Exercise 3.3.17* (Value of $s(a,p) =\sum\limits_{n=1}^p \genfrac{(}{)}{}{}{n(n+a)}{p}$)

Proof.

(a)

By definition, $\left(\frac{n^2}{p}\right)=1$ if $n\not\equiv 0(\mathit{mod}p)$, and $\left(\frac{p^2}{p}\right)=0$, thus $$\begin{array}{llll}\hfill s(0,p)& =\underset{n=1}{\overset{p}{\sum }}\left(\frac{n^2}{p}\right)& \hfill & \\ \hfill & =\underset{n=1}{\overset{p-1}{\sum }}1& \hfill & \\ \hfill & =p-1.& \hfill & \end{array}$$

Similarly,

$$s(p,p)=\underset{n=1}{\overset{p}{\sum }}\left(\frac{n(n+p)}{p}\right)=\underset{n=1}{\overset{p}{\sum }}\left(\frac{n^2}{p}\right)=p-1.$$

(b)

By Problem 16, $\underset{a=1}{\overset{p}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}\left(\frac{n+a}{p}\right)=0$, for every integer $n$. Therefore $$\begin{array}{llll}\hfill \underset{a=1}{\overset{p}{\sum }}s(a,p)& =\underset{a=1}{\overset{p}{\sum }}\underset{n=1}{\overset{p}{\sum }}\left(\frac{n(n+a)}{p}\right)& \hfill & \\ \hfill & =\underset{n=1}{\overset{p}{\sum }}\left(\frac{n}{p}\right)\underset{a=1}{\overset{p}{\sum }}\left(\frac{n+a}{p}\right)& \hfill & \\ \hfill & =0.& \hfill & \end{array}$$

(c)

Suppose now that $a\wedge p=1$. With the same notations as in Problem 16, with $\alpha ={[a]}_p\ne 0$, the map $$\phi \{\begin{array}{ccc}\hfill {𝔽}_p\hfill & \hfill \to \hfill & {𝔽}_p\hfill \\ \hfill x\hfill & \hfill \mapsto \hfill & \mathit{\alpha x}\hfill \end{array}$$

is bijective. Therefore the change of variable $y=\mathit{\alpha x}$ gives

$$\begin{array}{llll}\hfill s(a,p)& =\underset{n=1}{\overset{p}{\sum }}\left(\frac{n(n+a)}{p}\right)& \hfill & \\ \hfill & =\underset{y\in {𝔽}_p}{\sum }\left(\frac{y(y+\alpha )}{p}\right)& \hfill & \\ \hfill & =\underset{x\in {𝔽}_p}{\sum }\left(\frac{\mathit{\alpha x}(\mathit{\alpha x}+\alpha )}{p}\right)(y=\mathit{\alpha x})& \hfill & \\ \hfill & =\underset{x\in {𝔽}_p}{\sum }\left(\frac{{\alpha }^2}{p}\right)\left(\frac{x(x+1)}{p}\right)& \hfill & \\ \hfill & =\underset{n=1}{\overset{p}{\sum }}\left(\frac{n(n+1)}{p}\right)& \hfill & \\ \hfill & =s(1,p).& \hfill & \end{array}$$

(d)

By part (b), $\underset{a=1}{\overset{p}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}s(a,p)=0$. Using the results of part (c) and (a), $$\begin{array}{llll}\hfill 0& =\underset{a=1}{\overset{p}{\sum }}s(a,p)& \hfill & \\ \hfill & =\underset{a=1}{\overset{p-1}{\sum }}s(a,p)+s(p,p)& \hfill & \\ \hfill & =(p-1)s(1,p)+s(p,p)& \hfill & \\ \hfill & =(p-1)(s(1,p)+1).& \hfill & \end{array}$$

Therefore $s(1,p)=-1$, and by part (c)

$$s(a,p)=-1,a=1,2,\dots ,p-1.$$