Understand the problem

Given a prime number $p$ and let $\overline{v_1},\overline{v_2},\dotsc ,\overline{v_n}$ be $n$ distinct vectors of length $p$ with integer coordinates in an $\mathbb{R}^3$ Cartesian coordinate system. Suppose that for any $1\leqslant j<k\leqslant n$, there exists an integer $0<\ell <p$ such that all three coordinates of $\overline{v_j} -\ell \cdot \overline{v_k} $ is divisible by $p$. Prove that $n\leqslant 6$.

Source of the problem

Kürschák Competition 2018

Topic
Number Theory
Difficulty Level
Hard
Suggested Book

Start with hints

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Prove that any two vectors are either perpendicular or colinear.

Note that p^2| |\overline{v_j}-\ell\cdot\overline{v_k}|^2. This implies that p^2|\langle \overline{v_j},\overline{v_k}\rangle. Also, from the Cauchy-Schwarz inequality we get |\overline{v_j}|\cdot |\overline{v_k}|\ge |\langle\overline{v_j}.\overline{v_k}\rangle| hence -|\overline{v_j}|\cdot |\overline{v_k}|\le\langle\overline{v_j}.\overline{v_k}\rangle\le |\overline{v_j}|\cdot |\overline{v_k}|. Also, $latex | \overline{v_j}|\cdot |\overline{v_k}|=p^2$. As the dot product is also divisible by p^2, it has to be equal to \pm p^2 or 0. It cannot be p^2 because the vectors are distinct, hence it is either -p^2 or 0. Thus the two vectors are either perpendicular or colinear (adding to 0).

Let us plot the vectors in \mathbb{R}^3 and identify them with their tips (as the tails are at the origin). We join two tips by a straight line segment if they correspond to perpendicular vectors. Show that the number of straight lines is at least \frac{n(n-2)}{2}. Also prove that, there do not exist 4 vectors such that every possible pair of tips is joined by a line segment.

From the last assertion, show that the number of straight line segments in the figure is at most \frac{n^2}{3} (this is a special case of a result called Turan’s theorem). Thus, \frac{n^2}{3}\ge \frac{n(n-2)}{2}. Hence n\le 6

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