find integral value of equation 1/x+1/y=1/2007

1/x=1/2007-1/y

Since x,y>2007 we have x = 2007+a, y = 2007+b for some naturals a,b So, 1/(2007+a)+1/(2007+b) = 1/2007. Cross multiplying etc. we get, ab = 2007^2 = 3^4 X 223^2 For any ordered pair of naturals a,b satisfying ab = 2007^2, we have the corresponding pair (x,y) = (2007+a, 2007+b) as a solution. e.g. a=b=2007 gives x=y=4014 or a=1, b = 2007^2 gives x=2008,y=2007*2008 All in all we get 29 ordered pairs as solutions.