Every n has its own open question: what is s(n), the side of the smallest square that holds n unit squares? This page is about the questions that cover many n at once: patterns in the answers, how they behave as n grows, and what optimal packings look like. A few single values close the page.
Each item also says which n it concerns (all n, large n or one n) and who stated it. "Stated here" means we found no earlier statement; tell us if you know one. Most items are also stated precisely in Lean, as named propositions awaiting proofs: Conjectures.lean. Statements of method (lemmas about certificates, covering measures, Lean) are in the project's research wishlist, not here.
Take the k × k grid and remove c squares. For small c nothing better exists: s(k² − c) = k. For larger c, tilting some squares lets the rest squeeze into a smaller box. Write c*(k) for the largest c for which the grid is still optimal. It is the single number that says how far below k² the integer answers reach.
| k | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| c*(k) ≥ (proved) | 2 | 3 | 3 | 4 | 4 | 4 | 5 | 4 | 4 | 4 | 4 |
| c*(k) ≤ (best packings) | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 10 | 11 |
"Proved" row: s(6) = 3 (Kearney–Shiu 2002), s(13) = 4 (Bentz 2010), and the 2026 results on Proofs & results: s(k² − 4) = k for every k ≥ 5 (this project, unrefereed) and s(59) = 8 (wand125). "Best packings" row: the best known packings, which beat k at n = k² − c* − 1. Exact so far only for k = 2, 3; c*(4) is 4 exactly when s(12) = 4.
c*(k+1) ≥ c*(k): if removing c squares from the k × k grid never allows a smaller box, the same holds for every larger grid.
Erich Friedman's Conjecture 1 (Dynamic Survey DS7, 1998), "true of all the best known packings". It still is: the "best packings" row above never decreases, through k = 18. (The "proved" row can dip, as at k = 9, because it records what has been proved, not c* itself.) Proved for each c ≤ 4. Nagamochi (2005) covered c ≤ 2, but Karakuş found a gap in his proof in 2026. The 2026 results for c = 3 and 4 (unrefereed) imply the cases c ≤ 2 anyway.
Does c*(k) → ∞? Equivalently: for every c, is s(k² − c) = k for all large enough k?
Claimed in 2026: a preprint by Ryu Sungjoon (squarepacker/k2-minus-c, DOI; unrefereed) proves c*(k) ≥ 0.0353 log k − 1 for every k ≥ 2, partly by computer. It gives no usable thresholds for specific c. We have read parts of it and found no gap, but have not checked all of it. Before that we had found no attack on the question in print. Proved with explicit thresholds for c ≤ 4. It does not follow from the monotone conjecture (a monotone c* could stall), and it is probably the more tractable of the two.
Is c*(k+1) ≤ c*(k) + 1?
Stated here. The bound + 2 is easy: shrink a good packing for k and add an L-shaped border, which holds 2k − 1 squares, two fewer than the 2k + 1 that the grid's border holds. (Checked in Lean: cStarStepsOfTwo.) Together with the first two items, this says c* climbs by 0 or 1 at each step, forever. It cannot climb fast: a 2026 preprint by Ryu gives c*(k) = O(k3/8) with explicit constants, and two explicit bounds are checked in Lean here: c*(k) ≤ 4⌈√(2k − 7)⌉ − 1 for every k ≥ 7 and c*(k) < 18.9 k2/5 + 35.7 k1/5 + 122 for every k ≥ 3·105 (details).
Friedman's survey reports a conjecture that s(k² − k) = k "whenever k is small". It holds for k = 2, 3, fails for k = 11 (Cantrell 2025, s(110) ≈ 10.99679, unrefereed) and for every k ≥ 12 (Arslanov, Mustafin and Shangitbayev 2021). Which of k = 4, …, 10 hold?
So the last good k lies between 3 and 10. Nothing known rules out gaps (a failure at k = 4 would not decide k = 5); unit steps would rule them out. The smallest case is s(12) = 4 itself. On the packing side, s(90) < 10 is the natural next attempt (see §7).
In a box of side s, the best packing leaves some area W(s) uncovered. If s is an integer, W = 0. Otherwise the grid wastes about 2s times the fractional part, and tilting does much better.
