Let $f(x,y) = ax^2 + bxy - cy^2$ be an indefinite, irreducible, and primitive binary quadratic form. That is, we have $\gcd(a,b,c) = 1$ and $\Delta(f) = b^2 - 4ac > 0$ and not equal to a square integer.

Written in this way, it is clear that $f$ represents $a$ primitively, since $f(1,0) = a$. Suppose that $f$ also represents $-a^3$ primitively. What can be said about the algebraic properties of $f$? In particular, what can be said about the element in the narrow class group of $\mathbb{Q}(\sqrt{\Delta(f)})$ represented by $f$?

  • $\begingroup$ any particular examples in mind? The only thing that comes to mind is that when a form represents both $1$ and $-1,$ all forms inherit the $\pm$ and the discriminant is greatly restricted. I may have once proved that when any primitive form represents both some $w$ and $-w,$ then the principal form represents both $1,-1.$ If so, i don't recall how it went. $\endgroup$ – Will Jagy Jun 17 '19 at 23:03
  • $\begingroup$ @Will Jagy: It's a good thing you don't recall the proof since there are counterexamples! The principal form $x^2-34y^2$ takes the values $15$ at $(x,y)=(7,1)$ and $-15$ at $(11,2)$ but it does not take the value $-1$, as one can see from its topograph. $\endgroup$ – Allen Hatcher Jun 19 '19 at 15:34
  • $\begingroup$ @AllenHatcher thank you. I knew about $34$ and $-1,$ it did not occur to me to just find values of small absolute value and compare $\pm$ lists. Similar for $205$ and $221,$ but never for $pq$ with primes $p \equiv q \equiv 1 \pmod 4$ with $(p|q) = -1,$ then there is always an $x^2 - pq y^2 = -1.$ $\endgroup$ – Will Jagy Jun 19 '19 at 17:08

This question can be answered by general theory, at least when $\Delta$ is a fundamental discriminant (so it's not a square times a smaller discriminant). Assuming this, the general procedure for finding the forms of discriminant $\Delta$ that represent a number $n>1$ primitively is the following. Let the prime factorization of $n$ be $n=p_1^{e_1}\cdots p_k^{e_k}$ for distinct primes $p_i$. If $n$ is represented primitively by some form of discriminant $\Delta$ then each $p_i$ will also be represented (primitively). Let $Q_i$ be a form representing $p_i$. This is unique up to equivalence, giving an element in the narrow class group $CG(\Delta)$ that I will also call $Q_i$. Thus in $CG(\Delta)$ the two elements representing $p_i$ are $Q_i$ and $Q_i^{-1}$ (which coincide when $Q_i$ is symmetric, or "ambiguous" in Gauss's terminology). The statement is then that the elements of $CG(\Delta)$ representing $n$ primitively are precisely the products $Q_1^{\pm e_1}\cdots Q_k^{\pm e_k}$, with the restriction that $e_i$ must be $1$ for each $p_i$ dividing $\Delta$. The $2^k$ choices of signs for the exponents give $2^k$ potentially different elements of $CG(\Delta)$ representing $n$ primitively, though often the number is smaller due to relations that hold in $CG(\Delta)$ and to some forms $Q_i$ being symmetric.

For representing negative numbers $n$ when $\Delta>0$ one applies the preceding process to find the forms representing $|n|$ and then one takes the negatives of these forms to get the forms representing $n$.

Thus for the question of finding the forms representing $-a^3$ primitively one first applies the preceding remarks to find all the forms representing $a$ primitively, then one takes the cubes of these forms in $CG(\Delta)$ to get all the forms representing $a^3$ primitively, and finally one takes the negatives of these forms to get the forms representing $-a^3$ primitively. The condition that $e_i=1$ for primes $p_i$ dividing $\Delta$ means that $a$ must be relatively prime to $\Delta$ in order to have $-a^3$ represented primitively.

As an example consider discriminant $\Delta =136=8\cdot 17$. Here the class group is cyclic of order $4$ generated by the nonsymmetric form $Q=3x^2+2xy-11y^2$. In $CG(\Delta)$ we have $Q^2=34x^2-y^2$, $Q^3=Q^{-1}=3x^2-2xy-11y^2$, and $Q^4=x^2-34y^2$, the principal form giving the identity element of $CG(\Delta)$. It is not hard to see that $Q$ is equivalent to $-Q$, and even properly equivalent to it, say by drawing a small portion of its Conway topograph which has a 180 degree rotational "skew-symmetry" interchanging its positive and negative values. Thus if we take $a=3$, then $3$ is represented by $Q$ since $Q(1,0)=3$ and hence, passing to $CG(\Delta)$, $27$ is represented primitively only by $Q^{\pm3}=Q^{\pm1}$, hence $-27$ is represented primitively only by $-Q^{\pm3}=Q^{\pm1}$. All these forms are equivalent to $Q$. Explicitly we have $Q(7,4)=27$ and $Q(4,3)=-27$.

