58th Ukrainian National Mathematical Olympiad

Question

A sequence (x_n) satisfies the following conditions: x_1 = a, x_n+1 = 12(x_n - 1x_n), n N. Prove that there exists a number a such that the sequence (x_n) has exactly 2018 pairwise distinct elements. (If one of the elements of the sequence equals 0, then the sequence stops on that element.)

Accepted Answer

Let us denote x_1 = a = ctg . Then x_2 = 12(x_1 - 1x_1) = 12(ctg - tg ) = 12 ^2 - ^2 = 2 2 = ctg 2, In the same way we easily prove that x_n+1 = 12(x_n - 1x_n) = 12(ctg 2^n-1 - tg 2^n-1) = 12 ^2 2^n-1 - ^2 2^n-1 2^n-1 2^n-1 = ctg 2^n , n N. The statement of the problem will be satisfied if x_i 0, i = 1, 2017, and x_2018 = 0. Therefore, it is sufficient to choose such that x_2018 = ctg(2^2017) = 0. Thus we have 2^2017 = 2, and = 2^2018.