Can all three numbers $\ n\ \ n^2-1\ \ n^2+1\ $ be fine (as opposed to coarse)?

Question

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).

Answer 1

$n = 2673$ 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 $226981$.

Answer 2

For any $k\in\mathbb{N}$, the number $n=2^3\cdot3^{72k}$ works:

Obviously, $n$ itself is fine.

For $n^2-1 = 2^6\cdot3^{144k}-1 = (2^2\cdot 3^{48 k} + 2\cdot 3^{24 k} + 1)(2^2\cdot 3^{48 k}-2\cdot 3^{24 k} + 1)(2\cdot3^{24 k} + 1) (2\cdot3^{24 k} - 1)$, the first factor is divisible by $7$, the others are small, hence $n^2-1$ is fine.

For $n^2+1 = 2^6\cdot3^{144k}+1 = (2^2\cdot3^{48k}+2^2\cdot3^{36k}+2\cdot3^{24k}+2\cdot3^{12k}+1)(2^2\cdot3^{48k}-2^2\cdot3^{36k}+2\cdot3^{24k}-2\cdot3^{12k}+1)(2\cdot3^{24k}+2\cdot3^{12k}+1)(2\cdot3^{24k}-2\cdot3^{12k}+1)$, the first factor is divisible by $13$, the others are small, hence $n^2+1$ is fine.