Brocard's problem

As of 2026, Brocard's problem remains open: it is not even known unconditionally whether the number of solutions to $n! + 1 = m^2$ is finite. Under the $abc$ conjecture, Overholt's 1993 argument readily bounds $n$ because the radical $\mathrm{rad}(n!(m-1)(m+1))$ is at most $m \prod_{p \le n} p \approx m e^{(1+o(1))n}$, which is much smaller than $m^2 = n! + 1 \approx (n/e)^n$. Computationally, quadratic-residue sieves have ruled out any solution with $7 < n \le 10^{15}$, leaving little doubt that $(4,5)$, $(5,11)$, and $(7,71)$ are the only Brown numbers.