Exercise 5.19

Proof. Let $b$ such that $(b∕D)=-1$ and $b\wedge D=1$ : the existence of $b$ comes from Ex 5.18.

Let $S={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{a\in A}(a∕D)$, where $A$ is reduced residue system modulo $D$. As two reduced system modulo $D$ represent the same elements in $U(\mathbb{Z}∕\mathit{D\mathbb{Z}})$, the sum is independent of the reduced residue system $A$ : we can write

As $b\wedge D=1$, we know from Ex. 3.6 that $B=\mathit{bA}=\{\mathit{ba}|a\in A\}$ is also a reduced system modulo $D$. In other words, the application $U(\mathbb{Z}∕\mathit{D\mathbb{Z}})\to U(\mathbb{Z}∕\mathit{D\mathbb{Z}}),\bar{a}\mapsto \bar{a}\bar{b}$ is a bijection, so

As $(b∕D)=-1$, $-S=S$, so $S=0$.

Since $(a∕D)=\pm 1$, one half of the elements in $U(\mathbb{Z}∕\mathit{D\mathbb{Z}})$ satisfy $(a∕D)=1$, and one half of the elements in $U(\mathbb{Z}∕\mathit{D\mathbb{Z}})$ satisfy $(a∕D)=-1$. □