Let $x,y\ge 1$ be non-integer real numbers such that $\lfloor x^n\rfloor\lfloor y^n\rfloor$ is a perfect square for all natural numbers $n$. Does it follow that $x=y$?
From this question we know the condition under which $\lfloor a\rfloor\lfloor b\rfloor = \lfloor ab\rfloor$. I thought that this implies that for large $n$ it will not be the case that $\lfloor x^n\rfloor\lfloor y^n\rfloor = \lfloor x^ny^n\rfloor$, but now I do not see why.
If the conditions are slightly weakened in either of two ways, or ceiling is used instead of floor, the answer is "no". This is because of the existence of non-integral Pisot–Vijayaraghavan numbers (hereafter, PV numbers). These interesting algebraic real numbers have the following properties which are important for answering the question:
Given any $\epsilon > 0$, $\lfloor x^n+\epsilon\rfloor\lfloor y^n+\epsilon\rfloor$ can be a perfect square for all natural numbers $n$, for $x\ne y$.
Given any PV number $\alpha$, we can find natural numbers $p$ and $q$, where $q>1$, such that $x=\alpha^{p}$ and $y=q^2\alpha^p$ fulfill the condition. We need $p$ to make sure that the exponential decrease of the integral disparities of powers of $\alpha^p$ is sufficiently strong to overcome the exponential increase caused by multiplying it by powers of $q^2$ when $y$ is exponentiated. $p$ also needs to be chosen large enough so that when $n=1$ the integral disparity of both $x$ and $y$ are already smaller than $\epsilon$. We need $q$ simply to be sure that $x\ne y$.
$\lfloor x^{2n+1}\rfloor\lfloor y^{2n+1}\rfloor$ can be a perfect square for all natural numbers $n$, for $x\ne y$.
Some quadratic irrational PV numbers $\alpha$ have the property that the sequence $$\alpha^{2n+1} - \lfloor\alpha^{2n+1}\rfloor$$ decreases exponentially for all sufficiently large natural numbers $n$ (in other words, the sufficiently large odd powers of $\alpha$ are always larger than the integer closest to them, with exponentially decreasing integral disparity). The "golden ratio" $\phi={{1+\sqrt{5}}\over 2}$ is a well-known example of such a PV number. Given any such PV number $\alpha$ we can progress similarly as when constructing an example for Version 1: let $x=\alpha^{2p+1}$ and $y=q^2\alpha^{2p+1}$ for natural numbers $p$ and $q$, where $q>1$ and $p$ is sufficiently large.
$\lceil x^n\rceil\lceil y^n\rceil$ can be a perfect square for all natural numbers $n$, for $x\ne y$.
Some quadratic irrational PV numbers $\alpha$ have the property that the sequence $$\lceil\alpha^n\rceil - \alpha^n$$ decreases exponentially for all sufficiently large natural numbers $n$ (in other words, the sufficiently large powers of $\alpha$ are always smaller than the integer closest to them, with exponentially decreasing integral disparity). $2-\sqrt{2}$ is an example of such a PV number. Given any such PV number $\alpha$ we can progress similarly as when constructing an example for Version 1: let $x=\alpha^p$ and $y=q^2\alpha^p$ for natural numbers $p$ and $q$, where $q>1$ and $p$ is sufficiently large.
The three constructions above leave the original question unanswered. If there existed a PV number which has the property that the sequence $$\alpha^{n} - \lfloor\alpha^{n}\rfloor$$ decreases exponentially for all sufficiently large natural numbers $n$, then the answer to the original question would also be "no": however, as a consequence of the simultaneous version of Dirichlet's approximation theorem, no PV number has that property. In addition, this type of construction could be easily avoided by adding an extra constraint that at least one of $x$ or $y$ is transcendental.
