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.



Is there a real nonintegral number $x >1$ such that $\lfloor x^n \rfloor$ is a square integer for all $n \in \mathbb{N}$?
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](https://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number)). 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.


