Mordell’s Equation

The Diophantine equation $\displaystyle Y^2=X^3+k$ is called the Mordell equation. Here $k$ is a fixed nonzero integer, and we ask for integer or rational solutions $(X,Y)$ . This equation is simple to write down, but it already contains much of the arithmetic complexity of elliptic curves.

For $k\neq0$ , the curve

$\displaystyle E_k: Y^2=X^3+k$

is nonsingular. Indeed, the right-hand side has no repeated root unless $k=0$ . Equivalently, the discriminant of the short Weierstrass equation $Y^2=X^3+k$ is

$\displaystyle \Delta=-432k^2,$

which is nonzero exactly when $k\neq0$ . Thus $E_k$ is an elliptic curve over $\mathbb Q$ . That means its rational points form an abelian group, with a special point at infinity $\mathcal O$ as the identity element.

There are two natural questions. First, what are the rational points of finite order? These form the torsion subgroup $\displaystyle E_k(\mathbb Q)_{\mathrm{tors}}.$ Second, what are the rational points of infinite order? These form the free part of $E_k(\mathbb Q)$ . By the Mordell–Weil theorem,

$\displaystyle E_k(\mathbb Q)\simeq E_k(\mathbb Q)_{\mathrm{tors}}\oplus \mathbb Z^r,$

where $r$ is the rank of the elliptic curve. The torsion part can be described quite explicitly for Mordell curves. The infinite-order part is much harder. Even determining all integer points is difficult in general, although Siegel’s theorem tells us that there are only finitely many integer points on any elliptic curve.

Removing sixth powers

There is one simple normalization we should make at the beginning. If $k$ contains a sixth power factor, then the curve is isomorphic over $\mathbb Q$ to another Mordell curve with that sixth power removed.

Suppose $\displaystyle k=c^6k_0.$ Put $\displaystyle X=c^2X', Y=c^3Y'.$ Then $\displaystyle Y^2=X^3+k$ becomes $\displaystyle c^6(Y')^2=c^6(X')^3+c^6k_0.$ Dividing by $c^6$ , we get

$\displaystyle (Y')^2=(X')^3+k_0.$

So $E_k$ and $E_{k_0}$ are isomorphic over $\mathbb Q$ . Therefore, when studying the rational group structure, we may assume that $k$ is sixth-power-free: no prime exponent in the factorization of $k$ is at least $6$ .

The torsion subgroup

For sixth-power-free $k$ , the torsion subgroup of $\displaystyle E_k: Y^2=X^3+k$ is as follows:

$\displaystyle E_k(\mathbb Q)_{\mathrm{tors}}\simeq \begin{cases} \mathbb Z/6\mathbb Z, & \text{if } k=1,\\ \mathbb Z/3\mathbb Z, & \text{if } k \text{ is a square and } k\neq1,\\ \mathbb Z/3\mathbb Z, & \text{if } k=-432=-2^4 3^3,\\ \mathbb Z/2\mathbb Z, & \text{if } k \text{ is a cube and } k\neq1,\\ 0, & \text{otherwise.} \end{cases}$

Let us explain why this classification is natural.

The first tool is the Nagell–Lutz theorem. For an elliptic curve written in integral Weierstrass form, torsion points over $\mathbb Q$ have integer coordinates. More precisely, if $P=(X,Y)$ is a nontrivial torsion point, then $X,Y\in\mathbb Z$ , and either $Y=0$ or $Y^2$ divides the discriminant. For our curve, the discriminant is $-432k^2$ . This already tells us that rational torsion points are very special. We do not have to search among arbitrary rational points; torsion points must occur among integer points satisfying strong divisibility restrictions.

Before finding the actual $2$ – and $3$ -torsion, let us first see why no other prime torsion can occur.

We use a simple counting lemma: Suppose $p$ is a prime of good reduction for $E_k$ , meaning $p\nmid \Delta=-432k^2$ . Suppose also that $\displaystyle p\not\equiv1\pmod 3.$ Then

$\displaystyle |E_k(\mathbb F_p)|=p+1.$

Here $E_k(\mathbb F_p)$ means the group of points on the reduced curve $\displaystyle Y^2=X^3+k$ over the finite field $\mathbb F_p$ , including the point at infinity.

