### Example 15.3.1.

What is that solution? (Even if you don’t remember, you should be able to find it quickly.)

Let’s start by talking about \(x^3=y^2+2\) as a type of curve. Recall from Historical remark 3.5.3 that Bachet de Méziriac first asserted this had one positive integer solution in 1621, very early in the development of modern number theory.

What is that solution? (Even if you don’t remember, you should be able to find it quickly.)

Recall also that Fermat, Wallis, and Euler also studied this equation and gave various discussions and proofs of the uniqueness of its solution. As we first saw in Section 3.5, this equation is actually one of a more general class of equations called the *Mordell* equation:

\begin{equation*}
x^3=y^2+k\; , \;\; k\in\mathbb{Z}\text{.}
\end{equation*}

Louis Mordell was an early 20th-century American-born British mathematician. He proved some remarkable theorems about this class of equations. We have already seen that these are nontrivial, and that some have no solution (Proposition 7.6.3, or see below Fact 15.3.3). Even deciding whether there are no solutions or not turns out to be quite tricky; Helmut Richter has a somewhat old website^{ 4 }

with some tables of what *is* known about integer solutions.

`hr.userweb.mwn.de/numb/mordell.html`

Notice that Mordell’s set of curves are not quadratic/conic, but rather a set of cubic curves. Actually, as mentioned before, they are examples of a special type of elliptic curves, which makes them more mysterious (and, as it happens, more useful for cryptography – we allude to this briefly in Subsection 11.5.1).

One of Mordell’s remarkable theorems states that, for a given \(k\text{,}\) the equation can only have *finitely many* integer points (in fact, there are even useful bounds for how many that depend only on the prime factorization of \(k\)). At the same time, Mordell curves are apparently “simple” enough that they can still have infinitely many *rational* points (see Theorem 15.3.6). Gerd Faltings won a Fields Medal for proving that higher-degree curves cannot have infinitely many rational points. If you are online, see which points you can find in the interact below.

Proving things about Mordell’s equation is quite tricky, but once in a while there is something you can do. For instance, we can verify something we can see in the interact above.

There are no integer solutions to \(x^3=y^2-7\text{.}\)

Recall that we nearly finished the proof of this in Proposition 7.6.3! We had reduced to showing that

\begin{equation*}
y^2+1= (x+2)(x^2-2x+4)
\end{equation*}

was impossible if no prime of the form \(p=4n+3\) could divide \(y^2+1\text{.}\)

This is not possible, because Fact 13.3.2 implies there are no square roots of \(-1\) modulo \(p\) for this type of \(p\text{.}\)

Fact 15.3.3 is a simple version of the following far more general statement.

If the following hold:

- \(M\equiv 2\text{ (mod }4)\text{,}\)
- \(N\equiv 1\text{ (mod }2)\text{,}\) and
- all prime divisors \(p\) of \(N\) are of the form \(4k+1\text{.}\)

Then there is no solution to

\begin{equation*}
x^3=y^2-(M^3-N^2)\text{.}
\end{equation*}

The proof basically follows the same outline as Proposition 7.6.3 with Fact 15.3.3. See Exercise 15.7.8.

One can prove lots of similar statements using only congruence considerations^{ 5 }

. The previous theorem is [E.4.9, Theorem 14.1.2], and that text has several other interesting variants. See Conrad’s notes^{ 6 }

and [E.2.8, Theorem 7.4C.1] for even more special cases. See Subsection 17.5.4 for some other examples (without proof) of how knowing when square roots exist helps solve Mordell equations.

As one might expect, with more power more can be done. See [E.2.16, Section 11.6] or [E.4.9, Section 14.2] for results using the class number from Remark 13.3.4.

`www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf`

But there is a larger point to make, based on the very specific conditions on \(M\) and \(N\text{.}\) Namely, if we want to prove anything about such equations with methods we *currently* have access to in this text, we have no hope of getting any interesting *general* results.

Let’s see what I mean by “no hope” here by returning to Bachet’s original equation, \(x^3=y^2+2\text{.}\) What are some naive things we can say?

- It should be clear that \(x\) and \(y\) must have the same parity.
- If they are both even then \(x^3\) is divisible by 4, but \(y^2+2\equiv 2\text{ (mod }4)\text{,}\) which is impossible.
- So \(x\) and \(y\) are both odd.

That doesn’t really narrow things down much.

