Exercise 8.28

Proof.

(a)

$\chi (-1)={(-1)}^{(p-1)∕2}=-1$, and $\chi (p-x)=\chi (-x)=\chi (-1)\chi (x)=-\chi (x)$, thus $$\begin{array}{llll}\hfill \underset{x=1}{\overset{p-1}{\sum }}\mathit{x\chi }(x)& =\underset{x=1}{\overset{(p-1)∕2}{\sum }}\mathit{x\chi }(x)+\underset{x=(p-1)∕2+1}{\overset{p-1}{\sum }}\mathit{x\chi }(x)& \hfill & \\ \hfill & =\underset{x=1}{\overset{(p-1)∕2}{\sum }}\mathit{x\chi }(x)+\underset{y=1}{\overset{(p-1)∕2}{\sum }}(p-y)\chi (p-y)(x=p-y)& \hfill & \\ \hfill & =\underset{x=1}{\overset{(p-1)∕2}{\sum }}\mathit{x\chi }(x)-\left[p\underset{x=1}{\overset{(p-1)∕2}{\sum }}\chi (x)-\underset{x=1}{\overset{(p-1)∕2}{\sum }}\mathit{x\chi }(x)\right]& \hfill & \\ \hfill & =2\underset{x=1}{\overset{(p-1)∕2}{\sum }}\mathit{x\chi }(x)-p\underset{x=1}{\overset{(p-1)∕2}{\sum }}\chi (x).& \hfill & \end{array}$$

(b)

If we separate even and odd indices, we obtain, as $p$ is odd : $$\begin{array}{llll}\hfill \underset{x=1}{\overset{p-1}{\sum }}\mathit{x\chi }(x)& =\underset{k=1}{\overset{(p-1)∕2}{\sum }}2\mathit{k\chi }(2k)+\underset{k=0}{\overset{(p-1)∕2-1}{\sum }}(2k+1)\chi (2k+1)& \hfill & \\ \hfill & =\underset{x=1}{\overset{(p-1)∕2}{\sum }}2\mathit{x\chi }(2x)+\underset{x=1}{\overset{(p-1)∕2}{\sum }}(p-2x)\chi (p-2x)(2k+1=p-2x)& \hfill & \\ \hfill & =2\chi (2)\underset{x=1}{\overset{(p-1)∕2}{\sum }}\mathit{x\chi }(x)-\chi (2)\underset{x=1}{\overset{(p-1)∕2}{\sum }}(p-2x)\chi (x)& \hfill & \\ \hfill & =4\chi (2)\underset{x=1}{\overset{(p-1)∕2}{\sum }}\mathit{x\chi }(x)-\mathit{p\chi }(2)\underset{x=1}{\overset{(p-1)∕2}{\sum }}\chi (x).& \hfill & \end{array}$$

(c)

Let $S=\underset{x=1}{\overset{(p-1)∕2}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}\chi (x),T=\underset{x=1}{\overset{(p-1)∕2}{\sum }}\mathit{x\chi }(x)$. Then $$\begin{array}{llll}\hfill (a)\underset{x=1}{\overset{p-1}{\sum }}\mathit{x\chi }(x)& =2T-\mathit{pS},& \hfill & \\ \hfill (b)\underset{x=1}{\overset{p-1}{\sum }}\mathit{x\chi }(x)& =4\chi (2)T-\mathit{p\chi }(2)S.& \hfill & \end{array}$$

Subtracting theses equalities, we obtain

$$\begin{array}{llll}\hfill (4\chi (2)-2)T& =p(\chi (2)-1)S,& \hfill & \\ \hfill \frac{T}{p}& =\frac{\chi (2)-1}{4\chi (2)-2}S.& \hfill & \end{array}$$

$\chi (2)={(-1)}^{(p^2-1)∕8}=-1$ if $p\equiv 3(\mathit{mod}8)$, so $\frac{\chi (2)-1}{4\chi (2)-2}=\frac{1}{3}$, and $T∕p=(1∕3)S$.

$$\underset{x=1}{\overset{p-1}{\sum }}\mathit{x\chi }(x)∕p=2T∕p-S=-(1∕3)S,$$

$$\underset{x=1}{\overset{p-1}{\sum }}\mathit{x\chi }(x)∕p=-\frac{1}{3}\underset{x=1}{\overset{(p-1)∕2}{\sum }}\chi (x).$$

(d)

$\chi (2)={(-1)}^{(p^2-1)∕8}=1$ if $p\equiv 7(\mathit{mod}8)$ : $\frac{\chi (2)-1}{4\chi (2)-2}=0$. $$\underset{x=1}{\overset{p-1}{\sum }}\mathit{x\chi }(x)∕p=-\underset{x=1}{\overset{(p-1)∕2}{\sum }}\chi (x).$$

Verification : with $p=7$, the squares are $1,4,2=9$, so

$$\begin{array}{llll}\hfill \underset{x=1}{\overset{p-1}{\sum }}\mathit{x\chi }(x)∕p& =\frac{1}{7}\left[\left(\frac{1}{7}\right)+2\left(\frac{2}{7}\right)+3\left(\frac{3}{7}\right)+4\left(\frac{4}{7}\right)+5\left(\frac{5}{7}\right)+6\left(\frac{6}{7}\right)\right]& \hfill & \\ \hfill & =\frac{1}{7}\left(1+2-3+4-5-6\right)=-1& \hfill & \\ \hfill \underset{x=1}{\overset{(p-1)∕2}{\sum }}\chi (x)& =\left(\frac{1}{7}\right)+\left(\frac{2}{7}\right)+\left(\frac{3}{7}\right)& \hfill & \\ \hfill & =1+1-1=1.& \hfill & \end{array}$$

With $p=3$,

$$\begin{array}{llll}\hfill \underset{x=1}{\overset{p-1}{\sum }}\mathit{x\chi }(x)∕p& =\frac{1}{3}\left[\left(\frac{1}{3}\right)+2\left(\frac{2}{3}\right)\right]& \hfill & \\ \hfill & =\frac{1}{3}\left(1-2\right)=-\frac{1}{3}& \hfill & \\ \hfill \underset{x=1}{\overset{(p-1)∕2}{\sum }}\chi (x)& =\left(\frac{1}{3}\right)& \hfill & \\ \hfill & =1& \hfill & \end{array}$$

This confirms the misprints in the initial sentence.