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.

- May 20th 2009, 09:08 PMfardeen_genShow that all non integral solutions lie on exactly two lines
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. - May 21st 2009, 04:17 AMHallsofIvy
The left hand side

**is**an integer whether x and y are or not. If x and y are not integer, then the "fraction parts" must cancel. Either x= a+ r, y= b- r or x= a- r, y= b+ r, where a and b are integers, 0< r< 1. Those give the two lines. (And, of course, ab= a+b.) - May 21st 2009, 05:47 AMTheAbstractionist
- May 21st 2009, 05:52 AMpankaj
$\displaystyle x+y=0$ and $\displaystyle x+y=6$

- May 21st 2009, 06:01 AMIsomorphism
- May 21st 2009, 11:31 AMpankaj
[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,$\displaystyle x+y=6$