<s>[INST] show that $y^2 = x^5 - x +2 $ has no integer solutions I considered everything$\mod 3$:
$x,y \equiv 0 \mod 3 ,\;\;\;\; y^2 \equiv 0 \mod 3 ,\;\;\;\; x^5 \equiv 0 \equiv x \mod 3$
$x,y \equiv 1 \mod 3 ,\;\;\;\; y^2 \equiv 1 \mod 3 ,\;\;\;\; x^5 \equiv 1 \equiv x \mod 3$
$x,y \equiv 2 \mod 3 ,\;\;\;\; y^2 \equiv 1 \mod 3 ,\;\;\;\; x^5 \equiv 2 \equiv x \mod 3$
thus in particular $x^5 - x \equiv 0 \mod 3, \;\;\; \forall x \in \mathbb{Z},$ and furthermore
$x^5 - x + 2 \equiv 2 \mod 3, \;\;\; \forall x \in \mathbb{Z}$.
But, $y^2 \equiv 0,1 \mod 3$, and if we were considering that there would be an integer solution then both sides should have the same remainder modulo $3$, but LHS and RHS never have the same remainder thus there are no integer solutions?
This is my first time doing a proof of this form, I'm familiar with modular arithmetic, however I've never used it to prove that a polynomial equation didn't have integer solutions.
 [/INST] You can also use $\bmod 3$:
$$y^2-2=x(x^2-1)(x^2+1)$$
RHS: $x^2-1\equiv 0\bmod 3\Rightarrow RHS \equiv 0\bmod 3$
But :
LHS: $$y^2-1-1\equiv (0\bmod 3-1\bmod 3)\equiv -1\bmod 3$$
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