It's known that a cube can't be divided into smaller cubes of distinct sizes. So for fun, I defined a "wannabe cube" as a cuboid whose dimensions are (not equal, but) consecutive integers and looked to solve the analogous problem. So let's say the $k$-th wanna be cube is the rectangular prism of size $k\times(k+1)\times(k+2).$ Turns out the fifth wannabe cube can be divided into the first four. Or oppositely, we can build up the fifth wannabe cube from the first four as follows:
- Join the first and second wannabe cubes along a $2\times 3$ face producing a $2\times 3\times 5$ prism.
- Join this to the third wannabe cube along a $3\times 5$ face producing a $3\times5\times 6$ prism.
- Join this to the fourth wanna be cube along a $5\times6$ face produing a $5\times 6\times 7$ prism, which is the fifth wannabe cube.
This construction actually works in any dimension. Say if we define a wannabe $n$-cube as an $n$-dimensional orthotope of consecutive integral dimensions. Particularly, the $k$-th wannabe $n$-cube is the orthotope of size $k\times(k+1)\times...\times(k+n-1)$. Then the $(n+2)$-th wannabe $n$-cube can always be divided into the first $n+1$ such wannabe $n$-cubes.
In, say, 2 dimensions this amounts to fitting together the three rectangles of sizes $1\times 2$, $2\times 3$, and $3\times 4$ into a $4\times 5$ rectangle.
These constructions can be seen as physical embodiments of an identity on Pascal's triangle: $$\sum_{n=k}^{2k}\binom{n}{k}=\binom{2k+1}{k}$$ Namely, that the first $k+1$ entries of the $k$-th diagonal sum to the $(k+2)$-th entry.
I'm wondering if there are any divisions of wannabe $n$-cubes for $n>2$ that aren't given by this family of constructions. For 3 dimensions, I was able to show that the outlined construction is the only one which can be obtained by incrementally joining wannabe cubes to a growing cuboid along matching sides.
For clarity, I'm also considering two orthotopes the same regardless of their orientation. E.g. that a division into distinct wannabe cubes could not use both a $1\times2\times 3$ prism AND a $3\times 1\times 2$ prism.
Since the volumes of wanna be cubes are just 6 times the tetrahedral numbers, my current plan is to look for tetrahedral numbers which can be expressed as a sum of smaller distinct tetrahedral numbers in many ways, and then try building up a division in 3D space. But this hasn't turned up any new solutions. I know Smith diagrams and Kirchoff laws were used to solve squaring the square and thought maybe those laws could be used here. But I'm not sure how to get started with that approach.