P1 IMO-2024 (PENDING)

1. Problem Statement

Determine all real numbers \(\alpha\) such that, for every positive integer \(n\), the integer

\[\lfloor \alpha \rfloor + \lfloor 2\alpha \rfloor + \cdots + \lfloor n\alpha \rfloor\]

is a multiple of \(n\). (Note that \(\lfloor z \rfloor\) denotes the greatest integer less than or equal to \(z\). For example, \(\lfloor -\pi \rfloor = -4\) and \(\lfloor 2 \rfloor = \lfloor 2.9 \rfloor = 2\).)

2. Solution

It is clear that \(\alpha\) equals any even number posits a solution. For it becomes a routine sum of from \(\sum_i i(2m)\), where we substituted \(n\) for \(2m\). This is equal to \(m(n)(n+1)\), which is a clear multiple of \(n\). The converse can be said for odd numbers. For let \(n\) be a even number, the product is of form \(\alpha(n)(n+1)/2\). Now niether \(\alpha\) or \((n+1)\) offer the factor 2 needed for division, and so \(n\) has to offer. However now it is surely not a multiple of \(n\), as it has one less power of 2.

There is a little property: \(\lfloor i + \epsilon \rfloor = i + \lfloor \epsilon \rfloor\), where \(i\) is a integer and \(\epsilon \in (0,1)\).