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.

$a(a^{4n}-1)=a(a^{2n}-1)(a^{2n}+1)=a(a^{n}-1)(a^{n}+1)(a^{2n}+1)$.
Assume $30 | a(a^{4n}-1)$.
$30 | a(a^{4(n+1)}-1)$ if and only if $30 | a(a^{4(n+1)}-1)-a(a^{4n}-1)$.
Try factoring: $a(a^{4(n+1)}-1)-a(a^{4n}-1)$.  Think about how regardless of $a$, you can show that one of the terms of the factorization must be divisible by 2, another divisible by 3, and another divisible by 5.
