Set $z=\frac xy >0 \text{ and equation becomes } \lfloor z \rfloor \lfloor ny \rfloor = \lfloor nyz \rfloor \enspace \forall n>0$ Write $ny=p+\alpha_n, z=q+\beta \text{ where } 0\le\alpha_n,\beta<1$ and above is only true if $p\beta+q\alpha_n+\alpha_n\beta<1$ Since $p \ge 1$ and can be arbitrarily large $\beta=0, z=q$ is an integer and $q\alpha_n<1$ or $\alpha_n < \frac 1z$ with $z\ge 1$ If $z=1$ this is always true so in this case $x=y$ is a solution for all allowable reals If $z\ge 2$ then you require $\alpha_n < \frac 12 \enspace \forall n >0$ which is false unless $\alpha_1=0$ To prove this suppose that $0<\alpha_1<\frac 12$ and select the greatest $k$ such that $2^k\alpha_1< \frac 12$ Consequently, $\frac 12 \le 2^{k+1}\alpha_1 <1$ so $n=2^{k+1}$ gives $\alpha_n$ that violates $\alpha_n < \frac 12$ $\therefore \alpha_1=0$ and so $x,y$ are integers.