$\sqrt2 n^2$ is uniformly distributed.

Question

How to prove $(\sqrt2 n^2)_{n\in \mathbb{N}}$ is uniformly distributed mod 1.

I tried using Weyl's criterion but I seem to go nowhere.

Thank You very much.

Answer 1

$\textbf{Lemma}$(Van der Corput)

Let $a_n$ be a bounded sequence of complex numbers such that $\frac{1}{N}\sum_{n=1}^Na_{n}\overline{a_{n+m}} \to^{N \to +\infty} 0,\forall m \in \Bbb{N}$ then $\frac{1}{N}\sum_{n=1}^Na_n \to^{N \to +\infty}0$

Use this lemma with $a_n=e^{2\pi i \sqrt{2}{n^2}}$ and then Weil's theorem.

As an exercise instead of just simply invoking the lemma,you can prove that $a_n=e^{2\pi i \sqrt{2}{n^2}}$ satisfies the hypothesis of the lemma.