LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-26

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The leading coefficient of a reduced positive-definite form is minimal in its proper-equivalence class

Statement

Let $f = (a, b, c)$ be a reduced positive-definite binary quadratic form, and let $g$ be any form properly equivalent to $f$ . Then the leading coefficient of $g$ is at least $a$ .

Facts & Assumptions

Given: A reduced positive-definite form $f = (a, b, c)$ and a matrix $M = (\begin{array}{cc} p & q \\ r & s \end{array}) \in {S L}_{2} (Z)$ such that $g = f ∣ M$ .

[F1]

A reduced positive-definite form satisfies $∣ b ∣ \leq a \leq c$ (Reduced positive-definite binary quadratic forms).

[F2]

Proper equivalence means $g (x, y) = f (p x + q y, r x + s y)$ (Proper equivalence of binary quadratic forms).

[F3]

Primitive representation means representation by a pair of coprime integers (Integers represented, and primitively represented, by a binary quadratic form).

Proof

technique · direct

1.1F2givenalgebra

Since $f$ is positive definite, $a = f (1, 0) > 0$ . Also the leading coefficient of $g$ is $g (1, 0) = f (p, r) = a p^{2} + b p r + c r^{2}$ .

1.2F2F3algebra

Because $p s - q r = 1$ , every common divisor of $p$ and $r$ divides $1$ , so $\gcd (p, r) = 1$ . Thus the integer $f (p, r)$ is primitively represented by $f$ .

2.1F1step 1.1algebra

Using $∣ b ∣ \leq a \leq c$ from [F1], we have $f (p, r) \geq a p^{2} - a ∣ p r ∣ + a r^{2} = a (p^{2} - ∣ p r ∣ + r^{2})$ .

3.1step 2.1algebra

If $r = 0$ , then $p \neq 0$ because $p s - q r = 1$ , so $p^{2} - ∣ p r ∣ + r^{2} = p^{2} \geq 1$ . If $r \neq 0$ , then $p^{2} - ∣ p r ∣ + r^{2} = (∣ p ∣ - ∣ r ∣)^{2} + ∣ p r ∣ \geq 1$ . Hence in all cases $p^{2} - ∣ p r ∣ + r^{2} \geq 1$ .

4.1step 1.1step 2.1step 3.1∎

Combining steps 1.1, 2.1, and 3.1 gives $g (1, 0) = f (p, r) \geq a$ . So the leading coefficient of $g$ is at least $a$ .

Depends on

Used by

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

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

William Stein, Elementary Number Theory and Elliptic Curves, Theorem 9.3.2 (standard reference, not scraped)
Andrew Granville, Binary Quadratic Forms, Exercise 4.1f(a) (standard reference, not scraped)