W(s) = O(s1/2).
Friedman's Conjecture 2 (DS7, 1998 version only); Erdős and Graham (1975): "Perhaps the correct bound is O(α1/2)". known Upper bounds: O(s7/11) (Erdős–Graham 1975), improved several times to O(s3/5) (Chung–Graham 2020 claimed it, with an error found by Arslanov and Bui; proved by Bui 2025 and by McClenagan 2026, preprints). Lower bound: W(s) ≥ c·√(s·‖s‖) whenever s(s − ⌊s⌋) > 1/6, where ‖s‖ is the distance to the nearest integer (Roth–Vaughan 1978; their proof gives c = 10−100). So the exponent is between 1/2 and 3/5.
Not everyone agrees with the conjecture. Chung and Graham (2020) note one might guess O(x1/2) and add "The authors don't believe this." They offer prizes: $100 for W(x) = o(x3/5), $250 for O(x3/5−c), and $500 for W(N + ½) ≫ N1/2+c (a disproof also wins).
The lower bound fades as s approaches an integer, so it says nothing about c*(k): the staircase questions are about exactly the regime that the waste theory leaves open.
More squares never fit in less room, so s(n) ≤ s(n+1). Equality, a plateau, happens at every integer (the grid with a few holes). Away from integers it is rarer.
Is there an n with s(n) = s(n+1) and s(n) not an integer?
In the best-known table, yes: n = 147, 264, 290. In each case the best packing for n is the record for n + 1 with one square removed. None is proved. Two more, n = 232 and 295, were broken in October 2026 by packings of their own (Nate Chaoweeraprasit's of 232, jlevy/squares#481; ry-xu's of 295, #432). The proved bounds (2026 computer-checked floors, unrefereed) rule out a plateau for every n < 28; the first open case is s(28) = s(29). The last step was wand125's s(19) ≥ 193/40 = 4.825 (6 Oct 2026), above the side (7 + √7)/2 ≈ 4.82288 of the best packing of 18, so s(18) < s(19). That step rests on wand125's certificate as replayed by jlevy/squares (register-verified, partly sharing the producer's components), not on two separately written checkers.
The strongest answer to the question above is none. Below it is a ladder of weaker statements. Each is a step toward the ones above it, and a fallback if they fail.
Finitely many is exactly the last form plus one more statement: plateaus occur at only finitely many fractional parts. A different strengthening asks the same of every n, not just plateaus: each fractional part occurs for only finitely many n. That implies the per-fractional-part plateau form, but not finitely many, since infinitely many plateaus could each have their own fractional part.
Stated here. The implications are checked in Lean (ConjecturesProofs.lean). The evidence is structural, and it supports the per-fractional-part form directly. All five candidates (the three left and the two broken) belong to fixed-shape families: the same 45° construction repeated at side k + β for many k, with the same fractional part β. Each family's wasted area grows linearly in k, and the O(s3/5) constructions beat any such family eventually. So every family is optimal only finitely often. Ellsworth makes the same point from the data: "Even the known recurring patterns only extend finitely before becoming inoptimal." Getting from there to finitely many needs the plateaus to come from finitely many families, and that is the real gap. The all-n form is a real question too, because a family with sublinear waste is not obviously impossible.
| fractional part β | records (n ≤ 323) | n |
|---|---|---|
| √2/2 ≈ 0.7071 | 14 | 5, 10, 27, …, 262, 296: Göbel's strips (1979) |
| 0.5355 | 11 | 65, 82, …, 290, 291 |
| (√7 − 1)/2 ≈ 0.8229 | 7 | 18, 53, 151, 176, 204, 234, 299 |
| 0.6569 | 9 | 66, 124, 147, 148, …, 294 |
| 0.7782 | 6 | 150, 203, 233, 264, 265, 298 |
Göbel's strips are the best known packings for every strip size a < 44 except a = 3 (n = 17). The first to be beaten is a = 44 (n = 2043, David Ellsworth, 2024); the waste bound guarantees a last optimal one but gives no explicit place (see §5). Table computed from the best-known packings as of 10 Oct 2026 (script for the catalogue). In October 2026 new packings took six records out of these families: ry-xu's (jlevy/squares#432) at 84 and 295 (√2/2), 86 and 127 (√7) and 175 (0.7782), and Nate Chaoweeraprasit's (#481) at 232 (0.7782).
