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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.

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By considering polynomials of the form $x^3-3ax+2b$ (assuming you're working over the rationals) the discriminant takes the nice form $\delta^2=b^2-a^3$ – Riley E May 4 '11 at 15:40
@quanta: Actually I had a typo at first, so there's really a cube in there – Riley E May 4 '11 at 15:42
This is a good idea! We are left with $$(\delta-b)(\delta+b)=a^3$$ which may be possible to work with. – quanta May 4 '11 at 15:43
I think that $\text{gcd}(\delta-b,\delta+b)= (1\text{ or }2) \cdot \text{gcd}(\delta,b)$. – quanta May 4 '11 at 15:44
@Quanta: Yeah. It's $1$ if $b,\delta$ have different parity (odd/even) and $2$ if it's the same. When working with Diophantine equations, you can simplify things a bit by considering the "nontrivial" solutions: no variable is zero and all are relatively coprime. All other solutions can then be constructed from those. – Riley E May 4 '11 at 15:55
up vote 3 down vote accepted

An equation like $4a^3−27b^2=d^2$ for fixed $d$ defines a curve in the plane with coordinates $a,b$. It can be parametrised by rational functions if and only if it is singular, which is the case in your examples $d = 0$ and $d = b$. If not, it can be completed to an elliptic curve, which can not be parametrised by rational functions.

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doetoe, That is really interesting! Thanks for this result. I wonder if it would be possible to characterize all the $d$ that make this form singular? – quanta May 4 '11 at 22:22
Which I think would be $6a^2 = d \frac{\mathrm{d}d}{\mathrm{d}a} $, $27b = -d \frac{\mathrm{d}d}{\mathrm{d}b} $. – quanta May 4 '11 at 22:27

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.

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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).

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If you factor in the Eisenstein integers, you can get a parametrization though it's a bit messy.

Write $b + 3\sqrt{-3}d = b + 3d(2j+1) = (b+3d)+(3d)j = 2u$, so you can write $a^3 = u \bar{u}$. Since $u \bar{u}$ must be a cube, $u$ has to factor into bunches of product of primes of the form $p^3$ or $p^2 \bar{p}$, so you can find Eisenstein integers $v$ and $w$ such that $u = v^3w^2\bar{w}$ (though this decomposition is generally not unique)

This parametrization gives you $a = vw\bar{v}\bar{w}$ and $(b+3d)+(3d)j = 2v^3w^2\bar{w}$. If you want to get integer parameters, then you will have 4 parameters, and in order to get the equations for $b$ and $d$ you have to expand the product $v^3w^2\bar{w}$ to get its integer components, and then solve the system for $b$ and $d$.

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