Solve in integers $x^2 – x + 10 = y^3$. (You may use the fact that the class number of $\mathbb{Q}(\sqrt{-39})$ is $4$ and that its ring of integers is $\mathbb{Z}\left[\frac{1+\sqrt{-39}}{2}\right]$).
Approach: after factoring we get (in terms of ideals) $(x-\frac{1 + \sqrt{-39}}{2})(x – \frac{1 – \sqrt{-39}}{2}) = (y)^3$ and aim to obtain that the factors on the left are cubes of ideals (then since $3$ and the class number $4$ are coprime it would be easy to finish). If the factors are coprime we are fine, but if they are not, then a common prime factor ideal must divide $(1+\sqrt{-39}) = (2, \frac{1 + \sqrt{-39}}{2})^2(2,\frac{3 + \sqrt{-39}}{2})(5, \frac{1 + \sqrt{-39}}{2})$ (the factorization was done by Dedekind's theorem on $(2)$ and $(5)$). Any idea how to finish?
Any help appreciated!
EDIT: Just for the sake of curiosity, can anyone find an elementary way to solve this? It should be the case that this equation has no solutions, so it is not unlikely that an elementary proof exists.
Best Answer
Solved it now! Actually the norm of $(\sqrt{-39})$ is $39$ and the norm of the common factor $P$ also has to divide $N((y)^3) = y^6$, so $y$ must be divisible by $3$ or $13$ and this is impossible from moduli $3^2$ and $13^2$.
(Nothing new has been used here, just turned out I had fatal calculation errors, sorry!)