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 $x=0$: $$y^3+y^2=0⟹y^2(y+1)=0⟹y=0\text{ or }y=-1.$$ 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 $y=k{x}^2$ 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 $(x,y)=(0,0),(0,-1),(1,2)$.