Let $\ n\ $ be an arbitrary natural number ($\ 1\ 2\ \ldots).\ $ Then


* $\ n\ $ is coarse $\ \Leftarrow:\Rightarrow\ $ there exists a prime divisor $p$ of $\ n\ $ such that $\ p^3>n.$;
* $\ n\ $ is a p-cube $\ \Leftarrow:\Rightarrow\ $ the positive cubical root of $\ n\ $ is a prime number;
* $\ n\ $ is fine $\ \Leftarrow:\Rightarrow\ p^3<n\ $
for every prime divisor $\ p\ $ of $\ n$.


**Example:**   Natural $\ 64\ $ and
$$ 4095=64^2-1\ = 3^2\cdot 5\cdot 7\cdot13 $$
are both fine. However,


$$ 4097=64^2+1=17\cdot241 $$


is coarse.


**QUESTION**   Does there exist a fine natural number $\ n\ $ such that both $\ n^2-1\ $ and $\ n^2+1\ $ are fine too? (*My guess: perhaps NOT*).


Also, I don't expect that there is any p-cube $\ n\ $ such that both $\ n^2-1\ $ and $\ n^2+1\ $ are fine.


On the other hand, I believe that there are infinitely many coarse $\ n\ $ such that both $\ n^2-1\ $ and $\ n^2+1\ $ are fine (as rare as they may be).



Can all three numbers $\ n\ \ n^2-1\ \ n^2+1\ $ be fine (as opposed to coarse)?
$n = 2673$ has largest prime factor $11$ whose cube is $1331$.  

$n^2 - 1 = 7144928$ has largest prime factor $191$ whose cube is $6967871$.  

$n^2 + 1 = 7144930$ has largest prime factor $61$ whose cube is $226981$.


