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.

• One can get "close" to having $\lfloor x^{n} \rfloor$ being a square. For example, if $x = \frac{7+3\sqrt{5}}{2}$, then $\lfloor x^{n} \rfloor + 3$ is a square for every positive integer $n$. – Jeremy Rouse Nov 9 '20 at 2:14
• @Jeremy Rouse please put this as an answer I'd like to give you credit/upvote – Mike Nov 9 '20 at 2:44

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.
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.