Find maximum of n such that n^2 + n + 1 is factor of 19n^(2019) + 1 .


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Question Number 82514 by naka3546 last updated on 22/Feb/20

$F i n d m a x i m u m o f n s u c h t h a t$ $n^{2} + n + 1 i s f a c t o r o f 19 n^{2019} + 1 .$
Commented by naka3546 last updated on 22/Feb/20

$n \in N$
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