Solve $x^4 - 2x^3 + x = y^4 + 3y^2 + y \wedge (x,y) \in \mathbb{Z}^2$ I want to solve equation $x^4 - 2x^3 + x = y^4 + 3y^2 + y$ in integers. The task comes from the LXVI Polish Mathematical Olympiad. Series with this task ended twenty days ago, so it is legal to talk about it.

It is of course equal to $x(x-1)(x^2-x-1) = y(y^3 + 3y + 1) \wedge (x,y) \in \mathbb{Z}^{2}$.
It is obvious that $\left(\nexists x \in \mathbb{Z}\right)\left((x^2-x-1) = 0\right)$ and $\left(\nexists y \in \mathbb{Z}\right)\left( (y^3 + 3y + 1) = 0\right)$.
$$x(x-1)(x^2-x-1) = 0
\Longrightarrow x \in \lbrace 0;~1 \rbrace
\Longrightarrow y = 0$$
Therefore pairs $(0,0), (1,0)$ satisfy equation.
We can see
$|y| \neq 1 \Rightarrow y \nmid (y^3+3y+1)$
or
$\left(|x| \neq 1 \Rightarrow x \nmid (x-1) \wedge x \nmid (x^2 - x -1)\right)$, and lot similar.
Should note too, that any multiplier have to divide some multiplier from second side.
$$\frac{x(x-1)(x^2-x-1)}{y} = y^3+3y+1 \in \mathbb{Z} \Longrightarrow y \mid x(x-1)(x^2-x-1)\\
$$
And so on. With lot of transformations and special cases we can obtain, that there is no more solutions (if I didn't make mistake). But it looks badly, is so long and it is easy to confuse. Have anyone simple idea? I strongly believe here some exist.

Hint: $x^4-2x^3+x=y^4+3y^2+y\let\leq\leqslant\let\geq\geqslant$ is equivalent to $(2x^2-2x-1)^2-(2y^2+3)^2=4y-8$.

 We have $2x^2-2x-1+2y^2+3\mid4y-8$, hence $|2x^2-2x-1+2y^2+3|\leq|4y-8|$. As $x^2\geq x$ for $x\in\mathbb Z$ this gives $y^2+1\leq|2y-4|$. For $y\geq2$ this is $(y-1)^2\leq-5$; for $y\leq2$ this is $(y+1)^2\leq3$. It remains to check $y\in[-2,0]$.
$y=-2$: $x^2-x+y^2+1\mid 2y-4$ gives $x^2-x=3$, impossible.
$y=-1$: $x^2-x+2\mid6$ means $x^2-x=4$, impossible, or $x=0,1$. None of them works.
$y=0$: $x^2-x+1\mid4$ gives $x^2-x=0$, hence $x=0,1$. Both are solutions.

