<s>[INST] 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 $x, y \in \mathbb{Z}$?
$$x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0$$
 [/INST] $x^4 + 2x^3 -3x^2-4x-y^2-2y-6=0\iff(x^2+x)^2-4(x^2+x)=(y+1)^2+5$
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.
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