An integral binary quadratic form is a quadratic form (q.v.) in two variables over $\mathbb{Z}$, i.e. a polynomial

 $F(x,y)=ax^{2}+bxy+cy^{2},a,b,c\in\mathbb{Z}$

$F$ is said to be primitive if its coefficients are relatively prime, i.e. $\gcd(a,b,c)=1$, and is said to represent an integer $n$ if there are $r,s\in\mathbb{Z}$ such that $F(r,s)=n$. If $\gcd(r,s)=1$, $F$ is said to represent $n$ properly. The theory of integral binary quadratic forms was developed by Gauss, Lagrange, and Legendre.

In what follows, “form” means “integral binary quadratic form”.

Following the article on quadratic forms, two such forms $F(x,y)$ and $G(x,y)$ are equivalent if there is a matrix $M\in GL(2,\mathbb{Z})$ such that

 $G(x,y)=F(M(x,y)^{T})$

Matrices in $GL(2,\mathbb{Z})$ are matrices with determinant $\pm 1$. So if $\alpha,\beta,\gamma,\delta\in\mathbb{Z}$ and

 $det\left(\begin{array}[]{cc}\alpha&\gamma\\ \beta&\delta\end{array}\right)=\pm 1$

then if

 $G(x,y)=F(\alpha x+\beta y,\gamma x+\delta y)$

it follows that $G$ is equivalent to $F$. If $M\in SL(2,\mathbb{Z})$ (i.e. $\det M=1$), we say that $F$ and $G$ are properly equivalent, written $F\sim G$; otherwise, they are improperly equivalent.

Note that while both equivalence and proper equivalence are equivalence relations, improper equivalence is not. For if $F$ is improperly equivalent to $G$ and $G$ is improperly equivalent to $H$, then the product of the transformation matrices has determinant $1$, so that $F$ is properly equivalent to $H$. Since proper equivalence is an equivalence relation, we will say that two forms are in the same class if they are properly equivalent.

$GL(2,\mathbb{Z})$ is generated as a multiplicative group by the two matrices

 $\begin{pmatrix}1&0\\ 1&1\end{pmatrix},\quad\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$

so in particular we see that we can construct all equivalence transformations by composing the following three transformations:

Transformation Matrix Determinant
$(x,y)\mapsto(y,x)$ $\left(\begin{array}[]{cc}0&1\\ 1&0\end{array}\right)$ $-1$
$(x,y)\mapsto(y,-x)$ $\left(\begin{array}[]{cc}0&-1\\ 1&0\end{array}\right)$ $1$
$(x,y)\mapsto(x+dy,y)$ $\left(\begin{array}[]{cc}1&0\\ d&1\end{array}\right)$ $1$

Example: Let $F(x,y)=x^{2}+xy+6y^{2}$, $G(x,y)=82x^{2}+51xy+8y^{2}$. Then

 $G(x,y)=F(\alpha x+\beta y,\gamma x+\delta y)=F(4x+y,3x+y)$

so

 $\left(\begin{array}[]{cc}\alpha&\gamma\\ \beta&\delta\end{array}\right)=\left(\begin{array}[]{cc}4&3\\ 1&1\end{array}\right)$

The transformations to map $F$ into $G$ are

 $\begin{array}[]{rl}x^{2}+xy+6y^{2}&(x,y)\mapsto(x+y,y)\\ x^{2}+3xy+8y^{2}&(x,y)\mapsto(y,x)\\ 8x^{2}+3xy+y^{2}&(x,y)\mapsto(x+3y,y)\\ 8x^{2}+51xy+82y^{2}&(x,y)\mapsto(y,x)\\ 82x^{2}+51xy+8y^{2}\end{array}$

and

 $\left(\begin{array}[]{cc}4&3\\ 1&1\end{array}\right)=\left(\begin{array}[]{cc}0&1\\ 1&0\end{array}\right)\left(\begin{array}[]{cc}1&0\\ 3&1\end{array}\right)\left(\begin{array}[]{cc}0&1\\ 1&0\end{array}\right)\left(\begin{array}[]{cc}1&0\\ 1&1\end{array}\right)$
Title integral binary quadratic forms IntegralBinaryQuadraticForms 2013-03-22 16:55:44 2013-03-22 16:55:44 rm50 (10146) rm50 (10146) 10 rm50 (10146) Topic msc 11E16 msc 11E12 RepresentationOfIntegersByEquivalentIntegralBinaryQuadraticForms ReducedIntegralBinaryQuadraticForms