Solving the Diophantine equation $4x^3 + 27y^2 - z^2 = 0$

Question

This question is inspired by trying to find cubic polynomials whose discriminant is the negative of a perfect square.

Essentially the polynomial $t^3 + pt + q$ has discriminant $\Delta = -4p^3 - 27q^2$. So if $\Delta = -z^2$ then (after renaming the variables $p$ and $q$ to $x$ and $y$ respectively), we get the equation:

$$4x^3 + 27y^2 - z^2 = 0$$

I'm looking for integer solution of this equation, but I don't know how to tackle it.

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I know this equation does have non-trivial solutions, because $(-3n^2,4n^3,18n^3)$ is a solution for any $n \in \mathbb{Z}$. But can someone help me with generating more?

Answer 1

if $$ X= 2 x^3+ 162xy^2 $$ $$ Y = 6x^2 y + 54 y^3 $$ $$ Z = x^2 - 27 y^2, $$ then $$ X^2 - 27 Y^2 = 4 Z^3 $$

I see, I have permuted the variable names compared with the question. Sigh

Starting over...

A bit more generally,

if $$ X= x^3 -3Cxy^2 - BCy^3 $$ $$ Y = 3x^2 y +3Bxy^2 + (B^2-C) y^3 $$ $$ Z = x^2 + Bxy +C y^2, $$ then $$ X^2 +BXY+C Y^2 = Z^3 $$