Re: Mordell's equation $y^2 = x^3 + k$ and perfect numbers

Question

I have already tried a somewhat exhaustive search of the literature, but couldn't find anything close to the problem that I am working on.

My question is: When does Mordell's equation

$$Y^2 = X^3 + K$$

have only finitely many solutions over the field of rational numbers, if we allow $K$ itself to be a rational number?

I've seen a "criterion" (i.e. a set of sufficient conditions) related to the class number of the (real/imaginary) quadratic field $\mathbb{Q}(\sqrt{K})$, but it is limited only to $K$ being either $1$ or $2$ modulo $4$.

(The actual "criterion" (as stated in the Japanese[?] paper that I allude to) is:

Mordell's equation $Y^2 = X^3 + K$ has finitely many solutions in $\mathbb{Q}$ if

(1) $-K$ is not of the form $3t^2 + 1$ or $3t^2 - 1$; and

(2) $K \equiv 1 \pmod 4$ or $K \equiv 2 \pmod 4$; and

(3) $3$ does not divide the class number of the (real/imaginary) quadratic field $Q(\sqrt{K})$.)

Edit: Please refer to this hyperlink for more information as to the context of the previous "criterion". These have since been refuted by Kevin Buzzard (@Kevin - thank YOU!).

Thanks to Kevin for pointing out some of the subtle errors in my original post, particularly in the third condition. I was considering the case $K > 0$ (i.e. for real quadratic fields).

Now for the context:

Let $$Y = W + Z$$ and $$X = WZ$$ where $W$ and $Z$ are defined as: $$W = I(p^k) = \frac{\sigma_{1}(p^k)}{p^k}$$ $$Z = I(m^2) = \frac{\sigma_{1}(m^2)}{m^2}$$

Let $$N = {p^k}{m^2}$$ be a perfect number. (At this point, we don't have to distinguish between even or odd $N$ because the Euclid-Euler model for perfect numbers fits both cases. That is, the Eulerian form for an odd perfect number and the Euclidean form for an even perfect number have very similar multiplicative forms.)

We "know" that the exponent $k$ allows us to distinguish between even and odd $N$ in the sense that: (1) If $k$ = 1, then $N$ is even. (2) If $k$ > 1, then $N$ is odd. hyperlink

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Perhaps the important feature that should be used to distinguish between even perfect numbers $M=2^{q-1}(2^q - 1)$ and odd perfect numbers $N=p^k m^2$ (where $2^q - 1$ and $p$ are the Mersenne and special primes, respectively) is the index of a perfect number: (1) $\gcd(2^{q-1},\sigma(2^{q-1}))=1$ (2) $\gcd(m^2,\sigma(m^2))>1$ There is a known formula for computing $\gcd(m^2,\sigma(m^2))$. For example, if $k=1$, then it is equal to $2m^2 - \sigma(m^2)$.

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Thus, a (possibly) feasible and modern approach to the OPN problem (i.e. determining nonexistence or otherwise) will be to try establishing a finiteness result first (for particular values of $K$).

In other words, checking for finiteness of OPNs amounts to checking for finiteness of solutions for Mordell's equation

$$Y^2 = X^3 + K$$

for particular values of $K$.

And you will only have to check for values of $K$ in the range $[50, 399]$ (for a total of 350 elliptic curves), per the previous answer to this MathOverflow question.

$K$ falls in that range because the sum

$$Y = W + Z$$

is known to lie in the open interval $(57/20, 3)$.

Of course, the "juicy" implication is that: If you will be able to find a condition (e.g. equation, inequality, etc.) relating $k$ to $K$ and you are also able to further show that the number of solutions to the corresponding Mordell equation $Y^2 = X^3 + K$ is finite for all such $K$, then it would follow that there are only finitely many perfect numbers (odd and even).

Disclaimer: This is a "naive" approach based on my current understanding of elliptic curve theory. I am well-aware, of course, that the rationals are dense over the real numbers. (Edit: In addition, the abundancy indices and the abundancy outlaws are both dense over the rationals.) Which is why I was kind of surprised that there is no need to assume ("strict") rationality (i.e. $K \in \mathbb{Q} \setminus \mathbb{Z}$) for $K$ when checking for finiteness of solutions to Mordell's equation.

Answer 1

First of all, if you replace $k$ by $d^6k$ you get another equation such that the corresponding sets of rational solutions are in bijection. So, you might as well assume that $k$ is an integer. I don't think there is a simple, crisp criterion for the equation to have finitely many solutions. Birch-Swinnerton-Dyer predicts that this is the case if and only if the $L$-function of the elliptic curve does not vanish at $s=1$ and the "if" part is known (Coates-Wiles). There is no shortage of literature on that.

Answer 2

Here's another reason why there's unlikely to be a simple answer, unless you can find a simple answer to another question that appears to be very deep. You can try to do a descent to compute the $p$-Selmer group for some small prime $p$, and if that succeeds, you will have computed the rank. But in computing the $2$-Selmer group, you'll need to know the $3$-part of the ideal class group of $\mathbb{Q}(\sqrt{-k})$. Similarly, in computing the $3$-Selmer group, you'll need to know the $2$-part of the ideal class group of $\mathbb{Q}(\sqrt[3]{k})$. I'm not aware of any elementary conditions on $k$, for example, congruence conditions, which would let you determine the ranks of those parts of those ideal class groups. The situation is very different for the curves $E_k:y^2=x^3+kx$, since here there is a rational $2$-torsion point, so the $2$-Selmer group can be computed without working in an extension of $\mathbb{Q}$. That's why, for example, there are some special cases such as $k$ prime and congruent to 7 or 11 modulo 16 for which one knows that the Mordell-Weil rank is zero, so $E_k(\mathbb{Q})$ is finite.

Answer 3

A search for "Mordell's equation finiteness" shows:

www.math.uconn.edu/~kconrad/ross2008/mordell.pdf