TITLE: Is there a real nonintegral number $x >1$ such that $\lfloor x^n \rfloor$ is a square integer for all $n \in \mathbb{N}$?
QUESTION [15 upvotes]: This question was inspired by the following:
https://math.stackexchange.com/questions/3882691/lfloor-xn-rfloor-lfloor-yn-rfloor-is-a-perfect-square
Is there a real nonintegral $x>1$ s.t. $\lfloor x^n \rfloor$ is square integer for all positive integers $n$? I am asking because the question is interesting in and of itself, but also because the proof techniques should also be interesting.

REPLY [20 votes]: At the request of the OP, I am turning my comment into an answer. It is possible to have $\lfloor x^{n} \rfloor$ close to a square for all positive integers $n$. For example, if $x = \frac{7 + 3 \sqrt{5}}{2} = \phi^{4}$, where $\phi = \frac{1 + \sqrt{5}}{2}$, then $\lfloor x^{n} \rfloor + 3$ is a square for every positive integer $n$. For all integers $k$, $\phi^{k} + \left(-\frac{1}{\phi}\right)^{k} = L_{k}$, the $k$th Lucas number. Squaring this identity shows that $\phi^{2k} + 2 \cdot (-1)^{k} + \left(-\frac{1}{\phi}\right)^{2k} = L_{k}^{2}$ and so $L_{k}^{2} = L_{2k} + 2 \cdot (-1)^{k}$.
Thus
$$
  \lfloor x^{n} \rfloor + 3 = x^{n} + x^{-n} + 2 = L_{4n} + 2 = L_{2n}^{2}
$$
is the square of an integer.