Exercise 3.5.5 (Representations of $p$ by forms of discriminant $-20$)

Proof.

(a)

Let $f(x,y)=a{x}^2+\mathit{bxy}+c{y}^2$ be a reduced quadratic form of discriminant $-20$. The form $f$ is definite positive or negative, but we consider only the positive definite forms. By Theorem 3.19, $1\le a\le \sqrt{-d∕3}=\sqrt{20∕3}$, thus $a=1$ or $a=2$. By the definition of a reduced form, $|b|\le a$. Moreover, $b\equiv d\equiv 0(\mathit{mod}2)$. If $a=1$, then $b=0$, and if $a=2$, then $b\in \{-2,0,2\}$. $c$ is obtained by $b^2-4\mathit{ac}=-20$. For $a=2,b=\pm 2$, $c=3$, and for $a=2,b=0$, the equation $-20=b^2-4\mathit{ac}=-8c$ has no solution. Therefore $f=(1,0,5),f=(2,2,3)$, or $f=(2,-2,3)$. But $(2,-2,3)$ is not reduced, because $-a=b=-2$. This shows that $f_1(x,y)=x^2+5y^2$ and $f_2(x,y)=2x^2+2\mathit{xy}+3y^2$ are the only reduced quadratic forms positive definite of discriminant $-20$ (and these forms are primitive).

Moreover $2=f_2(1,0)$, but $f_1$ does not represent $2$, otherwise $2=a^2+5b^2$ for some integers $a,b$. Then $5b^2\le 2$, thus $b=0$, and $a^2=2$ has no solution. Since equivalent forms represent the same integers, $f_1$ and $f_2$ are not equivalent. Hence $H(-20)=2$.

(b)

Let $p$ be an odd prime. Then $p$ is represented by a form of discriminant $-20$ if and only if it is represented by a reduced form of discriminant $-20$, that is, if and only if $p$ is represented by $f_1$ or $f_2$. Moreover, by Corollary 3.14, $p$ is represented by a form of discriminant $-20$ if and only if $p\mid -20$ or $\left(\frac{-20}{p}\right)=1$.

Hence $p$ is represented by $x^2+5y^2$ or by $2x^2+2\mathit{xy}+3y^2$ if and only if $p=2$, $p=5$ or $\left(\frac{-20}{p}\right)=1$.

Using the law of quadratic reciprocity,

$$\begin{array}{llll}\hfill \left(\frac{-20}{p}\right)=1& ⟺\left(\frac{-5}{p}\right)=1& \hfill & \\ \hfill & ⟺\{\begin{array}{c}\left(\frac{-1}{p}\right)=\left(\frac{5}{p}\right)=1\hfill \\ \text{ or}\hfill \\ \left(\frac{-1}{p}\right)=\left(\frac{5}{p}\right)=-1\hfill \end{array}& \hfill & \\ \hfill & ⟺\{\begin{array}{c}\left(\frac{-1}{p}\right)=\left(\frac{p}{5}\right)=1\hfill \\ \text{ or}\hfill \\ \left(\frac{-1}{p}\right)=\left(\frac{p}{5}\right)=-1\hfill \end{array}& \hfill & \\ \hfill & ⟺\{\begin{array}{cc}p\equiv 1\hfill & (\mathit{mod}4)\hfill \\ p\equiv 1\hfill & (\mathit{mod}5)\hfill \end{array}\text{ or }\{\begin{array}{cc}p\equiv 1\hfill & (\mathit{mod}4)\hfill \\ p\equiv -1\hfill & (\mathit{mod}5)\hfill \end{array}& \hfill & \\ \hfill & \text{or }& \hfill & \\ \hfill & \{\begin{array}{cc}p\equiv -1\hfill & (\mathit{mod}4)\hfill \\ p\equiv 2\hfill & (\mathit{mod}5)\hfill \end{array}\text{ or }\{\begin{array}{cc}p\equiv -1\hfill & (\mathit{mod}4)\hfill \\ p\equiv -2\hfill & (\mathit{mod}5)\hfill \end{array}& \hfill & \\ \hfill & ⟺p\equiv 1,9,7,3(\mathit{mod}20).& \hfill & \end{array}$$

In conclusion, $p$ is represented by $x^2+5y^2$ or by $2x^2+2\mathit{xy}+3y^2$ if and only if $p=2$, $p=5$ or $p\equiv 1,3,7,9(\mathit{mod}20)$.

(See Problem 10 for more information.)