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The reduced primitive forms of discriminant $- 4$

Example

The only reduced primitive positive-definite binary quadratic form of discriminant $- 4$ is

$x^{2} + y^{2} = (1, 0, 1) .$

Consequently $h (- 4) = 1$ .

Facts & Assumptions

Given: A reduced primitive positive-definite form $(a, b, c)$ of discriminant $- 4$ .

[L1]

A reduced form of discriminant $Δ$ satisfies $a \leq \sqrt{∣ Δ ∣ / 3}$ (A reduced positive-definite form of discriminant $Δ$ satisfies $a \leq \sqrt{∣ Δ ∣ / 3}$ ).

[L2]

Each proper-equivalence class contains exactly one reduced form (Each proper-equivalence class of positive-definite integral binary quadratic forms contains exactly one reduced form).

[L3]

The class number $h (Δ)$ counts proper-equivalence classes of primitive positive-definite forms of discriminant $Δ$ (The class number of primitive positive-definite binary quadratic forms of discriminant $Δ$ ).

Verification

technique · direct

1.1L1givenalgebra

Here $a \leq \sqrt{4 / 3} < 2$ by [L1], so the positive integer $a$ must be $1$ .

2.1step 1.1algebra

The discriminant equation gives $- 4 = b^{2} - 4 c$ , so $4 c = b^{2} + 4$ . Since the form is reduced, $∣ b ∣ \leq 1$ ; the choices $b = \pm 1$ make $c = 5 / 4$ , not an integer, while $b = 0$ gives $c = 1$ . Thus the only reduced possibility is $(1, 0, 1)$ .

3.1L2L3step 2.1∎

The form $(1, 0, 1)$ is primitive, so by [L2] there is exactly one proper-equivalence class of primitive positive-definite forms of discriminant $- 4$ . Hence [L3] gives $h (- 4) = 1$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

William Stein, Elementary Number Theory and Elliptic Curves, Section 9.4 (standard reference, not scraped)

The reduced primitive forms of discriminant −4−4