<s>[INST] Prove that $n^{30}-n^{14}-n^{18}+n^2$ is divisible by $46410$ I have the following question.

Prove that $n^{30}-n^{14}-n^{18}+n^2$ is divisible by $46410$ for all positive integer $n$.

My attempt:
Firstly, notice that $$n^{30}-n^{14}-n^{18}+n^2$$
$$=n^2(n-1)^2(n+1)^2(n^2+n+1)(n^2-n+1)(n^2+1)^2(n^4-n^2+1)(n^4+1)(n^8+1).$$
Also, $$46410=2\times3\times5\times7\times13\times17$$
My first thought was to use induction, maybe substitute $n+1$ into $n$, then find terms that was already divisible by $46410$, but then I don't think this is a smart way to do it.
Are there any better ways to do it?
 [/INST] Because it divided by $n^2-n$, by $n^3-n$, by $n^5-n$, by $n^7-n$, by $n^{13}-n$ and by $n^{17}-n$ and use the Fermat's little theorem,
which gives that $n^2-n$ divided by $2$,  $n^3-n$ divided by $3$,$n^5-n$ divided by $5$,
$n^7-n$ divided by $7$, $n^{13}-n$ divided by $13$ and $n^{17}-n$ divided by $17$,
which says that our number divided by $2\cdot3\cdot5\cdot7\cdot13\cdot17=46410$
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