Consider for any integer $k\geqslant1$ : $$u_k=\lfloor k\sqrt2\rfloor$$ It's not difficult to prove that there are infinitely many $k$ such that $u_k$ is even and infinitely many $k$ such that $u_k$ is odd. Suppose, for example, that there exists $N\in\mathbb{N}^\star$ such that $u_k$ is odd for all $k\geqslant N$; then we have $u_{k+1}-u_k\geqslant2$ for all $k\geqslant N$ and thus $2k=O(u_k)$. But we know that $u_k=O(k\sqrt2)$ hence a contradiction, since $\sqrt2<2$.
And the same holds when replacing "odd" by "even".
Now my question ...
Consider, more generally, $\alpha\in\mathbb{R}-\mathbb{Q}$, $\alpha>0$ and, for any integer $k\geqslant1$ : $$u_k=\lfloor k\alpha\rfloor$$
Is it true that for any integer $n\geqslant2$ and any $r\in\{0,\cdots,n-1\}$, there are infinitely many $k$ such that $u_k\equiv r\pmod{n}$ ?
Since $\alpha$ is irrational, the number $\frac \alpha n$ is also irrational and hence the sequence $(\{\frac{ k\alpha} n\})_k$ is dense in the interval $[0, 1)$, where $\{x\}$ denotes the fractional part of $x$, i.e. $\{x\} = x - \lfloor x\rfloor$.
In particular, given any $r \in\{0, \dots, n - 1\}$, the interval $[\frac r n, \frac {r + 1} n)$ contains infinitely many numbers of the form $\{\frac {k\alpha} n\}$.
It only remains to note that, if $\{\frac {k \alpha} n\}$ lies in the interval $[\frac r n, \frac{r + 1} n)$, then we have $\frac r n + m \leq \frac {k\alpha} n < \frac{r + 1} n + m$ for some integer $m$, and hence $\lfloor k\alpha \rfloor = r + nm \equiv r \pmod n$.