Exercise 3.3.20* ($\sum\limits_{n=1}^p \genfrac{(}{)}{}{}{n^2 + a}{p} = -1$ if $p \nmid a$)

Proof. Put $\alpha ={[a]}_p\in {𝔽}_p^{\text{∗}}$. Then

(a)

Let $S=\{(x,y)\in {𝔽}_p^2|x^2-y^2=\alpha \}$.

Using the remark p. 123 (or the easy part of Problem 3.1.23),

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

(b)

Consider the two curves in the plane ${𝔽}_p^2$ defined by $$\begin{array}{llll}\hfill S& =\{(x,y)\in {𝔽}_p^2|x^2-y^2=\alpha \},& \hfill & \\ \hfill T& =\{(u,v)\in {𝔽}_p^2|uv=\alpha \}.& \hfill & \end{array}$$

Then $f:\{\begin{array}{ccc}\hfill S\hfill & \hfill \to \hfill & T\hfill \\ \hfill (x,y)\hfill & \hfill \mapsto \hfill & (x+y,x-y)\hfill \end{array}$ is well defined.

Indeed, if $(x,y)\in S$, $(x-y)(x+y)=\alpha$, so $(x+y,x-y)\in T$. Moreover $f$ is a bijection, with inverse $(u,v)\mapsto ((u+v)∕2,(u-v)∕2)$, so $|S|=|T|$.

We compute $|T|$.

Since $p\nmid a$, $\alpha \ne \bar{0}$. For $v=0$, there is no solution, and for each $v\ne 0$, we obtain the unique solution $(\alpha v^{-1},v)$. Therefore there exist $p-1$ solutions, so that $|S|=|T|=p-1$.

In conclusion, if $\alpha \ne 0$,

$$\begin{array}{llll}\hfill \mathrm{Card}\{(x,y)\in {𝔽}_p^2|x^2-y^2=\alpha \}& =p-1.& \hfill & \end{array}$$

(c)

We obtain in part (a), $|S|=\underset{n=1}{\overset{p}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}\left(1+\left(\frac{n^2+a}{p}\right)\right)$, and in part (b), $|S|=p-1$.

Therefore, if $p\nmid a$, $|S|-p=\underset{y=1}{\overset{p}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}\left(\frac{y^2+a}{p}\right)=-1.$

We have proved

$$\underset{n=1}{\overset{p}{\sum }}\left(\frac{n^2+a}{p}\right)=-1(\text{if }a\not\equiv 0(modp)).$$