# find tht integral solutions to the equation [x] [y] =x + y.Show that all the non -integral solution lie on exactly two lines.Determine the lines.

Swapnil Saxena
102 Points
11 years ago

Hi Manuj.

There will be 2 integral solutions to his question (2,2) and (0,0).The solution to the equation lie on the line x-y=0

Still trying to find an easy trick to find non integral solutions of the equation . so i l send the non integral solution later and the equation of the lines too . . .

manuj mittal
4 Points
11 years ago

did u find d integral solutions by just putting integers in equation......or by some mathematical approach.....?????????

Swapnil Saxena
102 Points
11 years ago

The non integral solutions of the equation lies on lines (x+y=0)---(1) and (x+y-6)=0------(2)

Swapnil Saxena
102 Points
11 years ago

The integral solutions are easy to find .

For x and y belong to integers, the equation will take form xy= x+y

or simply y(x-1)=x ==> y= x/(x-1) ==> y= 1+1/(x-1)

Solving the above equations (using graphs : Hypebolic Curve  ) , we can deduce that the integral soltuions are (0,0) and (2,2)

Swapnil Saxena
102 Points
11 years ago

The non integral solutions can be deduced through the following algorithm

Ttake x= Ix + Fx and y= Iy + Fy (I for Integral part and F for fractional part)

Then [x][y] = x+y

IxIy = Ix + Fx + Iy + Fy

==> IxIy = (Ix + Iy) + (Fx + Fy)

==> Iy( Ix-1 ) = (Ix) + (Fx + Fy)

==> Iy = ((Ix) + (Fx + Fy))/( Ix-1 )

For Iy to be integral, (Fx + Fy) should definitely be integral because (Ix-1) is itself integral

Since the only integral value that (Fx + Fy) can take = 1,except 0 (not possible as the solution are non integral and the fractional parts cannot take negative values or more than 0.9999)

Thus the equaion becomes ==> Iy = (Ix + 1)/( Ix-1 ) ==> Iy = 1+2/( Ix-1 )

Solving this equation using graphs(Hyperbolic Curves) ,

The only integral solution for Ix and Iy are (2,3) and (3,2) , (0,-1) , (-1 ,0) (Note: All these are solution of greatest integer function)

Using this result we can easily deduce that (2.99, 3.01), (2.98,3.02), (3.02, 2.98),(3.99, 2.01) are some of the solution of the equation.

In this way, the equation of the first line can be deduced, = x+y-6=0

Using this result we also deduced that (0.99,-0.99),(-0.99,0.99),(0.35,-0.35) are some of the solution of the equation.

In this way, the equation of the first line can be deduced, = x+y=0