Packing a 3xMxN-box with copies of a single tetracube

Question

There seems to be no solution for packing a 3xMxN-box with copies of this tetracube.

Does anybody know of a proof that this is impossible or is this still an open question?

Answer 1

Is it possible:

No. The reasoning is below:

First of all, $M$x$N$ must be divisible by 4 in a $3$x$M$x$N$ box because you're using shapes made up of 4 cubes. If we paint all the unit cubes like a 3D checkerboard, half of the "middle cubes" of the tetracubes must be of the opposite color of the rest. For this to be possible, $M$x$N$ must be divisible by 8.

Let's say there are $t_{bottom}$ tetracubes with three cubes at the bottom on the box's base, and $o_{bottom}$ with one. The base's area is $3t_{bottom}+o_{bottom}$. If $3t_{bottom}+o_{bottom} = MN$, then $t_{bottom}>o_{bottom}$ and $t_{top}>o_{top}$.

$3t_{bottom}+o_{bottom} = MN$
$3t_{top}+o_{top} = MN$
$t_{bottom}+t_{top}+3o_{top}+3o_{bottom} = MN$
$3(MN-3t_{bottom}) + 3(MN-3t_{top}) + t_{bottom}+t_{top} = MN$
$6MN-8t_{bottom}-8t_{top} = MN$
$5MN = 8t_{bottom}+8t_{top}$
$t_{top} = 5MN/8 - t_{bottom}$

Either the top or bottom layer has $5MN/16$ or more, and each has at least $5MN/8 - MN/3 = 7MN/24$. Both layers have a combined $MN/8$ lone legs of the tetracubes.

If two tetracubes combine into a cube, the latter can't be at a corner.

Let's look at the edges. There are $4*(N+1)*M + 4*N*(M+1) + 3*(N+1)*(M+1) = 11MN + 7N + 7M + 3$ in/on our prism.

Each tetracube has $48$ edges including the ones counted multiple times within the tetracube itself and there are $3MN/4$ tetracubes, so the product is $36MN$.

We also know that there are $5MN/8$ top- and bottom-centered ("justified") tetracubes in total and two edges at the top of each overlap with the surface.

Also, at least the 3 "deep" edges on every tetracube are counted only twice as well as the top/bottom 6 edges on each justified tetracube (the latter can overlap though). Assuming the rest are counted four times, the total count is:

4 times: $11MN + 7N + 7M + 3 - (23MN/4) = 21MN/4 + 7N + 7M + 3$
Twice: $18MN/4$
Once: $5MN/4$.

The total is at most $41MN/4 + 21MN + 28N + 28M + 12$.

$41MN/4 + 21MN + 28N + 28M + 12 >= 36MN$
$MN/4 + 28N + 28M + 12 >= 5MN$
$28N + 28M + 12 >= 19MN/4$
$112N + 112M + 48 >= 19MN$

Even assuming it could potentially work for sufficiently small $N$ and $M$ values, the fact that it's not applicable for any larger value (not even multiples) means it's impossible for any $N$ and $M$.