IMO 1967 LL ROM46

Origin: ROM

Problem

If x, y, z are real numbers satisfying the relations x+y+z = 1 and arctanx + arctan y + arctan z = \pi/4, prove that x2n+1 + y2n+1 + z2n+1 = 1 for all positive integers n.

Solution

Let us set arctan x = a, arctan y = b, arctanz = c. Then tan(a+b) = x+y 1−xy and tan(a + b + c) = x+y+z−xyz 1−yz−zx−xy = 1, which implies that (x −1)(y −1)(z −1) = xyz −xy −yz −zx + x + y + z −1 = 0. One of x, y, z is equal to 1, say z = 1, and consequently x + y = 0. Therefore x2n+1 + y2n+1 + z2n+1 = x2n+1 + (−x)2n+1 + 12n+1 = 1.