If we take $a=9$, this is represented primitively only by $Q^2=34x^2-y^2$, when $(x,y)=(1,5)$, so $9^3$ is represented primitively only by $Q^6=Q^2$ and $-9^3$ is represented primitively only by $-Q^2=x^2-34y^2$.

If we take $a=15=3\cdot 5$, then since $3$ and $5$ are both represented by $Q$, $15$ is represented primitively only by $Q^{\pm1}Q^{\pm1}$ with independent choices of signs, so $15$ is represented primitively only by $Q^0=x^2-34y^2$ (when $(x,y)=(7,1)$) and by $Q^2=34x^2-y^2$ (when $(x,y)=(2,11)$). So $15^3$ is represented primitively only by $Q^0$ and $Q^6=Q^2$, and $-15^3$ only by their negatives $Q^2$ and $Q^0$.

It is worth noting that $Q$ is equivalent to its negative but $x^2-34y^2$ is not equivalent to its negative $-x^2+34y^2$.

When $\Delta$ is not a fundamental discriminant, say $\Delta=d^2\Delta'$ with $\Delta'$ a fundamental discriminant, there are extra complications, although if one restricts attention to representing numbers relatively prime to $d$ then the theory described above still works.

[Added the next day.] The "wild conjecture" in Will Jagy's comment is true if by equivalent one means properly equivalent, i.e., equivalent by an element of $SL(2,Z)$. Thus a form is properly equivalent to its negative if and only if it is equivalent to a (reduced) form $ax^2+bxy-ay^2$. This is pretty easy to see from topographs. The topograph of a form of positive nonsquare discriminant has a periodic line separating the positive and negative values (Conway's "river") and the only way a form can be equivalent to its negative is for the topograph to have a "skew-symmetry" taking each value $n$ of the form to $-n$ and hence taking the separator line to itself. There are two types of such skew-symmetries: (a) a glide-reflection along the separator line interchanging positive and negative values, or (b) a 180 rotation about a point of the separator line switching its two ends and interchanging positive and negative values of the form. A type (a) skew-symmetry is an improper equivalence, in $GL(2,Z)-SL(2,Z)$, while a type (b) skew-symmetry is a proper equivalence in $SL(2,Z)$. Type (b) skew symmetries occur exactly at midpoints of edges in the separator line with values $a,-a$ on either side of the edge, so these correspond exactly to forms $ax^2+bxy-ay^2$. These forms are reduced, as is clear from the topograph since reduced forms occur exactly where the edges leading off the separator line switch from one side of the separator line to the other.

On the other hand glide-reflection skew-symmetries seem harder to detect. For the principal form they exist exactly when the form represents $-1$, i.e., when the fundamental unit has negative norm, and this is known to be hard to predict. In terms of topographs this is more of a global condition, not local as for rotational skew-symmetries.

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  • $\begingroup$ This is a good answer and almost answers my question. I guess the last remaining thing is to understand when is a form equivalent to its negative. $\endgroup$ – Stanley Yao Xiao Jun 19 '19 at 17:41
  • $\begingroup$ @StanleyYaoXiao just a suggestion for now, Gauss and Lagrange dealt with "reduced" indefinite forms $ax^2 + b xy + c y^2,$ reduced is equivalent to $ac < 0$ and $b > |a+c|,$ proof in Franz's book. Each form has a finite set of reduced representatives. Forms $\langle a,b,-a \rangle$ are equivalent to their negatives. For example, $Q$ above reduces to $\langle 3,10,-3 \rangle \; . \;$ So, wild conjecture, a form is equivalent to its negative if and only if it has a reduced representative $\langle a,b,-a \rangle.$ This does require that the discriminant be the sum of two squares...interesting $\endgroup$ – Will Jagy Jun 19 '19 at 18:52
  • $\begingroup$ for the discriminants I mentioned, 205 and 221, the principal form (represents $1$) does not also represent $-1,$ but the other genus has a pair of opposites, for 205 we have <3, 13, -3> and for 221 we have <5,11,-5> $\endgroup$ – Will Jagy Jun 19 '19 at 19:04
  • $\begingroup$ @Will Jagy: I've added somthing to my answer to explain why your conjecture is true if "equivalent" means "properly equivalent". $\endgroup$ – Allen Hatcher Jun 19 '19 at 20:01
  • $\begingroup$ This is very nice, Allen. $\endgroup$ – Will Jagy Jun 20 '19 at 2:23

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