n+1}+1)(a^{2n+2}+1)$ and again I'm stuck not being able to use my assumption.
Please note that I strictly need to use induction in this problem.

Here is how I would write up the main part of the induction proof (DeepSea and Bill handle the base case easily), in the event that you may find it useful:
\begin{align}
a^{4k+5}-a&= a^4(a^{4k+1}-a)+a^5-a\tag{rearrange}\\[1em]
&= a^4(30\eta)+a^5-a\tag{by ind. hyp.; $\eta\in\mathbb{Z}$}\\[1em]
&= a^4(30\eta)+30\ell\tag{by base case; $\ell\in\mathbb{Z}$}\\[1em]
&= 30(a^4\eta+\ell)\tag{factor out $30$}\\[1em]
&= 30\gamma.\tag{$\gamma\in\mathbb{Z}$}
\end{align}
