# Thread: Equation

1. ## Equation

I need to find the integer solutions to this equation.
$(x+y)^2 = x^2y^2$
Any Idea?
Thanks
everk

2. xy = x+y or x+y = -xy
the only solution is when x and y = 2 or x and y = -2

suppose y>x
then we can say y=kx+r k is an integer r<x

$x(kx +r ) = x + kx+r$

$x (kx +r) = x(k+1) +r$ the left side is a multiple of x so the right should be so should x divides r but x>r then r=0

so y = kx

$x(kx) = x + kx$ divde both sides by x

$kx = 1+k$

$x = \frac{1+k}{k}$ but x is an integer so k should equal 1

$x = 2$ and $y = 1x + 0 = x =2$

other case x+y = -xy same

so xy both 2 or both -2

3. Hi Amer,

Apart from x=y=2 and x=y=-2, wouldn't x=y=0 count as well?