Exercise 3.7.2

Question

Determine ${\cal F}(\Delta, 5)$ for all discriminants $\Delta$ with $|\Delta| \leq 10$.

Richard Ganaye · Accepted Answer

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\begin{proof} 
The discriminants $\Delta $ with $|\Delta| \leq 10$ are 
$$ -8, -7, -4, -3, 1, 4, 5, 8, 9.$$
(By the definition of section 3.3, $0$ is not a discriminant.)

Following Proposition 3.1.1, we search all $b \in \Z$ such that 
$$b^2 \equiv \Delta \pmod {20},  \qquad -5< b \leq 5.$$
The square numbers modulo $20$ for $-5< b \leq 5$ are given in the following array:
$$
\begin{array}{c|ccccccccc}
b & 0 & \pm 1 &\pm 2 & \pm 3 & \pm 4 & 5\
\hline
b^2 & 0 & 1 & 4 &9 & 16 & 5\
\end{array}
$$
\begin{enumerate}
\item[$\bullet$] If $\Delta  \in \{ -8,-7, -3, 8\}$, $\Delta$ is not the square modulo $20$ of some $b$ such that $-5<b\leq 5$, therefore
$$ {\cal F}(-8, 5) = {\cal F}(-7, 5) = {\cal F}(-3, 5) = {\cal F}(8, 5) =\varnothing.$$
\item[$\bullet$] If $\Delta =-4$, $\Delta \equiv 4^2 \pmod 20$, thus $b = \pm 4$, and $c =(b^2 + 4)/4 = 5$. There are two $\Gamma$-orbits in ${\cal F}(-4, 5)$:
$${\cal F}(-4, 5) = \{(5,4,5)\Gamma, (5,-4,5)\Gamma\}.$$
\item[$\bullet$] If $\Delta =1$, then $b = \pm 1$, $c = 0$.
$${\cal F}(1, 5) = \{(5,1,0) \Gamma, (5,-1,0)\Gamma\}.$$
\item[$\bullet$] If $\Delta =4$, $b = \pm 2$, $c = 0$.
$${\cal F}(4, 5) = \{(5,2,0) \Gamma, (5,2,0)\Gamma\}.$$
\item[$\bullet$] If $\Delta =5$, $b = 5$, $c = 1$.
$${\cal F}(5, 5) = \{(5,5,1) \Gamma\}.$$
\item[$\bullet$] If $\Delta =9$, $b = \pm 3$, $c = 0$
$${\cal F}(9, 5) = \{(5,3,0) \Gamma, (5,-3,0) \Gamma\}.$$
\end{enumerate}
\end{proof}

Exercise 3.7.2

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Exercise 3.7.2

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