State/Binary quadratic forms are uniquely determined by their values at a superbasis

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Proposition
Proof

Proposition: (Binary quadratic forms are uniquely determined by their values at a superbasis) Suppose that $\{ \pm \vec e, \pm \vec f, \pm \vec g \}$ is a lax superbasis. Suppose that $\alpha, \beta, \gamma \in \ZZ$. Then, there exists a unique integer-valued binary quadratic form $Q$, such that: $$Q(\vec e) = \alpha, Q(\vec f) = \beta, Q(\vec g) = \gamma.$$

Logical Connections

This statement logically relies on the following definitions and statements: Def/Lax superbasis, Def/Binary quadratic form, State/Arithmetic progression rule for binary quadratic forms

The following statements and definitions rely on the material of this page: State/The domain topograph has no circuits

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State/Binary quadratic forms are uniquely determined by their values at a superbasis

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	Suppose that $\{ \pm \vec e, \pm \vec f, \pm \vec g \}$ is a lax superbasis. Suppose that $\alpha, \beta, \gamma \in \ZZ$.
	By following a path from the lax superbasis $\{ \pm \vec e, \pm \vec f, \pm \vec g \}$, to the lax superbasis $\{ \pm (1,0), \pm (0,1), \pm (1,1) \}$, one may find unique integers $u,v,w \in \ZZ$, such that: If $Q$ is a binary quadratic form, then the following conditions are equivalent: $Q(1,0) = u$, $Q(0,1) = v$, and $Q(1,1) = w$. $Q(\vec e) = \alpha$, $Q(\vec f) = \beta$, $Q(\vec g) = \gamma$. Indeed, the arithmetic progression rule for binary quadratic forms allows us to determine the values of a quadratic form $Q$ at one lax superbasis from the values of $Q$ at any other lax superbasis. Hence, it suffices to find a quadratic form $Q$ such that $Q(1,0) = u$, $Q(0,1) = v$, and $Q(1,1) = w$. Solving a system of three linear equations in three variables, or a simple guess, yields a unique such quadratic form: $$Q(x,y) = ux^2 + (w-u-v) xy + v y^2.$$
	Hence there is a unique quadratic form $Q$ satisfying $Q(\vec e) = \alpha$, $Q(\vec f) = \beta$, and $Q(\vec g) = \gamma$.