x(y+1) = 3y -3
x = 3(y - 1)/(y+1)
daca (y+1) | 3 (y+1) ∈ D3 y+1 ∈ {-3, -1, 1, 3} y ∈ {-4, - 2, 0, 2}
x ∈ {3·(-5)/(-3), 3·(-3)/(-1), 3·(-1)/1 , 3·1/3 }
x ∈ {5, 9, -3, 1} (5,-4), (9, -2), (-3, 0), (1, 2)
daca (y+1) | (y-1) ⇒ (y+1) | (y+1-y+1) (y+1) | 2 (y+1) ∈ D2
(y+1) ∈ {-2, -1, 1, 2} y ∈ {-3, -2, 0, 1} x∈{6, 9, 3, 0}
(6,-3), (9, -2), (3, 0) , (0, 1)