Solutions to $\lfloor x\rfloor\lfloor y\rfloor=x+y$ 
Find all solutions to $$\lfloor x\rfloor\lfloor y\rfloor=x+y$$ and show that the non-Integral solutions lie on two unique lines. Also determine the equations of these 2 lines. 

I divided the problem into 2 cases:
$$\text{Case 1} : x,y \in \Bbb Z$$$$\text{Case 2}:x,y\notin \Bbb Z$$
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For Case 1, it can be shown that the solutions are those of the equation $xy=x+y$ and these are clearly $(0,0)$ and $(2,2)$.
$$$$
For Case 2, my approach was as follows:
$$\lfloor x\rfloor\lfloor y\rfloor-\lfloor x\rfloor-\lfloor y\rfloor=\{x\}+\{y\}$$
$$(\lfloor y\rfloor-1)(\lfloor x\rfloor-1)=\{x\}+\{y\}+1$$
Since $1<\{x\}+\{y\}+1<3$ (because both $x,y$ have fractional parts in Case 2)
$$1<(\lfloor y\rfloor-1)(\lfloor x\rfloor-1)<3$$
I don't know how to proceed further and would be grateful for any help with this problem. $$$$Many thanks in anticipation!
PS: I believe this is an old IITJEE question and is not homework.
 A: Let $u = \lfloor x \rfloor$, $s = x - u \in [0,1)$, $v = \lfloor y \rfloor$, $t = y - v \in [0,1)$.  Then $u v = u + s + v  + t$.  Now $s + t = u v - u - v$ is an integer, and since $0 \le s + t < 2$ this is either $0$ or $1$.


*

*If $s + t = 0$, $s = t = 0$, and $u v = u + v$.  We just saw that equation here, probably not by coincidence.

*If $s+t=1$, $uv = u + v + 1$ so $(u-1)(v-1) = 2$.  Looking at ways of factoring $2$, $(u, v)$ is either $(-1,0)$ or $(0,-1)$ or $(2,3)$ or $(3,2)$.
Correspondingly, $(x,y) = (u+s,v+t)$ is either on the line $x + y = 0$ or 
on the line $x+y = 6$.

A: Solution if $x$ and $y$ are not integer 
$1<(\lfloor y\rfloor-1)(\lfloor x\rfloor-1)<3 \Rightarrow (\lfloor y\rfloor-1)(\lfloor x\rfloor-1)=2$
now there can be two cases
case-1
$ (\lfloor x\rfloor-1)=1,(\lfloor y\rfloor-1)=2 \Rightarrow \lfloor x\rfloor=2 \lfloor y\rfloor=3 \Rightarrow x+y=6$ solution of this will infinte no of  sets lying on line $x+y=6$ , bounded between rectangle defined by $2\leqslant x <3$ and $3\leqslant y <4$
similarly solve for case $\lfloor x\rfloor=3 \lfloor y\rfloor=2$ and try to define rectangle
case-2
$ (\lfloor x\rfloor-1)=-1,(\lfloor y\rfloor-1)=-2 \Rightarrow \lfloor x\rfloor=0 \lfloor y\rfloor=-1 \Rightarrow x+y=0$ solution of this will infinte no of  sets lying on line $x+y=0$ , bounded between rectangle defined by $0\leqslant x <1$ and $-1\leqslant y <0$
similarly solve for case $\lfloor x\rfloor=-1 \lfloor y\rfloor=0$ and try to define rectangle
