Solve $y^2=x^4+x+2$ in $x,y\in \mathbb{Z}$ I have to solve $y^2=x^4+x+2$ in $x,y\in \mathbb{Z}$ and right now I am stuck. This is how far I came:
A little manipulation yields $y^2-2=x(x+1)(x^2-x+1)$. $x=1$ and $y=\pm 2$ are solutions. Assume from now on that $y$ is positive and $x>1$. I somehow want to find a prime $p=\pm 3\mod 8$ which  divides this expression so that $y^2=2\mod p$ and this is not possible then. But I really don't know how to do this. I really need some hints. Thanks.

hint: $x^4+x+2$ is between $(x^2)^2$ and $(x^2+1)^2$ for $x>2$.
