Saudi Arabian Mathematical Competitions

We rewrite the equation:
$$y^{3} + y^{2} - 2x^{3}y - 8x^{6} = 0.$$

Consider this as a cubic in $y$:
$$y^{3} + y^{2} - 2x^{3}y - 8x^{6} = 0.$$

Let us try small integer values for $x$.

If $x = 0$:
$$y^{3} + y^{2} = 0 \implies y^{2}(y + 1) = 0 \implies y = 0 \text{ or } y = -1.$$
So $(0, 0)$ and $(0, -1)$ are solutions.

If $x = 1$:
$$y^{3} + y^{2} - 2y - 8 = 0.$$
Try $y = 1$:
$1 + 1 - 2 - 8 = -8$.
Try $y = 2$:
$8 + 4 - 4 - 8 = 0$.
So $y = 2$ works. $(1, 2)$ is a solution.
Try $y = -2$:
$-8 + 4 + 4 - 8 = -8$.
Try $y = -1$:
$-1 + 1 + 2 - 8 = -6$.
Try $y = 4$:
$64 + 16 - 8 - 8 = 64$.
So only $y = 2$ works for $x = 1$.

If $x = -1$:
$$y^{3} + y^{2} + 2y - 8 = 0.$$
Try $y = 1$:
$1 + 1 + 2 - 8 = -4$.
Try $y = 2$:
$8 + 4 + 4 - 8 = 8$.
Try $y = -2$:
$-8 + 4 - 4 - 8 = -16$.
Try $y = -1$:
$-1 + 1 - 2 - 8 = -10$.
Try $y = 4$:
$64 + 16 + 8 - 8 = 80$.
So no integer solution for $x = -1$.

Try $x = 2$:
$8x^{6} = 8 \times 64 = 512$
$2x^{3}y = 2 \times 8y = 16y$
So:
$y^{3} + y^{2} - 16y - 512 = 0$
Try $y = 8$:
$512 + 64 - 128 - 512 = -64$
Try $y = 16$:
$4096 + 256 - 256 - 512 = 3584$
Try $y = -8$:
$-512 + 64 + 128 - 512 = -832$
Try $y = 4$:
$64 + 16 - 64 - 512 = -496$
Try $y = -4$:
$-64 + 16 + 64 - 512 = -496$
So no integer solution for $x = 2$.

Try $x = -2$:
$8x^{6} = 512$
$2x^{3}y = 2 \times (-8)y = -16y$
So:
$y^{3} + y^{2} + 16y - 512 = 0$
Try $y = 8$:
$512 + 64 + 128 - 512 = 192$
Try $y = -8$:
$-512 + 64 - 128 - 512 = -1088$
Try $y = 4$:
$64 + 16 + 64 - 512 = -368$
Try $y = -4$:
$-64 + 16 - 64 - 512 = -624$
So no integer solution for $x = -2$.

Now, for large $|x|$, the term $8x^{6}$ dominates, so $y^{3} \approx 8x^{6}$, so $y \approx 2x^{2}$.
Try $y = 2x^{2}$:
$y^{3} = 8x^{6}$
$2x^{3}y = 4x^{5}$
$y^{2} = 4x^{4}$
So:
$8x^{6} = 8x^{6} + 4x^{5} - 4x^{4}$
$0 = 4x^{5} - 4x^{4}$
$0 = 4x^{4}(x - 1)$
So $x = 0$ or $x = 1$.

If $x = 0$, $y = 0$ (already found).
If $x = 1$, $y = 2$ (already found).

Try $y = -2x^{2}$:
$y^{3} = -8x^{6}$
$2x^{3}y = -4x^{5}$
$y^{2} = 4x^{4}$
So:
$-8x^{6} = 8x^{6} - 4x^{5} - 4x^{4}$
$-16x^{6} + 4x^{5} + 4x^{4} = 0$
$4x^{4}(x + 1 - 4x^{2}) = 0$
So $x = 0$ or $x = -1$ or $x^{2} = 1/4$ (not integer).
If $x = 0$, $y = 0$ (already found).
If $x = -1$, $y = -2$.
Try $x = -1$, $y = -2$ in the original equation:
$(-2)^{3} = 8(-1)^{6} + 2(-1)^{3}(-2) - (-2)^{2}$
$-8 = 8 + 2(-1)(-2) - 4 = 8 + 4 - 4 = 8$
So $-8 = 8$ is false.
So $x = -1$, $y = -2$ is not a solution.

Try $y = kx^{2}$ for small integer $k$.
Try $y = x$:
$y^{3} = x^{3}$
$8x^{6} + 2x^{3}y - y^{2} = 8x^{6} + 2x^{4} - x^{2}$
So $x^{3} = 8x^{6} + 2x^{4} - x^{2}$
Try $x = 0$: $0 = 0$
Try $x = 1$: $1 = 8 + 2 - 1 = 9$
Try $x = -1$: $-1 = 8 + 2 - 1 = 9$
So only $x = 0$, $y = 0$ (already found).

Try $y = -1$:
$(-1)^{3} = 8x^{6} + 2x^{3}(-1) - (-1)^{2}$
$-1 = 8x^{6} - 2x^{3} - 1$
$0 = 8x^{6} - 2x^{3}$
$2x^{3}(4x^{3} - 1) = 0$
So $x = 0$ or $x^{3} = 1/4$ (not integer).
So $x = 0$, $y = -1$ (already found).

Therefore, the only integer solutions are $(x, y) = (0, 0), (0, -1), (1, 2)$.

Final answer:
All integer solutions are $(x, y) = (0, 0), (0, -1), (1, 2)$.