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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\ $ are both fine. However, $\ 4093=64^2-1\ $ is coarse (it is a prime).

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).

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).

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\ $ are both fine. However, $\ 4093=64^2-1\ $ is coarse (it is a prime).

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).

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\ $ are both fine. However, $\ 4093=64^2-1\ $ is coarse (it is a prime).

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).

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\ $ are both fine. However, $\ 4093=64^2-1\ $ is coarse (it is a prime).

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).

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\ $ are both fine. However, $\ 4093=64^2-1\ $ is coarse (it is a prime).

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).