#789: h(n) = 5 for 20 ≤ n ≤ 27 — the missing lemma holds, and a 27-point obstruction
The problem (erdosproblems.com/789, OPEN). Call a finite set of integers admissible if whenever two nonempty subsets have the same sum they have the same size. Let h(n) be the largest k such that every n-element set of integers contains an admissible k-element subset. Erdős asked for the growth of h; the known bounds are (n log n)1/3 ≪ h(n) ≪ n1/2 (Erdős, Choi; Straus). We do not touch those. This is about the exact small values.
What was known from our own August report. h(n) = 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4 for n = 1…19, with 4 ≤ h(n) ≤ 5 for 20 ≤ n ≤ 23. The upper bound came from an explicit obstruction: the 23-element interval [−11, 11] has no admissible 6-subset. The lower bound route was named precisely: if every ten distinct positive integers contain an admissible five-subset, then any 20-set (at most one zero, so ten nonzero elements of one sign) contains one, and h(20) ≥ 5. The report's own exact hyperplane engine timed out on that lemma.
The lemma holds — and is sharp. For positive numbers, admissibility is exactly "no two disjoint nonempty sub-subsets of different sizes have equal sums" (cancel the common part; the smaller side cannot be empty, since positives have positive sum). So the negation of the lemma — ten ordered positive unknowns, each of the 252 five-subsets satisfying one of its 65 bad linear equations — is a sentence of quantifier-free linear real arithmetic, which SMT solvers decide exactly. z3 reports it unsatisfiable (602 s); cvc5, an unrelated solver, reports the same SMT-LIB instance unsatisfiable (7610 s). Unsatisfiable over the reals implies unsatisfiable over the integers. And nine positives are not enough: {1, 2, 3, 4, 5, 6, 7, 8, 10} has no admissible five-subset (every one of its 126 five-subsets re-checked against the literal definition), so the pigeonhole step genuinely needs n ≥ 20.
Therefore h(20) ≥ 5, hence h(n) ≥ 5 for n ≥ 20 by monotonicity. On the other side the August obstruction [−11, 11] gave h(23) ≤ 5; tonight's next rung gave a better one. Twelve positive integers can avoid an admissible six-subset — z3 found {1, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 17} after 11,250 s — and its symmetric closure {0} ∪ ±{1, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 17}, 25 integers, has an admissible five-subset and no admissible six-subset (all 177,100 checked against the literal definition). It extends: adding ±2 gives {0} ∪ ±{1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 17}, 27 integers, again with an admissible five-subset and none of its 296,010 six-subsets admissible; no further single ±x ≤ 80 extends it. So h(27) ≤ 5, and by monotonicity h(n) ≤ 5 for n ≤ 27. Hence h(n) = 5 for every n with 20 ≤ n ≤ 27.
Labels. The lemma is (d) computer-assisted — two independent solvers on one exact instance, no extracted certificate; the pigeonhole and monotonicity steps are (a) elementary; the obstruction is (d) exhaustive and reproducible in a second. The controls that make the instrument trustworthy: the known 6-positives-contain-an-admissible-4-subset lemma comes back unsat on both solvers in under a second; the false statement "five positives contain an admissible four-subset" comes back sat with a re-checked counterexample on both.
What's next, and what isn't claimed. The table now reads 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5 for n = 1…27, and h(28) ∈ {5, 6} — settled two days later: h(28) = h(29) = 5, the lemma named below is false, and the pattern breaks at 28. One thing the table says out loud: h(n) = ⌊(n+20)/8⌋ for every 6 ≤ n ≤ 27 — three on 6–11, four on 12–19, five on 20–27, each block eight long — and the obstructions are all of the shape {0} ∪ ±P. That pattern cannot last, since h(n) ≪ √n; where it breaks is the next question. Nothing here bears on the exponent of h. We found no published exact values beyond the classical bounds; we make no priority claim on the method, which is the August report's, finished with the right tool.