Exercise 8.13

Proof. Suppose that $4p=A^2+27B^2=C^2+27D^2$, where $A\equiv C\equiv 1(\mathit{mod}3)$. We will show that $A=C$.

Let $\omega =e^{2\mathit{i\pi }∕3}=-1∕2+i\sqrt{3}∕2$. Then $i\sqrt{3}=2\omega +1$, and for all $x,y$, $x^2+3y^2=(x+i\sqrt{3}y)(x-i\sqrt{3}y)=(x+(2\omega +1)y)(x-(2\omega +1)y),$

With $x=A,y=3B$, we obtain

Note that $A,B$ are of same parity, since $4p=A^2+27B^2$.

So we can write $p=((A+3B)∕2+3\mathit{\omega B})((A-3B)∕2-3\mathit{\omega B})$ :

$\pi$ is a prime in $\mathbb{Z}[\omega ]$ : indeed $\pi =\mathit{uv},u,v\in \mathbb{Z}[\omega ]$ implies $p=N(\pi )=N(u)N(v)$, then $N(u)=1$ or $N(v)=1$, $u$ or $v$ is an unit, so $\pi$ is irreducible in the principal ideal domain $\mathbb{Z}[\omega ]$, thus $\pi$ is a prime in $\mathbb{Z}[\omega ]$.

As $\pi$ is a prime, it divides $\frac{C+3D}{2}+3\mathit{\omega D}$ or its conjugate. Since they have the same norm $p$, they are associated. The units of $\mathbb{Z}[\omega ]$ are $\pm 1,\pm \omega ,\pm {\omega }^2$, so there exists $12$ cases :

If we replace $D$ by $-D$, we obtain the 6 last cases from the 6 first cases, so it is sufficient to examine the first 6 cases. Recall that $(1,\omega )$ is a $\mathbb{Z}$-base of $\mathbb{Z}[\omega ]$.

1)

$A+3B+6\mathit{\omega B}=C+3D+6\mathit{\omega D}$.

Then $B=D$ and $A+3B=C+3D$, so $A=C$, which is the expected result. The five other cases are impossible :

2)

$A+3B+6\mathit{\omega B}=-C-3D-6\mathit{\omega D}$.

Then $B=-D,A=-C$. As $A\equiv C\equiv 1(\mathit{mod}3)$, this is impossible.

3)

$A+3B+6\mathit{\omega B}=\omega (C+3D+6\mathit{\omega D})=\omega (C+3D)+(-1-\omega )6D=-6D+\omega (C-3D)$.

Then $A+3B=-6D$, $A\equiv 0(\mathit{mod}3)$, this is impossible.

4)

$A+3B+6\mathit{\omega B}=-\omega (C+3D+6\mathit{\omega D})=-\omega (C+3D)+(1+\omega )6D=6D+\omega (-C+3D)$.

Then $A+3B=-6D$, $A\equiv 0(\mathit{mod}3)$, this is impossible.

5)

$A+3B+6\mathit{\omega B}={\omega }^2(C+D+6\mathit{\omega D})=(-1-\omega )(C+3D)+6D=-C+3D+\omega (-C-3D)$. Then $A+3B=-C+3D$, $A\equiv -C(\mathit{mod}3)$ , this is impossible.

6)

$A+3B+6\mathit{\omega B}=-{\omega }^2(C+3D+6\mathit{\omega D})=(1+\omega )(C+3D)-6D=(C-3D)+\omega (C+3D)$.

Then $6B=C+3D$, $C\equiv 0(\mathit{mod}3)$, this is impossible.

In conclusion $A=C$. □