Let $\alpha \in \mathbb{R}\setminus \mathbb{Q}$ be irrational. Does the arithmetic progression $(n\alpha )\_{n\in\mathbb{N}}$ becomes arbitrarily close to squares? 

More precisely, what can be said about the set of $\alpha$ such that for any $\varepsilon >0$ there are infinitely many $n,k\in \mathbb{N}$ such that 
$$|n\alpha- k^2|< \varepsilon \, . $$

Are there infinitely many such $\alpha$? Are they dense in $\mathbb{R}$? Are there any quantitative results about the size of this set? 

**Remarks:**

1. If $\alpha = \frac{a}{b}$ is rational, then the above sequence contains the subsequence $\{(a^{2m})\}\_{m\in\mathbb{N}}$, and so it coincides with infinitely many squares.
2. I found many references for a similar problem, numbers $\alpha$ for which there are inifinitley many pairs of reals $k,n$ such that $$|\alpha - \frac{k}{n^2}|$$ is arbitrarily small, but I don't see how that connects directly to my question.