IMO 1967 LL BUL5

Origin: BUL

Problem

Solve the system x2 + x −1 = y, y2 + y −1 = z, z2 + z −1 = x.

Solution

If one of x, y, z is equal to 1 or −1, then we obtain solutions (−1, −1, −1) and (1, 1, 1). We claim that these are the only solutions to the system. Let f(t) = t2 + t −1. If among x, y, z one is greater than 1, say x > 1, we have x < f(x) = y < f(y) = z < f(z) = x, which is impossible. It follows that x, y, z \leq1. Suppose now that one of x, y, z, say x, is less than −1. Since mint f(t) = −5/4, we have x = f(z) \in[−5/4, −1). Also, since f([−5/4, −1)) = (−1, −11/16) \subseteq(−1, 0) and f((−1, 0)) = [−5/4, −1), it follows that y = f(x) \in(−1, 0), z = f(y) \in[−5/4, −1), and x = f(z) \in(−1, 0), which is a contradiction. Therefore −1 \leqx, y, z \leq1. If −1 < x, y, z < 1, then x > f(x) = y > f(y) = z > f(z) = x, a contradiction. This proves our claim.