IMO 1971 SL 3

Origin: GDR

Problem

Knowing that the system x + y + z = 3, x3 + y3 + z3 = 15, x4 + y4 + z4 = 35, has a real solution x, y, z for which x2 + y2 + z2 < 10, find the value of x5 + y5 + z5 for that solution.

Solution

Let x, y, z be a solution of the given system with x2 + y2 + z2 = \alpha < 10. Then xy + yz + zx = (x + y + z)2 −(x2 + y2 + z2) = 9 −\alpha . Furthermore, 3xyz = x3 +y3+z3−(x+y+z)(x2+y2+z2−xy−yz −zx), which gives us xyz = 3(9 −\alpha)/2 −4. We now have

35 = x4 + y4 + z4 = (x3 + y3 + z3)(x + y + z) −(x2 + y2 + z2)(xy + yz + zx) + xyz(x + y + z) = 45 −\alpha(9 −\alpha)

• 9(9 −\alpha) −12. The solutions in \alpha are \alpha = 7 and \alpha = 11. Therefore \alpha = 7, xyz = −1, xy + xz + yz = 1, and x5 + y5 + z5 = (x4 + y4 + z4)(x + y + z) −(x3 + y3 + z3)(xy + xz + yz) + xyz(x2 + y2 + z2) = 35 \cdot 3 −15 \cdot 1 + 7 \cdot (−1) = 83.