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Baek 2024 note erdos conjecture about square packing
theorem_1_1: Baek, Koizumi and Ueoro's theorem that for all integers k and c with -k < c < k, the largest total side length of k^2+2c+1 axis-parallel squares packed in a unit square is k + c/k, which determines g(n) for every n.
theorem_2_1: Baek, Koizumi and Ueoro's theorem that for every positive integer k, the largest total side length of k^2+1 squares packed in a unit square with sides parallel to its sides is exactly k.
Jineon Baek, Junnosuke Koizumi, Takahiro Ueoro, A note on the Erdős conjecture about square packing. arXiv:2411.07274 (2024). The edition read is arXiv v2, dated November 19, 2024; labels and pages below are its own.
Let f(n) be the largest total side length of n squares packed in a unit square, and g(n) the same quantity when all squares are required to have sides parallel to the unit square (a modification the paper credits to Staton and Tyler). Erdős conjectured f(k^2+1) = k; Erdős-Soifer and Campbell-Staton independently proved f(k^2+2c+1) >= k + c/k for -k < c < k and conjectured equality, which Praton showed equivalent to the original conjecture. The paper says the conjecture for f remains unsolved. Theorem 1.1 (p. 1) proves the axis-parallel analogue in full: g(k^2+2c+1) = k + c/k for all integers k, c with -k < c < k. This determines g(n) for every n, since for k^2 < n < (k+1)^2 one of n-k^2 and (k+1)^2-n is odd. The key case is Theorem 2.1 (p. 2), g(k^2+1) = k for every positive integer k; Theorem 1.1 is restated as Corollary 2.2 (p. 4) and deduced from it by Praton's reduction, which the paper says carries over to g. The lower bound tiles the unit square by squares of side 1/k and replaces one tile by two squares of side 1/(2k). The upper bound is proved twice, by the second and third authors with k randomly shifted vertical lines spaced 1/k apart, and by the first author with a lattice-point count; the paper calls the two proofs essentially equivalent. The introduction also records, citing Staton-Tyler, that g(n) equals the tiling maximum h(n) when n is not 2, 3 or 5 and n+1 is not a square, and, citing Praton, that h(8) = 13/5 < 8/3 = g(8).
Source: https://arxiv.org/abs/2411.07274. The arXiv record names arXiv's non-exclusive distribution license (arXiv:2411.07274), every other right reserved.
Read status. Claims checked: Theorem 2.1 and Theorem 1.1 (with Corollary 2.2) were read clause by clause on the printed pages. The proofs (pp. 2--4) were read but not checked step by step; the reduction taken from Praton's paper was not checked here.
Bears on. #106: the problem asks whether f(k^2+1) = k. The paper proves the equality for g, where every square has its sides parallel to those of the unit square (Theorem 2.1), and the general formula g(k^2+2c+1) = k + c/k (Theorem 1.1). It does not address packings with a tilted square and does not claim to settle the question for f.
Results. Theorem 1.1 (p. 1; restated and proved as Corollary 2.2, p. 4); Theorem 2.1 (p. 2, with both proofs, pp. 2--4).
No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.