<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] 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 $3$.
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