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I am currently working with positive-definite, reduced, primitive, integral binary quadratic forms, and I have noticed something interesting.

Conjecture: Let $Q$ be a form of non-fundamental discriminant $\Delta$. Let $K=\mathbb{Q}(\sqrt{\Delta})$ be the associated quadratic field of discriminant $\Delta_K$ (which we note is fundamental). Let $R=\{m: m=Q(x,y), x,y\in\mathbb{Z}\}$, the set of integers represented by $Q$. Then $R\subseteq R_K$, where $R_K=\{m: m=Q_K(x,y), x,y\in\mathbb{Z}\}$ for a fixed form $Q_K$ of fundamental discriminant $\Delta_K$.

Example: Let $Q(x,y)=x^2+8y^2$. Then $\Delta=-32$. Let $R=\{m: m=Q(x,y), x,y\in\mathbb{Z}\}$. Then $R\subseteq\{m: m=x^2+2y^2, x,y\in\mathbb{Z}\}$.

This seems like a very basic fact about fundamental and non-fundamental discriminants, but I cannot find it anywhere in the literature. The only thing that has been suggested to me is the term maximal quadratic form, but I can't find that either. Is this a known theorem?

  • https://math.stackexchange.com/questions/1513733/solving-a-diophantine-equation-of-the-form-x2-ay2-byz-cz2-with-the-co – individ Feb 10 '19 at 14:46

1 Answers1

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tldr: characters.

I'm not sure I understand your remarks about quadratic fields. $\Delta$ is not a fundamental discriminant, so $4\mid\Delta$. But for a quadratic field $K=\mathbb{Q}(\sqrt\Delta)$, $\Delta$ must be square-free.

Following you I restrict to positive-definite reduced primitive integral binary quadratic forms.

We have the general result that, for any integer $m$ and discriminant $\Delta$, there is some quadratic form of discriminant $\Delta$ representing $m$ iff $\Delta$ is a square modulo $4m$.[1]

Your example is a bit trivial because $\langle 1,0,2\rangle=x^2+2y^2$ is the only quadratic form of discriminant $-8$, so if any quadratic form of discriminant $-8$ represents some integer $m$, that one does.

For any non-fundamental discriminant $\Delta$ define the corresponding fundamental discriminant $\Delta_K$ as that discriminant $\Delta/k^2$ with largest $k$. ($\Delta_K$ is fundamental by maximality of $k$.)

So let's consider instead the non-fundamental discriminant $\Delta=-80$. The corresponding fundamental discriminant is $\Delta_K=-20=\Delta/2^2$. So, for any $m$, if either $-20$ or $-80$ is a square modulo $m$, both are. So there is some quadratic form of discriminant $-20$ that represents $m$ iff there is some quadratic form of discriminant $-80$ that represents $m$. The question is: which one.

There are 2 quadratic forms of discriminant $-20$: $\langle 1,0,5\rangle$ and $\langle 2,2,3\rangle$. There are 3 of discriminant $-80$: $\langle 1,0,20\rangle$, $\langle 4,0,5\rangle$ and $\langle 3,2,7\rangle$.

If I understand your question correctly, then, applying it to my example, it is: given $Q$ of discriminant $-80$, how come all $m$ represented by $Q$ are represented by the same $Q_K$ of discriminant $-20$?

For any discriminant $\Delta$ (e.g. $\Delta=-80$), there are certain invariants of the quadratic forms of discriminant $\Delta$: characters.[2] Any $m, m'$ both represented by some quadratic forms of discriminant $\Delta$ are represented by the same one iff for each such character $\chi$ the respective values of $\chi$ for $m$ and $m'$ are equal. Crucially, each such character relates to which residue classes modulo some factor of $\Delta$ are represented. $\Delta_K\mid \Delta$ and so every character of $\Delta_K$ is a character of $\Delta$. So any such $m, m'$ which agree in all of $\Delta$'s characters agree in all of $\Delta_K$'s, and are thus represented by the same quadratic form of discriminant $\Delta_K$.

For example, for discriminants $-20$ and $-80$ alike, the characters are for $m\bmod 4$ and $m\bmod 5$. If $Q$ represents $m$, with either $Q=\langle 1,0,20\rangle$ or $Q=\langle 4,0,5\rangle$, then (because of $Q$'s values of $-80$'s characters) $m$ is a square both modulo 4 and modulo 5; this means that $Q_K=\langle 1,0,5\rangle$ represents $m$ because of $Q_K$'s values of $-20$'s characters. By contrast, if $Q=\langle 3,2,7\rangle$ represents $m$, then $m$ is 0 or a non-square both modulo 4 and modulo 5; this means that $Q_K=\langle 2,2,3\rangle$ represents $m$.


[1] See for example Andrew Granville. Binary quadratic forms, Prop. 4.1.

[2] See for example Gordon Pall. Arithmetic invariants of quadratic forms. Bulletin of the American Mathematical Society, vol.51, number 3 (March 1945), pp.185-197.

Rosie F
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