The 2018 South African Mathematics Olympiad — Problem 5
The final round of the South African Mathematics Olympiad will be taking place on Thursday, 28 July 2019. I have been writing about some of the problems from the senior paper from 2018. A list of all of the problems can be found here.
Today we will look at the fifth problem from the 2018 South African Mathematics Olympiad:
Determine all sequences \(a_1, a_2, a_3, \ldots\) of nonnegative integers such that \(a_1 < a_2 < a_3 < \ldots\), and \(a_n\) divides \(a_{n - 1} + n\) for all \(n \geq 2\).
Since the sequence \(a_1, a_2, \ldots\) is strictly increasing, we know that \(a_n \geq n - 1\) for all positive integers \(n\). (We could prove this rigorously by induction.) This means that \(a_{n - 1} + n \leq (a_n - 1) + (a_n + 1) = 2a_n\) for all \(n\), and so we know that \(a_{n - 1} + n\) is equal to either \(a_n\), or to \(2a_n\) for all positive integers \(n\). Perhaps we should try to figure out exactly when it is equal to \(a_n\), and when it is equal to \(2a_n\). If we knew, for example, that we always have that \(a_{n - 1} + n = a_n\), then we have reduced the problem to solving this recurrence relation.…
