# 2006 Indonesia MO Problems/Problem 7

## Problem

Let $a,b,c$ be real numbers such that $ab,bc,ca$ are rational numbers. Prove that there are integers $x,y,z$, not all of them are $0$, such that $ax+by+cz=0$.

## Solution

We divide the problem into three cases.

Case 1: At least one of $a,b,c$ equal zero

WLOG, let $a = 0$. Then $ax = 0$, no matter what integer $x$ is. We can set $y,z = 0$ and have a set $(x,0,0)$. This solution set works because not all of $x,y,z$ are equal to zero.

Similar reasoning can be used if all three variables are zero and if two of the three variables are zero.

Case 2: None of $a,b,c$ equal zero

Note that non-zero rational numbers are closed under multiplication or division. In other words, multiplying or dividing two rational numbers (unless when dividing by 0) will always result in a rational number.

Therefore, $\tfrac{ab}{ac} = \tfrac{b}{c}$ and $\tfrac{ab}{bc} = \tfrac{a}{c}$ are also rational numbers. In the original equation, dividing both sides by $c$ results in $\tfrac{ax}{c} + \tfrac{by}{c} + z = 0$.

Since $\tfrac{a}{c}$ and $\tfrac{b}{c}$ are rational, let $\tfrac{a}{c} = \tfrac{m_1}{n_1}$ and $\tfrac{b}{c} = \tfrac{m_2}{n_2}$, where $m_1, m_2, n_1, n_2$ are integers. Therefore, we can let $x = n_1$ and $y = n_2$ to have $\tfrac{ax}{c} + \tfrac{by}{c}$ be an integer. Since addition and subtraction are closed under integers, $z$ would also be an integer, so there is an integral solution that satisfies the equation

Thus, in all two cases, there are solutions $(x,y,z)$ (where all of $x,y,z$ are integers) that satisfy the equation.