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Conjectures:

For all prime numbers $p>5$ there exist a prime number $q<p$ such that $q\equiv m!\!\pmod p$, for some $m$ with $2<m<p$.

and

For all primes $p$ there is a prime $q<p^2$ such that $q\equiv m!\pmod {p^2}$, for some $m$ with $2<m<p^2$. (Even $2<m<p$ as pointed out in comment).

Both tested for all $p<100,000,000$.

There are reasons to believe that the conjectures are true. If $p$ is a large prime there are a lot of nonzero solutions to $n\equiv m!\!\pmod p$ and to $n\equiv m!\!\pmod {p^2}$. The probability for one of those $n$ to be a prime increase with p.

https://math.stackexchange.com/questions/1838931/conjecture-about-primes-and-the-factorial

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  • $\begingroup$ If $p$ is prime then $m!\equiv0\bmod{p^2}$ for all $m\ge2p$, indeed $m!$ is divisible by $p$ for all $m\ge p$, and there is no prime $q$ satisfying either condition except $q=p$. $\endgroup$
    – Gerry Myerson
    Commented Jul 6, 2016 at 5:43
  • $\begingroup$ @GerryMyerson: yes, the condition for $m$ in the second conjecture could be $2<m<p$. The statement "It exist a prime $q<n$ so that $q\equiv m!\pmod n$, for some $m$ with $2<m<n$" is true for almost 1/3 of all numbers $n$, as it seems, and I wanted to expose the general form. $\endgroup$
    – Lehs
    Commented Jul 6, 2016 at 6:06

1 Answer 1

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You might consider the inverse problem: for small m and prime q, factorize m! - q. Note that when q is greater than m all such prime factors of the difference must also be greater than m. Thus for many prime factors p, m is a solution of the desired congruence. Your question now becomes how many prime factors p are covered this way for a small assortment of m. For example, for m=5, all the primes p between 73 and 113 have 120 equal a prime modulo p, since 120-p has to be less than 49 and have all prime factors greater than 7. This reasoning should be generalizable to the point of showing that exceptional primes p are either less than 7 or have density 0 in the primes.

Gerhard "Turns It Over To Professionals" Paseman, 2016.07.06.

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