IMO 1993 SL 25

Origin: VIE

Problem

Solve the following system of equations, in which a is a given number satisfying |a| > 1: x2 1 = ax2 + 1, x2 2 = ax3 + 1, \cdot \cdot \cdot \cdot \cdot \cdot x2 999 = ax1000 + 1, x2 1000 = ax1 + 1.

Solution

We need only consider the case a > 1 (since the case a < −1 is reduced to a > 1 by taking a′ = −a, x′ i = −xi). Since the left sides of the equations are nonnegative, we have xi \geq−1 a > −1, i = 1, . . . , 1000. Suppose w.l.o.g. that x1 = max{xi}. In particular, x1 \geqx2, x3. If x1 \geq0, then we deduce that x2 1000 \geq1 ⇒x1000 \geq1; further, from this we deduce that x999 > 1 etc., so either xi > 1 for all i or xi < 0 for all i. (i) xi > 1 for every i. Then x1 \geqx2 implies x2 1 \geqx2 2, so x2 \geqx3. Thus x1 \geqx2 \geq\cdot \cdot \cdot \geqx1000 \geqx1, and consequently x1 = \cdot \cdot \cdot = x1000. In this case the only solution is xi = 1 2(a + \sqrt a2 + 4) for all i. (ii) xi < 0 for every i. Then x1 \geqx3 implies x2 1 \leqx2 3 ⇒x2 \leqx4. Similarly, this leads to x3 \geqx5, etc. Hence x1 \geqx3 \geqx5 \geq\cdot \cdot \cdot \geqx999 \geqx1 and x2 \leqx4 \leq\cdot \cdot \cdot \leqx2, so we deduce that x1 = x3 = \cdot \cdot \cdot and x2 = x4 = \cdot \cdot \cdot . Therefore the system is reduced to x2 1 = ax2 + 1, x2 2 = ax1 + 1. Subtracting these equations, one obtains (x1 −x2)(x1 + x2 + a) = 0. There are two possibilities: (1) If x1 = x2, then x1 = x2 = \cdot \cdot \cdot = 1 2(a − \sqrt a2 + 4). (2) x1 + x2 + a = 0 is equivalent to x2 1 + ax1 + (a2 −1) = 0. The discriminant of the last equation is 4 −3a2. Therefore if a > \sqrt 3, this case yields no solutions, while if a \leq \sqrt 3, we obtain x1 = 2(−a − \sqrt 4 −3a2), x2 = 1 2(−a + \sqrt 4 −3a2), or vice versa.