Prove that $x^3-x = y^2+1$ has no integer solution

Question

Prove that $x^3-x = y^2+1$ has no integer solution:

I began the proof by case distinction considering the cases if x,y are both even, if x,y both odd, if x even, y odd and the last one if x odd and y even.

If x,y are even: then $x^3$ and $x$ are always even, so is $x^3-x$ even too. $y^2$ is even and the term $y^2+1$ is always odd. Hence the equality in this case is false.

If x,y are odd: then $x^3$ and $x$ are odd, so is $x^3-x$ always even. $y^2$ is odd, but $y^2+1$ is always even. This means that the equality could possibly be true.

(The other two cases are then analogously proven)

How should I continue?

Answer 1

HINT :

Instead of mod 2, considering in mod 3 helps.

The LHS is divisible by 3, but the RHS isn't.