Integral solutions

take simple values of x and y
and see if ti satisfies the relation

for eg x=y=0 will satisfy

as [x],[y] are integers. so, x+y should be integer or x and y should be integer.

CASE I (x,y are both integers)
then [x]=x, [y]=y
So, xy=x+y
==> y=x/(x-1)

but y is an integer
Putting diff values of x
we have x=0,x=2 as the only solutions
So the solution set is (0,0) and (2,2)

CASE II (x+y is an integer)
then {x}+{y}=1
So, [x][y]=[x]+[y]+1
==> [y]=([x]+1)/([x]-1)

Again putting various values of [x] and [y]
we have [x]=0,[x]=2,[x]=-1,[x]=3 as the only solutions
So the solution set is ([x],[y])=(0,-1) and (-1,0) and (2,3) and (3,2)

So the graph luks like this for non-integral solution
Hence the two lines are y=x+1 and y=x-1

Intagral solutions are(0,0)&(2,2)
Lines are coming x+y=6 and x+y=0 :(
Not the same as asish.