Illustration of Theorem 10.11 With Modulus 8

$$\begin{array}{rlrl}{5}^1& \equiv \phantom{0}5(\mathrm{mod}8),& -5^1& \equiv 3(\mathrm{mod}8),\\ 5^2& \equiv \phantom{1}1(\mathrm{mod}8),& -5^2& \equiv 7(\mathrm{mod}8).\end{array}$$

For every odd integer $n$ there is a unique $b(n)$ such that $$n\equiv (-1{)}^{(n-1)/2}{5}^{b(n)}(\mathrm{mod}8),\text{ with }1\le b(n)\le 2.$$

$$\begin{array}{rlrlrl}n& =1:& 5^2& \equiv 1(\mathrm{mod}8),& b(1)& =2,\\ n& =3:& -5^1& \equiv 3(\mathrm{mod}8),& b(3)& =1,\\ n& =5:& 5^1& \equiv 5(\mathrm{mod}8),& b(5)& =1,\\ n& =7:& -5^2& \equiv 7(\mathrm{mod}8),& b(7)& =2.\end{array}$$

---

$$f(n)=\{\begin{array}{ll}(-1{)}^{(n-1)/2}& \text{if }n\text{ is odd},\\ 0& \text{if }n\text{ is even}.\end{array}$$

$$g(n)=\{\begin{array}{ll}{e}^{i2\pi b(n)/2}& \text{if }n\text{ is odd},\\ 0& \text{if }n\text{ is even}.\end{array}$$

$$\begin{array}{rrrrr}n& 1& 3& 5& 7\\ f(n)& (-1{)}^{(1-1)/2}& (-1{)}^{(3-1)/2}& (-1{)}^{(5-1)/2}& (-1{)}^{(7-1)/2}\\ g(n)& e^{i2\pi b(1)/2}& e^{i2\pi b(3)/2}& e^{i2\pi b(5)/2}& e^{i2\pi b(7)/2}\end{array}$$

$$\begin{array}{rrrrr}n& 1& 3& 5& 7\\ f(n)& 1& -1& 1& -1\\ g(n)& 1& -1& -1& 1\end{array}$$

---

$${\chi }_{a,c}=f(n{)}^{a}g(n{)}^c$$

$$\begin{array}{rrrrr}n& 1& 3& 5& 7\\ {\chi }_{1,1}(n)& f(1{)}^{1}g(1{)}^1& f(3{)}^{1}g(3{)}^1& f(5{)}^{1}g(5{)}^1& f(7{)}^{1}g(7{)}^1\\ {\chi }_{1,2}(n)& f(1{)}^{1}g(1{)}^2& f(3{)}^{1}g(3{)}^2& f(5{)}^{1}g(5{)}^2& f(7{)}^{1}g(7{)}^2\\ {\chi }_{2,1}(n)& f(1{)}^{2}g(1{)}^1& f(3{)}^{2}g(3{)}^1& f(5{)}^{2}g(5{)}^1& f(7{)}^{2}g(7{)}^1\\ {\chi }_{2,2}(n)& f(1{)}^{2}g(1{)}^2& f(3{)}^{2}g(3{)}^2& f(5{)}^{2}g(5{)}^2& f(7{)}^{2}g(7{)}^2\end{array}$$

$$\begin{array}{rrrrr}n& 1& 3& 5& 7\\ {\chi }_{1,1}(n)& 1& 1& -1& -1\\ {\chi }_{1,2}(n)& 1& -1& 1& -1\\ {\chi }_{2,1}(n)& 1& -1& -1& 1\\ {\chi }_{2,2}(n)& 1& 1& 1& 1\end{array}$$