Is there a pattern to the signs of the norms of quadratic fundamental units?

Introductory Algebraic Number Theory by Alaca & Williams deals with this in a fair level of detail in Chapter 11.

• Theorem 11.5.4: given prime $d \equiv 1 \pmod 4$, the fundamental unit has norm $-1$. e.g., $d = 5$, $N(\phi) = -1$.
• Theorem 11.5.5: if $d$ is divisible by some prime $p \equiv 3 \pmod 4$, the fundamental unit has norm 1. e.g., $N(2 + \sqrt 3) = N(5 + 2 \sqrt 6) = 1$.
• Theorem 11.5.6: if $d = 2p$ and $p \equiv 5 \pmod 8$, the fundamental unit has norm $-1$. e.g., $N(3 + \sqrt{10}) = N(5 + \sqrt{26}) = -1$.
• Theorem 11.5.7: if $d = pq$ with distinct primes $p \equiv q \equiv 1 \pmod 4$ and $$\left(\frac{p}{q}\right) = \left(\frac{q}{p}\right) = -1$$ (I know that by quadratic reciprocity that's a bit redundant) then the fundamental unit has norm $-1$. e.g., $N(8 + \sqrt{65}) = -1$.

These are from my notes, not directly from the book; hopefully I haven't made any mistakes of transcription. I don't remember if they say anything about the $d$ not covered by these four theorems. Proofs are included for all four of them, I think one or two of them is given more than one proof.

There are some partial results; if $d\equiv3\pmod 4$ then the fundamental unit has norm $+1$. The same is true if $d$ has a positive factor congruent to $3$ modulo $4$. The proof is just working modulo $4$.

On the other hand if $d=p\equiv1\pmod 4$ is a prime, then the fundamental unit has norm $-1$. To prove this one can show that if $\varepsilon>0$ is a unit of norm $1$ in $K=\Bbb Q(\sqrt d)$ then $\sqrt{\varepsilon}\in K$ also.

Things get more complicated when there are more prime factors. If $d=pq$ where $p$ and $q$ are primes $\equiv1\pmod 4$ then one can have both $+1$ and $-1$ as the norm of the fundamental unit.