 Obstruction to norm-minus-one units from a prime congruent to three modulo four

If a squarefree positive $d$ has a prime divisor $p\equiv 3(\mathrm{mod}4)$, a unit of norm $-1$ in $\mathbb{Q}(\sqrt{d})$ would give $a^2-d{b}^2=-4$ for integers $a,b$. Modulo $p$ this makes $-1$ a quadratic residue, which is impossible. The obstruction is sufficient, not a full criterion for negative-norm units.