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. 
 A: 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.
A: 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.
A: A search for "Mordell's equation finiteness" shows:
www.math.uconn.edu/~kconrad/ross2008/mordell.pdf
