[x][y]=x+y
[x][y]=[x]+[y]+{x}+{y} ({x} and {y} denote the fractional parts of x and y respectively)
0 {x}+{y}<2
0 ([x][y]-[x]-[y])<2
1 (1+[x][y]-[x]-[y])<3
1 ([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 and will be integers
Case 2
{x}+{y}=0-0-(-1)=1 and thus
Case 3
{x}+{y}=0-0-0=0
{x}={y}=0,but then and are integers
Case 4
{x}+{y}=6-2-3=1
Therefore,