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How many points can one can place in $\mathbb{R}^n$, with the requirement that no $n+1$ points lie in the same $\mathbb{R}^{n-1}$-plane, and the euclidean distance between every two points is an integer?

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@J.M. if $n+1=3$ then $n-1=1$ so the condition reduces to "no three points in a line". – Gerry Myerson Nov 24 '11 at 23:03
up vote 4 down vote accepted

Take $n=2$. By an old result of Erdős and Anning, there is an infinite collection of points on the unit circle such that all mutual distances are rational. By scaling up by $K!$ for large $K$, this means that for any integer $k$, we can find a circle, and $k$ points on that circle, such that all mutual distances are integers.

In the same paper, Erdős and Anning show that if we have an infinite collection of points in the plane such that all mutual distances are integers, then these points must lie in a line.

The paper referred to above is quite accessible.

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It is for this reason that one usually also imposes the restriction that no $4$ points lie on a circle. In this case there known examples of 7 points in the plane. – Gerry Myerson Nov 24 '11 at 23:02

With the restriction no three points in a line, no four points on a circle, there is a 7 point configuration on the plane. I came here to ask about the upper bound in 3-space, with the added restriction no 4 points on a plane.

[1] Tobias Kreisel and Sascha Kurz, There are integral heptagons, no three points on a line, no four on a circle, Discrete Comput. Geom. 39 (2008), no. 4, 786–790, MR2413160 (2009d:52021) labeled diagram

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As a lower bound: n+1, arranged in a simplex (triangle/tetahedron/etc.)

In the case n=2, you can do four points: (0,0) (0,3) (4,0) (4,3)... or you can use any Pythagorean triple {a,b,c} and place points at (0,0) (0,a) (b,0) (b,a). However, that approach cannot be extended beyond n=2, as there are no known cuboids with integer sides and all-integer diagonals.

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