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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.
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    $\begingroup$ Isn't your problem equivalent to making $k^2/\alpha$ close to an integer? (For which you found references.) $\endgroup$ – Lucia Dec 20 '19 at 21:59
  • $\begingroup$ @Lucia , I don't see it. It is making $\alpha$ close to $k^2 /n$, and it does not seem equivalent. $\endgroup$ – Amir Sagiv Dec 20 '19 at 23:36
  • $\begingroup$ If $1/\alpha$ is close to $n/k^2$, then $\alpha$ is close to $k^2/n$? $\endgroup$ – user6976 Dec 21 '19 at 0:02
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    $\begingroup$ What is an example of a number that does not have this property? $\endgroup$ – Pietro Majer Dec 21 '19 at 10:19
  • $\begingroup$ @PietroMajer, I don't know that there is (or that there isn't). $\endgroup$ – Amir Sagiv Dec 22 '19 at 2:21
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This holds for all $\alpha \in \bf R$. If $\alpha \in \bf Q$ it's easy, so we may assume $\alpha$ irrational. Divide by $\alpha$ to get $$ |\alpha^{-1} k^2 - n| < \alpha^{-1} \epsilon. $$ So, we want to show that $\alpha^{-1} k^2$ comes arbitrarily close to integers. This is a special case of Weyl's equidistribution theorem: if $P \in {\bf R}[x]$ is a nonconstant polynomial with an irrational leading term then the sequence $P(1),P(2),P(3),\ldots$ is equidistributed $\bmod 1$.

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  • $\begingroup$ A quick follow up - if this is true for a polynomial $P$, wouldn't it also be true for functions of slower growth, e.g., $ax+bx^{1/2}$? $\endgroup$ – Amir Sagiv Oct 19 at 18:01
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The set of real numbers for which the property is not satisfied has Lebesgue measure zero :

Assume $\alpha>0$ and write $\alpha$ in the form $\alpha=1/\beta^{2}$, $\beta>0$. Then, the approximation problem becomes $$ \left|\beta^{2}-\frac{n}{k^{2}}\right|<\frac{\epsilon}{k^{2}},\qquad\text{(with a different }\epsilon). $$ By Khinchin's theorem, for almost all real number $\beta$ (in the sense of Lebesgue measure) there exists an infinite number of rational approximations $p/q$ satisfying $$ \left|\beta-\frac{p}{q}\right|<\frac{1}{q^{2}\log q}. $$ Hence, for $q$ large, $$ \left|\beta^{2}-\frac{p^{2}}{q^{2}}\right|<\frac{3\beta}{q^{2}\log q}<\frac{\epsilon}{q^{2}}. $$ Since the above inequality is satisfied for almost all real numbers $\beta$, the original approximation property is also satisfied for almost all $\alpha=1/\beta^{2}$.

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  • $\begingroup$ Thank you, considering the above answer by @noam D. Elkies, it does not seem that there are any such numbers $\alpha$ (which of course does not contradict your statement). $\endgroup$ – Amir Sagiv Dec 22 '19 at 2:25

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