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

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Properly equivalent binary quadratic forms represent the same integers, with primitive representations in bijection

Statement

Let $f$ and $g$ be integral binary quadratic forms. If $f$ and $g$ are properly equivalent, then:

$f$ and $g$ represent exactly the same integers.
Primitive representations correspond bijectively: for each integer $n$ , a pair $(x, y)$ with $\gcd (x, y) = 1$ satisfies $g (x, y) = n$ if and only if the pair $(p x + q y, r x + s y)$ has relatively prime coordinates and satisfies $f (p x + q y, r x + s y) = n$ , where $g = f ∣ (\begin{array}{cc} p & q \\ r & s \end{array})$ .

Facts & Assumptions

Given: Integral binary quadratic forms $f$ and $g$ , an integer $n$ , 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 all integers $x, y$ (Proper equivalence of binary quadratic forms).

[F2]

The form $h$ represents $n$ when $n = h (u, v)$ for some integers $u, v$ , and it primitively represents $n$ when moreover $\gcd (u, v) = 1$ (Integers represented, and primitively represented, by a binary quadratic form).

[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

If $g (x, y) = n$ , then $n = f (p x + q y, r x + s y)$ by [F1], so every representation of $n$ by $g$ yields a representation of $n$ by $f$ .

F1F2

2.1

Since $p s - q r = 1$ , the inverse matrix is $M^{- 1} = (\begin{array}{cc} s & - q \\ - r & p \end{array}) \in {S L}_{2} (Z)$ . By [L1], $(f ∣ M) ∣ M^{- 1} = f$ , so $f = g ∣ M^{- 1}$ . Applying step 1.1 to $g$ and $M^{- 1}$ gives the converse implication. Therefore $f$ and $g$ represent exactly the same integers.

F1L1step 1.1algebra

3.1

Let $u = p x + q y$ and $v = r x + s y$ . If $\gcd (x, y) = 1$ and an integer $d$ divides both $u$ and $v$ , then $d$ divides $s u - q v = (p s - q r) x = x$ and $- r u + p v = (p s - q r) y = y$ , so $\gcd (u, v) = 1$ . The same argument with $M^{- 1}$ gives the converse, so the correspondence of steps 1.1 and 2.1 restricts to a bijection on primitive representations.

F2step 2.1algebra∎

Depends on

Used by

Forms of discriminant -24 need not be properly equivalent Counterexample
Odd primes congruent to 1 or 3 modulo 8 are represented by x²+2y² Example

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, Proposition 9.2.4 (standard reference, not scraped)
Andrew Granville, Binary Quadratic Forms, Exercise 4.1d (standard reference, not scraped)