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The reduced primitive forms of discriminant $- 20$

Example

The reduced primitive positive-definite binary quadratic forms of discriminant $- 20$ are

$x^{2} + 5 y^{2} = (1, 0, 5), 2 x^{2} + 2 x y + 3 y^{2} = (2, 2, 3) .$

Consequently $h (- 20) = 2$ .

Facts & Assumptions

Given: A reduced primitive positive-definite form $(a, b, c)$ of discriminant $- 20$ .

[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

1.1

Here $a \leq \sqrt{20 / 3} < 3$ , so $a$ is $1$ or $2$ .

L1givenalgebra

2.1

If $a = 1$ , then $- 20 = b^{2} - 4 c$ , so $4 c = b^{2} + 20$ . Reducedness gives $∣ b ∣ \leq 1$ . The values $b = \pm 1$ make $c = 21 / 4$ , not an integer, while $b = 0$ gives $c = 5$ , yielding $(1, 0, 5)$ .

step 1.1algebra

2.2

If $a = 2$ , then $- 20 = b^{2} - 8 c$ , so $8 c = b^{2} + 20$ . Reducedness gives $∣ b ∣ \leq 2$ . The values $b = 0, \pm 1$ give no integer $c$ , while $b = \pm 2$ gives $c = 3$ ; the boundary rule forces $b = 2$ . Thus the only reduced possibility with $a = 2$ is $(2, 2, 3)$ .

step 1.1algebra

3.1

The two forms of steps 2.1 and 2.2 are primitive and distinct, and [L2] shows that no other reduced primitive form of discriminant $- 20$ exists. Hence [L3] gives $h (- 20) = 2$ .

L2L3step 2.1step 2.2∎

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 −20−20