Suppose the origin is one of the points, we can take $(3,0)$ as the second point. If $(x,y)$ is some other point in the set I think we can use the fact that $x^2+y^2$, for $x$ and $y$ odd, is never an integer. There is also a lot of lines that the points certainly cannot belong to. But I could not go beyond.
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I don't think your question is phrased right. What does 'greatest number of point(s) mean? And distance between any of them is odd? Are you talking about a set of points or something else? – sato Dec 24 '20 at 09:46
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3@Mastermind817I think the question is asking " A set of points such that the distance between any two points you pick is an odd natural number." – Dec 24 '20 at 09:52
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@Mastermind817 Forester is right. Sorry the bad translation – Gabriela Da Silva Dec 24 '20 at 09:57
1 Answers
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The answer is $3$.
Assume the contrary, let's say we can find four points $O,A,B,C$ such that following six distances are all odd integers: $$(a,b,c,a_1,b_1,c_1) = (BC,CA,AB,OA,OB,OC)$$
Since $O, A, B, C$ lies in the plane, they form a degenerate tetrahedron with volume $V = 0$. Express $V$ in terms of corresponding Cayley Menger determinant, the distances satisfy
$$\left|\begin{matrix} 0 & 1 & 1 & 1 & 1\\ 1 & 0 & a_1^2 & b_1^2 & c_1^2\\ 1 & a_1^2 & 0 & c^2 & b^2 \\ 1 & b_1^2 & c^2 & 0 & a^2\\ 1 & c_1^2 & b^2 & a^2 & 0 \\ \end{matrix}\right| = 288V^2 = 0\tag{*1}$$
Recall if $u$ is an odd integer, then $u^2 \equiv 1 \pmod 8$.
Taking modulo $8$ on both sides of $(*1)$, we arrive at a contradiction.
$$4 = \left|\begin{matrix} 0 & 1 & 1 & 1 & 1\\ 1 & 0 & 1 & 1 & 1\\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1\\ 1 & 1 & 1 & 1 & 0 \\ \end{matrix}\right| \equiv 0 \pmod 8$$
This means it is impossible to find four points in the plane whose pairwise distances are all odd integers.
Since it is trivial to find three points at odd integral distances from each other (eg. the vertices of an equilateral triangle of side $1$), the greatest number we seek is $3$.
achille hui
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