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:** &nbsp; 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** &nbsp; 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).