Is it true that there are infinitely many square numbers of the form $\lfloor n\sqrt{2}\rfloor$?

Question

Is it true that there are infinitely many square numbers in the sequence $\left \lfloor n \sqrt{2} \right \rfloor $ with $n$ is a whole number? If not, how can we prove it?

Answer 1

Interesting question. Beatty sequences give that every positive natural number is either of the form $\lfloor \sqrt{2}\,a\rfloor$ or of the form $\lfloor (2+\sqrt{2})\,b\rfloor$, for some $a,b\in\mathbb{N}$. If we assume that only a finite number of squares belong to the first sequence, we have that every large enough square is of the form $\lfloor (2+\sqrt{2})\,b\rfloor$ for some $b\in\mathbb{N}$. Now we may use the general properties of continued fractions to derive a contradiction. We have that $$ \sqrt{2}=[1;2,2,2,2,\ldots] $$ and if $\left\{\frac{p_k}{q_k}\right\}$ is the sequence of convergents of such continued fraction, $$ q_k^2 \sqrt{2}-p_k q_k = \frac{(-1)^k}{2\sqrt{2}}+o(1) $$ as $k\to +\infty$. Similarly, $$ \frac{1}{2+\sqrt{2}}=[0;3,2,2,2,\ldots] $$ and if $\left\{\frac{p_k}{q_k}\right\}$ is the sequence of convergents of such continued fraction, $$ \frac{q_k^2}{2+\sqrt{2}}-p_k q_k = \frac{(-1)^k}{2\sqrt{2}}+o(1) $$ as $k\to +\infty$. Assuming that $q_k^2$ belongs to the complementary Beatty sequence we get $$ b \leq \frac{q_k^2}{2+\sqrt{2}} \leq b+\frac{1}{2+\sqrt{2}},\qquad b\in\mathbb{N} $$ which contradicts the previous statement if $k$ is large and odd.

The same argument works also by switching the roles of the original Beatty sequence and its complementary sequence. In particular we have that

There are infinite squares of the form $\lfloor \sqrt{2}\,a\rfloor$ and infinite squares of the form $\lfloor \left(2+\sqrt{2}\right)\,b\rfloor$. Every non-zero square is represented by exactly one of such forms.

A more advanced approach is proving that the set $\{\lfloor n\sqrt{2}\rfloor:n\geq 0\}$ contains a difference set $S-S$ where $S\subset\mathbb{N}$ has a positive density. Indeed, by the Furstenberg-Sárközy theorem there are infinite squares in any difference set of the previous form. This is the quantitative version of the fact that the squares form a Heilbronn set, i.e. a neighbourhood of zero in the Bohr topology.

The first positive squares of the form $\lfloor \sqrt{2}\,a\rfloor$ are $$1, 4, 9, 16, 25, 36, 49, 100, 121, 144, 169, 196, 289, 400.$$ The first positive squares of the complementary form $\lfloor \left(2+\sqrt{2}\right)\,b\rfloor$ are $$64, 81, 225, 256, 324, 361, 484, 529.$$