How do you prove that rational points on $y^2 = x^3 - 2$ are of the form $(A/B^2, C/B^3)$, where are $A, B, C$ are coprime?

Question

I was only browsing this book on number theory and the author shows how the solution $(3, 5)$ can be used to generate other exotic rational solutions and then in the end leaves the problem I'm asking as an exercise. I don't have quite the background to prove this, can someone maybe come up with an elementary proof?

Answer 1

Let $x=A/p$ and $y=C/q$ in lowest terms. Cross-multiplying then gives us $$ p^3C^2 = q^2(A^3-2p^3)$$

Since $A/p$ is in lowest terms, $A$ and $p$ are coprime, and thus so are $A^3$ and $p^3$, and thus so are $p^3$ and $A^3-2p^3$. Therefore all of the prime factors in $p^3$ on the left must come from $q^2$ on the right. Similarly, $C^2$ and $q^2$ are coprime, so all of the prime factors of $p^2$ on the left must come from $A^3-2p^3$ on the right. Thus, $$ p^3 = q^2 \qquad\text{and}\qquad C^2 = A^3-2p^3 $$ Now each prime factor of this common value of $p^3$ and $q^2$ must have an exponent that is both a multiple of $3$ and of $2$ -- that is, $p^3=q^2$ is a sixth power, and the sixth root will be the desired $B$.

Finally, $A^3-C^2=2p^3$ so any common prime factor of $A$ and $C$ must divide $2p^3$. It cannot divide $p^3$ (because $A/p$ was in lowest terms), but it cannot be $2$ either, because if $A$ and $C$ were even then $A^3-C^2$ would be a multiple of $4$, which $2p^3$ isn't. So $A$ and $C$ are coprime.