# Distribution of 'square classes' of binary quadratic forms

Let $$f$$ be a binary quadratic form with integer coefficients and non-zero discriminant $$D$$. Suppose for simplicity that $$D$$ is a fundamental discriminant (which in particular implies that $$f$$ is primitive). We say that $$f$$ represents a square mod $$D$$ if for any $$a$$ representable by $$f$$ we have

$$a \equiv \square \pmod{p}$$

for each prime $$p | D$$. Note that this condition only depends on the $$\text{GL}_2(\mathbb{Z})$$-equivalence class of $$f$$. We can thus say that the $$\text{GL}_2(\mathbb{Z})$$-class of $$f$$, $$[f]$$, is a square class if for a prime $$q$$ representable by $$f$$ and co-prime to $$D$$ satisfies $$q \equiv \square \pmod{p}$$ for $$p | D$$, and if $$D \equiv 0 \pmod{4}$$ (respectively $$8$$), then $$q \equiv 1 \pmod{4}$$ (respectively $$8$$). By our earlier discussion, the choice of $$q$$ representable by $$f$$ does not matter.

How are 'square classes' distributed in the ideal class group of $$\mathcal{O} = \mathcal{O}_{\mathbb{Q}(\sqrt{D})}$$? Heuristically, for each (odd) prime $$p | D$$, half of the classes of the ideal class group should be square mod $$p$$ and the other half should not, so the number of square classes ought to be $$h_2(D) 2^{-\omega(D)}$$ where $$h_2(D)$$ is the class number of $$\mathcal{O}$$ and $$\omega(n)$$ denotes the number of prime divisors of $$n$$.

• Are you expecting an answer going beyond genus theory? Or is the principal genus theorem (Gauss, DA 1801) the answer? – Franz Lemmermeyer Oct 20 '18 at 9:12
• @FranzLemmermeyer yes, it seems that 'square classes' are precisely the classes in the principal genus. If you want to write that as an answer I will accept it. – Stanley Yao Xiao Oct 26 '18 at 0:59

The standard definition for a form to belong to the principal genus of forms with fundamental discriminant $$d$$ is that the primes $$p$$ coprime to $$d$$ that the form $$Q$$ represents satisfy $$(d_1/p) = \ldots = (d_t/p)$$, where $$d =d_1 \cdots d_t$$ is the factorization of $$d$$ into prime discriminants. In particular, $$Q$$ is allowed to represent primes $$p \equiv \pm 1 \bmod 8$$ if $$d_1 = 8$$ occurs in the factorization, and primes $$p \equiv 1, 3 \bmod 8$$ if $$d_1 = -8$$ occurs (for example, $$40 = 5 \cdot 8$$ and $$-40 = 5 \cdot (-8)$$ are factorizations into prime discriminants, $$40 = -5 \cdot (-8)$$ is not since $$-5$$ is not a discriminant).
Gauss's principal genus theorem states that a form is in the principal genus if and only if the equivalence classof $$Q$$ is a square in the class group. The genus of a form is the sign vector $$((d_1/p), \ldots, (d_t/p))$$, and there are forms for every sign vectors whose entries have product $$+1$$.