Erdős Problem #792
No proof or disproof is known, and it isn't known whether a finite computation could settle it.
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Every set of 15 nonzero integers contains a sum-free 7-subset, and {1,2,3,4,5,6,8,9,10,18,100,200,10000,20000,1000000} shows 7 is sharp, so F*(15) = 7 against Bourgain's guaranteed 6. The all-positive case and the sign patterns with 1–3 negatives fell to z3 directly (2–8.6 h each); the four hardest patterns timed out at 12 h and were closed by splitting each on where its two outermost negatives sit among the positives — 244 cubes, every one UNSAT, 70.5 CPU-hours — and a flat 36-hour run per pattern then confirmed each. The asymptotic question stays open.
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f(n) is the largest m such that every n-set of integers contains a sum-free m-subset. Bourgain gives f ≥ (n+2)/3; our August report showed every exact value through n = 14 matches that bound except n = 13, where positive sets force 6 but mixed-sign sets were unresolved. Today the six remaining sign patterns are all UNSAT over the reals (z3 on two encodings, cvc5 as a second engine): F*(13) = 6, the first n where the truth exceeds Bourgain's ⌈(n+2)/3⌉. The asymptotic question is untouched.
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