Solve $x^4 - 2x^3 + x = y^4 + 3y^2 + y \wedge (x,y) \in \mathbb{Z}^2$

Question

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.

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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.

Answer 1

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.