Consider the equation $\displaystyle \lfloor x \rfloor \lfloor y \rfloor = x + y$.
Show that all non-integral solutions of this equation lie on exactly two lines.
[x][y]=x+y
[x][y]=[x]+[y]+{x}+{y} ({x} and {y} denote the fractional parts of x and y respectively)
0$\displaystyle \leq ${x}+{y}<2
0$\displaystyle \leq $([x][y]-[x]-[y])<2
1$\displaystyle \leq $(1+[x][y]-[x]-[y])<3
1$\displaystyle \leq $([x]-1)([y]-1)<3
([x]-1)([y]-1)=1,2
Case 1
([x]-1)([y]-1)=1
[x]-1=1 and [y]-1=1
[x]=[y]=2
{x}+{y}=(2)(2)-2-2=0
{x}={y}=0,but then $\displaystyle x$ and $\displaystyle y$ will be integers
Case 2
$\displaystyle ([x]-1)([y]-1)=2$
$\displaystyle
[x]-1=-1 and [y]-1=-2
$
$\displaystyle [x]=0 and [y]=-1$
{x}+{y}=0-0-(-1)=1 and thus $\displaystyle x+y=0$
Case 3
$\displaystyle ([x]-1)([y]-1)=1$
$\displaystyle [x]-1=-1 and [y]-1=-1$
$\displaystyle [x]=0 and [y]=0$
{x}+{y}=0-0-0=0
{x}={y}=0,but then $\displaystyle x$ and $\displaystyle y$ are integers
Case 4
$\displaystyle ([x]-1)([y]-1)=2$
$\displaystyle [x]-1=1 and [y]-1=2$
$\displaystyle [x]=2 and [y]=3$
{x}+{y}=6-2-3=1
Therefore,$\displaystyle x+y=6$