There do not exist points

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

Solution

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.