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.
Best Answer
Your reasoning is correct.
Here a similar way using little Fermat for $p=5$:
$$x^5\equiv x \mod 5$$
Hence,
$$y^2\equiv 2 \mod 5$$
but the quadratic residues $\mod 5$ are $0,1,4$.
So, no solution in integers exists.