Problem Statement
Let\[A = \left\{ \sum_{n\in S}n! : S\subset \mathbb{N}\textrm{ finite}\right\}.\]If $k\geq 2$, then does $A$ contain only finitely many $k$th powers? Does it contain only finitely many powerful numbers?
Categories:
Number Theory Factorials
Progress
Asked by Erdős at Oberwolfach in 1988. It is open even whether there are infinitely many squares of the form $1+n!$ (see [398]).This was motivated in part by a problem of Mahler which he discussed with Erdős a few days before his death in 1988: if $k\geq 5$ and\[A_k= \left\{ \sum_{n\in S}k^n : S\subset \mathbb{N}\textrm{ finite}\right\}\]then does $A_k$ contain only finitely many squares? Mahler showed that there are infinitely many squares in $A_k$ for $k\leq 4$, and found only one square for $k\geq 5$, namely\[1+7+7^2+7^3=400.\]Brindza and Erdős [BrEr91] proved that, for any $r$, if $n_1!+\cdots+n_r!$ is powerful then $n_1\ll_r 1$.
Source: erdosproblems.com/1108 | Last verified: January 19, 2026