MathDB
a_{n+1} = a^2_{n} + 2018 for n>=1, exists at most one perfect cube

Source: Irmo 2018 p2 q9

September 16, 2018
perfect cubeSequencerecurrence relationnumber theory

Problem Statement

The sequence of positive integers a1,a2,a3,...a_1, a_2, a_3, ... satisfies an+1=an2+2018a_{n+1} = a^2_{n} + 2018 for n≥1n \ge 1. Prove that there exists at most one nn for which ana_n is the cube of an integer.
Back to ProblemsView on AoPS