= Solution
For the real quadratic field $K=\mathbb Q(\sqrt d)$ with squarefree $d>1$, <Dirichlet unit theorem> gives
$$
\boxed{\mathcal O_K^\times=\{\pm\epsilon^n:n\in\mathbb Z\}}
$$
for a <fundamental unit> $\epsilon>1$, since the <roots of unity> in a real field are $\pm1$.
If $p\mid d$ and $p\equiv3\pmod4$, a <unit> of norm $-1$ would have the form $(a+b\sqrt d)/2$ with <integers> $a,b$ and satisfy $a^2-db^2=-4$. Reducing modulo the odd <prime> $p$ gives $(a/2)^2\equiv-1\pmod p$, impossible because $-1$ is not a <quadratic residue> for such a <prime>. Hence \b[every <unit> has norm $+1$ in this case]. In contrast, $\boxed{d=2,\ u=1+\sqrt2}$ has norm $-1$.
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