s(n) is always an algebraic number: the set of valid packings is described by polynomial inequalities (Tarski–Seidenberg). Small cases are simple: integers, or numbers like 2 + 1/√2. Some records are even rational: s(50) ≤ 7 + 4/7 tilts squares by the 3-4-5 angle, and s(293) ≤ 17 + 26/41 (Ellsworth, 2024–25) uses the 20-21-29 triangle. The questions here ask whether that simplicity ever runs out for good.
Does the algebraic degree of s(n) tend to infinity along the non-integer values? Is there a last n whose optimal side is built from square roots alone? Is any optimal non-integer side rational?
Stated here. Two data points. The value s(11), settled in 2026 by a computer-assisted proof, uses a single tilt angle (49.8°), and its side is a root of an irreducible degree-8 polynomial. The best known s(28) (Ellsworth, 2025) also uses one tilt angle (59.7°), and its degree-6 polynomial has Galois group S6, so that side cannot be written with radicals at all. Yet values built from square roots alone still hold records as late as n = 626 (27/2 + 17√2/2). Rational records exist (n = 50, 230, 261, 293, all with Pythagorean tilt angles), but none is proved optimal. We found no literature on any of these questions.
Must optimal packings use more and more distinct tilt angles as n grows? A cleaner large-n form: let Wm(s) be the least waste using at most m distinct angles. Axis-parallel packings alone give W1 ≈ 2s·frac(s), linear in s. At a fixed fractional part, is Wm linear in s for every fixed m?
Stated here. If yes, the general O(s3/5) bound forces optimal packings to use ever more angles. Every sublinear construction we know of (Erdős–Graham onward) varies the tilt from stack to stack, and we found no result about packings with a bounded number of angles. §5 has what the catalogue shows so far.
If s(n) is not an integer, then s(k²n) < k·s(n) for every k ≥ 2. In words: lay out a k × k array of copies of an optimal packing that beats the grid, and the result is never optimal for k²n squares.
Stated here. We found it in neither Friedman's survey, nor Ellsworth's catalogue, nor the wasted-space papers. Tiling always gives s(k²n) ≤ k·s(n) (checked in Lean: tilingBound), and so does scaling the packing by k and cutting each square into k², so the conjecture is that this inequality is strict. The hypothesis is needed: when s(n) is an integer, a tiled grid can be optimal: s(32) = 6 = 2·s(8) (see Proofs & results). known True for all k ≥ K(n). The tiling wastes area k²(s(n)² − n), which grows like k², while the best packings of side k·s(n) waste O(k3/5) (§2; Erdős–Graham's 7/11 already suffices). For large k the saving even grows linearly in k. In Lean this is a conditional theorem, tilingsEventuallyNotOptimal_of: the waste bound and positive waste at non-integer sides imply it. K(n) is effective but large, so the content is at small k, above all k = 2.
The best-known table agrees wherever it can check. For each of the 28 non-integer records with 4n ≤ 323, the record for 4n beats twice the record for n, by at least 0.41 (at n = 5 and 65). For k = 3 and 4 the smallest margins are 1.04 and 1.59. These compare best-known values, not optima. The test cases lie just past the table. The current records for n = 241, 273 and 307 (Francisco Couzo, 2026, in jlevy's register) double to sides 31.9763, 33.9679 and 35.9621 for 964, 1092 and 1228 squares. Neither catalogue lists these n. All three doublings can be beaten (October 2026, this project): starting from 2 × 2 tilings of the records, a short search gives s(964) ≤ 31.8595, s(1092) ≤ 33.8406 and s(1228) ≤ 35.8474, each checked exactly by two separately written verifiers (write-up). The gains are 0.115–0.127. The best start scales the packing by 2 and cuts each square into a 2 × 2 block, and those blocks can shear apart along the cuts. Plain and pinwheel tilings gain 0.008–0.038, as their seams relax. Mirrored copies, whose seams meet their own reflections, barely move. This is evidence about best-known values, not a proof: it shows that s(4n) is below twice the record for n, while the conjecture is about the true s(n).
