<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] By the Chinese remainder theorem, you just have to prove that $n^5\equiv n\mod 5$, $\bmod 3$ and $\bmod2$.
The case $n=5$ is exactly the statement of Little Fermat's theorem.
Also, the same theorem says that $n^3\equiv n\mod 3$, so $\;n^5=n^3\cdot n^2\equiv n\cdot n^2=n^3\equiv n$.
Similarly, modulo $2$, $\;n^5=\bigl(n^2\bigr)^2n\equiv n^2n\equiv n\cdot n=n^2\equiv n.$
</s>