How to solve the Diophantine equation $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$? How can one solve the following Diophantine equation in $x, y \in \mathbb{Z}$?
$$x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$$
 A: Since this is a Diophantine equation, we only seek for integer solutions. Notice that $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$ implies $(y-x^2-x+3)(y+x^2+x-1)=-9$, we only have $6$ cases: $((y-x^2-x+3),(y+x^2+x-1))$ must be one of $(1,-9),(9,-1),(3,-3),(-1,9),(-9,1),(-3,3)$. However, after checking all these cases there are no solutions.
A simpler way to do this: modulo 4, we have
$$x^4 + 2x^3 -3x^2-4x-y^2-2y-6\equiv x^4 + 2x^3 +x^2-y^2-2y+2\equiv (x(x+1))^2-(y+1)^2+3\equiv 3-(y+1)^2$$
However, $3-(y+1)^2$ won't be zero modulo $4$ since the squares must be $0$ or $1$ modulo $4$, not $3$.
A: $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0\iff(x^2+x)^2-4(x^2+x)=(y+1)^2+5$
Putting $X=x^2+x$ we get $(X-2)^2=(y+1)^2+9$ whose only solutions are clearly
$$(X-2,y+1)=(\pm3,0),(\pm5,\pm4)$$ in both cases, since $X=x^2+x$, we have the equations for the unknown $x$
$$x^2+x-5=0\text{ and } x^2+x+1=0\\x^2+x-7=0\text{ and } x^2+x+3=0$$ these four equations have not integral roots, consequently the diophantine equation $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$ has no solution.
