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.

Don't factor entirely.  Just factor enough to realize
$a^{4n+1} - a = a(a^{4n} - 1)=a(a - 1)(a^{4n-1} + a^{4n-2} + .... + a + 1)$.
Assume for $n = k$ that
$a^{4k + 1}-a= a(a^{4n} - 1)=a(a - 1)(a^{4k-1} + a^{4k-2} + .... + a + 1)= 30M$
then $a^{4(k+1) + 1} - a = a(a-1)(a^{4k + 3} + ...)=$
$a(a-1)(a^{4k + 3} + a^{4k + 2} + a^{4k+1} + a^{4k}) +a(a - 1)(a^{4k-1} + a^{4k-2} + .... + a + 1)$
$= a(a-1)(a^{4k + 3} + a^{4k + 2} + a^{4k+1} + a^{4k}) + 30M$.
$= a(a-1)(a^3 + a^2 + a + 1)a^{4k} + 30M$
$= (a^{4*1 + 1} - a)a^{4k} + 30M$.
But $30|(a^{4*1 + 1} - a) $ By our initial step.
So $30|(a^{4*1 + 1} - a)a^{4k} + 30M$.
====
In hind sight it's easy to see that $a^{4(n+1) + 1} - a = a^{4(n+1) + 1} - a^{4n+1} + a^{4n + 1} - a = (a^{4*1 + 1} - a)a^{4n} - (a^{4n+1} - a)$.
So $a^{4*1 + 1} -a|a^{4*m + 1} - a;m \ge 1$ via induction.