Related: Ellsworth calls a packing primitive if it is not a smaller packing with L-shaped borders added. Several records are built from two copies of a smaller record plus grid filler, which is a different construction from tiling (s(171) from two s(50); s(290) and s(291) from two s(65)).
Are optimal packings eventually asymmetric? The Explore page reports each record's symmetry.
Small n is full of patterns: explicit families, packings with two or three angles, simple side lengths. The large-n theory says most of these patterns must end, usually without saying where. The questions here ask where. They matter less than the ones above, but they are concrete, the catalogue gives data for them, and a single good packing can answer one. The last k with s(k² − k) = k (§1) and the fixed-shape families (§3) are questions of the same kind.
A Göbel strip wastes area about equal to its side length (waste ≈ s). Any construction wasting less than (1 − ε)·s at fractional part near 0.707 beats it. So the sublinear constructions of §2 beat every large enough strip, but their constants were never written down, and the crossover they give is presumably astronomical. In the catalogue, the first strip to fall (n = 2043) was beaten by an optimised one-off, not by a family. The question: what is the best waste coefficient κ (waste ≈ κ·s) of a fixed-shape family at a given fractional part? Göbel's strips have κ ≈ 1 and the plain grid at fractional part 0.707 has κ ≈ 1.41. Is there an explicit family with κ < 1 that beats the strips from some modest n, ideally below 2043? And are the strips optimal among families with two angles (axis and 45°)?
Stated here. The existence of some explicit beating family is not in doubt; computing the crossover for a published construction is bookkeeping, not research.
In the catalogue, packings with five or more angle classes sit mostly just below a perfect square (fractional part ≳ 0.87), where the small-tilt constructions live. Exceptions are n = 68, 103, 123 and 297. Packings with at most three classes still hold about half of each interval up to k = 17, with no visible decline. Does the many-angle regime spread down from the top of each interval as k grows, and how fast?
Two examples. Göbel's strips hold the records up to a = 43 (except a = 3; §3). The first strip to be beaten is n = 2043, but some larger strips (around n = 2938) are still listed unbeaten, so where the best-known strips stop is unsettled. Ellsworth's table also notes that the best known s(n² + 1) has fractional part at least ½ for every n < 42; Hajba's record for n = 42 (s(1765) ≈ 42.488) is the first below ½. Patterns like these are only as good as the search behind them, so each break is partly a fact about packings and partly a fact about where searching got hard.
Call a packing locally optimal if no nearby arrangement fits in a smaller box. Two related notions come from Ellsworth. A packing is rigid if it "cannot be continuously transformed into any other valid packing without changing the size of its enclosing square"; he calls this rare, and Explore reports rigidity for every record. Two packings in the same box are alternative packings if neither "can be reached by continuously translating and/or rotating the squares" from the other. Jamming theory, which studies such notions for disks and spheres, seems not to have been applied to squares.
At an integer side the box size is constant on a huge set of packings: everything that slides around inside the k × k box. A point of that set can be locally optimal while the same flat set leads, by large motions, to a smaller box. A full row and a full column are enough for local optimality: tilting a row of k squares together by a small angle θ widens it to k cos θ + sin θ ≈ k + θ − kθ²/2, which stays above k until θ ≈ 2/k. So 7 squares in an L inside a 4 × 4 box are locally optimal, yet they slide into a 3 × 3 corner. The same goes for a grid with c > c*(k) holes that leaves a row and a column full.
Judging the whole flat set fixes this. Call a packing plateau-optimal if no continuous motion that never enlarges the box reaches a smaller box. Equivalently, every packing reachable from it inside the same box is locally optimal in the plain sense. (Packings form a semi-algebraic set, so the curve selection lemma turns "a smaller box arbitrarily close by" into "a path to a smaller box".) So plateau-optimality depends only on the connected piece of the plateau, which plain local optimality does not. It is the weakest condition with that property that still implies plain local optimality. This is not global optimality in disguise, because the packings in a box need not form one connected set. The best known packings for n = 38 and n = 149 (side k + √2/2) come in alternatives: Göbel's 1979 strips, Cantrell's 2005 alternative for 38, and Ellsworth's rigid 2024 alternative for 149 (Göbel strips page).
