Use the result of Ex. 0.16 to deduce another proof of the fact that any prime $p\equiv_6 1$ is of the form $3a^2+b^2$.
Exercise 0.16: Prove that the ring $\Bbb Z[\omega]$ is a UFD, where $\omega^2+\omega+1=0$.
This is what I've tried:
Let $N(a+b\omega)=(a+b\omega)(a+b\omega^2)=a^2-ab+b^2$, where $a+b\omega\in\Bbb Z[\omega]$. If $p\equiv_6 1$, then $(\frac{-3}{p})=1$ so there exists an $s\in\Bbb Z$ such that $p\mid s^2+3$. We have $$s^2+3=(s+1+2\omega)(s+1+2\omega^2)\tag{1}$$
Also, $p\not\mid s+1+2\omega,s+1+2\omega^2$ so $p$ is not prime. Therefore $p=\alpha\beta$, where $\alpha,\beta\in\Bbb Z[\omega]$ and non of them are units. Taking the norm we get $p^2=N(\alpha\beta)=N(\alpha)N(\beta)$. This gives $N(\alpha)=N(\beta)=p$ since $N(\alpha),N(\beta)\neq 1$. Therefore $$p=a^2-ab+b^2\tag{2}$$
The rest is manipulating eq. (2) with algebra:
$$4p=(2a-b)^2+3b^2\\p=\left(a-\frac{b}{2}\right)^2+3\left(\frac{b}{2}\right)^2$$
Both $a,b$ cannot be even, otherwise $2\mid p$ by eq. (2). Setting $2x=b, y=a-x$ yields $$p=y^2+3x^2$$
Is this correct?