Let $p \in \mathbb{N}$ be prime. Show that if there is an element in $\mathbb{Z}[\sqrt{3}]$ with norm $p$, then there is no element of norm $-p$ Let $p \in \mathbb{N}$ be prime Show that if there is an element in $\mathbb{Z}[\sqrt{3}]$ with norm $p$, then there is no element of norm $-p$.
What I have so far is:

Suppose $N(P) = \pm p$. The other elements of norm $\pm p$ are UP, UP¯, where U is a unit. But no element has norm
$−1$.
(Else, $N(x + y\sqrt{3}) = x^2 − 3y^2.$ If this is equal to $−1$ then $x^2 \equiv −1$ mod $3$. However, this
is impossible since $−1$ is not a quadratic residue modulo $3$.)
So $N(U) = 1$.
Hence $$N(UP) = N(UP¯) = N(P).$$

From here, how can I show $N(P)$ is not $-p$?

Suppose you have two elements $x,y \in \mathbb{Z}[\sqrt{3}]$. I will write $x=\alpha + \beta \sqrt{3}$ and $y=\gamma + \delta\sqrt{3}$, $N(x)=p$ and $N(y)=-p$, where $N(x)$ denotes the norm. We know $N(x)=\alpha^2-3\beta^2$ and of course $N(y)=\gamma^2-3\delta^2$.
Now we have
$N(x)=-N(y) \iff \alpha^2-3\beta^2 = 3\delta^2-\gamma^2 \iff \alpha^2+\gamma^2 = 3(\beta^2+\delta^2)$.
From this we can deduce $\frac{\alpha^2+\gamma^2}{3}\in \mathbb{Z}$ or in other words $\alpha^2+\gamma^2 \equiv 0 \mod 3$. Since only $0,1$ are quadratic resiudes mod 3, this can only happen if $\alpha^2 \equiv \gamma^2 \equiv 0 \mod 3$. But now we have $3\mid \alpha^2-3\beta^2 =N(x)$. This is a contradiction, because we assumed $N(x)=p$ to be prime in $\mathbb{N}$.
