TITLE: Help to find cubics with square discriminant
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: The discriminant of an irreducible cubic is a square if and only if it is a cyclic cubic, that is, if and only if its Galois group is the cyclic group of order $3$. There are tabulations in the literature of all cyclic cubics with discriminant less than $n$, for various values of $n$. These can be found by web-searching the phrase "cyclic cubic" (or "abelian cubic", which amounts to the same thing, since if the Galois group isn't the cyclic group of order $3$, it must be the symmetric group on three letters, which group is not abelian).