# Re: Mordell's Equation $y^2 = x^3 + k$ and Perfect Numbers

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) $K \equiv 1 (mod 4)$ or $K \equiv 2 (mod 4)$; 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. For more details regarding this, please refer to this link.)

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. (Again, refer to the link for more details. There is also a related MathOverflow post here.)

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 kinda surprised that there is NO need to assume ("strict") rationality (i.e. $K \in \mathbb{Q}$ but not in $\mathbb{Z}$) for $K$ when checking for finiteness of solutions to Mordell's equation.

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@Arnie: Diophantine equations are "hard". There will be other criteria, some unconditional, some conditional on the truth of some parts of the Birch--Swinnerton-Dyer conjecture. There may well be some easy criteria when $k$ is prime; one can do the descent and computer Selmer ranks sometimes. But the bottom line is that all these equivalent, or conditionally equivalent, criteria, for general $k$, will involve mathematics that you cannot explain to a high-school kid. You can get as fancy as you like but the bottom line is that you can ask a computer algebra package for the rank of the... – Kevin Buzzard Dec 4 '10 at 8:21
...curve for any given explicit small k and you'll get an answer very quickly. You ask "when does the equation have finitely many solutions"---but this isn't really a mathematics question because you don't give a criterion which makes any given answer acceptable to you: e.g. clearly "it has finitely many solutions iff it has finitely many solutions" is not an acceptable answer, even though it's logically correct. How about "it has finitely many solns if the L-function doesn't vanish at 1". Is that any good? It sounds useless in practice! Why not just compute the solutions! – Kevin Buzzard Dec 4 '10 at 8:26
Arnie: you are living in a dream world. You cannot expect to find a wonderful simple criterion for this equation to have finitely many solutions. There are plenty of methods for computing class numbers of real quadratic fields. There are also plenty of methods for computing ranks of elliptic curves such as the one in your question. You are asking these questions but it's hard to believe that the answers can help you. Say for example that I just tell you some explicit formula whose answer is the class group---such formulas exist. What will you do now? – Kevin Buzzard Dec 4 '10 at 12:26
@Arnie: you're barking up the wrong tree. I don't care what the class number is. $7$ isn't $1$ or $2$ mod 4, so (2) fails, so "(1) and (2) and (3)" fails regardless of the truth status of (3). So according to the "criterion" you're claiming, the curve should have infinitely many points. And my computer says it doesn't. Let me again stress that I am a bit skeptical about a genuine provable "iff" criterion of this form. – Kevin Buzzard Dec 5 '10 at 13:13
Arnie: please stop using MO as a place to announce your work – Yemon Choi Apr 13 '11 at 6:45

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.

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@Felipe, thanks a lot! On a side note, BSD is one of the "deepest" conjectures in number theory (i.e. Millenium Problem) but I am actually working on something which is somewhat "unrelated" -- and it's the OPN conjecture. – Arnie Bebita Dris Dec 4 '10 at 4:08
You might as well also assume that $k \not\equiv 0 \bmod 27$, because the curves $y^2 = x^3 + k$ and $y^2=x^3−3^{-3}k$ are isogenous and thus have the same rank. – Noam D. Elkies Jun 10 '11 at 2:39
Thanks Noam! Any references in the literature for that? (My apologies, I am still in the process of self-studying elliptic curve theory...) – Arnie Bebita Dris Jun 12 '11 at 21:28
@Arnie: if you ever need some non trivial consequence of BSD conjecture in your research, you might be interested in mathoverflow.net/questions/149815/… where I give a few ideas about the weak version thereof (equality of the two kinds of ranks). Hope it helps. – Sylvain JULIEN Apr 5 '15 at 19:11
Thanks @SylvainJULIEN, will check that out! =) – Arnie Bebita Dris Apr 5 '15 at 19:16

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.

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