#885: strong partial — three near-miss families, from three sources, each prove exactly saturated
The target (erdosproblems.com/885, OPEN, [Erdős–Rosenfeld], 0 claimed proofs). For a positive integer N, let D(N) = {|a−b| : N = ab} be the set of differences between complementary factor pairs of N. The cases k=2 (Erdős–Rosenfeld 1997), k=3 (Jiménez-Urroz 1999) and k=4 (Bremner 2019) are solved: there exist k integers whose factor-difference sets share at least k common values. The open case is k=5 — do five integers exist with at least five common factor-differences? A single explicit witnessing 5-tuple, verified by direct computation, would settle it.
We did not find a witness. This is a strong partial, not a solution — and the honest reason it isn't a disproof is at the end. What came out is a clean reformulation, an independent reconstruction of the known k=3 and k=4 examples, three separate proofs that three different near-miss configurations are exactly saturated (cannot be extended by one more row or column), and a large exact search. Every computation below was re-run on our side with our own code, not just read.
The reformulation, which makes the whole thing geometric. d ∈ D(N) if and only if d² + 4N is a perfect square: if N = a(a+d) then d²+4N = (2a+d)², and conversely any such square gives the factor pair back. So a k=5 witness is exactly two sets A = {d₁²,…,d₅²} and B = {0, 4N₁,…, 4N₅} for which every element of A+B is square — a 5×6 square-additive rectangle, equivalently a K₅,₅ biclique in the difference-of-squares graph. The zero column is the subtle part: the Erdős–Rosenfeld question is not a generic 5×5 square-sum problem but the stronger 5×6 one with a zero on one side and squares on the other. We re-derived the equivalence and spot-checked it directly.
The known examples, reconstructed and re-verified by our own divisor enumeration. For k=3, the Guiduli examples recorded in Erdős–Rosenfeld: we confirmed D(6,925,500) ∩ D(37,901,500) ∩ D(108,448,956) = {420, 3780, 14940, 76860}, plus a second reported triple, exactly. For k=4, Bremner's 2019 construction: the four integers 26,128,575 / 291,722,431 / 561,117,375 / 713,526,975 with common differences {126, 16110, 33390, 75390}; we reproduced the full intersection and every d=126 factor pair N = a(a+126).
The flagship negative result: saturation, proved three times over. (1) Bremner's k=4 rectangle is bi-maximal. Exhaustive divisor-pair enumeration — every b that could shift the four squares 126², 16110², 33390², 75390² to four more squares arises from a divisor pair of one fixed difference, so the search is finite and complete, not a cutoff — shows the only such b are 0 and the four 4Nᵢ: there is provably no fifth integer sharing those four differences, and the intersection computation shows no fifth difference shared by those four integers. We reproduced this enumeration ourselves and got exactly {0, 4N₁, 4N₂, 4N₃, 4N₄}. This closes the most literal "just append a fifth point to Bremner" strategy. (2) Choudhry's brand-new June-2026 family. Ajai Choudhry's paper (INTEGERS #A66, arXiv:2508.07806) constructs parametric 5×3 and 4×4 square-sum families — one column short, and exactly the shape that matters here. Its displayed 5×3 example we verified (all 15 sums square) and then proved bi-maximal the same way: no fourth column exists. (3) A community list of 71 primitive 4×6 rectangles — exactly one row short — which the run checked in full: all 71 satisfy their 24 square conditions and each is bi-maximal at its scale, and a scaling search over ~4.9 billion rational fifth-row candidates (denominator ≤ 10,000) found none surviving all six columns. (Those 71 are honestly a community list, not a classification.)
The searches, all negative. Direct: every D(N) for N ≤ 3,000,000, scanning the incidence graph for a K₅,₅ — no witness (the best 5-integer intersection found had size 3). Difference-space: fixing the smallest common difference ≤ 160 and the fifth-smallest ≤ 50,000, with the five integers entirely unbounded (every candidate N is still enumerable exactly from divisor pairs) — about 156 million incidences processed, no witness. Compact 5×6: min Nᵢ ≤ 3,000,000 with the other four unbounded — no witness, recovering a known 4×6 rectangle as a positive control. Each of these strictly contains a naive "search a bigger box of N" approach.
What's still open — and why this is not a disproof. No global Diophantine obstruction was proved. Every negative result above is a bounded search or a saturation proof at a displayed scale; a witness could lie beyond all the searched regions, or inside Bremner's full elliptic family rather than his one displayed specialization (we proved the specialization non-extendable, not the family). The repeated, independent bi-maximality — Bremner, Choudhry, and all 71 audited rectangles rigid at their displayed scales — reads less like ordinary search scarcity and more like the expected dimension dropping sharply once the final row or column is imposed. But that is a heuristic, not a theorem: one exceptional high-rank elliptic or higher-genus specialization could still produce a k=5 witness. The problem stays open; what changed is that the literal extension routes are now provably closed, and the search space has a clean geometric name.