Exercise 3.7.5 (Number of automorphisms)

Proof. Let $\mathrm{Aut}(f)$ denote the group of automorphisms of a form $f$. Then $w(f)=|\mathrm{Aut}(f)|$ is the order of this group. Consider two primitive positive definite quadratic forms $f=(a,b,c)$ and $g=(a^{\prime },b^{\prime },c^{\prime })$ of same discriminant $d$, so that

• If $w(f)=4$, then by Theorem 3.26, $a=c$ and $b=0$. Since $f$ is primitive and positive definite, $a=c=1$, so $f(x,y)=x^2+y^2$. Then $d=-4$. Since $H(-4)=h(-4)=1$, all forms of discriminant $-4$ are properly equivalent to $f$, so $g\stackrel{+}{\sim }f$. By Theorem 2.36,

  $$w(f)=w(g)=4.$$

• If $w(f)=6$, then the same theorem shows that $a=b=c$. Since $f$ is primitive and positive definite, $a=b=c=1$, so $f(x,y)=x^2+\mathit{xy}+y^2$, and $d=-3$. By Problem 3, $H(-3)=h(-3)=1$, so $g\stackrel{+}{\sim }f$, and by Theorem 3.26

  $$w(f)=w(g)=6.$$

• If $w(f)\ne 4$ and $w(f)\ne 6$, then $w(f)=2$. Assume, for the sake of contradiction, that $w(g)=4$ or $w(g)=6$. Then, by the first two items applied to $g$, $w(f)=w(g)\in \{4,6\}$, which is false. Therefore

  $$w(f)=w(g)=2.$$

For any given $d<0$, the primitive positive definite quadratic forms of discriminant $d$ all have the same number of automorphisms. □