32nd IMO shortlist 1991/19 solution

32nd IMO 1991 shortlist

Problem 19

Show that the only rational solution q in the range 0 < q < 1 to the equation cos(3πq) + 2 cos(2πq) = 0 is 2/3.

Solution

We have cos 2A = 2 cos²A - 1, cos 3A = 4 cos³A - 3 cos A. So put x = cos(πq). Then 4x³ - 3x + 4x² - 2 = 0, or (2x + 1)(2x² + x - 2) = 0, so x = -½ or (-1 ± √17)/4. For q to be real we must have |x| <= 1, so x = -½ or (√17 - 1)/4. Since q is in the range 0 to 1, x = -½ implies q = 2/3. Thus we have to show that if cos(πq) = (√17 - 1)/4, then q is irrational.

The idea is to show that cos(πq), cos(2π), cos(4πq), cos(8πq), cos(16πq), ... are all distinct. But for any given positive integer d, nπ/d = one of 0, π/d, 2π/d, ... , (2d-1)π/d mod 2π, so there are at most 2d possible values for cos(nπ/d), where n is an arbitrary integer. Hence if q is rational, the there are only finitely many possible values for cos(2ⁿπq).

We show by induction that cos(2ⁿπq) = (a_n + b_n√17)/4 for odd integers a_n, b_n. We have already shown that a₁ = -1, b₁ = 1. Suppose the result is true for n. Then cos(2ⁿ⁺¹πq) = 2cos²(2ⁿπq) - 1 = 2(a_n + b_n√17)/16 - 1. So a_n+1 = (a_n² + 17b_n²)/2 - 4, b_n = a_nb_n. Odd squares = 1 mod 4 and 17 = 1 mod 4, so a_n² + 17b_n² = 2 mod 4, and hence a_n+1 is odd. b_n+1 is obviously odd. So the result is true for n+1 and hence for all n. Also a_n+1 > ½ a_n² + ½ ≥ a_n, so all the a_n are different. Hence all the cos(2ⁿπq) are different, as claimed.

Note that although it might seem more natural to find cos(nπq) for all integers n, it is much harder work, because the recurrence is more complicated. All we need is an infinite number of distinct values, so we pick a subsequence that is easy to deal with.

32nd IMO shortlist 1991