<s>[INST] Equality involving greatest integer 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?
 [/INST] 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*}
</s>