Why is this true? Since $p\not\equiv1\pmod3$ , the cubing map $\displaystyle x\mapsto x^3$ is a bijection on $\mathbb F_p$ . Therefore, for each value of $Y\in\mathbb F_p$ , the equation $\displaystyle X^3=Y^2-k$ has exactly one solution $X\in\mathbb F_p$ . Thus there are exactly $p$ affine points, one for each choice of $Y$ . Including the point at infinity gives $\displaystyle |E_k(\mathbb F_p)|=p+1.$

Now use the reduction map. For primes of good reduction, rational torsion injects into $E_k(\mathbb F_p)$ away from the characteristic $p$ . In particular, if $\ell\neq p$ is a prime dividing the rational torsion order, then $\ell$ must divide $|E_k(\mathbb F_p)|$ . But for the primes in the lemma, $|E_k(\mathbb F_p)|=p+1$ . This gives strong restrictions.

Choose primes $p\equiv5\pmod{12}$ avoiding the finite set of bad primes. Then $p\not\equiv1\pmod3$ , so the lemma applies, and $\displaystyle |E_k(\mathbb F_p)|=p+1.$ But if $p\equiv5\pmod{12}$ , then $p+1\equiv6\pmod{12}$ , so $4\nmid p+1$ . Therefore the rational torsion cannot have a subgroup of order $4$ .

Similarly, choose primes $p\equiv2\pmod9$ avoiding the bad primes. Then $p\not\equiv1\pmod3$ , and $\displaystyle p+1\equiv3\pmod9,$ so $9\nmid p+1$ . Hence the rational torsion cannot have order divisible by $9$ .

Finally, if $q>3$ is a prime, choose a prime $p\equiv2\pmod{3q}$ avoiding the bad primes. Then $p\not\equiv1\pmod3$ , but $\displaystyle p+1\equiv3\pmod q,$ so $q\nmid p+1$ . Therefore no prime $q>3$ can divide the torsion order.

Thus the torsion order can only involve the primes $2$ and $3$ , and the previous arguments rule out high powers. So the torsion subgroup has order dividing $6$ . Now we only need to decide when $2$ -torsion and $3$ -torsion occur.

Two-torsion

A point $P=(X,Y)$ has order $2$ precisely when $P=-P$ . On a Weierstrass equation, negation sends $\displaystyle (X,Y)\mapsto(X,-Y).$ Thus $P=-P$ exactly when $Y=0$ .

So the $2$ -torsion points are exactly the solutions of $\displaystyle 0=X^3+k.$ Equivalently, $\displaystyle X^3=-k.$ Therefore a rational $2$ -torsion point exists exactly when $-k$ is a rational cube. Since $k$ is an integer, this is the same as saying that $k$ is an integer cube up to sign. In the convention above, this is recorded by saying that the cube case gives $\mathbb Z/2\mathbb Z$ , except when $k=1$ , where additional $3$ -torsion appears and the torsion becomes $\mathbb Z/6\mathbb Z$ .

Three-torsion

Now let us look for points of order $3$ . If $P$ has order $3$ , then $\displaystyle 3P=\mathcal O.$ This means $\displaystyle 2P=-P.$ Since $P=(X,Y)$ and $-P=(X,-Y)$ have the same $X$ -coordinate, the condition $2P=-P$ says that the $X$ -coordinate of $2P$ must equal $X$ .

For the curve $\displaystyle Y^2=X^3+k,$ the duplication formula gives

$\displaystyle X(2P)=\frac{X^4-8kX}{4(X^3+k)}.$

Therefore the condition $X(2P)=X$ becomes

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

Multiplying through, $\displaystyle X^4-8kX=4X(X^3+k).$ So

$\displaystyle X^4-8kX=4X^4+4kX.$

Rearranging, $\displaystyle 3X^4+12kX=0.$ Dividing by $3$ ,

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

Thus $\displaystyle X(X^3+4k)=0.$ So there are two possibilities. First, $X=0$ . Then the curve equation gives $\displaystyle Y^2=k.$ Thus if $k$ is a square, we get points $\displaystyle (0,\pm\sqrt{k})$ of order $3$ . Second, we may have $\displaystyle X^3=-4k.$ Then the curve equation gives $\displaystyle Y^2=X^3+k=-4k+k=-3k.$ So we need both

$\displaystyle -4k \text{ to be a cube}, -3k \text{ to be a square}.$

Because $k$ is sixth-power-free, these two conditions force

$\displaystyle k=-2^4 3^3=-432.$

Indeed, comparing prime exponents: for primes other than $2$ and $3$ , the exponent in $k$ must be divisible by both $2$ and $3$ , hence by $6$ , so it must be zero because $k$ is sixth-power-free. At the prime $2$ , the conditions force exponent $4$ . At the prime $3$ , they force exponent $3$ . The sign must be negative so that $-3k$ is positive. Hence $k=-432$ .

