IMO 1965 · Problem 4

Find all sets of four real numbers \(x_1, x_2, x_3, x_4\) such that the sum of any one and the product of the other three is equal to 2.

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

Answer

The solutions are \[ (x_1,x_2,x_3,x_4)=(1,1,1,1) \] and all permutations of \[ (3,-1,-1,-1). \]

Solution

Write the four numbers as \(a,b,c,d\). The conditions are \[ a+bcd=b+acd=c+abd=d+abc=2. \]

If all four numbers equal \(t\), then \[ t+t^3=2, \qquad (t-1)(t^2+t+2)=0. \] Since \(t^2+t+2=(t+\tfrac12)^2+\tfrac74>0\), we obtain \(t=1\).

Otherwise, by relabelling, assume \(a\ne b\). Subtracting the equations for \(a\) and \(b\) gives \[ (a-b)(1-cd)=0, \] so \(cd=1\). The equation for \(a\) therefore yields \(a+b=2\). Consequently, \[ 4=(a+b)^2=(a-b)^2+4ab>4ab, \] and hence \(ab<1\).

Subtracting the equations for \(c\) and \(d\), we obtain \[ (c-d)(1-ab)=0. \] Because \(ab<1\), this forces \(c=d\). Together with \(cd=1\), it follows that either \(c=d=1\) or \(c=d=-1\).

If \(c=d=1\), the equation \(c+abd=2\) gives \(1+ab=2\), contradicting \(ab<1\). Thus \(c=d=-1\), and the same equation gives \[ -1-ab=2, \qquad ab=-3. \] Since \(a+b=2\), the numbers \(a,b\) are the roots of \[ u^2-2u-3=(u-3)(u+1)=0. \] Thus they are \(3\) and \(-1\), in either order.

Finally, \((1,1,1,1)\) satisfies every equation. For \((3,-1,-1,-1)\), selecting \(3\) gives \(3+(-1)^3=2\), while selecting any \(-1\) gives \(-1+3(-1)(-1)=2\). Hence all listed solutions work, and the cases above exhaust every possibility.

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