# The divisor bound for binary quadratic forms

The divisor bound asserts that for a large integer $n\in \mathbb{Z}$, the number of divisors of $n$ is at most $n^{o(1)}$ as $n\rightarrow\infty$. See here for a discussion of proofs using elementary analysis.

This MO topic asked the same question with $\mathbb{Z}$ replaced by the ring of integers in a number field.

My question is whether a similar bound holds in the case of the number of representations by a binary, positive, integral quadratic form.

More precisely, let $Q(x,y) = a x^2+ b x y + c y^2$ be a binary quadratic form with $a,b,c$ integers and discriminant $\Delta = b^2 - 4 a c <0$.

For any positive integer $n$, let $r_Q(n)$ be the number of integer solutions $(x,y)$ of the equation $$n = Q(x,y).$$

Question: For $n$ large, is it known that $r_Q(n) = n^{o(1)}$?

When $\Delta$ is a fundamental discriminant, then one can use the theory of quadratic fields and the divisor bounds therein to deduce the same for our case. When $\Delta$ is a general discriminant, the cases seem tricky (especially when the associated fundamental discriminant $\Delta_0$ is $1 \mod 4$).
From D.A.Buell, Binary Quadratic Forms, Section 4.4, it seems that the classical theory of quadratic forms doesn't impose any condition on $\Delta$. In fact it seems to proceed along the lines of factoring $n$ into prime factors, getting at most $2$ classes of forms which represent each prime (depending on $\Delta$, of course) and composing them to obtain the classes of forms representing $n$. Then, within each class, it is up to the modular group to generate more solutions (the orbit of $(\pm 1,0)$ under $SL_2\mathbb{Z}$). The cardinality of this orbit seems to be bounded by an absolute constant.
Let $r_Q^*(n)$ be the number of primitive representations of $n$ by $Q$. Let $\mathcal{C}$ be a set of representatives of the classes of binary quadratic forms of discriminant $\Delta$. We can assume that $Q\in\mathcal{C}$. It is known that $\sum_{Q'\in\mathcal{C}}r_{Q'}^*(n)$ equals the number of residue classes $b\pmod{2n}$ such that $b^2\equiv\Delta\pmod{4n}$, see e.g. Page 66 in Zagier: Zetafunktionen und quadratische Körper (Springer 1981). By the Chinese Remainder Theorem, we can calculate this as a product of local densities, in fact this is an instance of Siegel's mass formula. Estimating crudely we get $$\sum_{Q'\in\mathcal{C}}r_{Q'}^*(n) \ll_\Delta 2^{\omega(n)},$$ where $\omega(n)$ is the number of prime factors of $n$, whence $r_Q^*(n)\ll_{\Delta,\epsilon} n^\epsilon$ readily follows. Finally, $$r_Q(n)=\sum_{d^2\mid n}r_Q^*(n/d^2) \ll_{\Delta,\epsilon} n^\epsilon.$$
• @Krishnan: I don't think there is an English translation, but the quoted result should be available in many textbooks. Also, I think it is very simple to show that $r_Q^*(n)$ is at most the number of solutions of the congruence, you don't need a reference for that. – GH from MO Apr 26 '13 at 13:07