59th Ukrainian National Mathematical Olympiad

Question

Given a fixed natural number n > 1, 2019 natural numbers are placed around the circle in such a way that the product of any two neighboring numbers is a perfect n-th power. Is it always the case that the product of any (not necessarily neighboring) two numbers is also a perfect n-th power? (A. Nikolaev, B. Rublyov)

Accepted Answer

For the odd n = 2l + 1 one can prove the statement as follows: P^2 = (a_1a_2)(a_2a_3)(a_3a_4)(a_2018a_2019)(a_2019a_1) = m^2l+1, where P = a_1a_2 a_2019. From here it follows that P = a_1a_2 a_2019 = s^2l+1. Therefore, a_1 = P(a_2a_3)(a_4a_5)(a_2018a_2019) = s^2l+1u^2l+1 = v_1^2l+1, and the same holds for all a_i = v_i^2l+1, i = 2, , 2019. From here it directly follows that a_i a_j = v_i^2l+1 v_j^2l+1 = v^2l+1. For even n = 2l one can write P^2 = (a_1a_2)(a_2a_3)(a_3a_4)(a_2018a_2019)(a_2019a_1) = m^2l, where P = a_1a_2 a_2019. From here it follows that P = m^l. Therefore, a_1^2 = P^2(a_2a_3)(a_4a_5)(a_2018a_2019) = m^2lu^2l = v_1^2l, and the same holds for all a_i^2 = v_i^2l, i = 2, , 2019. From here it directly follows that r^2l = (a_1a_2)(a_2a_3)(a_3a_4)(a_k-1a_k) = a_1a_2^2a_3^2 a_k-1…