Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

If $x$ is a real number, let $\{x\} = x - \lfloor x \rfloor$ denote the fractional part of $x$. Given an integer number $m \ge 3$, there exists a positive integer number $n$ such that $\{n\pi\} < 1/m$, for the set $\{k\pi : k = 0, 1, 2, 3, \dots\}$ is dense in the closed unit interval $[0, 1]$. Consequently,
$$\lfloor kn\pi \rfloor = \lfloor k \lfloor n\pi \rfloor \rfloor + \lfloor k\{n\pi\} \rfloor = k \lfloor n\pi \rfloor + \lfloor k\{n\pi\} \rfloor = k \lfloor n\pi \rfloor, \quad k = 1, 2, \dots, m,$$
so $\lfloor n\pi \rfloor, \lfloor 2n\pi \rfloor, \dots, \lfloor mn\pi \rfloor$ are $m$ numbers in $S$ in arithmetic progression with ratio $\lfloor n\pi \rfloor$.

Suppose, if possible, that $S$ contains an infinite arithmetic progression $\lfloor n_k \pi \rfloor$, $k = 0, 1, 2, 3, \dots$, with (integral) ratio $r$, where the $n_k$ form a strictly increasing sequence of positive integer numbers. Write $n_k \pi = \lfloor n_0 \pi \rfloor + kr + \{n_k \pi\}$ to deduce that $r$ is positive and
$$n_{k+1} - n_k = \frac{r + \{n_{k+1}\pi\} - \{n_k\pi\}}{\pi} \in \left( \frac{r-1}{\pi}, \frac{r+1}{\pi} \right).$$
The length of this interval is less than $1$, so $n_{k+1} - n_k = n$ for some positive integer $n$ and all indices $k$. Hence, $n_k = n_0 + kn$, so $n_k/k \xrightarrow{k \to \infty} n$. On the other hand, $n_k/k \xrightarrow{k \to \infty} r/\pi$, so $\pi = r/n$ which contradicts irrationality of $\pi$.