I need to find the integer solutions to this equation.
$\displaystyle (x+y)^2 = x^2y^2$
Any Idea?
Thanks
everk
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
$\displaystyle x(kx +r ) = x + kx+r $
$\displaystyle 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
$\displaystyle x(kx) = x + kx $ divde both sides by x
$\displaystyle kx = 1+k $
$\displaystyle x = \frac{1+k}{k} $ but x is an integer so k should equal 1
$\displaystyle x = 2 $ and $\displaystyle y = 1x + 0 = x =2 $
other case x+y = -xy same
so xy both 2 or both -2