A rigid packing is a reachable set of one, so it is automatically plateau-optimal. So rigid packings that are not optimal answer the bare question. Ellsworth's rigid alternative for n = 18 has side 2 + 2√2 ≈ 4.8284, above the best known 4.8229. The open question is about packings that can move: is there a plateau-optimal packing with free motion whose side is larger than s(n)? Natural candidates are the Göbel strip for n = 2043, which no longer holds the record (Ellsworth notes that Göbel's arrangement is not rigid at n = 149), and grids with c > c*(k) holes.
How many mutually unreachable packings can share the best side, and how fast can that number grow with n?
Stated here. n = 149 already gives hundreds. Ellsworth's rigid alternative uses just five rotated squares, each in an s(5) block (four axis-parallel squares around one 45° square, side 2 + √2/2). We modelled such packings exactly as an integer program, with every axis-parallel square shifted by 0 or √2/2 in each coordinate. In that model, every packing of 149 squares uses exactly five blocks. Their x-corners form a chain from 0 to 10 in steps of 2 and 3, and so do their y-corners, giving 6 chains each. The blocks pair the two chains by a permutation. Not every pairing works (no simple rule about neighbouring blocks predicts which), but over 470 do. Each of them fixes the remaining squares uniquely in the model, which suggests each is rigid and is its own alternative. At n = 38, by contrast, two blocks have room to slide, which is why those alternatives have free motion. (Exploratory computation, not yet a theorem: goebel_alt.)
For c > c*(k), how large a change to the grid is needed to beat side k? Is the winner a small-tilt perturbation, or a different structure built from 45° blocks?
Is every packing with side close to s(n) close to an optimal one, apart from squares that are free to move? Computer proofs that work by subdividing the space of placements are, in effect, proofs of this. Cases where it fails (several distant packings of nearly equal side) are where such proofs struggle. Alternatives at equal side, as at n = 149, are the extreme case: such a proof must handle each one separately.
The most wanted exact value: the smallest open case of s(k² − k) = k, and it decides whether c*(4) is 3 or 4. The best proved floor is 3.9715 (squarepacker, 5 Oct 2026, re-weighting the points of our 3.9686 certificate). See twelve squares.
The questions above also point at specific n where a better packing is plausible:
| k − s(k² − c) | c = 9 | 10 | 11 | 12 | 13 |
|---|---|---|---|---|---|
| k = 8 | .054 | .153 | .177 | .293 | .310 |
| 9 | grid | .056 | .119 | .173 | .201 |
| 10 | grid | .050 | .118 | .161 | |
| 11 | grid | .003 | .050 | .095 | |
| 12 | grid | .014 | .050 | ||
| 13 | grid | .018 | |||
| 14 | grid |
"grid": no known packing beats side k (blank cells to its left are grids too, with fewer holes). Best-known packings as of 10 October 2026 (the catalogue and the records posted since; see Sources §1). All other single values, with their proved floors, are on Bounds.
For an a × b grid, let c*(a, b) be the most squares one can remove while the grid stays optimal (no smaller a′ × b′ box). Is c*(a, b) nondecreasing in each of a and b separately? This is half a step of Friedman's conjecture. A related, easier-looking case keeps the height fixed and lets the width grow.
Stated here. Rectangles that can be squeezed are known (Arslanov, Mustafin and Shangitbayev 2021). Nagamochi's lower bound for rectangles (2005) has a gap (Karakuş 2026).
Erdős Problem #106 asks for the largest total side length f(n) of n squares packed in a unit square. Its conjecture f(k² + 1) = k was disproved in 2026 (f(17) > 4, Lean-checked; erdosproblems.com/106). Erdős also asked when f(n+1) = f(n), the analogue of our plateaus.
Is s₃(k³ − c) = k for all large k, for each fixed c? Even c = 1 seems not to be in print for unit cubes in a cube. Friedman's "Cubes in Cubes" page lists packings only.
Last revised 2026-10-07. References are on Sources. The working list behind this page, with proof pieces and their status, is the project's wishlist.