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: $$ \operatorname{hull}(\{(0, 3, 2), (1, 1, 4), (2, 2, 0), (2, 5, 3), (3, 0, 2), (3, 3, 5), (4,4, 1), (5, 2, 3)\}). $$

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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.)

  • $\begingroup$ 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 $\endgroup$
    – donaastor
    Feb 25, 2021 at 22:17
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    $\begingroup$ 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. $\qquad$ $\endgroup$ Feb 26, 2021 at 1:18
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    $\begingroup$ That this is possible in sufficiently large dimension follows more generally from the answers to this question. $\endgroup$
    – M. Winter
    Mar 2, 2021 at 16:28
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    $\begingroup$ @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. $\endgroup$ Mar 2, 2021 at 19:41
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    $\begingroup$ @M.Winter I found a proof of course. I will write an answer as soon as I get to a computer. $\endgroup$
    – donaastor
    Mar 6, 2021 at 15:55

1 Answer 1


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 $s$.

enter image description here

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.

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

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