Let $P$ be a prime ideal of $\mathcal O_K$ ($K$ a quadratic field) and let $P$ have norm $p$ where $p$ is a split prime. Is it possible for the ideal class $[P]$ to have order less than three? I feel like this must happen sometimes, as e.g. $3$ splits in $\mathcal O_K$ where $K=\mathbb Q(\sqrt{-5}),$ but the class group $Cl(K)$ is $C_2$. So if $P_3$ is a prime ideal of norm $3$ here, it's class must have order 1 or 2. However, at the same time, $P_3 \overline{P_3}=(3)$ is principal, and so $[\overline{P_3}]=[P_3]^{-1}$. But $P_3 \neq \overline{ P_3}$. So must we have $[\overline{P_3}]=[P_3]$, in this case, since the order of the class divides 2? Is there any other way of telling when $[\overline{P}]=[P]$ in general?
In a similar vein, can ideal classes of prime ideals over ramified primes $p$ ever be of order 1 or are they always order 2? We would need $(a+b\sqrt d)=P \implies a^2-db^2=p$ for some $a, b \in \mathbb Z$ (or $\frac{1}{2}\mathbb Z$, if $d \equiv 1 \pmod 4$) and $K=\mathbb Q(\sqrt d)$. Are there simple examples when this happens? Attempt 1: since $p$ ramifies, either $p|d$ or $p=2, d \equiv 3 \pmod 4$. In the first case, we get $a^2=p-db^2$ which is divisible by $p$. Hence $p|a$ and so $d>0$. In the second case, $a^2-3b^2 \equiv 2 \pmod 4 \implies a^2+b^2 \equiv 2 \pmod 4 \implies 2 \nmid a,b$.
Attempt 2: For ramified primes, $P=\overline P$ and so $a+b\sqrt d \sim a-b\sqrt d $. If $d<-1, d \equiv 2,3 \pmod 4$ then the units are the elements of norm 1, so $\pm 1$. So either $a+b\sqrt d=a-b\sqrt d $ or $-a+b\sqrt d$. So either $b=0$ or $a=0$, but then $p=a^2$ or $-db^2$, which gives a contradiction unless $b=1,p=-d$. (is this argument correct?)