IMO 1963 · Problem 4

Find all solutions \(x_1,x_2,x_3,x_4,x_5\) of the system \[ \begin{aligned} x_5+x_2&=yx_1\\ x_1+x_3&=yx_2\\ x_2+x_4&=yx_3\\ x_3+x_5&=yx_4\\ x_4+x_1&=yx_5, \end{aligned} \] where \(y\) is a parameter.

Official text of the problem. Official problem paper (PDF)

Answer

All solutions are \[ (x_1,x_2,x_3,x_4,x_5)= \begin{cases} (t,t,t,t,t),&y=2,\\[2mm] \bigl(a,b,yb-a,-y(a+b),ya-b\bigr), &y=\dfrac{-1+\sqrt5}{2}\ \text{or}\ \dfrac{-1-\sqrt5}{2},\\[2mm] (0,0,0,0,0),&\text{otherwise}, \end{cases} \] where \(a,b,t\) are arbitrary real numbers.

Solution

Read indices cyclically, so that \(x_{i+5}=x_i\). The equations become \[ x_{i-1}+x_{i+1}=yx_i. \] Let \(S=x_1+x_2+x_3+x_4+x_5\). Adding all five equations gives \[ (y-2)S=0. \] Also, for every \(i\), \[ y^2x_i =y(x_{i-1}+x_{i+1}) =x_{i-2}+2x_i+x_{i+2}. \] The five cyclic indices \(i-2,i-1,i,i+1,i+2\) represent all five variables. Therefore \[ \begin{aligned} S &=x_i+(x_{i-1}+x_{i+1})+(x_{i-2}+x_{i+2})\\ &=x_i+yx_i+(y^2-2)x_i\\ &=(y^2+y-1)x_i. \end{aligned} \]

If \(y=2\), this identity gives \(S=5x_i\) for every \(i\), so all five variables are equal. Conversely, every tuple \((t,t,t,t,t)\) satisfies the system when \(y=2\).

Suppose now that \(y\ne2\). Then \(S=0\), hence \[ (y^2+y-1)x_i=0 \] for every \(i\). Unless \(y^2+y-1=0\), all five variables must vanish, and this tuple indeed satisfies the system.

It remains to consider \[ y^2+y=1, \qquad\text{that is,}\qquad y=\frac{-1\pm\sqrt5}{2}. \] Set \(x_1=a\) and \(x_2=b\). The second, third, and fourth equations successively force \[ \begin{aligned} x_3&=yb-a,\\ x_4&=yx_3-b =(y^2-1)b-ya =-y(a+b),\\ x_5&=yx_4-x_3 =(1-y^2)a-(y^2+y)b =ya-b. \end{aligned} \] Thus every solution has the stated form.

Conversely, for arbitrary \(a,b\), these definitions satisfy the second, third, and fourth equations by construction. The remaining two equations hold because \[ x_5+x_2=(ya-b)+b=ya=yx_1 \] and \[ x_4+x_1 =(1-y)a-yb =y^2a-yb =y(ya-b) =yx_5. \] This proves both the completeness and the validity of all the listed solutions.

The solution was checked against John Scholes' IMO solution archive (Kalva).