5 points on a plane with rational distances

Question

Can you find 5 points on a plane whose Euclidean distances between them are all rational numbers and no 3 points out of them are co-linear?

If the answer is yes, can we find a construction for any number of points?

I thought about this for fun. It is trivial for 3 points or co-linear points. For 4 points, @peterwhy suggested a rectangle of size 3*4.

One easy construction for any number of points where almost all points are colinear except 2 points. First, consider three points (0,0), (0,1), (0,-1). For each rational triplet $(1,a,b)$ where $a^2+ 1 = b^2$, add two points to $(\pm a,0)$. There are infinitely many such triplets.

Answer 1

Mathworld says you can get six and cites Richard K. Guy, "Unsolved Problems in Number Theory", D20, p. 185

He shows a hexagon with opposite sides parallel, $85, 68, 80$ long. The diagonals that do not pass through the center are $127,131,133,158$

Answer 2

http://www.famosweb.org/arkiv/15-3.pdf (in danish, I'll try to find time to produce a translation later, but it might not until the weekend) contains a proof of the following:

For $n\geq 2$, there exists a limited, enumerably infinite set $A_n \subset \mathbb R^n$ such that any pair of points from $A_n$ has rational distance and $A_n$ spans $\mathbb R^n$.

As for "limited", the proof seems to construct the points so the all have norm $1$.

It's not exactly the same, as this doesn't say that no three points are co-linear, but as I remember it (I read in in 2002, it was in a student magazine I edited, and kind of a follow-up to an exercise I had posted in a previous issue) no three points are co-linear.

Answer 3

Based on Henrik's finding, here's a summarized translation of Ansgar Grunnet, Om punkter med rational afstand:

Consider $z=a+bi\in\mathbb Q[i]$ with $|z|=1$. Then $$\begin{align}|z^2-1|^2&=|z-1|^2|z+1|^2\\&=((a-1)^2+b^2)((a+1)^2+b^2)\\&=(2-2a)(2+2a)\\&=4(1-a^2)\\&=4b^2\end{align}$$ hence $|z^2-1|=|2b|\in\mathbb Q$. Since for all $n\in\mathbb N$, we also have $z^n\in\mathbb Q[i]$ and $|z|^n=1$, we conclde that $|z^{2n}-1|$ is rational for all $n$. If we let $a_n=z^{2n}$, this implies that the distance between $a_n$ and $a_m$ is rational for all $n,m\in\mathbb N$, $n>m$: $$ |a_n-a_m|=|z^{2m}|\cdot|z^{2(n-m)}-1|=|z^{2(n-m)}-1|\in\mathbb Q.$$

From the Pythagorean triple $(3,4,5)$ we can pick $z=\frac35+\frac45i$, for example. Remains to show that the $a_n$ are pairwise distinct, which amounts to $z^{2n}\ne1$ for all $n\in\mathbb N$. This can be seen by noticing that $(3+4i)^2=-7+24i\equiv 3+4i\pmod 5$, hence $(3+4i)^2\equiv 3+4i\pmod 5$ for all $n\ge 1$ and thus finally $\Im (5z)^n\ne 0$ for all $n\ge 1$.