Exercise 3.5.15* (Fundamental discriminants)

Proof. :

• Case $d\equiv 1(\mathit{mod}4)$.

  $\text{(⇐)}$

  If $d$ is not a fundamental discriminant, then there exists a non-primitive form $(a,b,c)$ of discriminant $d$. Thus, there exists $e>1$ such that $a=\mathit{eA},b=\mathit{eB},c=\mathit{eC},A,B,C\in \mathbb{Z}$. Then $d=e^2(B^2-4\mathit{AC}),e>1$, has a square factor.

  $\text{(⇒)}$

  Suppose that $d$ has a square factor $e^2$: $d=e^{2}m,e>1$. Then $e$ is odd, so $e^2\equiv 1(\mathit{mod}4)$. Since $d\equiv 1(\mathit{mod}4)$, then $m\equiv 1(\mathit{mod}4)$. So $m$ is a discriminant, of the principal form $f=(1,1,\frac{1-m}{4})$. Thus $d$ is the discriminant of the form $\mathit{ef}=(e,e,e\frac{1-m}{4})$, which is not primitive, and so $d$ is not a fundamental discriminant.

  If $d\equiv 1(\mathit{mod}4)$ then $d$ is a fundamental discriminant if and only if $d$ is square-free.

• Case $d\equiv 0(\mathit{mod}4)$.

  $\text{(⇐)}$

  Suppose that $d∕4\equiv 2,3(\mathit{mod}4)$, and that $d∕4$ is square-free. Reasoning by contradiction, if $d$ is not a fundamental discriminant, then as in the previous case, $d$ is the discriminant of a form $(a,b,c)$ where $a=\mathit{eA},b=\mathit{eB},c=\mathit{eC},e>1$.

  Then $d=e^2(B^2-4\mathit{AC})$. If $e$ is even, then $e=2e^{\prime }$, so $d∕4={e{}^{\prime }}^2(B^2-4\mathit{AC})\equiv {(e^{\prime }B)}^2\equiv 0$ or $1(\mathit{mod}4)$, which is contrary to the hypothesis. So $e$ is odd, which means that $B^2-4\mathit{AC}$ is even, so $B$ is even: $B=2B^{\prime }$. Thus $d∕4=e^2({B{}^{\prime }}^2-\mathit{AC})$ (where $e>1$), so $d∕4$ has a square factor, which is contrary to the hypothesis. It is thus proven that the hypotheses " $d∕4\equiv 2,3(\mathit{mod}4)$, and $d∕4$ is without a square factor" indeed imply that $d$ is a fundamental discriminant.

  $\text{(⇒)}$

  If $d∕4\equiv 0(\mathit{mod}4)$, then $d=4^{2}m,m\in \mathbb{Z}$, and $4m\equiv 0(\mathit{mod}4)$, so $4m$ is a discriminant, of the form $f=(1,0,-m)$, so $d$ is the discriminant of the form $2f=(2,0,-2m)$, which is not primitive.

  If $d∕4\equiv 1(\mathit{mod}4)$, then $d=4m,m\equiv 1(\mathit{mod}4)$, so $m$ is a discriminant, of the form $f=(1,1,\frac{1-m}{4})$. So $d$ is the discriminant of the form $2f=(2,2,\frac{1-m}{2})$, and $d$ is not a fundamental discriminant.

  This proves that a fundamental discriminant $d\equiv 0(\mathit{mod}4)$ satisfies $d∕4\equiv 2,3(\mathit{mod}4)$. It remains to prove that $d∕4$ is square-free. If this were not the case, then $d=4e^{2}m,$ where $e>1$, and $4m\equiv 0(\mathit{mod}4)$ is a discriminant, of the form $f=(1,0,-m)$, and $d$ is the discriminant of the form $\mathit{ef}=(e,0,-\mathit{em})$, which contradicts the fact that $d$ is a fundamental discriminant.

  If $d\equiv 0(\mathit{mod}4)$ then $d$ is a fundamental discriminant if and only if $d∕4$ is square-free and $d∕4\equiv 2$ or $3(\mathit{mod}4)$.