Is there a real number $x\geq 1$ that is not an integer, such that $\lfloor x^n\rfloor$ is a perfect square for all positive integers $n$?
There are non-integer values of $x$ which give infinitely many perfect squares, for example $x=\sqrt{2}$ gives $\lfloor x^n\rfloor=(2^k)^2$ for all $n=4k$. We can also find $x$ that will be a perfect square for all $n \leq N$ for some fixed integer $N$, for example $x=1+\frac{1}{m}$ with $m>\frac{1}{\sqrt[N]{2}-1}$ gives $\lfloor x^n\rfloor=1$ for $n \leq N$. However the condition to hold for all $n$ seems to be too strict, I suspect there is not even an example that holds for almost all $n$, but I have no idea how to prove that.
This has been recently asked here but was removed by the author. They came up with the problem when thinking about $\lfloor x^n\rfloor\lfloor y^n\rfloor$ is a perfect square.
Remark: Interestingly enough, if we expect $\left\lfloor \frac{x^n}{2}\right\rfloor$ to be a perfect square instead then there are examples such as $x=17 +12 \sqrt2$ with $\left\lfloor \frac{x^n}{2}\right\rfloor= \left ( \frac{(3+2 \sqrt2)^n – (3-2 \sqrt2)^n}{ \sqrt2} \right )^2$, see this aops thread.
Best Answer
Assume there's a non-integral $x$ which satisfies the stated criteria. For $n = 1$, for some integer $m \ge 1$, we'd have
$$\lfloor x\rfloor = m^2 \;\;\to\;\; x = m^2 + r, \;\; 0 \lt r \lt 1 \tag{1}\label{eq1A}$$
With $n = 2$, we get
$$(m^2)^2 \lt x^2 = (m^2 + r)^2 \lt (m^2 + 1)^2 \tag{2}\label{eq2A}$$
Thus, for $\lfloor x^2\rfloor$ to be a perfect square requires that
$$\lfloor x^2\rfloor = (m^2)^2 \;\;\to\;\; (m^2+r)^2 \lt (m^2)^2 + 1 \tag{3}\label{eq3A}$$
Have $n$ double each time, so $n = 2^k$. Using the first $2$ terms of the binomial theorem expansion, and with large enough $k$, we have
$$\begin{equation}\begin{aligned} & (m^2+r)^{2^{k}} - (m^{2^k}+1)^2 \\ & \gt (m^2)^{2^k} + 2^{k}(m^2)^{2^k-1}r - (m^{2^k})^2 - 2(m^{2^k}) - 1 \\ & = m^{2^{k+1}} + 2^{k}rm^{2^{k+1}-2} - m^{2^{k+1}} - 2m^{2^k} - 1 \\ & = 2m^{2^k}\left(2^{k-1}rm^{2^{k}-2}-1\right) - 1 \\ & \gt 0 \end{aligned}\end{equation}\tag{4}\label{eq4A}$$
This shows $\left\lfloor x^{2^k}\right\rfloor \ge (m^{2^k}+1)^2$ for all sufficiently large $k$. Note that, for any $k$, we have $x^{2^k}\gt \left(m^{2^{k}}\right)^2$, and $\left\lfloor x^{2^k}\right\rfloor$ can't be any value between $\left(m^{2^k}\right)^{2}+1$ and $\left(m^{2^k}+1\right)^{2}-1$, inclusive. Thus, for some $k$, there must be a switch between the floor value being $(m^{2^k})^2$, and then for $k+1$ it being at least $(m^{2^{k+1}}+1)^2$, i.e., that
$$(m^2+r)^{2^{k}} \lt (m^{2^k})^2+1, \;\; (m^2+r)^{2^{k+1}} \ge \left((m^{2^k})^2+1\right)^2 \tag{5}\label{eq5A}$$
However, squaring the left side above gives $(m^2+r)^{2^{k+1}} \lt \left((m^{2^k})^2+1\right)^2$, contradicting the right side above. This shows the original assumption that there's such a non-integral $x$ must be false.