 # 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, there exists an integer $0<\ell  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

Number Theory
Hard
##### Suggested Book

Do you really need a hint? Try it first!

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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