One way to attack, say, $y^2 + 65 = x^3$ in integers is to factor as $(y+\sqrt{-65})(y-\sqrt{-65})= x^3$, show that the ideals $(y+\sqrt{-65})$, $(y-\sqrt{-65})$ are coprime - hence both cubes of an ideal; then using that the class number is $8$ (and thus coprime with $3$), we get that $(y+\sqrt{-65})$ is a cube of a principal ideal, then $y+\sqrt{-65}=(a+b\sqrt{-65})^3$ for integers $a,b$ and it's easy to finish.

But what do we do for, say, $y^2 + 79 = x^3$? Can we factor this appropriately in the ring of integers, which is now $\mathbb{Z}[\frac{1+\sqrt{-79}}{2}]$ and proceed as above?

Any help appreciated!

  • $\begingroup$ $\mathbb{Z}[(1+\sqrt{-7})/2]$ is Euclidean, so obviously it works even better in this case. $\endgroup$ – user10354138 May 31 at 0:30
  • $\begingroup$ Ok, what about $\mathbb{Z}[\frac{1+\sqrt{-79}}{2}]$ or some other random case (for which it turns out that the class number is coprime to $3$, in case we need it). $\endgroup$ – DesmondMiles May 31 at 0:32
  • $\begingroup$ The integers in the ring of integers of $\mathbb Q(\sqrt{-79})$ are of the form $\dfrac{a+b\sqrt{-79}}{2}$ where $a$ and $b$ have same parity so you can act likely your first example. $\endgroup$ – Piquito Jun 4 at 15:35

The Mordell equation is discussed in Chapter 6 of these notes. See in particular Theorems 6.7 and 6.8, which together give sufficient conditions on $k \in \mathbb{Z}$ for the Mordell equation $y^2 = x^3 -k$ to have either no integral solutions (if $k \neq 3a^2 \pm 1$) or precisely the integral solutions $(a^2+k,\pm a(a^2-3k))$ (if $k = 3a^2 \pm 1$).

One of the conditions is indeed that the class number of $\mathbb{Q}(\sqrt{-k})$ is prime to $3$, and the table on p. 85 confirms that this condition holds for $k = 79$. However, the other condition is that $k \equiv 1,2 \pmod{4}$, so that $\mathbb{Z}[\sqrt{-k}]$ is the full ring of integers of $\mathbb{Q}(\sqrt{-k})$. On that same page the example of $k = 47$ is discussed: here the class number is $5$, which is prime to $3$, but the Mordell equation has more than $2$ solutions. This shows that the condition that $k \equiv 3 \pmod{4}$ cannot simply be ignored.

These results are only the beginning of the study of the Mordell equation. See for instance this page, which records that $y^2 = x^3 - 79$ has two integral solutions (even though $79 \neq 3 a^2 \pm 1$).


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