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Forms of discriminant $- 24$ need not be properly equivalent

Statement refuted

Two integral binary quadratic forms with the same discriminant need not be properly equivalent. The forms

$f = (1, 0, 6) and g = (2, 0, 3)$

both have discriminant $- 24$ , but they are not properly equivalent.

Facts & Assumptions

Given: The forms $f = (1, 0, 6)$ and $g = (2, 0, 3)$ .

[F1]

The discriminant of $(a, b, c)$ is $b^{2} - 4 a c$ (The discriminant of a binary quadratic form).

[L1]

Properly equivalent forms represent exactly the same integers (Properly equivalent binary quadratic forms represent the same integers, with primitive representations in bijection).

[L2]

A form represents $n$ when it takes the value $n$ at some integer pair (Integers represented, and primitively represented, by a binary quadratic form).

Counterexample

technique · direct

1.1F1givenalgebra

The discriminants are $0^{2} - 4 \cdot 1 \cdot 6 = - 24$ and $0^{2} - 4 \cdot 2 \cdot 3 = - 24$ .

1.2L2givenalgebra

The form $f$ represents $1$ , since $f (1, 0) = 1$ .

1.3L2givenalgebra

The form $g$ does not represent $1$ : if $g (x, y) = 1$ , then $2 x^{2} + 3 y^{2} = 1$ , but $y \neq 0$ would force the left-hand side to be at least $3$ , while $y = 0$ would give $2 x^{2} = 1$ , impossible in integers.

2.1L1step 1.2step 1.3∎

Since $f$ and $g$ do not represent the same integers, [L1] shows that they are not properly equivalent.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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, discussion after Proposition 9.2.8 (standard reference, not scraped)

Forms of discriminant −24−24 need not be properly equivalent