Exercise 9.39

Proof. Let $p$ be a rational prime, $p\equiv 1(\mathit{mod}6)$. As $p\equiv 1(\mathit{mod}3)$, we know from Theorem 2, Chapter 8, that there are integers $A$ and $B$ such that $4p=A^2+27B^2,A\equiv 1(\mathit{mod}3)$, and that $A$ is uniquely determined by these conditions.

Then $A,B$ have same parities. If we take $a=\frac{A+3B}{2},b=3B$, then $A=2a-b,B=\frac{b}{3}$, and $4p={(2a-b)}^2+3b^2$, so $p=a^2-\mathit{ab}+b^2$. If $\pi =a+\mathit{b\omega }$, then $N(\pi )=p$. Since $A=2a-b\equiv 1[3]$, and $b=3B\equiv 0[3]$, then $a\equiv -1[3]$, so $\pi$ is a primary prime.

$\text{∙}$ Suppose that $2\mid B$. Since $p=a^2-\mathit{ab}+b^2$ is odd, and $b=3B$,

By Proposition 9.6.1,

Therefore

Here ${\chi }_{\pi }$ is of order 3, so ${\chi }_{\pi }^2\ne 𝜀$. By Exercise 8.6,

where $\rho$ is the Legendre’s character.

In this case, $2\mid B$, ${\chi }_{\pi }(2)=1$, so $J({\chi }_{\pi },{\chi }_{\pi })=J({\chi }_{\pi },\rho )$, and by Lemma 1 section 4, where $p\equiv 1[3]$ and $p=N(\pi )$,

By Exercise 8.15,

and the Exercise 8.27(b) gives

thus

Moreover, since $J({\chi }_{\pi },\rho )=\pi =a+\mathit{b\omega }$, by Exercise 8.27(c),

Therefore

where $m=(p-1)∕6.$ Since $A\equiv 1(\mathit{mod}3)$,

$\text{∙}$ Conversely, suppose that $\left(\genfrac{}{}{0.0pt}{}{3m}{m}\right)\equiv -1(\mathit{mod}p)$. Then $A=2a-b\equiv -\left(\genfrac{}{}{0.0pt}{}{3m}{m}\right)(\mathit{mod}p)$. Write $J({\chi }_{\pi },\rho )=c+\mathit{d\omega }$. By Exercise 8.27(c), $2c-d\equiv -\left(\genfrac{}{}{0.0pt}{}{3m}{m}\right)(\mathit{mod}p)$. thus

Since $|J({\chi }_{\pi },\rho )|=\sqrt{p}$,

thus $d\equiv \pm b(\mathit{mod}p)$.

By Exercise 8.6,

Here ${\chi }_{\pi }$ is of order 3, therefore ${\chi }_{\pi }{(2)}^{-2}={\chi }_{\pi }(2)\in \{1,\omega ,{\omega }^2\}$, so

If ${\chi }_{\pi }(2)=\omega$, then $a+\mathit{b\omega }=\omega (c+\mathit{d\omega })=-d+\omega (c-d)$. Then $a=-d\equiv \pm b(\mathit{mod}p)$. As $a\equiv -\mathit{b\omega }(\mathit{mod}\pi )$, we would have $-\mathit{b\omega }\equiv \pm b(\mathit{mod}\pi )$. Here $\pi \nmid b$, otherwise $p=N(\pi )\mid N(b)=b^2$, so $p\mid b$, and $p=a^2-\mathit{ab}+b^2$, so $p\mid a$, and $p^2\mid p$, which is a nonsense. Therefore $\pi \mid \omega \pm 1$, where $\pi$ is a primary prime: it’s impossible. Indeed $\omega +1$ is a unit and $\omega -1$ is prime, so $\pi \mid \omega -1=-\lambda$ implies that $\pi$ and $\lambda$ are associate, in contradiction with $N(\pi )=p\ne 3=N(\lambda )$.

If ${\chi }_{\pi }(2)={\omega }^2$, then $a+\mathit{b\omega }={\omega }^2(c+\mathit{d\omega })=(d-c)-\mathit{\omega c}$, so $a=d-c,b=-c$.

Reasoning modulo $\bar{\pi }=a+b{\omega }^2=(a-b)+\mathit{b\omega }$, where $\bar{\pi }\mid \pi \bar{\pi }=p$, we obtain

where $d\equiv \pm b(\mathit{mod}\bar{\pi })$, so $-\mathit{b\omega }\equiv \pm b(\mathit{mod}\bar{\pi })$. Since $N(\bar{\pi })=p$, we obtain the same contradiction as above.

So ${\chi }_{\pi }(2)=1$, and the previously proved equivalence $2\mid B⟺{\chi }_{\pi }(2)=1$ show that $2\mid B$.

Conclusion:

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