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Find all positive integers $n$ such that $n^5 - 5n^3 + 5n + 1 | n!$

I know that $ n^5-5n^3+5n+1=(n+1)(n^4-n^3-4n^2+4n+1)$, but I have no idea where to go from here.

This was from a local contest.

If there are an infinite amount, I would like to know like a "general" solution.

Adola
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    Well, a natural first step is to simply search. Maybe you can learn something by examining small $n$. – lulu Jun 02 '20 at 16:27
  • There's an old post here, with something similar like $n^9+1 \mid n!$, and it got answered. I can't find the link. – Bart Michels Jun 02 '20 at 16:32
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    $693, 4613, 5587, 5873, 6000, 6239, 7293, 7660, 7964, ...$. Sequence doesn't seem to be in OEIS (yet). – Robert Israel Jun 02 '20 at 16:40
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    @BartMichels May be you were thinking about this? Joni Teräväinen's answer in particular. I'm afraid Schinzel's theorem is above my paygrade. – Jyrki Lahtonen Jun 02 '20 at 17:14
  • Yes! Great you found it! – Bart Michels Jun 02 '20 at 17:18
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    The zeros of that quartic are $2\cos(2^k\pi/15)$ with $k=1,2,3,4$. In other words, it splits in the real subfield of the fifteenth cyclotomic field. I betcha that plays a role. – Jyrki Lahtonen Jun 02 '20 at 17:35
  • For its part, that comes from the connection to Chebyshev polynomials. If $P(x)=x^5-5x^3+5x$, then $$P(2\cos\alpha)=2\cos5\alpha$$ for all $\alpha$. – Jyrki Lahtonen Jun 02 '20 at 17:44
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    I suspect the contest screwed this up. The quintic polynomial $f(n)$ given should be better than $n$-smooth infinitely often and, with a little work, one should be able to show that for many of these values, we have $f(n) \mid n!$. An explicit characterization of such $n$ would be surprising to me. – Mike Bennett Jun 02 '20 at 17:57

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Possibly, the explicit despiprion of such $n$ is beyond our possibilities. But for infiniteness of such $n$ user Math1Zzang on mathlinks proved the next generalization of your problem (post #3 here: https://artofproblemsolving.com/community/c6h1793837p11880596):

For every positive integer $m$, there exists polynomial $P_{m}(x)$ such that $P_{m}\left(x+\frac{1}{x}\right)=x^{m}+\frac{1}{x^m}$. Then there exists infinitely many positive integers $n$ for which $P_{m}(n)+1|n!$

Pavel Kozlov
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