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. 
Question: 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.

If $a=\sqrt2$ and $x^2-2y^2=1(x,y\in\mathbb N)$ then $x-\sqrt2y=\dfrac{1}{x+\sqrt2y},$
$$9xy-9\sqrt{2}y^2=\dfrac{9y}{x+\sqrt2y}\in(3,4)$$
so $\lfloor{(3y)^2\sqrt2}\rfloor=9xy-4$ is even since $y$ is even.
If $k\in\mathbb N,\sqrt k \notin \mathbb N$ then we can prove it for $a=u+v\sqrt k,u,v\in\mathbb Q$ as above.
I know that if $a$ is an counterexample and the continued fraction representations of$a=[a_0; a_1,a_2,....]$ then $a_{k},a_{k+2},a_{k+4},\cdots$ must be all even for some $k\in \mathbb N$.

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.
