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Let $\alpha \in I$ where $I$ is some closed interval that does not contain $0$. I am interested in upper bound for $$ M(\alpha) = \#\{ 1 \leq n \leq N: \| \alpha n^2/N \| < 1/N \} $$ where $N$ is a sufficiently large positive integer with respect to $I$. I was wondering is it reasonable to expect $$ \sup_{\alpha \in I} M(\alpha) \ll N^{\varepsilon} ? $$ or would one expect this to be not true? (or rather the statement holds for $\sup_{\alpha \in I - B_N}$ and some subset $B_N$ that depends on $N$. Is it possible to bound the measure of $B_N$ in terms of $N$?).

The question is related to my question from couple years ago (Distribution of $\alpha n^2/q$ modulo $1$?) in which a reference to Lemma 1 of http://www.numdam.org/item/ASNSP_1999_4_28_4_591_0.pdf was given. Here an upper bound is given, but it wasn't clear to me if it was sharp or not. Any input would be appreciated. Thank you.

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    $\begingroup$ Hmm, if $\alpha = \frac{1}{4}$, $N = k^2$, then all $n$ divisible by $2k$ should be in $M(\alpha)$, and there are about $\frac{\sqrt{N}}{2}$ such numbers, if I understood the question correctly. $\endgroup$
    – Aleksei Kulikov
    Commented Apr 14 at 21:29
  • $\begingroup$ Thank you, the question is fixed. as you point out, I think it's necessary to remove some subset. I was hoping it can be achieved with $B$ being some very small set, like measure $0$ or $\ll N^{-1}$ or so... $\endgroup$
    – Johnny T.
    Commented Apr 14 at 21:55
  • $\begingroup$ What is the order of quantifiers on $N$ and $B$? Is it "For every $N$ there is $B$" or "There is $B$ such that for every $N$"? $\endgroup$
    – fedja
    Commented Apr 15 at 17:55
  • $\begingroup$ I was thinking the former. Let me fix it. thank you $\endgroup$
    – Johnny T.
    Commented Apr 15 at 21:06

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