Find all pairs (x, y) of integers that satisfy this equation:

1 + x²y = x² + 2xy + 2x + y

Solving for y gives 
y = 1 + 4x/(x^2 - 2x -1)

X is relatively prime to x^2 - 2x -1, so either 
(a) x = 0 or 
(b) 4 is divided by x^2 - 2x -1

In the 2nd case, x^2 - 2x - 1 = p where p is in (1, 2, 4, -1, -2, -4)

Solving that quadratic equation gives
x = 1 +/- sqrt(p+2)

if p = -1, then x is 0 or 2
if p = -2 then x is 1
if p = 2  then x is -1 or 3

So, the only solutions are:
(0,1), (2,-7),  (1,-1), (-1,-1) and (3,7) 

Posted by Steve Herman on 2015-06-14 11:24:56