 Solution

For the real quadratic field $K=\mathbb{Q}(\sqrt{d})$ with squarefree $d>1$, Dirichlet unit theorem givesfor a fundamental unit $ϵ>1$, since the roots of unity in a real field are $\pm 1$.

If $p\mid d$ and $p\equiv 3(\mathrm{mod}4)$, a unit of norm $-1$ would have the form $(a+b\sqrt{d})/2$ with integers $a,b$ and satisfy $a^2-d{b}^2=-4$. Reducing modulo the odd prime $p$ gives $(a/2{)}^2\equiv -1(\mathrm{mod}p)$, impossible because $-1$ is not a quadratic residue for such a prime. Hence every unit has norm $+1$ in this case. In contrast, $\boxed{d=2,u=1+\sqrt{2}}$ has norm $-1$.