Properly equivalent reduced forms with the same leading coefficient are equal

The leading coefficient of $g$ is $g(1,0)=f(p,r)=a{p}^2+bpr+c{r}^2$, and because $\mathrm{\mid }b\mathrm{\mid }\le a\le c$ one has $g(1,0)\ge a(p^2-\mathrm{\mid }pr\mathrm{\mid }+r^2)\ge a$. Since $g$ has leading coefficient exactly $a$, equality holds throughout.

Equality in step 1.1 forces $p^2-\mathrm{\mid }pr\mathrm{\mid }+r^2=1$, so $(\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }r\mathrm{\mid })$ is one of $(1,0)$, $(0,1)$, or $(1,1)$.

If $(\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }r\mathrm{\mid })=(1,0)$, then $r=0$ and $p=\pm 1$. The determinant condition gives $s=p$ and $M=\left(\begin{array}{cc}p& q\\ 0& p\end{array}\right)$. The transformed middle coefficient is $b^{\mathrm{\prime }}=b+2aq$ when $p=1$ and $b^{\mathrm{\prime }}=b-2aq$ when $p=-1$, so $\mathrm{\mid }{b}^{\mathrm{\prime }}\mathrm{\mid }\le a$ and $\mathrm{\mid }b\mathrm{\mid }\le a$ force $q=0$ unless $\mathrm{\mid }b\mathrm{\mid }=a$ and $b^{\mathrm{\prime }}=-b$. But the boundary rule in [F2] forbids $b^{\mathrm{\prime }}=-a$ for a reduced form, so $q=0$, hence $b^{\mathrm{\prime }}=b$ and $c^{\mathrm{\prime }}=c$.

If $(\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }r\mathrm{\mid })=(0,1)$, then $p=0$ and $r=\pm 1$. Equality in step 1.1 gives $c=a$, so reducedness of $f$ yields $0\le b\le a$. The determinant condition gives $q=-r$, so $M=\left(\begin{array}{cc}0& -r\\ r& s\end{array}\right)$, and direct substitution gives $g=(a,-b+2ars,a-brs+a{s}^2)$. Since $g$ is reduced, $\mathrm{\mid }-b+2ars\mathrm{\mid }\le a$. If $s=0$, then $g=(a,-b,a)$, and reducedness of $g$ with $a=c^{\mathrm{\prime }}=a$ forces $-b\ge 0$; together with $b\ge 0$ this gives $b=0$, hence $g=f$. If $s\mathrm{\ne }0$, then the same bound implies $2a\mathrm{\mid }s\mathrm{\mid }-b\le a$, so $\mathrm{\mid }s\mathrm{\mid }=1$ and $b=a$; because $0\le b\le a$, this means $b=a$, and the sign of $-b+2ars$ shows $rs=1$. Then $g=(a,a,a)=f$.

If $(\mathrm{\mid }p\mathrm{\mid },\mathrm{\mid }r\mathrm{\mid })=(1,1)$, equality in step 1.1 forces $c=a$ and $\mathrm{\mid }b\mathrm{\mid }=a$. By reducedness, $b=a$. Replacing $M$ by $-M$ if necessary does not change the substitution, so we may assume $(p,r)=(1,-1)$. Then $s+q=1$, and the transformed middle coefficient is $b^{\mathrm{\prime }}=a(2q-1)$. Since $\mathrm{\mid }{b}^{\mathrm{\prime }}\mathrm{\mid }\le a$, one has $q=0$ or $q=1$; reducedness excludes $b^{\mathrm{\prime }}=-a$, so $q=1$ and $b^{\mathrm{\prime }}=a$. The discriminant identity $b^{\mathrm{\prime }2}-4a{c}^{\mathrm{\prime }}=b^2-4ac$ then gives $c^{\mathrm{\prime }}=a=c$, so again $g=f$.

The three cases of step 2.1 are exhaustive, and each yields $g=f$. Therefore properly equivalent reduced forms with the same leading coefficient are equal.