Timeline for Find every prime $p$ such that $p^3$ divides $(p-1)!+1$ [duplicate]
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Sep 10, 2019 at 12:23 | history | closed | Gerry Myerson nt.number-theory Users with the nt.number-theory badge or a synonym can single-handedly close nt.number-theory questions as duplicates and reopen them as needed. | Duplicate of Stronger versions of Wilson's Theorem | |
Sep 10, 2019 at 11:26 | history | edited | apple | CC BY-SA 4.0 |
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Sep 10, 2019 at 7:53 | comment | added | Greg Martin | The usual heuristic is that the "probability" that a prime $p$ satisfies $p^3 \mid \big( (p-1)!+1\big)$ is $1/p^2$. Since there are no such primes up to $2\times10^{13}$ (as per the Wikipedia article on Wilson primes), our heuristic is that there is a (much) less than $1$ in $2\times10^{13}$ chance that any such primes exist. But yes, a proof seems hopeless at present. | |
Sep 10, 2019 at 7:26 | comment | added | Gerry Myerson | I suspect the answer to all three of your questions is, nobody knows. | |
Sep 10, 2019 at 6:08 | history | asked | apple | CC BY-SA 4.0 |