IMO 2004 SL A3

Origin: CAN | Category: Algebra

Problem

Does there exist a function s: Q \to{−1, 1} such that if x and y are distinct rational numbers satisfying xy = 1 or x + y \in{0, 1}, then s(x)s(y) = −1? Justify your answer.

Solution

The answer is yes. Every rational number x > 0 can be uniquely expressed as a continued fraction of the form a0 + 1/(a1 + 1/(a2 + 1/(\cdot \cdot \cdot + 1/an))) (where a0 \inN0, a1, . . . , an \inN). Then we write x = [a0; a1, a2, . . . , an]. Since n depends only on x, the function s(x) = (−1)n is well-defined. For x < 0 we define s(x) = −s(−x), and set s(0) = 1. We claim that this s(x) satisfies the requirements of the problem. The equality s(x)s(y) = −1 trivially holds if x + y = 0. Suppose that xy = 1. We may assume w.l.o.g. that x > y > 0. Then x > 1, so if x = [a0; a1, a2, . . . , an], then a0 \geq1 and y = 0 + 1/x = [0; a0, a1, a2, . . . , an]. It follows that s(x) = (−1)n, s(y) = (−1)n+1, and hence s(x)s(y) = −1. Finally, suppose that x + y = 1. We consider two cases: (i) Let x, y > 0. We may assume w.l.o.g. that x > 1/2. Then there exist natural numbers a2, . . . , an such that x = [0; 1, a2, . . . , an] = 1/(1 + 1/t), where t = [a2, . . . , an]. Since y = 1 −x = 1/(1 + t) =

[0; 1 + a2, a3, . . . , an], we have s(x) = (−1)n and s(y) = (−1)n−1, giving us s(x)s(y) = −1. (ii) Let x > 0 > y. If a0, . . . , an \inN are such that −y = [a0; a1, . . . , an], then x = [1 + a0; a1, . . . , an]. Thus s(y) = −s(−y) = −(−1)n and s(x) = (−1)n, so again s(x)s(y) = −1.