Exercise 8.15

Proof. $x\mapsto -x$ is a bijection between the set of roots of $x^3=b$ and the set of roots of ${(-x)}^3=b$, so $N(x^3=b)=N({(-x)}^3=b)=N(x^3=-b)$.

As $\chi$ is a character of order 3, the characters whose order divides 3 are $𝜀,\chi ,{\chi }^2$. Using Prop. 8.1.5, we obtain, if $D\ne 0$,

We know (Theorem 1) that $J(\chi ,𝜀)=J({\chi }^2,𝜀)=J(𝜀,\rho )=0,J(𝜀,𝜀)=p$, so

As ${\chi }^2(D)={\chi }^{-1}(D)=\overline{\chi (D)}$, and as $\overline{\rho (D)}=\rho (D)$, then $J({\chi }^2,\rho )=J(\bar{\chi },\bar{\rho })=\overline{J(\chi ,\rho )}$, and

If $\chi (2)=1$, then from Exercise 8.6 we have

With $D=1$ (if $\chi (2)=1$), we obtain

From Prop. 8.3.4 we know that $J(\chi ,\chi )=a+\mathit{b\omega },b\equiv 0(\mathit{mod}3),a\equiv -1(mod3)$.

$\pi +\bar{\pi }=2\mathrm{Re}J(\chi ,\chi )=2a-b\equiv 1(\mathit{mod}3)$, and $p=|J(\chi ,\rho ){\text{|}}^2=a^2-\mathit{ab}+b^2$, so $4p={(2a-b)}^2+3b^2$.

Writing $A=2a-b,B=b∕3$, we obtain $4p=A^2+27B^2,A\equiv 1(\mathit{mod}3)$ (the unicity of $A$ if proved in Exercise 8.13).

Conclusion : $N(y^2=x^3+1)=p+A$, where $4p=A^2+27B^2,A\equiv 1(\mathit{mod}3)$.

If $p=31$, 3 is a primitive element, and $2=3^{24}={(3^8)}^3$ in ${𝔽}_{31}$, therefore $\chi (2)=1$.

$31=4+27$, $4\times 31=124=4^2+27\times 2^2$, and $4\equiv 1(\mathit{mod}3)$, so

if $p=31$, $N(y^2=x^3+1)=35$. □