<s>[INST] Showing "$30$ divides $n^5-n$ for all $n\in\Bbb N$" using induction 
Prove that $(n^5 - n)$ divides by $30$ for every $ n\in N$, using induction only. 

How on earth do I do that? Thing is  $(n^5 - n)$  can't be opened using any known formula...
 [/INST] Let $f(m)=m^5-m$
$\displaystyle\implies f(1)=1^5-1=0$
Let $f(m)$ is divisible by $30$ for $m=n$
Now $\displaystyle f(n+1)-f(n)=(n+1)^5-(n+1)-(n^5-n)=5n^4+10n^3+10n^2+5n$
$\displaystyle=5n(n^3-n)+10(n^3-n)+15n^2+15n$
$\displaystyle=5(n+2)n(n+1)(n-1)+15n(n+1)$
Observe that $n(n+1)(n-1),$ being product of $3$ consecutive integers is divisible by $3$ 
and $n(n+1)$ is divisible by $2$
$\displaystyle\implies n(n+1)(n-1)$ is divisible by lcm$(2,3)=6$
$\displaystyle\implies f(n+1)-f(n)$ is divisible by $30$
But, $f(n)$ is divisible by $30,\implies f(n+1)$ will be divisible by $30$
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