Now, Euler *nearly* proves the following fact.

The only positive solution to the Bachet equation is \(x=3,y=5\text{.}\)

Proving this is already a little sophisticated, and is closely connected to the use of complex numbers in Section 14.1. Here we will give the idea behind Euler’s ‘proof’.

In examining \(a^2+b^2\text{,}\) we factored it as \((a+bi)(a-bi)\) using a square root of negative 1 (relative to \(\mathbb{Z}\)). Similarly, we would like to factor the \(y^2+2\text{.}\) But it can’t be done in \(\mathbb{Z}[i]\text{.}\)

Instead, we could try to use the square root of \(-2\text{,}\) and define

\begin{equation*}
\mathbb{Z}[\sqrt{-2}]=\{a+b\sqrt{-2} \mid a,b\in\mathbb{Z}\}
\end{equation*}

Then

\begin{equation*}
x^3=\left(y-\sqrt{-2}\right)\left(y+\sqrt{-2}\right)
\end{equation*}

We haven’t done anything with cubes yet …

Here is the tricky bit. In the integers, if \(x^3=pq\) and \(\gcd(p,q)=1\text{,}\) then \(p\) and \(q\) must both be perfect (integer) cubes. So Euler *assumes this works* in \(\mathbb{Z}[\sqrt{-2}]\) as well, and that the factors of \(y^2+2\) are “coprime” (whatever that means in this new number system). (A very nice discussion of this is in [E.4.14], including a full proof in its appendix.)

Then some basic algebraic manipulation of

\begin{equation*}
y-\sqrt{-2}=\left(a+b\sqrt{-2}\right)^3
\end{equation*}

and divisibility considerations end up showing that \(b \mid 1\) and \(a=\pm b\text{,}\) which ends up implying \(y=\pm 5\) and \(x=3\text{.}\) (We will not take this further; see Exercise 15.7.10.)

Where’s the problem? It turns out you *can* say that if a product of coprime numbers is a cube, then the factors are cubes in this situation; however, it requires some (geometrically motivated) proof, just like with \(\mathbb{Z}[i]\text{.}\) In his 1765 “Vollständige Anleitung zur Algebra”, sections 187-188 and 191, Euler explicitly says that this just works – in *any* number system with \(\mathbb{Z}[\sqrt{c}]\text{.}\) He solves the original Bachet equation in section 193, and solves \(x^3=y^2+4\) using the same technique in section 192, without realizing he had not proved this implicit assumption. (This is the same assumption he tacitly made in examining Fermat’s Last Theorem for the case \(n=3\text{.}\))

But we shouldn’t be too hard on Euler! He was one of the first people to even consider some essentially random new number system obtained by adjoining \(\sqrt{c}\) (for some integer \(c\)) to the integers. And as noted in Example 3.5.4, in 1738 he gave a correct *and full* proof of the observation that \(8\) and \(9\) is the only time a perfect square is preceded by a perfect cube, which is Mordell’s equation for \(k=-1\text{.}\) (See also Question 3.5.5.)

If you are interested in more information about how to prove cases of Mordell’s equation, there are many good resources, including a nice one on Keith Conrad’s website^{ 7 }

. But even finding a bound on the *size* of solutions to Mordell’s equation for a given \(k\) is tricky.

`www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf`

- Mordell, Siegel, and Thue all had a part after World War I in showing there are finitely many solutions for a given \(k\text{,}\) but said nothing about how big \(x\) and \(y\) might be.
- An early bound on the size of the numbers was that\begin{equation*} |x| < e^{10^{10}|k|^{10^4}} \end{equation*}which is of course ridiculously huge.
- More recent conjectures are that \(x\) has absolute value less than \(e^C |k|^{2+\epsilon}\text{,}\) where \(\epsilon\) is as small as you want and \(C\) seems to pretty close to one, probably less than two.

We cannot close discussion of this topic without a final very famous result carrying Mordell’s name. Recall that these curves can have infinitely many *rational* points, even if they have finitely many (or zero) integer points. The following is a bit of a surprise, then; the rational points can still be *described* finitely.

Essentially, the set of (rational) points on a Mordell curve is a combination of finitely many “cyclic” (recall Fact 14.2.7) groups (in a very specific way I will not describe), and so it can be described using finitely many of the rational points.

If you like, the number of rational points might be infinite, but not *too* infinite.