Putting the $2$ – and $3$ -torsion analyses together gives the torsion classification.

The special curve $E_1^2=X^3+1$

Let us work out the most famous case: $\displaystyle E_1: Y^2=X^3+1.$

First, $Y=0$ gives $\displaystyle X^3=-1,$ so $X=-1$ . Thus $(-1,0)$ is a $2$ -torsion point. Next, because $k=1$ is a square, the points $\displaystyle (0,1), (0,-1)$ are $3$ -torsion points. The points $\displaystyle (2,3), (2,-3)$ also lie on the curve.These points have order $6$ . In fact the torsion subgroup is

$\displaystyle E_1(\mathbb Q)_{\mathrm{tors}}=\{\mathcal O,(-1,0),(0,\pm1),(2,\pm3)\}\simeq\mathbb Z/6\mathbb Z.$

This curve has rank zero, so these torsion points are all the rational points on $E_1$ .

Integral points versus torsion points

The torsion points are always very special, and Nagell–Lutz tells us that rational torsion points must have integer coordinates. But the converse is false: not every integer point is torsion.

For example, on $\displaystyle Y^2=X^3-2,$ we have the integer point $\displaystyle (X,Y)=(3,5).$ This point is not torsion. In fact, for $k=-2$ , the torsion classification gives $\displaystyle E_{-2}(\mathbb Q)_{\mathrm{tors}}=0,$ and the point $(3,5)$ has infinite order. Indeed, it generates the free part of $E_{-2}(\mathbb Q)$ , so this curve has rank $1$ .

Thus torsion is only one part of the story. Integral points can be torsion, but they can also have infinite order. Determining all integral points is a separate and difficult problem. Siegel’s theorem says there are finitely many integral points on any elliptic curve, but the theorem is not constructive in an elementary way. For specific Mordell equations, however, we can sometimes solve the integer problem using unique factorization in quadratic rings.

Example 1: $Y^2=X^3-1$

Consider $\displaystyle Y^2=X^3-1.$ Equivalently, $\displaystyle X^3=Y^2+1.$ It is clear that $(X,Y)=(1,0)$ is a solution. We claim it is the only integer solution.

Write $\displaystyle Y^2+1=(Y+i)(Y-i)$ in the Gaussian integers $\mathbb Z[i]$ . Thus $\displaystyle X^3=(Y+i)(Y-i).$ The ring $\mathbb Z[i]$ is a unique factorization domain. The idea is that if the two factors $Y+i$ and $Y-i$ are coprime, and their product is a cube, then each factor must itself be a cube up to a unit. One first checks that the two factors are coprime up to units. Indeed, any common divisor must divide their difference $\displaystyle (Y+i)-(Y-i)=2i.$ A parity check shows that no nonunit Gaussian prime divides both factors. Therefore they are coprime. Hence $\displaystyle Y+i=u(a+bi)^3$ for some unit $u\in{\pm1,\pm i}$ and integers $a,b$ . Expanding, $\displaystyle (a+bi)^3=(a^3-3ab^2)+b(3a^2-b^2)i.$ Since the imaginary part of $Y+i$ is $1$ , up to a unit we are forced into the equation $\displaystyle b(3a^2-b^2)=\pm1.$ Thus $b=\pm1$ . Checking the resulting possibilities gives only $a=0$ , and hence $Y=0$ . Then $\displaystyle X^3=1,$ so $X=1$ .

