Find all integers $x, y$ such that $y^{2}=x^{3}-3 x+2$.
Let $x, y$ be integers.

$$
y^{2}=x^{3}-3 x+2 \Longleftrightarrow y^{2}=(x-1)\left(x^{2}+x-2\right)=(x-1)^{2}(x+2)
$$

If $y^{2}=x^{3}-3 x+2$, then $x+2$ is a perfect square (if $x \neq 1$, then $x+2 \geq 0$ and every prime factor in the factorization of $x+2$ appears with an even exponent (2 times that in $y$ minus 2 times that in $x-1)$ ). By setting $x=k^{2}-2$ for $k \in \mathbb{N}, y= \pm k\left(k^{2}-3\right)$.

Conversely, we verify that the pairs $\left(k^{2}-2, \pm k\left(k^{2}-3\right)\right)$ with $k \in \mathbb{N}$ are valid:

$$
( \pm k)^{2}\left(k^{2}-3\right)^{2}=\left(k^{2}-2+2\right)\left(k^{2}-2-1\right)^{2}
$$