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.


Too long for a comment:

Assume w.l.o.g. that $a<1$ and write $a=0.a_1 a_2a_3\ldots$. If there are infinitely many $i$ such that $a_{2i}=0$, then we are done, because we can take the corresponding even powers of 2 as the squares. Otherwise, there is a $k$ such that $2^{2k}a$ has only ones in its binary rep after the point, which means that the fractional part of $2^{2k}a$ is at least $1/3$.

Thus, it is probably sufficient to show that there are infinitely many $n$ such that the fractional part of $n^2a$ is less than 1/3.


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