Therefore the only integer solution is $\displaystyle (X,Y)=(1,0).$

Example 2: $Y^2=X^3-2$

Now consider $\displaystyle Y^2=X^3-2.$ Rearrange: $\displaystyle X^3=Y^2+2.$ Factor in the ring $\mathbb Z[\sqrt{-2}]$ :

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

The ring $\mathbb Z[\sqrt{-2}]$ is also a unique factorization domain. As before, one checks that the two factors are coprime up to units. Since their product is a cube, each factor must be a cube up to a unit. Thus $\displaystyle Y+\sqrt{-2}=(a+b\sqrt{-2})^3$ up to sign. Expand: $\displaystyle (a+b\sqrt{-2})^3=(a^3-6ab^2)+b(3a^2-2b^2)\sqrt{-2}.$ The coefficient of $\sqrt{-2}$ on the left is $1$ , so $\displaystyle b(3a^2-2b^2)=1.$ This forces $b=\pm1$ . If $b=1$ , then $\displaystyle 3a^2-2=1,$ so $a^2=1$ . Thus $a=\pm1$ . The norm of $a+b\sqrt{-2}$ is $\displaystyle a^2+2b^2.$ Since $X^3$ is the norm of the cube, we get $\displaystyle X=a^2+2b^2=1+2=3.$ The real part gives $\displaystyle Y=a^3-6ab^2.$ For $a=1$ , this gives $Y=-5$ . For $a=-1$ , this gives $Y=5$ . Thus the integer solutions are

$\displaystyle (X,Y)=(3,\pm5).$

Example 3: $Y^2=X^3-3$

Consider $\displaystyle Y^2=X^3-3.$ We show that there are no integer solutions. First suppose $X$ is even. Then $X^3\equiv0\pmod8$ , so $\displaystyle Y^2=X^3-3\equiv -3\equiv5\pmod8.$ But a square modulo $8$ is only $0,1,4$ . So this is impossible. Hence $X$ must be odd. If $X$ is odd, then $X^3-3$ is even, so $Y$ is even. Write $Y=2Z$ . Then

$\displaystyle 4Z^2=X^3-3.$

Add $4$ to both sides: $\displaystyle 4(Z^2+1)=X^3+1.$ Factor: $\displaystyle X^3+1=(X+1)(X^2-X+1).$ So $\displaystyle 4(Z^2+1)=(X+1)(X^2-X+1).$ Since $X$ is odd, the factor $X^2-X+1$ is odd. Therefore the factor $4$ on the left must divide $X+1$ : $\displaystyle 4\mid X+1.$ Now look at the other factor $X^2-X+1$ . Apart from a harmless common factor issue that can be checked separately, its prime divisors must divide a number of the form $Z^2+1$ . An odd prime dividing $Z^2+1$ is congruent to $1\pmod4$ . Therefore $\displaystyle X^2-X+1\equiv1\pmod4.$ But for odd $X$ , the congruence $\displaystyle X^2-X+1\equiv1\pmod4$ forces $\displaystyle X\equiv1\pmod4.$ That is, $\displaystyle 4\mid X-1.$

We have obtained both $4\mid X+1$ and $4\mid X-1$ , which is impossible. Hence there are no integer solutions.

Example 4: $Y^2=X^3-4$

Now consider $\displaystyle Y^2=X^3-4.$ Rearrange: $\displaystyle X^3=Y^2+4.$ Factor in $\mathbb Z[i]$ : $\displaystyle Y^2+4=(Y+2i)(Y-2i).$ Using unique factorization in $\mathbb Z[i]$ , one obtains $\displaystyle Y+2i=(a+bi)^3$ up to multiplication by a unit. Expanding, $\displaystyle (a+bi)^3=(a^3-3ab^2)+b(3a^2-b^2)i.$ The imaginary part must be $2$ , so $\displaystyle b(3a^2-b^2)=2.$ Thus $b$ must divide $2$ , so $\displaystyle b=\pm1,\pm2.$ Checking these cases gives the solutions. If $b=1$ , then $\displaystyle 3a^2-1=2,$ so $a^2=1$ . Then $\displaystyle X=a^2+b^2=2,$ and $\displaystyle Y=a^3-3ab^2=\pm2.$ This gives $\displaystyle (X,Y)=(2,\pm2).$ If $b=-2$ , then $\displaystyle -2(3a^2-4)=2,$ so $3a^2-4=-1$ , hence $a^2=1$ . Then $\displaystyle X=a^2+b^2=1+4=5,$ and $\displaystyle Y=a^3-3ab^2=a(1-12)=-11a.$ Thus we get

