PropositionStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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Proper equivalence preserves discriminant and primitivity of the form

Statement

Let $f$ and $g$ be properly equivalent integral binary quadratic forms. Then:

$f$ and $g$ have the same discriminant.
$f$ is primitive if and only if $g$ is primitive.

Facts & Assumptions

Given: Integral binary quadratic forms $f = (a, b, c)$ and $g = (A, B, C)$ , and a matrix $M = (\begin{array}{cc} p & q \\ r & s \end{array}) \in {S L}_{2} (Z)$ with $g = f ∣ M$ .

[F1]

Proper equivalence means $g (x, y) = f (p x + q y, r x + s y)$ for a determinant-one integer matrix (Proper equivalence of binary quadratic forms).

[F2]

The discriminant of $(u, v, w)$ is $v^{2} - 4 u w$ (The discriminant of a binary quadratic form).

[F3]

The form $(u, v, w)$ is primitive when the only integers dividing all three coefficients are $1$ and $- 1$ (Primitive binary quadratic forms).

[L1]

Integral substitution defines a right action of ${S L}_{2} (Z)$ on integral binary quadratic forms (Integral substitution defines a right action of ${S L}_{2} (Z)$ on integral binary quadratic forms).

Proof

technique · direct

1.1

Expanding $g (x, y) = f (p x + q y, r x + s y)$ gives $A = a p^{2} + b p r + c r^{2}$ , $B = 2 a p q + b (p s + q r) + 2 c r s$ , and $C = a q^{2} + b q s + c s^{2}$ . A direct simplification yields $B^{2} - 4 A C = (p s - q r)^{2} (b^{2} - 4 a c) = b^{2} - 4 a c$ , since $p s - q r = 1$ . Thus $f$ and $g$ have the same discriminant.

F1F2algebra

2.1

Suppose an integer $d$ divides $a$ , $b$ , and $c$ . Then the formulas of step 1.1 show that $d$ also divides $A$ , $B$ , and $C$ .

step 1.1algebra

3.1

Since $p s - q r = 1$ , the inverse matrix $M^{- 1} = (\begin{array}{cc} s & - q \\ - r & p \end{array})$ is integral and lies in ${S L}_{2} (Z)$ . By [L1], $f = g ∣ M^{- 1}$ , so the same argument as in step 2.1 with $M^{- 1}$ shows that every common divisor of $A$ , $B$ , and $C$ also divides $a$ , $b$ , and $c$ .

F1L1step 2.1algebra

4.1

Steps 2.1 and 3.1 show that $(a, b, c)$ and $(A, B, C)$ have exactly the same common divisors. Therefore one form is primitive exactly when the other is, by [F3].

F3step 2.1step 3.1∎

Depends on

Used by

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Sources

William Stein, Elementary Number Theory and Elliptic Curves, Proposition 9.2.8 (standard reference, not scraped)
Andrew Granville, Binary Quadratic Forms, Exercise 4.1d (standard reference, not scraped)