Exercise 3.5.7 (There are finitely many classes if the discriminant is a positive perfect square)

Proof.

(a)

Let $f(x,y)=A{x}^2+\mathit{Bxy}+C{y}^2$ be a quadratic form whose discriminant $d\ne 0$ is a perfect square, so that $d={\delta }^2>0$ for some positive integer $\delta$. By Problem 3.4.9, there are integers $h_1,k_1,h_2,k_2$ such that $$f(x,y)=(h_{1}x+k_{1}y)(h_{2}x+k_{2}y)$$

(read again the second solution of Problem 3.4.9).

Then $(h_1,k_1)\ne (0,0)$, otherwise $f=0$ and $d=0$. Thus $h_1\wedge k_1\ne 0$, and $f(-k_1,h_1)=0$, therefore

$$f\left(-\frac{k_1}{h_1\wedge k_1},\frac{h_1}{h_1\wedge k_1}\right)=0.$$

Put $x_0=-\frac{k_1}{h_1\wedge k_1},y_0=\frac{h_1}{h_1\wedge k_1}$. Then $f(x_0,y_0)=A{x}_0^2+B{x}_0{y}_0+C{y}_0^2=0$, and $x_0\wedge y_0=1$, so that $0$ is properly represented by $f$.

By Problem 3, there are integers $u,v$ such that $A=\left(\begin{array}{cc}\hfill x_0\hfill & \hfill u\hfill \\ \hfill y_0\hfill & \hfill v\hfill \end{array}\right)\in \Gamma$. Put $g=f\cdot A$, i.e.

$$\begin{array}{llll}\hfill g(x,y)& =f(x_{0}x+\mathit{uy},y_{0}x+\mathit{vy})& \hfill & \\ \hfill & =A{(x_{0}x+\mathit{uy})}^2+B(x_{0}x+\mathit{uy})(y_{0}x+\mathit{vy})+{(y_{0}x+\mathit{vy})}^2& \hfill & \\ \hfill & =(A{x}_0^2+B{x}_0{y}_0+C{y}_0^2)x^2+\mathit{rxy}+s{y}^2,& \hfill & \end{array}$$

where $r,s$ are integers.

Since $f(x_0,y_0)=A{x}_0^2+B{x}_0{y}_0+C{y}_0^2=0$, we obtain $g(x,y)=\mathit{rxy}+s{y}^2$, that is $g=(0,r,s)$.

(Alternatively, we can apply the result of Problem 4 to $n=0$.)

For any integer $k$, $g\cdot T^k=(0,r,\mathit{rk}+s)$ (recall that $T=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$ and $(a,b,c)\cdot T^k=(a,b+2\mathit{ak},a{k}^2+\mathit{bk}+c)$).

Since $d\ne 0$, then $r\ne 0$, thus we can choose $k$ such that $0\le \mathit{rk}+s<|r|$. Put $a=\mathit{rk}+s$, and $b=-r$. For this value of $k$, $h=g\cdot T^k=(0,-b,a)$, and $0\le a<|b|$.

Finally put $q=h\cdot S$, where $S=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{array}\right)$. Then $q=(a,b,0)$, i.e.

$$q(x,y)=a{x}^2+\mathit{bxy},$$

and $q$ is properly equivalent to $f$.

In conclusion, $f$ is properly equivalent to a form $q(x,y)=a{x}^2+\mathit{bxy}+c{y}^2$ for which $c=0$ and $0\le a<|b|$.

(b)

The discriminant of $q$ is $b^2$, and $b$ is equivalent to $f$, thus $b^2=d={\delta }^2$, so $b=\pm \delta$. Moreover $0\le a<|b|=\delta$. Hence there are finitely many forms $(a,b,0)$ of discriminant $d$ satisfying $0\le a<|b|$. Since every form of discriminant $d$ is properly equivalent to such a form by part (a), there are only finitely many equivalence classes of form of this discriminant.