$\displaystyle (X,Y)=(5,\pm11).$

The other cases do not give integer solutions. Therefore the integer solutions are

$\displaystyle (2,\pm2), (5,\pm11).$

Example 5: $Y^2=X^3-5$

Finally consider $\displaystyle Y^2=X^3-5.$ We show there are no integer solutions. If $X$ is even, then $X^3\equiv0\pmod8$ , so $\displaystyle Y^2=X^3-5\equiv -5\equiv3\pmod8,$ which is impossible because squares modulo $8$ are $0,1,4$ . Thus $X$ is odd. Then $Y$ is even. Write $Y=2Z$ . Then $\displaystyle 4Z^2=X^3-5.$ Add $4$ : $\displaystyle 4(Z^2+1)=X^3-1.$ Factor: $\displaystyle X^3-1=(X-1)(X^2+X+1).$ Thus $\displaystyle 4(Z^2+1)=(X-1)(X^2+X+1).$ Since $X^2+X+1$ is odd, the factor $4$ must divide $X-1$ : $\displaystyle 4\mid X-1.$ On the other hand, the prime divisors of the odd factor $X^2+X+1$ are forced, through the equation above, to divide a sum of two squares. Hence they are congruent to $1\pmod4$ , so $\displaystyle X^2+X+1\equiv1\pmod4.$ For odd $X$ , this forces $\displaystyle X\equiv -1\pmod4,$ or $\displaystyle 4\mid X+1.$

Again we get both $4\mid X-1$ and $4\mid X+1$ , impossible. Therefore there are no integer solutions.

Rational points of infinite order

The integral-point examples above solve some equations completely over $\mathbb Z$ , but rational points are a different question. The group $E_k(\mathbb Q)$ can contain infinitely many rational points even if there are only finitely many integer points.

For example, $\displaystyle E_{-2}: Y^2=X^3-2$ has the integer points $\displaystyle (3,\pm5).$ The torsion classification says that $E_{-2}(\mathbb Q)_{\mathrm{tors}}=0$ , because $-2$ is neither a square nor a cube, and it is not $-432$ . Thus $(3,5)$ is not torsion. In fact, it has infinite order, and one can show using descent that $\displaystyle E_{-2}(\mathbb Q)\simeq \mathbb Z,$ generated by $(3,5)$ . So the rank is $1$ .

On the other hand, $\displaystyle E_1: Y^2=X^3+1$ has torsion subgroup $\displaystyle E_1(\mathbb Q)_{\mathrm{tors}}\simeq\mathbb Z/6\mathbb Z,$ generated by a point such as $(2,3)$ . In this case the rank is zero, so the rational points are exactly the torsion points:

$\displaystyle E_1(\mathbb Q)=\{\mathcal O,(-1,0),(0,\pm1),(2,\pm3)\}.$

So two curves that look very similar can behave quite differently. The curve $Y^2=X^3+1$ has only finitely many rational points, all torsion. The curve $Y^2=X^3-2$ has no torsion but has infinitely many rational points generated by one point of infinite order.

The Mordell curves $\displaystyle E_k: Y^2=X^3+k$ are among the simplest elliptic curves, but they already show the main features of the arithmetic of elliptic curves. The torsion subgroup is quite rigid. After removing sixth powers from $k$ , the rational torsion is usually zero. It can only be $\mathbb Z/2\mathbb Z$ , $\mathbb Z/3\mathbb Z$ , or $\mathbb Z/6\mathbb Z$ , with the special cases described above. The proof combines Nagell–Lutz, reduction modulo primes, and explicit $2$ – and $3$ -torsion calculations. Integral points are finite in number by Siegel’s theorem, but finding them requires more work. In small examples, unique factorization in quadratic rings such as $\mathbb Z[i]$ and $\mathbb Z[\sqrt{-2}]$ can solve the equation completely. Rational points of infinite order are harder still. They are governed by the rank of the elliptic curve. The torsion subgroup can be computed explicitly in these Mordell cases, but the free part usually requires descent or deeper methods. This is why even the simple-looking equation $\displaystyle Y^2=X^3+k$ opens the door to the full arithmetic theory of elliptic curves.

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