# 1976 IMO Problems/Problem 5

## Problem

We consider the following system with $q = 2p$:

$\begin{matrix} a_{11}x_{1} + \ldots + a_{1q}x_{q} = 0, \\ a_{21}x_{1} + \ldots + a_{2q}x_{q} = 0, \\ \ldots , \\ a_{p1}x_{1} + \ldots + a_{pq}x_{q} = 0, \\ \end{matrix}$

in which every coefficient is an element from the set $\{ - 1,0,1\}$$.$ Prove that there exists a solution $x_{1}, \ldots,x_{q}$ for the system with the properties:

a.) all $x_{j}, j = 1,\ldots,q$ are integers$;$

b.) there exists at least one j for which $x_{j} \neq 0;$

c.) $|x_{j}| \leq q$ for any $j = 1, \ldots ,q.$

## Solution

First of all note that we have $(q + 1)^q - 1$ possible nonzero vectors $(x_1,\cdots,x_q)$ such that $0\leq x_i\leq q$ are integers.

But $a_{j1}x_1 + \cdots + a_{jq}x_q$ can only assume $q^2 + 1$ different values, because if it is maximized/minimized by $(M_1,M_2,\cdots,M_q)/(m_1,m_2,\cdots,m_q)$, we have that $\sum_{i = 1}^{q}a_{ji}(M_i - m_i)\leq q\times q = q^2$ (if $a_{ji} = 0$, it doesn't affect the sum, if it is $1$, $M_i = q,m_i = 0$, and if it is $- 1$, $M_i = 0,m_i = q$).

From this we conclude that there are at most $(q^2 + 1)^p = (q^2 + 1)^{q/2}$ possible values for the vector $(a_{11}x_{1} + \ldots + a_{1q}x_{q},\cdots,a_{p1}x_{1} + \ldots + a_{pq}x_{q})$. But we have that: $(q + 1)^q - 1 = (q^2 + 1 + 2q)^{q/2} - 1 =$ $= (q^2 + 1)^{q/2} + \left( - 1 + \sum_{j = 1}^{q/2}{{q/2}\choose j}(q^2 + 1)^{q/2 - j}(2q)^j\right) > (q^2 + 1)^{q/2}$

We conclude that by the pigeonhole principle there are two distinct vectors being mapped to the same vector. Taking their difference we have a vector with the desired properties.

The above solution was posted and copyrighted by Jorge Miranda. The original thread for this problem can be found here: [1]