OPEN
This is open, and cannot be resolved with a finite computation.
Are\[2^n\pm 1\]and\[n!\pm 1\]powerful (i.e. if $p\mid m$ then $p^2\mid m$) for only finitely many $n$?
Cushing and Pascoe
[CuPa16] have shown the answer to the second question is yes assuming the abc conjecture - in fact, for any fixed $k\geq 0$, there are only finitely many $n$ and powerful $x$ such that $\lvert x-n!\rvert \leq k$.
CrowdMath
[Cr20] has shown that the answer to the first question is yes, again assuming the abc conjecture.
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This page was last edited 31 October 2025.
Additional thanks to: seewoo5
When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:
T. F. Bloom, Erdős Problem #936, https://www.erdosproblems.com/936, accessed 2025-11-16
