# Mordell curve when $d\equiv 3 \pmod 4$

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!

• $\mathbb{Z}[(1+\sqrt{-7})/2]$ is Euclidean, so obviously it works even better in this case. – user10354138 May 31 at 0:30
• 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). – DesmondMiles May 31 at 0:32
• 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. – Piquito Jun 4 at 15:35

## 1 Answer

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 $$79 \neq 3 a^2 \pm 1$$).