Exercise 8.27

Proof.

(b)

By Exercise 8.15,

$$N(y^2=x^3+D)=p+2\mathrm{Re}(\pi ),\mathrm{where}\pi =(\mathit{\rho \chi })(D)J(\chi ,\rho ).$$

Moreover, with $D=1$, we obtain

$$\begin{array}{llll}\hfill N(y^2+x^3=1)& =N(y^2+{(-x)}^3=1)& \hfill & \\ \hfill & =p+2\mathrm{Re}J(\chi ,\rho ).& \hfill & \end{array}$$

$J(\chi ,\rho )=\underset{a+b=1}{\sum <!--\; FUNCTION\; APPLICATION\; -->}\chi (a)\rho (b)$, where $\rho (b)\in \{-1,1\},\chi (a)\in \{1,\omega ,{\omega }^2\}$, so $J(\chi ,\rho )\in \mathbb{Z}[\omega ]$.

$J(\chi ,\rho )=a+\mathit{b\omega },a\in \mathbb{Z},b\in \mathbb{Z}$ and $2\mathrm{Re}(J(\chi ,\rho ))=2a-b$.

(a)

By Exercise 8.8, $$\begin{array}{llll}\hfill \underset{x}{\sum }\rho (1-x^3)& =\underset{{\lambda }^3=𝜀}{\sum }J(\rho ,\lambda )& \hfill & \\ \hfill & =J(\rho ,𝜀)+J(\rho ,\chi )+J(\rho ,{\chi }^2)& \hfill & \\ \hfill & =2\mathrm{Re}J(\chi ,\rho ).& \hfill & \end{array}$$

So

$$\begin{array}{llll}\hfill N(y^2=1-x^3)& =p+\underset{x}{\sum }\rho (1-x^3)& \hfill & \\ \hfill & =p+2a-b.& \hfill & \end{array}$$

(c)

Reducing modulo $p$, we obtain in ${𝔽}_p$ : $$\begin{array}{llll}\hfill 2\bar{a}-\bar{b}& =\underset{x}{\sum }\rho (1-x^3)& \hfill & \\ \hfill & =\underset{t\in {𝔽}_p}{\sum }{(1-t^3)}^{\frac{p-1}{2}}& \hfill & \\ \hfill & =\underset{t\in {𝔽}_p}{\sum }\underset{k=0}{\overset{(p-1)∕2}{\sum }}\left(\genfrac{}{}{0.0pt}{}{\frac{p-1}{2}}{k}\right){(-1)}^k{t}^{3k}& \hfill & \\ \hfill & =\underset{k=0}{\overset{(p-1)∕2}{\sum }}{(-1)}^k\left(\genfrac{}{}{0.0pt}{}{\frac{p-1}{2}}{k}\right)\underset{t\in {𝔽}_p}{\sum }{t}^{3k}& \hfill & \\ \hfill & =\underset{k=1}{\overset{(p-1)∕2}{\sum }}{(-1)}^k\left(\genfrac{}{}{0.0pt}{}{\frac{p-1}{2}}{k}\right)\underset{t\in {𝔽}_p}{\sum }{t}^{3k}.& \hfill & \end{array}$$

Let $S_k=\underset{t\in {𝔽}_p}{\sum <!--\; FUNCTION\; APPLICATION\; -->}{t}^{3k}(0<k\le \frac{p-1}{2})$, and $g$ a primitive root in ${𝔽}_p$ : $t=g^i,0\le i\le p-2$.

$S_k=\underset{i=0}{\overset{p-2}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{(g^i)}^{3k}=\underset{i=0}{\overset{p-2}{\sum }}{(g^{3k})}^i$

If $g^{3k}\ne 1,S_k=\frac{g^{3k(p-1)}-1}{g^{3k}-1}=0$, if not $S_k=p-1=-1$.

$$g^{3k}=1⟺p-1\mid 3k⟺\frac{p-1}{3}\mid k⟺k=\frac{p-1}{3}$$

and $\frac{p-1}{3}$ even, so

$$2a-b\equiv -\left(\genfrac{}{}{0.0pt}{}{\frac{p-1}{2}}{\frac{p-1}{3}}\right)(\mathit{mod}p)(\mathrm{where}J(\chi ,\rho )=a+\mathit{b\omega }).$$