prove that $a^{25}-a$ is divisible by 30 I'm trying to prove that $30|a^{25}-a$.
First of all I said that it equals
$a(a^3-1)(a^3+1)(a^6+1)(a^{12}+1)$, so that it must be divisible by 2.
Now I want to show it can be divided by $3,6$ using Fermat's little theorem.
I need a bit of direction since I'm kind of lost.

You need divisibility by $2,3,5$.
$a$ and $a^3-1$ are of different parity, so one of them is odd, one is even. Thie product is of course even.
$a^3\equiv a\pmod{3}$, so one of numbers $a^3-1$, $a$, $a^3+1$ is always divisible by $3$.
Now, for $5$.


*

*The case $5|a$ is trivial.

*$a\equiv 1\pmod{5}\implies a^3\equiv 1\pmod{5}\implies 5|a^3-1$

*$a\equiv 2\pmod{5}\implies a^6+1\equiv 65\equiv 0\pmod{5}\implies 5|a^6+1$

*$a\equiv 3\pmod{5}\implies a^6\equiv 9^3\equiv (-1)^3\equiv -1\pmod{5}\implies 5|a^6+1$

*$a\equiv 4\pmod{5}\implies a\equiv -1\pmod{5}\implies a^3\equiv -1\pmod{5}\implies 5|a^3+1$
