This is about the frequency of integral solutions to $$ b^2 - 4 a^3 = \Delta, $$ when $\Delta < 0$ is a discriminant of positive binary quadratic forms such that the class number is divisible by 3. My observation is that integral solutions are pretty frequent when $|\Delta|$ is small. For instance, there are 25 discriminants with class number exactly 3. Of these, in 19 cases we can write the non-principal, order three form as $$ \langle a, \pm b, a^2 \rangle, $$ the failures seeming to be $$ \Delta = -124, \; -172, \; -307, \; -547, \; -652, \; 907,$$ with the first two proved impossible in integers, maybe all six.

Note that there may be multiple solutions for a given $\Delta,$ which usually means the forms reduce to the same thing or the principal form, as in $$ \langle 2, 1, 4 \rangle = \langle 4, 15, 16 \rangle $$ but $$ \langle 10, 63, 100 \rangle = \langle 1, 1, 8 \rangle. $$ On the other hand, in the relatively rare event of 3-rank larger than one, we may have genuinely distinct cube roots of 1, and for $\Delta = -3299$ we find
$$ \langle 11,45, 121 \rangle = \langle 11, 1, 75 \rangle, $$ $$ \langle 15, 101, 225 \rangle = \langle 15, 11, 57 \rangle, $$
$$ \langle 23, 213, 529 \rangle = \langle 23, -17, 39 \rangle. $$

So, the question is, how likely is it that a positive form of order three in the class group can be written as $$ f(x,y) = a x^2 + b x y + a^2 y^2 ? $$

EDIT: of course this becomes Mordell's equation when $\Delta$ is even, which is how I know there are proofs for $-124, -172$ but a rational solution $a = 1177/36, \; b = 40355/108$ for $-172.$ Hope I got the factor of 2 in the right place.

EDIT TOOOOO: forgot. by Dirichlet's "united forms" description of composition, $$ \langle a,b,a^2 \rangle^2 = \langle a^2, b,a \rangle, $$ so such a form is either the principal form or of order three, as long as $$ \gcd(a,b) = 1. $$

EDIT THREE, Saturday: The importance of this for my other recent questions appeared when I found that $ \langle 4,1,9 \rangle$ integrally represents 77, but $ \langle 4,1,9 \rangle + \langle 4,1,9 \rangle,$ or $g(x,y,z,w) = 4 x^2 + x y + 9 y^2 + 4 z^2 + z w + 9 w^2,$ does not represent $77^2 = 5929.$ Now, $ \langle 4,1,9 \rangle$ has order five, so the set of values it represents is not required to be multiplicative. Similarly, if $f$ is of order 2 "ambiguous," it seems always true that $f$ represents some $n$ (often prime) while $f(x,y) + f(z,w)$ fails to represent $n^2.$ Note that the original form in question 88905 is of order three, indeed $$ \langle 4,2, 3 \rangle = \langle 3,-2,4 \rangle = \langle 3,-8,9 \rangle. $$ So three seems to be the order of the day.


1 Answer 1


If you treat $b^2-4a^3=\Delta$ as an elliptic curve, say $E_\Delta$, then of course there are only finitely many integer points on $E_\Delta$ (effectively by Baker). But since you're interested in the number of solutions, then possibly it's useful to know that under the assumption that $\Delta$ is 6'th power free, there exists an absolute constant $C$ such that $$\#E_\Delta(\mathbb{Z})\le C^{1+\hbox{rank }E_\Delta(\mathbb{Q})}.$$ [J. Reine angew. Math 378 (1987), 60-100.] Actually, you can probably relax the 6'th power free assumption on $\Delta$ if you restrict to integer solutions such that $\gcd(a^2,b^3)$ is 6'th power free.

  • $\begingroup$ Thanks, Joe. I realized at some point that there were various possible subquestions, and that I may not have hit quite the right note. As $n \rightarrow -\infty,$ both the count of possible values of $0 \geq b^2 - 4 a^3 \geq n$ and the number of discriminants with class number divisible by 3 with $0 \geq \Delta \geq n$ go to infinity. What I had in mind was, given class number divisible by 3, is there a nonzero likelihood that there is any solution at all to $b^2 - 4 a^3 = \Delta?$ I admit, this does not end things, for example you may just be rewriting the principal form... $\endgroup$
    – Will Jagy
    Mar 4, 2012 at 0:31
  • $\begingroup$ @W.Jagy: On average one expects zero integral solutions as $|\Delta| \rightarrow \infty$, and I think this can be proved under the ABC conjecture, or just Hall's conjecture; more precisely, there should be only $O(n^{5/6 + \epsilon})$ solutions of $|b^2 - 4a^3| \leq n$ (and it's easy to show the number of solutions is $\gg n^{5/6}$). $\endgroup$ Mar 4, 2012 at 2:42

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