Exercise 11.12

Proof.

$\text{∙}$ Assume first that $p\equiv 3(\mathit{mod}4)$. First, we count the number of affine points on $C_2$.

In this case, there is no character of order $4$, and the only characters whose order divides $4$ are $𝜀$ and $\rho$, where $\rho$ is the Legendre’s character. Then Exercises 8.1, 8.2, with $d=4\wedge (p-1)=2$, and Proposition 8.1.5 show that $N(x^4=a)=N(y^2=a)=1+\rho (a).$ Therefore

= $\rho (-1){\sum <!--\; FUNCTION\; APPLICATION\; -->}_{a^{\prime }+c^{\prime }=1}\rho {(4D)}^2\rho (a^{\prime })\rho (b^{\prime })(a=4D{a}^{\prime },c=4D{b}^{\prime })=\rho (-1)J(\rho ,\rho ).$

Moreover, by Theorem 1(c), Chapter 8, since ${\rho }^2=𝜀$,

Putting all together, we obtain

Then Exercise 11 gives

The projective closure of $C_1$ has equation $y^{2}t=x^3-\mathit{Dx}{t}^2$. For $t=0$, $x=0$, thus $[0,1,0]$ is the only point at infinity. The number of projective points on $C_1$ is

$\text{∙}$ Now we assume that $p\equiv 1(\mathit{mod}4)$. Then there is a character $\chi$ of order $4$ on $F_p$.

The inner sum for each fixed pair $(i,j)$ is

Since ${\chi }^2$ is of order 2, $\rho ={\chi }^2$, thus

and, using $J(𝜀,𝜀)=p,J(𝜀,{\chi }^j)=0$ if $j\ne 0$,

Since $J({\chi }^2,{\chi }^2)=J({\chi }^2,{\chi }^{-2})=-{\chi }^2(-1)=-1$, and ${\chi }^3=\bar{\chi }$, we obtain

Since $\chi {(4)}^2=\chi (2^4)={\chi }^4(2)=1$, $\chi (4)=\pm 1$ is real, therefore

We must add one to obtain the number of affine points of $C_1$, and one more to the point at infinity. Thus the number of projective points on $C_1$ is

But $\chi (-1)={(-1)}^{\frac{p-1}{4}}$. To prove this equality, take $g$ a generator of $F_p^{\text{∗}}$ such that $\chi (g)=i$ (such a generator exists, since $\chi (g)=\pm i$: if $\chi (g)=-i$, replace $g$ by $g^{-1}$). Since $g^{p-1}=1$, and $g^{(p-1)∕2}\ne 1$, we obtain $g^{(p-1)∕2}=-1$, thus $\chi (-1)=\chi {(g)}^{(p-1)∕2}=i^{(p-1)∕2}={(-1)}^{(p-1)∕4}$. Moreover $\chi (4)={\chi }^2(2)=\rho (2)={(-1)}^{(p^2-1)∕8}$. Thus, for $p=4k+1$,

Alleluia! We conclude

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