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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.

Will Sherwood
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user342209
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2 Answers2

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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 $x+y = 6$.
Community
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Robert Israel
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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

vikas meena
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