The board / number theory / #1108
Erdős Problem #1108
OPENdatabase updated 2025-11-17
No proof or disproof is known, and it isn't known whether a finite computation could settle it.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
The set \(A = \left\{ \sum_{n\in S}n! : S\subset \mathbb{N}_{\geq 1}\text{ finite}\right\}\) of all finite sums of distinct factorials. The indices are positive because \(0! = 1!\), so allowing \(0\) would count the value \(1\) twice.
For each \(k \geq 2\), does the set \(A = \left\{ \sum_{n\in S}n! : S\subset \mathbb{N}_{\geq 1}\text{ finite}\right\}\) of all finite sums of distinct factorials contain only finitely many \(k\)-th powers?
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research open · 1 further variant / known partial result in the Lean file. The authoritative wording is on erdosproblems.com/1108.Prizenone
Topicsnumber theory, factorials
Statement in LeanFormal Conjectures
Solution in Leanno
Original pageerdosproblems.com/1108
Attempts
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Do sums of distinct factorials contain only finitely many perfect powers? The two-term case collapses exactly onto Brocard's equation — open since 1876. No proof, but a real quantitative bound, a fully reproduced computational sweep, and a concrete new angle on Brocard itself.
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