Mordell’s Equation

The equation $\displaystyle E_k: Y^2=X^3+k$ is called Mordell’s equation. For $k \neq 0$ , it is non-singular and defines an elliptic curve (genus 1, group structure).

Rational Points: We first describe the torsion part of the rational solutions. By Nagell-Lutz, these solutions should have integer coordinates.

Torsion: Assume that $k$ is free of sixth powers. If not we can use the change of variables $X \to c^2X, Y \to c^3 Y$ to remove the sixth powers.

$\displaystyle c^6 Y^2 = c^6X^3+k \implies Y^2=X^3 +\frac{k}{c^6}$

The torsion is zero in most cases. With $k$ sixth power free we have

$\displaystyle E(\mathbb{Q})_{t o r s} \simeq \begin{cases}\mathbb{Z} / 6 \mathbb{Z} & \text { if } k=1 \\ \mathbb{Z} / 3 \mathbb{Z} & \text { if } k=-432=-2^{4} 3^{3}, \text { or if } k=\square \text { and } k \neq 1 \\ \mathbb{Z} / 2 \mathbb{Z} & \text { if } k=\text { cube and } k \neq 1 \\ 0 & \text { otherwise } \end{cases}$

Let us first consider $\displaystyle E_{1}: Y^{2}=X^{3}+1$ .

If $Y=0$ , then $X=-1$ , so there is one 2-torsion element. Using Nagell-Lutz, the square of the Y-coordinate dividies the discriminant. In this case we have $Y^2|27$ which implies that $Y \in \{\pm 1, \pm 3\}$ . Now check that the points $(X, Y) =(0, \pm 1), (2, \pm 3)$ actually are torsion, are of order 3 and 6. So we have

$\displaystyle E(\mathbb{Q})_{\text {tors }}={\mathcal{O},(-1,0),(2, \pm 3),(0, \pm 1)} \simeq \mathbb Z/6$ .

We use the following lemma to get constraints on the size of the torsion.

Lemma: If $p$ is a prime such that $p \neq 1 \mod 3$ where $E_k$ has good reduction that is $p \nmid \Delta$ , then $|E(F_{p})|=p+1$ (including point at infinity).

This is because the cubing map is an isomorphism $x \to x^3 \mod p$ and for every value of $Y^2-k$ has a unique cuberoot $X$ , giving $p$ solutions. If you include the point at infinity, we get $p+1$ points.

But the modulo p map lets us see the torsion as a subgroup of $E(\mathbb F_p)$ , hence the size of the torsion divides $p+1$ .

Choosing a large prime $p = 5 \mod 12$ we can see that $4 \nmid |E(\mathbb{Q})_{\text {tors }}|$ , because otherwise $4| |E(\mathbb{Q})_{\text {tors }}| | p+1 \implies 4|p+1$

Similarly choose a large prime $p = 2 \mod 9$ (which satisfies the conditions of the lemma), to see that $9 \nmid |E(\mathbb{Q})_{\text {tors}}$ , and finally $p = 2 \mod 3q$ for any prime $q>3$ to get $q \nmid |E(\mathbb{Q})_{\text {tors}}|$

Thus $|E(\mathbb{Q})_{\text {tors}}|$ can only be divisible by $2$ and $3$ and nothing else.

$\displaystyle Y=0 \implies x^3+k =0$ which has solution only when $k$ is a cube, hence cubes are the only cases with two torsion.

Let’s now consider 3-torsion. $3P=\mathcal O \implies 2P =-P$ , so using the duplication formula we get

$\displaystyle \frac{X^{4}-8 k X}{4\left(X^{3}+k\right)}=X$

$\displaystyle X^4+4kX=0$

$X=0 \implies Y^2=k$ , therefore when $k$ is a square, we have 3-torsion,

$X^3=-4k \implies Y^2=-3k$ . These two equation together with the fact that k is free of sixth powers implies that $k =-2^4 3^3$ , in which case there is 3-torsion.

Putting all these together proves the determines the torsion in all the cases.

We saw that the torsion points are integer points, but not all integer points are torsion.

Integral Points: There are finitely many integral points on any elliptic curve (Siegel’s theorem). But determining the integer points is hard.

Here are some examples to determine all the integer points:

a) $\displaystyle Y^2=X^3-1$

It’s easy to see that $(1, 0)$ is a solution. In fact, it’s the only integer solution. To see this write

$\displaystyle X^3=Y^2+1 =(Y+i)(Y-i),$

and use the unique factorization in $\mathbb Z[i]$ to get

$\displaystyle Y+i =(a+bi)^3$

which implies that $Y=0$ . (We have argue that the factors are coprime)

b) $\displaystyle Y^2=X^3-2$

Rearranging the equation to

$\displaystyle X^3=Y^2+2 =(Y+\sqrt{-2})(Y-\sqrt{-2})$

and use the unique factorization in $\mathbb Z[\sqrt{-2}]$ to see that

$\displaystyle Y+\sqrt{-2} =(a+b\sqrt{-2})^3$

which implies that $b=\pm 1$ . And $b=1$ gives the solutions $\displaystyle (X,Y)=(3, \pm 5)$ , while $b=-1$ is not possible.

c) $\displaystyle Y^2=X^3-3$

If $X$ is even, $Y^2 = -3 \mod 8$ which is not possible. If $X$ is odd, $Y$ is even and we have

$\displaystyle 4((Y/2)^2+1) =X^3+1=(X+1)(X^2-X+1) \implies 4|X+1$ .

$X^2-X+1$ is an odd number with prime factors only of the form $p =1 \mod 4$ , but this implies $X^2-X+1 =1 \mod 4 \implies 4|X-1$ . Therefore there are no solutions.

d) $\displaystyle Y^2=X^3-4$

$\displaystyle X^3=Y^2+4 =(Y+2i)(Y-2i)$ and use the unique factorization in $\mathbb Z[i]$ to see that

$Y+2i =(a+bi)^3$ which implies $Y=a\left(a^{2}-3 b^{2}\right), \quad 2=b\left(3 a^{2}-b^{2}\right)$

$b= \pm 1, \pm 2 \implies (X, Y) =(2, \pm 2), (5, \pm 11)$

e) $\displaystyle Y^2=X^3-5$

The argument is similar to $\displaystyle Y^2=X^3-3$ , we get that $Y$ is even and writing

$4((Y/2)^2+1) = (X-1)(X^2+X+1)$

we get that $4|X-1$ and looking the prime factors of $X^2+X+1$ , we get that $4|X+1$ , which is a contradiction.

But what about rational points? We determined the rational points of finite order (torsion), but what about point of infinite order?

For instance $Y^2=X^3-2$ , the integer points we found $(3, \pm 5)$ are not torsion. In fact, this curve has no torsion from the computation in the beginning. $(3, 5)$ generates all the rational points and hence the rank is $1.$