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I am trying to find the integral solutions to the equation $$[x][y]=x+y$$ and show that all non-integral solutions lie on exactly two lines. Here $[x]$ denotes greatest integer function.

For integer solutions I solved the equation $xy=x+y.$ Upon solving it I got $y=\frac{x}{x-1}$ and was not able to proceed further. For non integral values, I got two inequalities $y>1$, $x>1$ but these aren't lines. These I got by breaking greatest integer of $x$ and $y$ into their subsequent fractional and integer parts and using the concept of basic inequalities but my answer is still incomplete. How do I proceed?

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  • $\begingroup$ $y > 1$ and $x > 1$ are not lines. $\endgroup$
    – quasi
    May 5, 2017 at 5:07
  • $\begingroup$ Right, I would correct my post. $\endgroup$
    – Bai Li
    May 5, 2017 at 5:08

2 Answers 2

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First suppose $x,y$ are integers. \begin{align*} \text{Then}\;\;&\lfloor x \rfloor \lfloor y \rfloor = x + y\\[4pt] \iff\; &xy = x + y\\[4pt] \iff\; &xy - x - y = 0\\[4pt] \iff\; &xy - x - y + 1 = 1\\[4pt] \iff\; &(x-1)(y-1) = 1\\[4pt] \iff\; &x-1 = 1\;\;\text{and}\;\;y-1=1\\[0pt] &\;\;\;\;\text{or}\\[0pt] &x-1 = -1\;\;\text{and}\;\;y-1= -1\\[4pt] \iff\; &(x,y)=(2,2)\;\;\text{or}\;\;(x,y)=(0,0)\\[4pt] \end{align*} Next suppose at least one of $x,y$ is not an integer.

\begin{align*} \text{Let}\;\; &a=\lfloor x \rfloor,\;\;r=x-\lfloor x \rfloor\\[4pt] &b=\lfloor y \rfloor,\;\;s=y-\lfloor y \rfloor\\[4pt] \end{align*} Thus we have $x = a + r$ and $y = b + s$, where $a,b$ are integers, and $0 \le r,s < 1$.

By assumption, $r,s$ are not both equal to $0$, hence $0 < r + s < 2$. \begin{align*} \text{Then}\;\;&\lfloor x \rfloor \lfloor y \rfloor = x + y\\[4pt] \implies\; &ab = (a + r) + (b + s)\\[4pt] \implies\; &ab - a - b = r + s\\[4pt] \implies\; &r + s \in \mathbb{Z}\\[4pt] \implies\; &r + s = 1\;\;\text{and}\;\;r,s>0\\[4pt] \implies\; &ab - a - b = 1\\[4pt] \implies\; &ab - a - b + 1 = 2\\[4pt] \implies\; &(a-1)(b-1) = 2\\[4pt] \end{align*}

From the equation$\;(a-1)(b-1) = 2,\;$you get $4$ cases for the pair $(a,b)$, one for each of the factorizations $$(1)(2)=2 ,\;\;\; (2)(1)=2 ,\;\;\; (-1)(-2)=2 ,\;\;\; (-2)(-1)=2 $$ Noting that $r+s=1$ and $r,s>0$, for each valid pair $(a,b)$, the pair $(x,y)$ must satisfy \begin{align*} &\begin{cases} x=a+r\\[2pt] y=b+(1-r)\\ \end{cases} \\[2pt] &\;\;\;\,0 < r < 1\\[2pt] \end{align*} which is the segment of the line $x+y = a + b + 1$, strictly between the points $(a,b+1),\;(a+1,b)$.

Conversely, if $a,b$ are integers such that $ab - a - b = 1$, then for any point $(x,y) \in \mathbb{R}^2$ satisfying \begin{align*} &\begin{cases} x=a+r\\[2pt] y=b+(1-r)\\ \end{cases} \\[2pt] &\;\;\;\,0 < r < 1\\[2pt] \end{align*} \begin{align*} \qquad\qquad\text{we have}\; \lfloor x \rfloor \lfloor y \rfloor &= ab\\[4pt] &= a + b + 1\\[4pt] &=(a+r)+\left(b+(1-r)\right)\\[4pt] &=x + y\\[4pt] \end{align*} At this point, I'll let you finish it.

If you work it out correctly, the set $S$ of all points $(x,y) \in \mathbb{R}^2$ satisfying $$\lfloor x \rfloor \lfloor y \rfloor = x + y$$ has $4$ components$\,-\,$an isolated point, and $3$ open line segments.

More precisely, $S = \{P\} \cup L_0 \cup L_1 \cup L_2$, where \begin{align*} {\small{\bullet}}\;\,&P\;\text{is the point}\;(2,2)\text{.}\\[5pt] {\small{\bullet}}\;\,&L_0\;\text{is the segment of the line}\;x+y=0\;\text{strictly between}\\[-0.5pt] &\text{the points}\;(-1,1),\;(1,-1)\text{.}\\[5pt] {\small{\bullet}}\;\,&L_1\;\text{is the segment of the line}\;x+y=6\;\text{strictly between}\\[-0.5pt] &\text{the points}\;(2,4),\;(3,3)\text{.}\\[5pt] {\small{\bullet}}\;\,&L_2\;\text{is the segment of the line}\;x+y=6\;\text{strictly between}\\[-0.5pt] &\text{the points}\;(3,3),\;(4,2)\text{.}\\[5pt] \end{align*}

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  • $\begingroup$ The two lines are x+y=6 and x+y=0. Solution completed! $\endgroup$
    – Bai Li
    May 5, 2017 at 16:26
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Write $x=a+\alpha$ and $y=b+\beta$ where $a,b\in\mathbb{Z}$ and $0\leq \alpha, \beta<1$.

Then your equation boils down to $$ab = a+\alpha+b+\beta.$$

Now since $\alpha+\beta$ must be an integer and $0\leq \alpha+\beta < 2$, either $\alpha+\beta=0$ or $\alpha+\beta=1$.

In the former case, $$(a-1)b = a$$ and $a-1$ must divide $a$. The only way this can happen is if $a-1=1$ or $a=0$, so the solutions are $(x,y) = (2,2)$ or $(0,0)$.

In the latter case, $ab+a+b+1$ and $$(a-1)b = a+1.$$

For what values of $a$ will $a-1$ divide $a+1$?

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  • $\begingroup$ Already done that. Two inequalities are coming as I have mentioned in the question. Please be more clear. $\endgroup$
    – Bai Li
    May 5, 2017 at 5:27

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