Let's see what I mean by “no hope” here by returning to Bachet's original equation, \(y^3=x^2+2\). 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 \(y^3\) is divisible by 4 but \(x^2+2\equiv 2\text{ mod }(4)\), which is impossible.
• So \(x\) and \(y\) are both odd.
• That doesn't really narrow things down too much, really.

Now, Euler nearly proves that \(x=5,y=3\) is the only positive solution. This is already a little sophisticated, and is closely connected to the use of complex numbers in the previous chapter.

• In examining \(a^2+b^2\), we factored it as \((a+bi)(a-bi)\) for a square root of negative 1.
• Just as there, we try to factor the \(x^2+2\). But it can't be done in \(\mathbb{Z}[i]\).
• So we instead use the square root of \(-2\), and define \[\mathbb{Z}[\sqrt{-2}]=\{a+b\sqrt{-2} \mid a,b\in\mathbb{Z}\}\]
• Then \[y^3=(x-\sqrt{-2})(x+\sqrt{-2})\] And now what do we do? We haven't done anything with cubes yet.

Here is the tricky bit. In the integers, if \(y^3=pq\) and \(gcd(p,q)=1\), 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 \(x^2+2\) are “coprime” (whatever that means in this new number system).

Then some basic algebraic manipulation of \[x-\sqrt{-2}=\left(a+b\sqrt{-2}\right)^3\] and and divisibility considerations end up showing that \(b \mid 1\) and \(a=\pm b\), which ends up showing \(x=\pm 5\) and \(y=3\). (We will skip these details, since we will not take this further.)

Where's the problem? It turns out you can say that a product which is a cube is a product of cubes in this situation, but it requires some (geometrically motivated) proof, just like with \(\mathbb{Z}[i]\). In his 1765 "Vollständige Anleitung zur Algebra," sections 187-188 and 191, Euler explicitly says that this just works – in any number system with \(\mathbb{Z}[\sqrt{c}]\). He solves this one in section 193, and solves \(x^2+4=y^3\) using the same technique in section 192, without realizing the problem.

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 of this type. And, in 1738, he gives 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 \(k=-1\).

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 here.

In case you are wondering, even finding a bound on the size of solutions to Mordell's equation for a given \(k\) is tricky.

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

Finally, we have to mention a very famous result, called Mordell's Theorem. Recall that these curves can have infinitely many rational points, even if they have finitely many (or zero) integer points. Nonetheless, the rational points on such a curve can be described finitely, using group theory. Essentially, the set of points is a combination of finitely many “cyclic” groups (in a very specific way I will not describe), and so it can be described using finitely many of the rational points. So basically, the rational points might be infinite, but not too infinite.