Prove that $\forall a\in \mathbb{R\smallsetminus Q}$, there exist infinitely many $n\in \mathbb N$ such that $\lfloor{an^2}\rfloor$ is even.

We use the following lemma, which can be proved by using van der Corput's theorem:

If $\alpha$ is an irrational number, then the sequence $\alpha n^2$ is equidistributed modulo 1.

Assume $\alpha > 0$. If there is only finitely many $n$ such that $\lfloor \alpha n^2 \rfloor$ is even. Then for sufficiently large $n$, we can write $$\alpha n^2 = 2N_n+1+r_n$$ where $0<r_n <1$ and $N_n$ is an integer. Hence $$\frac{1}{2} < \left\{\frac{\alpha}{2}n^2 \right\} = \frac{1+r_n}{2} < 1 $$

$\{x \}$ means the fractional part of $x$. Thus, the sequence $\dfrac{\alpha}{2}n^2$ is not equidistributed modulo 1, a contradiction.

In the same way, we can prove there are infinitely many $n$ such that $\lfloor \alpha n^2 \rfloor$ is odd.