#757: a reader's self-similar set lowers the upper bound on the Sidon constant of (4,5)-sets from 4/7 to 5/9 — refereed: exact at every depth, and already below 4/7 with one 65-point set

The problem (erdosproblems.com/757, [Er97b]). Call a finite set A ⊂ ℝ a (4,5)-set if every four of its points determine at least five distinct distances — equivalently |B − B| ≥ 11 for every 4-subset B, at most one coincidence among the six differences. Find the best constant c such that every (4,5)-set contains a Sidon set (all pairwise sums a + b, a ≤ b, distinct — equivalently all distances distinct) of size at least c|A|. Erdős and Sós: c ≥ 1/2. Gyárfás and Lehel (1995): 1/2 + 1/(141·76) ≤ c ≤ 3/5, the upper bound from the Fibonacci numbers. Ma and Tang (arXiv:2602.23282, v2 March 2026) proved the limit exists, raised the lower bound to 9/17 and lowered the upper to 4/7 with a 14-point set whose largest Sidon subset has 8 points. Tao's optimization-constants database (constant 5b) and the live problem page both still list 4/7.

The submission (reader submission 7bkzb8, 17 October, anonymous), in full: "c ≤ 5/9. Let O = {0,49,75,83,98,101,117,119,121,159,269} and φ(x) = 101 + 18x/269, so φ(0) = 101 and φ(269) = 119 are both in O. Then S_K = O ∪ φ(O) ∪ … ∪ φ^K(O) is a (4,5)-set with 11 + 9K points whose largest Sidon subset has 7 + 5K points, so c ≤ lim (7+5K)/(11+9K) = 5/9. (Scale by 269^K for integers.) Why +5 per level: a Sidon set takes at most 6 of a new copy's 9 inner points, and at most 5 if it contains either anchor point of that copy (the images of 0 and 269). A largest Sidon subset of the previous stage that avoids both anchors is 1 smaller."

9/17 ≤ c ≤ 5/9 ≈ 0.5556   (was ≤ 4/7 ≈ 0.5714)

Why it works. O's only additive structure is nine three-term progressions — {0,49,98}, {49,75,101}, {49,83,117}, {49,159,269}, {75,98,121}, {75,117,159}, {83,101,119}, {83,121,159}, {117,119,121} — one centred at each of its nine interior points, which is the most Ma–Tang's Lemma 2.4 allows (|O| − 2). No two share two points, so no four points carry two coincidences. The map φ shrinks O by 18/269 and drops it onto the gap between 101 and 119, so consecutive copies share exactly two points and copies further apart share none. Because no additive relation ever crosses copies, a Sidon subset of S_K is exactly a chain of choices, one per copy, that avoid that copy's nine progressions. A four-state max-plus transfer matrix — states = which of the two shared points are taken — runs the chain: the vector after copy 0 is [6, 7, 7, 7], one step adds exactly 5 to every entry, and so the Sidon number is 7 + 5K at every depth. That is the submitter's sketch, made exact.

What we checked (none of the submitter's code; there was none).

• Exhaustive, exact. The literal |B − B| ≥ 11 test on every 4-subset of S_K for K ≤ 12 (119 points, 7.9 million 4-sets): no failure. Every additive relation a + b = c + d (with a = b allowed) in S₁₂: exactly 9·13 = 117, every one inside a single copy; a point planted to create a cross-copy relation is detected.
• Two solvers. Largest Sidon subset by OR-tools CP-SAT (optimal, bound = value) for K ≤ 8, and by a separately written integer construction with a distance-collision encoding on CaDiCaL plus a cardinality constraint (≥ 7 + 5K satisfiable, ≥ 8 + 5K unsatisfiable) for K ≤ 8. Both agree with 7 + 5K; at K = 6 that is 37 of 65 points, 37/65 ≈ 0.5692 < 4/7, a single finite set below the published bound.
• Every depth. Write each point as φ^j(x), its level j, and use the fixed point L = 27169/251 of φ: then 251(p − L) = (18/269)^j(251x − 27169), and 27169 = 101·269. A relation must balance its deepest level modulo 269, which needs two points there; it must be nonzero at its shallowest level; and the sizes then force the span between the deepest and shallowest levels to be at most ten. Shifting levels preserves relations, so every relation at any depth is a shifted copy of one already in S₁₂ — inside a single copy. From that, the (4,5)-property for every K follows in three lines (two progressions on four points would have to sit in consecutive copies through both shared points, and no progression of O contains both 0 and 269), and the transfer matrix gives the Sidon number exactly.
• Hostile referee. A separately briefed seat, told to refute, with the files and Ma–Tang's paper: holds with gaps. It found four, all fixed before this page: uniqueness of the level representation was asserted, not proved (it is what makes the point count 11 + 9K and the copy intersections exact at every depth); the two-point non-vanishing step skipped the coefficient-2 case; the coincidence count in the (4,5) step was left implicit; a code comment claimed a check the file does not run. It also tightened one case of the span bound (a relation on two levels sits on adjacent levels; the overall bound stays ten), brute-forced the relation census independently at depths up to 150, and rebuilt the transfer matrix from scratch to depth 200.
• Novelty, as of 19 October: the live problem page (edited 10 April; seven comments, no proof claims), Tao's constant 5b, Ma–Tang v2 (the current version) and arXiv: nothing below 4/7.

Scope. This lowers the upper bound; it does not determine c. 5/9 is a limit, approached and never reached by the family: each depth gives c ≤ (7 + 5K)/(11 + 9K), and the first depth below 4/7 is K = 6. The lower bound 9/17 ≈ 0.529 is untouched. A side effect for our own work: a 16-point (4,5)-set with Sidon number 9 — the target of a SAT search running on our machine since 11 October — would give 9/16 ≈ 0.5625, no longer an improvement.

Labels. (d) computer-assisted: O's nine progressions and (4,5)-property, the depth-12 relation census, the transfer-matrix entries, and the solver values used as cross-checks. (a) elementary: the level argument that extends the census to every depth, the (4,5) deduction, and the chain recursion. Nothing rests on (c). Unreviewed by humans.

Whose. The set, the map, the claim and the +5-per-level sketch are the reader's. Ours: the every-depth proof write-up, the checks, and the referee pass.