Exercise 3.5.12 (A positive semidefinite quadratic form is equivalent to $gx^2$)

Proof. If $a=0$, then $b=0$, because $b^2=4\mathit{ac}=0$. Then $f(x,y)=c{y}^2$, where $c\ge 0$ since $f$ is positive semi definite. Then $g=\gcd (a,b,c)=\gcd (0,0,c)=c$, so $f(x,y)=g{y}^2\stackrel{+}{\sim }g{x}^2$, because $(0,0,g)\cdot S=(g,0,0)$, where $S=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{array}\right)\in \Gamma$.

We suppose now that $a\ne 0$. Since $d=b^2-4\mathit{ac}=0$, by equation (3.3),

where $a>0$ since $f$ is positive semidefinite. This shows that $f(-b,2a)=0$, and so $f(x_0,y_0)=0$, where $(x_0,y_0)=\left(-\frac{b}{b\wedge 2a},\frac{2a}{b\wedge 2a}\right)$ satisfies $x_0\wedge y_0=1$.

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 $h=f\cdot A$. Then

where $B$ and $C$ are integers.

Since $f$ and $h$ have the same discriminant, $B^2=0$, so $B=0$, and $h(x,y)=C{y}^2$. Moreover, by Problem 11,

thus $h(x,y)=g{y}^2$.

Finally, $(0,0,g)\cdot S=(g,0,0)$, thus $f$ is properly equivalent to the form $g{x}^2$, where $g=\gcd (a,b,c)$. □