=> (y + 2).x2 + 1 - 4 = y2 - 4
=> (y+2).x2 - 3 = (y - 2)(y+2)
=> (y+2)x2 - (y+2).(y - 2) = 3
=> (y+2)(x2 - y + 2) = 3
=> y + 2 \(\in\) Ư(3) = {3;-3;1;-1}
y+2 | 3 | -3 | 1 | -1 |
y | 1 | -5 | -1 | -3 |
x2 -y + 2 | 1 | -1 | 3 | -3 |
x2 | 0 | -8 | 0 | -8 |
x | 0 | loại | 0 | loại |
Vậy (x;y) = (0;1); (0;-1)