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Prove that there do not exist four points in $\mathbb{R}^2$ whose pairwise distances are all odd integers.

Tolaso J Kos wrote:Prove that there do not exist four points in $\mathbb{R}^2$ whose pairwise distances are all odd integers.

Here is a solution I have seen for this exercise in a magazine.

Suppose these four points do exist. We can translate them so that one of them is at the origin. Let the four points be $0, a, b, c \in \mathbb{R}^2$ Then:


$$|a|, \; |b|, \; |c|, \; |d|, \; |a-b|, \; |c-a|, \; |b-c|$$


are all odd integers so their squares are all $1 \pmod 8$. It follows that:


$$2a \cdot b= |a|^2+|b|^2- |a-b|^2 \equiv 1 \pmod 8$$


Let $V$ be the $2\times 3$ matrix whose columns are $a, b, c$. Consider the Gram Matrix:


$$B=V^t V= \begin{pmatrix}

a\cdot a & a\cdot b & a\cdot c\\

b\cdot a& b\cdot b &b\cdot c \\

c \cdot a &c \cdot b & c \cdot c

\end{pmatrix}$$


We have:


$$2B \equiv \begin{pmatrix}

2 &1 &1 \\

1& 2 & 1\\

1&1 &2

\end{pmatrix} \pmod 8$$


Thus $\det (2B)= 4 \pmod 8$ and hence $\det B \neq 0$. However, this is impossible since $${\rm rank} B = {\rm rank} V^t V < {\rm rank }V <2$$ as $V$ is a $2\times 3$ matrix.

Awesome result!