# Elementary properties of integral binary quadratic forms

Let $f = ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{Z}$. $D = b^2 - 4ac$ is called the discriminant of $f$. We say $f$ is positive definite if $a \gt 0$ and $D \lt 0$(cf. this question). We say $f$ is primitive if gcd$(a, b, c) = 1$.

Let $\sigma = \left( \begin{array}{ccc} p & q \\ r & s \end{array} \right)$ be an element of $GL_2(\mathbb{Z})$. This means that $p,q,r,s$ are integers and det $\sigma = \pm 1$. We denote the quadratic form $f(px + qy, rx + sy)$ by $f^{\sigma}$.

My question Is the following proposition correct? If yes, how do you prove it?

Proposition Let $f$ and $\sigma$ be as above.

1. The discriminant of $f^{\sigma}$ is the same as that of $f$.

2. If $f$ is positive definite, $f^{\sigma}$ is also so.

3. If $f$ is primitive, $f^{\sigma}$ is also so.

## 1 Answer

1) A binary quadratic form $f(x, y)=ax^2+bxy+cy^2$ can be written

$$f(x, y)=\begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} a & \frac{b}{2} \\ \frac{b}{2} & c \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$$

so $f(x, y)$ corresponds to the $2 \times 2$ matrix $M$ in the center, with discriminant $D=-4\det M$.

Note that

$$\begin{pmatrix} px+qy \\ rx+sy \end{pmatrix}=\begin{pmatrix} p & q \\ r & s \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$$

so in fact $f^{\sigma}$ corresponds to the $2 \times 2$ matrix $\sigma ^T M \sigma$, with discriminant $$D'=-4\det(\sigma ^T M \sigma)=-4\det M (\det \sigma)^2=-4 \det M=D$$

2) For $\begin{pmatrix} x \\ y \end{pmatrix} \not =\begin{pmatrix} 0 \\ 0 \end{pmatrix}$, we have $\begin{pmatrix} px+qy \\ rx+sy \end{pmatrix} \not =\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ as well. Then since $f$ is positive definite, $$f^{\sigma}(x, y)=f(px+qy, rx+sy)>0$$, so $f^{\sigma}$ is positive definite.

3)We prove the contrapositive statement. Suppose that $f^{\sigma}$ is not primitive, and the $\gcd$ of its coefficients is $d>1$. Put $f^{\sigma}(x, y)=g(x, y)d$, where $g(x, y)$ is also a binary quadratic form, then $f=(f^{\sigma})^{\sigma^{-1}}=(gd)^{\sigma^{-1}}=dg^{\sigma^{-1}}$, which is clearly not primitive.

• Beautiful. Thanks. – Makoto Kato Nov 17 '13 at 15:10