= Solution
The <Dirichlet unit theorem> states that for a number field of signature $(r_1,r_2)$,
$$
\mathcal O_K^\times\cong\mu(K)\times\mathbb Z^{r_1+r_2-1}.
$$
For $K=\mathbb Q(\sqrt5)$,
$$
\mathcal O_K=\mathbb Z[\phi],
\qquad \phi=\frac{1+\sqrt5}{2},
\qquad N(\phi)=-1.
$$
The field is real quadratic, so the unit rank is one and its only roots of unity are $\pm1$. To see that $\phi$ is fundamental, suppose a positive unit satisfies $1<u<\phi$. If its norm is $1$, its <integral> trace $u+u^{-1}$ lies strictly between $2$ and $\sqrt5$; if its norm is $-1$, its trace $u-u^{-1}$ lies strictly between $0$ and $1$. Neither interval contains an integer. Reducing any positive unit by a suitable power of $\phi$ now proves the <units of the quadratic field Q square root of five> formula
$$
\boxed{\mathcal O_K^\times=\{\mathord\pm\phi^n:n\in\mathbb Z\}}.
$$
If $\mathcal O_L^\times/\mathcal O_K^\times$ is finite, the two unit <groups> have the same rank, namely one. Put $d=[L:K]$ and let $(r_1,r_2)$ be the signature of $L$. Then
$$
r_1+r_2-1=1,
\qquad
r_1+2r_2=[L:\mathbb Q]=2d.
$$
Therefore $r_1=4-2d$ and $r_2=2d-2$. For a proper extension, nonnegativity forces
$$
\boxed{d=2},
$$
with signature $(0,2)$.
This degree occurs: take $L=K(i)$. It is a totally imaginary quadratic extension of $K$, so the unit ranks agree and the quotient is finite. It is nontrivial because $i\in\mathcal O_L^\times$ but $i\notin\mathcal O_K^\times$ (and its coset has order two). This is the <finite relative unit quotient over a real quadratic field> example.
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