$x^2+k=y^3$ no integral solution

Question

Let $k > 0$ be a squarefree positive integer and $k = 1, 2 \mod(4)$. Suppose there exists no integer $a$ s.t. $k = 3a^2 \pm 1$. Additionally, $3$ should not divide the class number of $\mathbb Q (\sqrt{-k})$.

Why does then the equation $x^2 + k = y^3$ have no integral solution?

See for Mordell curve on this site. See also here. A solution is given here. In Theorem $I$, part $(B)$ is the assumption that the class number is not divisible by $3$. — Dietrich Burde, Dec 07 '19 at 14:38

score 3 · Accepted Answer · answered Dec 07 '19 at 14:42

This is well-known, and can be found on the web. See for example http://faculty.csuci.edu/brian.sittinger/mordell.pdf

I only sketch the proof. Let $I=(x-\sqrt{-k})$ and $J=(x+\sqrt{-k})$, so that $IJ=(y)^3$. One may show that $I$ and $J$ are coprime (this is a bit of work...)

By uniqueness of decomposition into produts of powers of prime ideals, $I$ and $J$ are cubes of some ideals. Write $I=(x-\sqrt{-k}))=K^3$. Since the class group has no element of order $3$ by assumption, $K$ is principal ideal, i.e. $K=(z)$. We then have $z^3=\pm(x-\sqrt{-k})$ (units here are equal to $\pm 1$). A bit of algebra and the assumption on $k$ then shows that you have no solution.

$x^2+k=y^3$ no integral solution

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