Solving the Diophantine equation $y^2 = x^4+x+ 2$ for $x,y \in \mathbb{Z}$

Question

I want to solve the Diophantine equation $y^2 = x^4+x+ 2$ for $x,y \in \mathbb{Z}$.

I already found 4 solutions: $(x,y) = (1,\pm2)$ and $(x,y)=(-2,\pm4)$. It can probably be solved using some factorization argument, but I don't know how.

Answer 1

If $x\geq -1$, then $$x^4<y^2<(x^2+2)^2\,.$$ Therefore, for $x\geq -1$, there exists a solution iff $x^4+x+2=(x^2+1)^2$. The only integer root of the last equation is $x=1$, yielding the solutions $(x,y)=(1,\pm2)$.

If $x\leq -3$, then $$(x^2-2)^2<y^2<x^4\,.$$
Therefore, for $x\leq -3$, there exists a solution iff $x^4+x+2=(x^2-1)^2$. The last equation does not have an integer root, whence there does not exist a solution in this case.

If $x=-2$, then we have the solutions $(x,y)=(-2,\pm 4)$. Therefore, there are only four solutions $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ to $y^2=x^4+x+2$, which are $(1,\pm2)$ and $(-2,\pm4)$.

Answer 2

There is a "simple method" to solve the Diophantine equation $$ Y^2=X^4+aX^3+bX^2+cX+d $$ in general, see here. The second section does it for the example of $Y^2=X^4-8X^2+8X+1$, but $Y^2=X^4+X+2$ should be even easier.