plz solve

one thing is obvious.. (if it is not do tell me..)

{x}+{y} = 1 or 0

if it is zero then it is simple that x and y are both integers..

if it is 1 then x=[x]+f

and y=[y]+1-f

we get

[x][y] = [x]+[y]+1

now may be u have a way to move ahead?

basically take 2 cases

[x]>[y]
[x]=[y]

and [x]<[y]

u will get the solution :)

ok.. now when
Case 1
[x]=[y] = k let

[x][y] = [x]+[y]+1
k² = 2k+1

no integer solution to this case!

Case 2
[x]>[y]
[x][y] = [x]+[y]+1

RHS <=2[x] (simple to see!)

hence, [y]<=2

now this gets far more simple! (doesnt it?)

If it still doesnt make sense.. let me know.. i will post the full solution :)

Case 1
[x]=[y] = k let
[x][y] = [x]+[y]+1
k² = 2k+1
no integer solution to this case!

Case 2
[x]>[y]
[x][y] = [x]+[y]+1

RHS <=2[x] (simple to see!)

hence, [y]<=2

sub-case 1: [y]=0

then [x][y]=[x]+[y]+1
0=[x]+1
so for this having a solution, [x]=-1

now for this case u have respective solutions in x and y

again take subcase [y]=1 and the other remaining cases :)

dude this is a real long solution...

I am sure u have the complete idea now..

do it for individual cases.. or i for one am not sure how to do it in a smaller way :)