# Difference between revisions of "1963 IMO Problems/Problem 4"

## Problem

Find all solutions $x_1,x_2,x_3,x_4,x_5$ of the system $\begin{eqnarray*} x_5+x_2&=&yx_1\\ x_1+x_3&=&yx_2\\ x_2+x_4&=&yx_3\\ x_3+x_5&=&yx_4\\ x_4+x_1&=&yx_5,\end{eqnarray*}$ where $y$ is a parameter.

## Solution

Notice: The following words are Chinese.

$2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)$

$x_1+x_2+x_3+x_4+x_5=0$时,因为$x_1,x_2,x_3,x_4,x_5$关于原方程组轮换对称,所以

$x_1=x_2=x_3=x_4=x_5=0$

$x_1=x_2=x_3=x_4=x_5$

The solution in English (translated by Google Translate):

First of all, we can add the five equations to get:

$2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)$

When $x_1+x_2+x_3+x_4+x_5=0$, Because $x_1,x_2,x_3,x_4,x_5$ is symmetric in the original equations,

$x_1=x_2=x_3=x_4=x_5=0$

Otherwise, dividing both sides by $(x_1+x_2+x_3+x_4+x_5$, we get $y=2$, and clearly

$x_1=x_2=x_3=x_4=x_5$

Summarizing, if $y=2$, then the answer is of the form $x_1=x_2=x_3=x_4=x_5$. Otherwise, $x_1=x_2=x_3=x_4=x_5=0$.