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!).

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.)

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.

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!).

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

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 $2m^2 - \sigma(m^2)$.

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.

(2) If $k$ > 1, then $N$ is odd. (Again, refer to the link for more details. There is also a related MathOverflow post here.)

(2) If $k$ > 1, then $N$ is odd. (Again, refer to the link for more details. There is also a related MathOverflow post here.)