What are the rational solutions to the equation
$$
y^3 = x^4 + x,
$$
in particular, are there any (finite) solutions other than $(x,y)=(0,0)$ and $(-1,0)$?


Context: This is the simplest-looking example of a Picard curve, that is, curve $y^3=P(x)$, where $P(x)$ is a polynomial of degree 4. The rank of its Jacobian is 0, the simplest case possible. Hashimoto and Morrison <https://arxiv.org/abs/2002.03291> treated more difficult case of rank $1$. For rank $0$, Jackson Morrow [$y^3 = x^4 + x + 2$, and existence of rational points on rank 0 Picard curves](https://mathoverflow.net/questions/407677) presented some Magma code that can prove that there are no rational points (if this is indeed the case). However, the curve $y^3 = x^4 + x$ has some rational points, so deeper analysis seems to be required. Hence the question.



$y^3=x^4+x$, and computing all rational points on rank $0$ Picard curves
Let $x = \frac{a}{d}$, $y = \frac{b}{d}$, $\gcd(a, d, b) = 1$.
$$b^3d = a^4 + ad^3$$
Suppose $p \mid a$ and $p \mid d$ for prime $p$. Then $b$ is not divisible by $p$.
$$\nu\_p(d) = \nu\_p(a^4 + ad^3) = \nu\_p(a) + \nu\_p(a^3 + d^3) = 4\nu\_p(a)$$
The last equality is true because $\nu\_p(a) < \nu\_p(d)$. Hence, we can make the substitution $a = a\_1c$, $d = d\_1c^4$; $(a\_1, d\_1, c)$ are pairwise coprime.


$$b^3d\_1 = a\_1^4 + a\_1d\_1^3c^9$$
$d\_1 \mid a\_1$. But this is only possible if $d\_1 = \pm 1$.
$(\pm b)^3 = a\_1(a\_1^3 \pm c^9)$
Whence $a\_1^3 \pm c^9$ must be a cube, which contradicts Fermat's theorem.


