prove that $a^{25}-a$ is divisible by 30

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.

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

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

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

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

Hint. By Fermat's little theorem, for any prime $p$, $a^{p}-a$ is divisible by $p$. Moreover, the polynomial $P(a)=a^{25}-a$ is divisible by $(a^r-a)$ for $r=2,3,5$.

As $(a^n-a)$ divides $a^{25}-a$ for $n=2,3,5,7,13$. For Little Fermat Theorem we know that $p$ divides $a^p-a$ for primes $p$, so $a^{25}-a$ is divisible by $2730$ for any integer $a>1$

Hope this helps