What are the rational solutions to $y^4=x^3+x+1$?

Question

What are the rational solutions to $y^4=x^3+x+1$?

This equation is interesting because it has substitution $y^2=z$ that reduces it to elliptic curve $z^2=x^3+x+1$. Sometimes, the existence of such substitution makes an equation much easier. For example, if the resulting elliptic curve has rank $0$, then it has finitely many rational points, and it is easy to check for which of them $z$ is a perfect square. As another example, if the original equation has no rational solutions at all, then we can search for a prime $p$, or for a set of primes, for which the Mordell-Weil Sieve gives a contradiction. It is even conjectured that for any equation with no solutions there exists a set of primes that works.

However, in this example the resulting elliptic curve has rank $1$ and therefore infinitely many rational points. Also, the original equation has solutions $(x,y)=(0,\pm 1)$, hence no set of primes gives a contradiction via the Mordell-Weil Sieve. Still, I have a hope that the existence of the above reduction to elliptic curve should make the equation easier than a "random" genus $3$ equation. Hence the question.

Answer 1

This is not an answer, just an account of an attempt to solve the problem that didn't work. I think it sheds some light on the techniques that might be needed, which is why I am sharing it.

Although the plane quartic $C : y^{4} = x^{3} z + xz^{3} + z^{4}$ only has two automorphisms over $\mathbb{Q}$, it has $16$ automorphisms over $\overline{\mathbb{Q}}$. In particular, there is an order $2$ automorphism $\iota$ defined over $K = \mathbb{Q}[w]$, where $w^{6} - 10w^{4} + 31w^{2} - 31 = 0$. The quotient curve $C / \langle \iota \rangle$ is an elliptic curve with $j$-invariant $1728$, namely $$ E : y^{2} = x^{3} + (2w^{3} + 3w^{2} - 4w - 6)x. $$ This suggests the possibility of trying elliptic curve Chabauty. The idea here is that the restriction of scalars ${\rm Res}_{K/\mathbb{Q}} E$ is a $6$-dimensional abelian variety, and there is a map from $J(C) \to {\rm Res}_{K/\mathbb{Q}} E$. Nils Bruin has written code that will attempt Chabauty's method on the image of $J(C)$ inside ${\rm Res}_{K/\mathbb{Q}} E$. This could only work if the dimension of the image is less than its rank.

We have that $E(K) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}^{3}$, and Magma is capable of finding three points of infinite order (although it's not clear if Magma is able to prove that these are generators). The final ingredient we need is a map from $C \to \mathbb{P}^{1}$ defined over $\mathbb{Q}$ that factors through $C \to C / \langle \iota \rangle$. One such map is $$ \varphi(x,y,z) = (-44x^{4} + 232x^{3}z + 88x^{2}z^{2} + 584xz^{3} + 188z^{4}, 33x^{3}z + 33xz^{3} + 33z^{4}). $$ From this, one can compute the map $\psi : E \to \mathbb{P}^{1}$ so that $\varphi$ is the composition of $C \to C / \langle \iota \rangle$ with $\psi$. Using this, I attempted elliptic curve Chabauty using Nils Bruin's code, and it was not successful.

This suggests to me that the image of $J(C)$ inside ${\rm Res}_{K/\mathbb{Q}} E$ has dimension $2$ or $3$, and rank equal to the dimension. If this is true, $J(C)$ would have rank $3$ over $\mathbb{Q}$ and some other technique might be needed (like étale descent or quadratic Chabauty).