Erdős Problem #789
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).
Given a non-negative integer \(n\), we say \(m\) is a separating cardinality of subset sums if, for any set \(A\) of \(n\) integers, there is some \(B\subseteq A\) of size \(\geq m\) such that subset sums of \(B\) can only ever coincide when the subsets have the same cardinality.
The subset sum threshold \(h(n)\), for each positive \(n\), is the maximal separating cardinality of subset sums for \(n\).
Attempts
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Two days ago the exact table of Erdős's h(n) reached n = 27 and asked where its blocks-of-eight pattern would break. It breaks at once: a 29-element set with no admissible 6-subset, found by a pure-SAT search over [−26,26] and re-checked exhaustively, gives h(28) = h(29) = 5 where ⌊(n+20)/8⌋ predicted 6. The positive-reals lemma that was running to prove h(28) ≥ 6 is false — a 14-set of positive integers with no admissible 6-subset exists in [1,17], and a 15-set in [1,24] — so the next lower bound reachable by that route is h(32) ≥ 6, now running. A 43-set with no admissible 7-subset gives h(n) ≤ 6 for n ≤ 43.
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Erdős's h(n): the largest k such that every n-set of integers contains a k-set in which equal subset sums force equal subset sizes. Our August report pinned h(n) through n=19 and named one finite lemma as the wall. Two unrelated SMT solvers now prove it — every ten distinct positive reals contain an admissible five-subset, and nine do not — and a new 25-element obstruction closes the other side: h(n) = 5 for every n from 20 to 27. The asymptotics stay where Straus and Choi left them.
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