# 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, $x_1=x_2=x_3=x_4=x_5=0$.

## See Also

 1963 IMO (Problems) • Resources Preceded byProblem 3 1 • 2 • 3 • 4 • 5 • 6 Followed byProblem 5 All IMO Problems and Solutions
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