Is there a real nonintegral number $x >1$ such that $\lfloor x^n \rfloor$ is a square integer for all $n \in \mathbb{N}$? 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.
 A: There is no such number.  Suppose $\alpha>1$ is a real number such that $\lfloor \alpha^n \rfloor$ is a square for all $n\in {\Bbb N}$.  Put $\beta=\sqrt{\alpha}$.
Now for each $n$ we have
$$ 
m^2 + 1 > \alpha^n \ge m^2
$$
for some integer $m$, so that taking square-roots
$$ 
m + \frac{1}{2m} > \beta^n \ge m. 
$$
In other words $\beta^n$ has exponentially small fractional part, indeed at most $1/(2m) \approx 1/(2\beta^n)$.
A theorem of Pisot now states that $\beta$ must be a Pisot--Vijayaraghavan (or PV) number (see for example the Wikipedia page on PV numbers).  That is $\beta$ is an algebraic integer $>1$ such that all its Galois conjugates are $<1$ in absolute value.  Suppose that $\beta$ has degree $k$ and that $\beta_1$, $\ldots$, $\beta_{k-1}$ are its Galois conjugates.  Now for all $n$, we must have
$$ 
\beta^n + \beta_1^n + \ldots +\beta_{k-1}^{n} \in {\Bbb Z},   
$$
and from our assumption it follows that for large $n$
$$ 
|\beta_1^n + \ldots +\beta_{k-1}^n |\le \frac{3}{4\beta^n}.  
$$
Write each $\beta_j$ in polar coordinates as $r_j e^{2\pi i\theta_j}$.  Note that
$$ 
\beta r_1 \cdots r_{k-1} \ge 1, 
$$
since this is the absolute value of the norm of $\beta$.  Moreover by Dirichlet's theorem we may find arbitrarily large $n$ with $\Vert n\theta_j \Vert \le 1/10^6$ for all $1\le j\le k-1$.  Then, for such $n$,
\begin{align*}
Re(\beta_1^n + \ldots + \beta_{k-1}^n) &\ge 0.99 (r_1^n +\ldots +r_{k-1}^n) \ge 0.99 (k-1) (r_1\cdots r_{k-1})^{n/(k-1)} \\
&\ge 0.99 (k-1) \beta^{-n/(k-1)},
\end{align*}
by AM-GM.  That gives a contradiction.
The argument also shows that if $\alpha^n$ is within a bounded distance of a square, then $\beta$ again is a PV number, and moreover its degree $k$ must be $2$ and its norm must be $1$ in size.  In other words, we must have $\beta = (r+ \sqrt{r^2 \pm 4})/2$ for some natural number $r$.  This is in keeping with Jeremy Rouse's comment above.
A: 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.
