This should be well-known, but I can't find a reference (or a proof, or a counter-example...). Let $d$ be a positive square-free integer. Suppose that there is no element in the ring of integers of $\mathbb{Q}(\sqrt{d})$ with norm $-1$. Then I believe that no element of $\mathbb{Q}(\sqrt{d})$ has norm $-1\ $
(in fancy terms, the homomorphism $H^2(G,\mathscr{O}^\*)\rightarrow H^2(G,\mathbb{Q}(\sqrt{d})^\*)$, with $G:=\operatorname{Gal}(\mathbb{Q}(\sqrt{d})/\mathbb{Q})=$ $\mathbb{Z}/2 $, is injective). Is that correct? If yes, I'd appreciate a proof or a reference.



Norms in quadratic fields
This is false. The smallest counterexample is $d = 34$. Let $K = \mathbb{Q}(\sqrt{34})$. The fundamental unit in $\mathcal{O}\_{K} = \mathbb{Z}[\sqrt{34}]$ is $35 + 6 \sqrt{34}$, which has norm $1$, and therefore, there is no element in $\mathcal{O}\_{K}$ with norm $-1$.


However, $\frac{3}{5} + \frac{1}{5} \sqrt{34}$ has norm $-1$, so there is an element of norm $-1$ in $K$.


