<s>[INST] Help to find cubics with square discriminant If the discriminant $b^2-4c$ of the quadratic $x^2 + bx + c$ is a square then it factors. For every discriminant $d^2$ we have can parametrize them all $(b,c) = (d + 2 h,h(d+h))$. edit I realized now that the quadratic case is trivial because it the discriminant is a square iff it factors, so the two variable parametrization is $(x-a)(x-b)$ so it might not represent what is happening with the cubic.
I was hoping to do a similar parametrization for the cubics $x^3 - ax + b$ with square discriminant $4a^3 - 27b^2 = d^2$ but factoring in the Eisenstein integers does not seem to make the problem any easier.
Are there any other promising approaches I could try?

I noticed the problem is simple when $d=0$. In that case we have $a = 3 m^2$, $b = 2 m^3$. Also when $d = b$ we also get a simple parametrization, $a = 7m^2$, $b = 7m^3$ but I don't think these will help to get the general case.
 [/INST] I would not focus on parametrization as such. Instead, note that $4 a^3 = d^2 + 27 b^2.$ This gives an easily stated restriction on the prime factorization of $a.$ It is necessary and sufficient that $a \geq 0$ and, whenever any prime $p | a$ and $p \equiv 2 \pmod 3,$ then the exponent of $p$ must be even. So $a = 2$ or $a = 5$ or $a=10$ are impossible. Without the cube, there would be a restriction on the prime 3 as well, but it turns out not to matter because you have $a^3.$ Without the $4$ in $4a^3,$ there would be a competition between $d^2 + 27 b^2$ and the other forms in the genus, $4 u^2 \pm 2 u v + 7 v^2.$ But
$$ 
\left(  \begin{array}{cc}
  4 & 0 \\
  1 & 1  
\end{array} 
  \right) 
\; \cdot \; 
\left(  \begin{array}{cc}
  1 & 0 \\
  0 & 27  
\end{array} 
  \right) 
\; \cdot \;
\left(  \begin{array}{cc}
  4 & 1 \\
  0 &  1  
\end{array} 
  \right) =  \; \;\; \; 
 \left(  \begin{array}{cc}
  16 & 4 \\
  4 & 28  
\end{array} 
  \right)  
 $$
Now, try $a = 4 = 2^2,$ so the exponent on 2 is even, you get $4a^3 = 256,$ and you get $d = 16, b = 0,$ which seems too easy. i will put some more below, maybe skip squares... But, given a legal $a$ as described, the number of pairs $d,b$ is finite. 
$$a = 1,4a^3 = 4, ( d = \pm 2, b = 0), $$
$$a = 3, 4a^3 = 108, ( d = \pm 9, b = \pm 1), ( d = 0, b = \pm 2), $$
 $$a = 7, 4a^3 = 1372, ( d = \pm 20, b = \pm 6), ( d = \pm 7, b = \pm 7), $$
  $$a = 12, 4a^3 = 6912, ( d = \pm 72, b = \pm 8), ( d = 0, b = \pm 16), $$
 $$a = 13, 4a^3 = 8788, ( d = \pm 70, b = \pm 12), ( d = \pm 65, b = \pm 13). $$
Well, given all possible representations $a = s^2 + s t + t^2,$ one may construct all $d,b,$ an annoying task unless $a$ is prime with $a \equiv 1 \pmod 3.$ This is, essentially, what factoring in the Eisenstein integers will give you.
Actually, that last bit was needlessly pessimistic. With a little special treatment of the primes 2,3, I can see how to create all possible representations of $a = j^2 + 3 k^2,$ create all possible representations of $4a^3 = m^2 + 3 n^2,$ then just keep the ones when $3 | n.$ Rather involved but easy enough to program. 
</s>