Problem Statement
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 $n$?
Categories:
Number Theory Powerful
Progress
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.
Source: erdosproblems.com/936 | Last verified: January 19, 2026