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The answer is most probably "It is always the set of all prime numbers or empty". Indeed, assume that $S$ is not empty. First of all, $S$ must contain $2$ (if it contains $p>2$, then $p^2+1$ is even). Then it also contains $2\cdot 2+1=5, 2\cdot 5+1=11, 7\mid 5\cdot 11+1, 3\mid 2\cdot 7+1,$ etc. It looks like then $S$ contains all prime numbers. I cannot prove it yet, but the proof should not be that difficult.

Update

Here is a stronger conjecture. Let $S$ be the set of all primes from $2$ to $p>17$. Let $q$ be the next prime after $p$. Then $q$ divides $rs+1$ for some $r,s\in S$. This of course implies the statement above. I wonder if my conjecture is already known. I checked it for all $p<35,000$.

The answer is most probably "It is always the set of all prime numbers or empty". Indeed, assume that $S$ is not empty. First of all, $S$ must contain $2$ (if it contains $p>2$, then $p^2+1$ is even). Then it also contains $2\cdot 2+1=5, 2\cdot 5+1=11, 7\mid 5\cdot 11+1, 3\mid 2\cdot 7+1,$ etc. It looks like then $S$ contains all prime numbers. I cannot prove it yet, but the proof should not be that difficult.

Update Here is a stronger conjecture. Let $S$ be the set of all primes from $2$ to $p>17$. Let $q$ be the next prime after $p$. Then $q$ divides $rs+1$ for some $r,s\in S$. This of course implies the statement above. I wonder if my conjecture is already known. I checked it for all $p<100,000$. I stop checking and will wait for an opinion of a real number theorist.

The answer is most probably "It is always the set of all prime numbers or empty". Indeed, assume that $S$ is not empty. First of all, $S$ must contain $2$ (if it contains $p>2$, then $p^2+1$ is even). Then it also contains $2\cdot 2+1=5, 2\cdot 5+1=11, 7\mid 5\cdot 11+1, 3\mid 2\cdot 7+1,$ etc. It looks like then $S$ contains all prime numbers. I cannot prove it yet, but the proof should not be that difficult.

Update

Here is a stronger conjecture. Let $S$ be the set of all primes from $2$ to $p>17$. Let $q$ be the next prime after $p$. Then $q$ divides $rs+1$ for some $r,s\in S$. This of course implies the statement above. I wonder if my conjecture is already known. I checked it for all $p<1500$.

The answer is most probably "It is always the set of all prime numbers or empty". Indeed, assume that $S$ is not empty. First of all, $S$ must contain $2$ (if it contains $p>2$, then $p^2+1$ is even). Then it also contains $2\cdot 2+1=5, 2\cdot 5+1=11, 7\mid 5\cdot 11+1, 3\mid 2\cdot 7+1,$ etc. It looks like then $S$ contains all prime numbers. I cannot prove it yet, but the proof should not be that difficult.

Update

Here is a stronger conjecture. Let $S$ be the set of all primes from $2$ to $p>17$. Let $q$ be the next prime after $p$. Then $q$ divides $rs+1$ for some $r,s\in S$. This of course implies the statement above. I wonder if my conjecture is already known. I checked it for all $p<35,000$.

The answer is most probably "It is always the set of all prime numbers or empty". Indeed, assume that $S$ is not empty. First of all, $S$ must contain $2$ (if it contains $p>2$, then $p^2+1$ is even). Then it also contains $2\cdot 2+1=5, 2\cdot 5+1=11, 7\mid 5\cdot 11+1, 3\mid 2\cdot 7+1,$ etc. It looks like then $S$ contains all prime numbers. I cannot prove it yet, but the proof should not be that difficult.

The answer is most probably "It is always the set of all prime numbers or empty". Indeed, assume that $S$ is not empty. First of all, $S$ must contain $2$ (if it contains $p>2$, then $p^2+1$ is even). Then it also contains $2\cdot 2+1=5, 2\cdot 5+1=11, 7\mid 5\cdot 11+1, 3\mid 2\cdot 7+1,$ etc. It looks like then $S$ contains all prime numbers. I cannot prove it yet, but the proof should not be that difficult.

Update

Here is a stronger conjecture. Let $S$ be the set of all primes from $2$ to $p>17$. Let $q$ be the next prime after $p$. Then $q$ divides $rs+1$ for some $r,s\in S$. This of course implies the statement above. I wonder if my conjecture is already known. I checked it for all $p<1500$.