Values of 'n' for $n^{2019}+20$ to be divisible $n^2+n+1$ For how many integers $n$, $n^2+n+1$ is a divisor of $n^{2019}+20$?
Could someone give me some hint to proceed in this question? $n=-1,0,1$ are the values obtained by hit and trial but how to get all values?

$$n^{2019}+20=n^{2019}-1+21=(n^3)^{673}-1+21=$$
$$=(n^3-1)\left(n^{2016}+...+1\right)+21=(n-1)(n^2+n+1)\left(n^{2016}+...+1\right)+21,$$
which says that $\frac{21}{n^2+n+1}\in\mathbb Z$ and an the rest is  easy checking.
I got $\{-5,-3,-2,-1,0,1,2,4\}$.
