<s>[INST] How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but I don't know how.
 [/INST] Let $f(n)=n^p-n$
Clearly, f(1) =1-1=0 is divisible by p.
Now, $f(n+1)-f(n)=((n+1)^p-(n+1))-(n^p-n)= \sum _{1≤r≤p-1} pC_rn^r$.
But prime p| $pC_r$  for 1≤r≤p-1.
So, the difference is divisible by p.
So, using mathematical induction, we can say p|($n^p-n$)
This comes from Fermat's Little theorem.
So,  $n^5-n$ is divisible by 5.
$n^5-n = n(n^4-1) =  n(n^2-1)(n^2+1) = (n^3-n)(n^2+1)$
Now,  $n^3-n$ is divisible by 3, so is $n^5-n$.
$n^5-n = n(n^4-1) =  n(n^2-1)(n^2+1) = n(n-1)(n+1)(n^2+1) = (n^2-n)(n+1)(n^2+1)$
Now,  $n^2-n$ is divisible by 2, so is $n^5-n$.
So, $n^5-n$ is divisible by 2,3,5 so, it is divisible by lcm(2,3,5)=30
</s>