Can a regular $n$-simplex have vertices in $\mathbb Z^n$ for $n > 1$?

Asked Feb 20 '19 at 01:25

Active Feb 21 '19 at 06:52

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Trivially, a regular $0$-simplex (point) and $1$-simplex (line segment) can have integer vertices in $0$ and $1$ dimensional Euclidean space respectively. On the other hand, a regular $2$-simplex (equilateral triangle) cannot have vertices in $\mathbb Z^2$.

Is it true more generally that an $n$-simplex cannot have vertices in $\mathbb Z^n$ for $n > 1$? Is there an easy way to see this?

Edit:

Eugen J. Ionascu provides an example of a regular tetrahedron with vertices in $\mathbb Z^3$: $(0,0,4),(7,0,3),(3,5,0),(4,5,7)$

Do such regular $n$-simplices exist for all $n \neq 2$?

edited Jun 12 '20 at 10:38

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asked Feb 20 '19 at 01:25

Peter Kagey

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5

What about $(0,0,0),(1,1,0),(1,0,1),(0,1,1)$? – Michael Biro Feb 20 '19 at 01:29
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See one of the answers in https://mathoverflow.net/questions/38724/coordinates-of-vertices-of-regular-simplex. The statement is that it's possible iff n+1 is the sum of one, two, four or eight odd squares! – Michael Biro Feb 20 '19 at 01:36
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It looks like the question is answered here on MathOverflow. – Peter Kagey Feb 20 '19 at 01:45

Can a regular $n$-simplex have vertices in $\mathbb Z^n$ for $n > 1$?

Edit:

0 Answers0