IMO 1963 - Problem 4

Problem: Find all solutions $x_1,...,x_5$ to the system of equations:
$\left\{\begin{array}{I} x_5+x_2=yx_1,\\ x_1+x_3=yx_2,\\x_2+x_4=yx_3,\\ x_3+x_5=yx_4,\\x_4+x_1=yx_5, \end{array}\right.$ where $y$ is a real parameter.
Solution: Summing up all the equations yields $2(x_1+x_2+x_3+x_4+x_5)=y(x_1+x_2+x_3+x_4+x_5)$. If $y=2$, then the given equations imply $x_1-x_2=x_2-x_3=...=x_5-x_1$; hence $x_1=x_2=...=x_5$, which is clearly a solution. If $y\ne 2$, then $x_1+...+x_5=0$, and summing the first three equalities gives $x_2=y(x_1+x_2+x_3)$. Using that $x_1+x_3=yx_2$ we obtain $x_2=(y^2+y)x_2,i.e,(y^2+y-1)x_2=0$. If $y^2+y-1\ne 0$ then $x_2=0$ and similarly $x_1=...=x_5=0$. If $y^2+y-1=0$, it is easy to prove that the last two equations are the consequence of the first three. Thus choosing any values for $x_1$ and $x_5$ will give exactly one solution for $x_2,x_3,x_4$.