# 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:

$$\text{Case 1} : x,y \in \Bbb Z$$$$\text{Case 2}:x,y\notin \Bbb Z$$  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.

• Here $(\lfloor y\rfloor-1)(\lfloor x\rfloor-1)=\{x\}+\{y\}+1\in \mathbb{Z}$, So $\{x\}+\{y\}+1 = 2$ – juantheron Jun 9 '16 at 5:46
• @juantheron Sorry Sir, I couldn't understand. What is $\in \Bbb Z$? Please could you elaborate your comment? – user342209 Jun 9 '16 at 5:47
• bcz $(\lfloor x \rfloor -1)(\lfloor y \rfloor -1)\in \mathbb{Z}$ – juantheron Jun 9 '16 at 5:48
• Oh, alright. But how would we proceed to find those values of $x$ and $y$ which satisfy Case 2? – user342209 Jun 9 '16 at 5:50
• @juantheron Thanks Sir, I was able to solve the question now! – user342209 Jun 9 '16 at 5:51

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 $1$.

1. If $s + t = 0$, $s = t = 0$, and $u v = u + v$. We just saw that equation here, probably not by coincidence.
2. 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 $x+y = 6$.

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