<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] Hint:
Consider the factors of $n^{30}-n^{18}-n^{14}+n^2$
$$(n-1)^2,n^2,(n+1)^2,\left(n^2+1\right)^2,n^2-n+1,n^2+n+1,n^4+1,n^4-n^2+1,n^8+1$$
for any $n>1$ we have an even number and a multiple of $3$ in the first three factors.
With similar arguments it can be shown that at least one of the factor is a multiple of $5,7,11,13,17$ which are the other factors of $46410$.
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