There do not exist points
- Tolaso J Kos
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There do not exist points
Prove that there do not exist four points in $\mathbb{R}^2$ whose pairwise distances are all odd integers.
Imagination is much more important than knowledge.
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
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Re: There do not exist points
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.
Imagination is much more important than knowledge.
Re: There do not exist points
Awesome result!
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