Exercise 3.7.4

Proof. Assume that $\mathrm{\Delta }$ is a fundamental discriminant. Then $\mathrm{\Delta }$ is a discriminant, thus $\mathrm{\Delta }\equiv 0,1(\mathit{mod}4)$.

$\text{∙}$

Assume first that $\mathrm{\Delta }\equiv 1(\mathit{mod}4)$. Reasoning by contradiction, assume that $\mathrm{\Delta }$ is not square free. Then $\mathrm{\Delta }=d^{2}m$, where $d>1$. Since $\mathrm{\Delta }$ is odd, $d$ is odd, thus $d^2\equiv 1(\mathit{mod}4)$, and this implies $m\equiv 1(\mathit{mod}4)$. Therefore $m$ is a discriminant, of the principal form $f=\left(1,1,\frac{1-m}{4}\right)$. Thus $\mathrm{\Delta }$ is the discriminant of the form $\mathit{df}=\left(d,d,d\frac{1-m}{4}\right)$, which is not primitive. By exercise 3.7.3, $\mathrm{\Delta }$ is not a fundamental discriminant. This contradiction proves that $\mathrm{\Delta }$ is square free.

$\text{∙}$

Now assume that $\mathrm{\Delta }\equiv 0(\mathit{mod}4)$. First, we show that $\mathrm{\Delta }∕4\equiv 2,3(\mathit{mod}4)$.

If $\mathrm{\Delta }∕4\equiv 0(\mathit{mod}4)$, then $\mathrm{\Delta }=4^{2}m,m\in \mathbb{Z}$, where $4m$ is the discriminant of the form $f=(1,0,-m)$. Therefore $\mathrm{\Delta }$ is the discriminant of the form $4f=(4,0,-4m)$, which is not primitive. By Exercise 3.7.3, this is a contradiction, because $\mathrm{\Delta }$ is a fundamental discriminant.

If $\mathrm{\Delta }∕4\equiv 1(\mathit{mod}4)$, then $\mathrm{\Delta }=4m$, where $m\equiv 1(\mathit{mod}4)$. Therefore $m$ is the discriminant of the form $f=\left(1,1,\frac{1-m}{4}\right)$, and so $\mathrm{\Delta }$ is the discriminant of the form $2f=\left(2,2,2\frac{1-m}{4}\right)$. This proves that $\mathrm{\Delta }$ is not a fundamental discriminant.

The only possibilities are $\mathrm{\Delta }∕4\equiv 2,3(\mathit{mod}4)$. Now we prove that $\mathrm{\Delta }∕4$ is square free. If not, $\mathrm{\Delta }=4d^{2}m,d>1$, where $4m$ is the discriminant of the form $f=(1,0,-m)$, so $\mathrm{\Delta }$ is the discriminant of the form $\mathit{df}=(d,0,-\mathit{dm})$, which is not primitive. This is impossible, since $\mathrm{\Delta }$ is a fundamental discriminant.

Conversely, assume that (1) or (2) is true.

$\text{∙}$

First, assume that $\mathrm{\Delta }\equiv 1(\mathit{mod}4)$, and that $\mathrm{\Delta }$ is square free. Since $\mathrm{\Delta }$ is square free, the conductor of $\mathrm{\Delta }$ is 1, thus $\mathrm{\Delta }$ is a fundamental discriminant.

$\text{∙}$

Now, assume that $\mathrm{\Delta }\equiv 0(\mathit{mod}4),\mathrm{\Delta }∕4\equiv 2,3(mod4)$, and $\mathrm{\Delta }∕4$ is square free.

Reasoning by contradiction, suppose that the conductor $f$ of $\mathrm{\Delta }$ is such that $f>1$. Then $\mathrm{\Delta }∕f^2$ is the discriminant of some form $g=(a,b,c)$, and $\mathrm{\Delta }=f^2(b^2-4\mathit{ac})$.

If $f$ is even, then $f=2f^{\prime }$, thus $\mathrm{\Delta }∕4={f{}^{\prime }}^2(b^2-4\mathit{ac})\equiv {(f^{\prime }b)}^2\equiv 0,1(\mathit{mod}4)$, and this contradicts the hypothesis $\mathrm{\Delta }∕4\equiv 2,3(\mathit{mod}4)$. Therefore $f$ is odd, thus $b^2-4\mathit{ac}$ is even, so $b$ is even. Using $b=2b^{\prime }$, for some integer $b^{\prime }$, we obtain $\mathrm{\Delta }∕4=f^2({b{}^{\prime }}^2-\mathit{ac})$, where $f>1$, so $\mathrm{\Delta }∕4$ is not square free, in contradiction with the hypothesis. This proves that $\mathrm{\Delta }$ is a fundamental discriminant.