Sign of fundamental unit in real quadratic number fields with 1 mod 4 discriminant factors

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Let $K$ be a real quadratic number field of discriminant $D$ with fundamental unit $\varepsilon$. Further, I want to assume that each positive factor $n$ of $D$ satisfies $n \equiv 1 \pmod 4$. (For other cases I have enough information concerning my question, hence we do not need to discuss them here.)

Now let us write $D=p_1 p_2 \cdots p_k$ as product of primes. Is the sign of $N(\varepsilon)$ well understood in terms of the primes? What I know is that

  • for $k=1$ we have $N(\varepsilon)=-1$,
  • for $k=2$ and $(p_1/p_2)=-1$ we have $N(\varepsilon)=-1$.

What happens if $k=2$ and $(p_1/p_2)=1$? What happens for $k>2$?

Edit: Some examples for the case $D=pq$ with $(p/q)=1$ ordered by discriminant:

  • $D = 5 \cdot 29$, $N(\varepsilon)=-1$
  • $D = 5 \cdot 41$, $N(\varepsilon)=+1$
  • $D = 13 \cdot 17$, $N(\varepsilon)=+1$
  • $D = 5 \cdot 61$, $N(\varepsilon)=+1$
  • $D = 13 \cdot 29$, $N(\varepsilon)=+1$
  • $D = 5 \cdot 89$, $N(\varepsilon)=-1$
  • $D = 5 \cdot 101$, $N(\varepsilon)=+1$