# Approximation of a square with an irrational arithmetic progression

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.
• Isn't your problem equivalent to making $k^2/\alpha$ close to an integer? (For which you found references.) – Lucia Dec 20 '19 at 21:59
• @Lucia , I don't see it. It is making $\alpha$ close to $k^2 /n$, and it does not seem equivalent. – Amir Sagiv Dec 20 '19 at 23:36
• If $1/\alpha$ is close to $n/k^2$, then $\alpha$ is close to $k^2/n$? – user6976 Dec 21 '19 at 0:02
• What is an example of a number that does not have this property? – Pietro Majer Dec 21 '19 at 10:19
• @PietroMajer, I don't know that there is (or that there isn't). – Amir Sagiv Dec 22 '19 at 2:21

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 $$\bmod 1$$.
• 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., $ax+bx^{1/2}$? – Amir Sagiv Oct 19 '20 at 18:01
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 $$\alpha=1/\beta^{2}$$.
• 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). – Amir Sagiv Dec 22 '19 at 2:25