# Do there exist uniform triangular prisms with all vertices in $\mathbb Z^3$?

It's quite easy to find a regular square prism (cube) or a regular triangular antiprism (octahedron) with vertices in $$\mathbb Z^3$$. Take for instance, take the convex hulls \begin{align*} &\operatorname{hull}(\{(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1)\}), \text{ and} \\ &\operatorname{hull}(\{(0,0,1),(0,1,0),(1,0,0),(0,0,-1),(0,-1,0),(-1,0,0)\}) \end{align*} respectively.

For a much more interesting example of embedding the cube, see Table 1 from Ionascu and Obando: $$\operatorname{hull}(\{(0, 3, 2), (1, 1, 4), (2, 2, 0), (2, 5, 3), (3, 0, 2), (3, 3, 5), (4,4, 1), (5, 2, 3)\}).$$

### Question

Do there exist uniform triangular prisms with all vertices in $$\mathbb Z^3$$ (equivalently $$\mathbb Q^3$$)? If not, can it be done in $$\mathbb Z^4$$, $$\mathbb Z^5$$, etc?

(I've checked all examples in $$\mathbb Z^3$$ where the triangular faces have side lengths up to $$\sqrt{700}$$. This did not produce any examples, but there are plenty of near-misses.)

• I hesitate to write an answer again (because it seems rocky and ugly, so I might wait a little longer to try to tidy it up), but you can be sure that there isn't such thing in $\mathbb{Z}^3$, I've checked. This time correctly :D Feb 25, 2021 at 22:17
• I instantly thought of this: \begin{align} & (1,0,0),\quad(0,1,0),\quad(0,0,1) \\ {} \\ & (2,1,1), \quad (1,2,1), \quad (1,1,2) & & \phantom{mmmmmmmmmmmmmmmmmmmmmmmmmm} \end{align} But of course that doesn't work. $\qquad$ Feb 26, 2021 at 1:18
• That this is possible in sufficiently large dimension follows more generally from the answers to this question. Mar 2, 2021 at 16:28
• @M.Winter: I think it actually does work - the post relies on a rational squared distance, which is true of the distance between any two points in $\mathbb{Q}^n$, so things should be OK. Mar 2, 2021 at 19:41
• @M.Winter I found a proof of course. I will write an answer as soon as I get to a computer. Mar 6, 2021 at 15:55

Suppose that there exists a triangular prism in $$\mathbb{Z}^3$$ of side length $$s$$. From a given vertex $$V$$, let $$a$$ and $$b$$ be the vectors on the triangular face containing $$V$$ and $$c$$ the vector from $$V$$ to its pair on the other triangular face. Note that $$a,b$$, and $$c$$ are all of length $$s$$.

Now, consider the integer vector $$a\times b$$; since $$a$$ and $$b$$ are $$60^\circ$$ apart, it has length $$s^2\cdot\frac{\sqrt{3}}2$$. Up to a change of sign, this vector is parallel to $$c$$, so $$c$$ is a rational multiple of $$a\times b$$ (since both lie in $$\mathbb{Z}^3$$). Thus, $$s$$ and $$\frac{s^2\sqrt{3}}2$$ differ by a rational factor, so we conclude that $$s$$ is a rational multiple of $$\sqrt{3}$$.

This means that we can scale the prism so that its side length is an integer multiple of $$\sqrt{3}$$. Then scale it by a further factor of $$2$$, so that the midpoints of each edge also have integer coordinates. Let $$k\sqrt{3}$$ be the integer distance from a vertex to a midpoint of an edge.

But now consider the integer vector from the midpoint of a non-triangle edge to a non-adjacent vertex: it has length $$(k\sqrt{3})\cdot \sqrt{5}$$.

So, the squared distance $$15k^2$$ between these two integer points is the sum of three squares. Note that $$k^2$$ is of the form $$4^n(8a+1)$$, so $$15k^2$$ is of the form $$4^n(8b+7)$$. But, by Legendre's three-square theorem, such integers are not expressible as the sum of three squares! So no such prism exists.

There do exist triangular prisms in $$\mathbb{Z}^5$$: Take the first three coordinates to be any permutation of $$(0,0,1)$$, and the last two to be either $$(0,0)$$ or $$(1,1)$$. This gives a triangular prism of side length $$\sqrt{2}$$.

I'm not sure yet about $$\mathbb{Z}^4$$ - my guess is no, but checking some small examples seems worthwhile.

• Your example for $\mathbb{Z}^5$ is very nice Feb 25, 2021 at 22:47