Can a rectangle be tiled with $6$ smaller rectangles, such that no smaller set of those rectangles forms a rectangle?

Terminology: A compound is a proper subset of the smaller rectangles whose union is a rectangle. The problem is to prove there is no compound-free tiling by 6 rectangles.

First, we note that the four corners of R must be occupied by four separate rectangles:

• If the same rectangle occupies two adjacent corners, it splits R by a straight line, which is not allowed according to the partial results.
• If it occupies two diagonally opposite corners, it is R and no further rectangles can be placed.

After placing the rectangles at the corners:

.----1----.
.  |   |  .
.  |   ---.
.---      4
2     ----.
.--   |   .
. |   |   .
.---3-----.

how can we assign the remaining two rectangles to the edges 1, 2, 3 and 4 and the interior? We note that unless they are assigned to two opposite edges (the boxcar case), there are two adjacent edges with only three distinct rectangles on them, which can occur in two essentially different configurations:

.-------.
.   |   .
.   |---.
.---- B . (1)
.  |A   .
.-------.
.------.
.  | X .
.  |---.
.----  . (2)
. Y |  .
.------.

In configuration (1), to prevent the formation of compounds, at least two rectangles each must sit on edges A and B. But that makes a minimum of seven rectangles – and we only have six.

The remaining space in configuration (2) is in the shape of an ell, which must be filled by three rectangles. It is quite easy to show that up to rotations/reflections the only compound-free tiling of an ell by three rectangles is

.---.
.   .
.   .
.   .
.   .---.
.   | Y .
.-------.
.   X   .
.-------.

There are only two ways to place this ell tiling into configuration (2). In one of them, the rectangles marked X form a compound; in the other, the rectangles marked Y form a compound (try it).

Hence there is no compound-free 6-rectangle tiling of R with two adjacent edges sharing three distinct rectangles.

---

For the boxcar case, here are the two essentially different arrangements of the four corner squares:

.----------.
.   |   |  .
.   |   |  .
.   |   |  .
.   |   |  .
A---B   C--D (cis)
. |      | .
. |      | .
. |      | .
. |      | .
.----------.
.----------.
.   |   |  .
.   |   |  .
.   |   |  .
.   |   |  .
A---B   |  . (trans)
. |    C---D
. |    |   .
. |    |   .
. |    |   .
.----------.

In both configurations, the only time when the remaining space can be filled by two rectangles is when ABCD forms a straight line, but that causes AD to split R, which is again not allowed by the partial results.

We have covered all possible arrangements of the two remaining rectangles, and shown that none of them can give rise to a compound-free tiling. In conclusion, the tiling asked for in the original question cannot be done for n = 6.