$y^3=x^4+x$, and computing all rational points on rank $0$ Picard curves

Question

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 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.

Answer 1

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.

Answer 2

I think there are no more rational points: Write the equation as $y^3z=x^4+xz^3$ for integers $x,y,z$ where $z\ge1$ and $\operatorname{gcd}(x,y,z)=1$. We consider two cases:

• $\operatorname{gcd}(x, z)=1$. As $z$ divides $x^4$, we get $z=1$. From $y^3=x(x^3+1)$ and $\operatorname{gcd}(x,x^3+1)=1$ we get that $x^3+1$ is a cube, and we are readily done.
• $t=\operatorname{gcd}(x, z)>1$. As $t$ is relatively prime to $y$, we get $t^4\mid z$. Write $z=at^4$, $x=bt$ for integers $a$ and $b$. This yields $y^3a=b(b^3+a^3t^9)$. So $a$ divides $b^4$. But $a$ and $b$ are relatively prime, hence $a=\pm1$. But $b$ and $t$ are relatively prime too, so we arrive at $\pm y^3=b^3\pm t^9$, and we finish in view of the Fermat equation for exponent $3$.