I am attempting to compute the ideal class group of the real quadratic field $K = \Bbb Q(\sqrt{65})$, which has ring of integers $\mathcal{O}_K = \Bbb Z\left[\frac{1 + \sqrt{65}}{2}\right]$.
The discriminant $d_K$ of $K$ is
\begin{align} d_K &= \begin{vmatrix} 1 &\frac{1 + \sqrt{65}}{2}\\ 1 &\frac{1 - \sqrt{65}}{2}\end{vmatrix}^2\\ &= 65 \end{align}
and, as a real quadratic field, there are no complex embeddings so the Minkowski bound $M_K$ is
\begin{align} M_K &= \frac{2!}{2^2}\left( \frac{2}{\pi}\right)^0 \sqrt{65}\\ &\approx 4.03 \end{align}
so the ideal class group is generated by primes lying above $2$ and $3$. Checking the factorisation of $X^2 - X - 16 \bmod p$ for $p = 2, 3$ we get
$$X^2 - X - 16 \equiv X(X+1) \bmod 2\\ X^2 - X - 16 \equiv X^2 + 2X + 2 \bmod 3$$
so $2$ splits and $3$ is inert, so that $(2) = \mathfrak{p}_2\bar{\mathfrak{p}}_2$ and $(3) = \mathfrak{p}_3$ with $N(\mathfrak{p}_2) = N(\bar{\mathfrak{p}}_2) = 2$ and $N(\mathfrak{p}_3) = 9$.
Since $\mathfrak{p}_3$ is principal, it does not generate the ideal class group so we need only think about $\mathfrak{p}_2$.
What follows might seem like a silly quesion so brace yourselves!
If the norm equation $a^2 + ab - 16b^2 = 2$ has solutions in the integers, should it have solutions modulo every prime?
I ask because the equation has no solutions modulo $5$, and this is perhaps a way of showing that there is no element of norm $2$ and so $\mathfrak{p}_2$ cannot be principal (the same computation shows there are no elements of norm $-2$, so this is also covered).
The confusion stems from the fact that the book I'm using quotes the ideal class $[(5, \sqrt{65})]$ as a generator for the class group, though this ideal has norm $5 > 4.03$, hence the confusion. In addition, following an argument by Keith Conrad in his expository papers, it seems that since $(5) = \mathfrak{p}_5^2$, if $\mathfrak{p}_5$ were principal we should be able to write
$$5 = (a + b\sqrt{65})^2 u$$
for some $u \in \mathcal{O}_K^\times$ and $a, b \in \frac{1}{2}\Bbb Z$. The units of $\mathcal{O}_K$ look like $\pm(8 + \sqrt{65})^n$ for $n \in \Bbb Z$, and since $N(8 + \sqrt{65}) = -1$, the units of norm $1$ have the form $(8 + \sqrt{65})^{2k}$ for some $k \in \Bbb Z$. In particular, this means that the $u$ term in the above equation can be absorbed into the $(a+b\sqrt{65})^2$ term, but this implies that $\sqrt{5} \in \mathcal{O}_K$, and so $\mathfrak{p}_5$ is not principal, which seems to contradict the Minkowski bound.
The fact that $(5,\sqrt{65})$ has a norm above the Minkowski bound does not imply it must be principal. Rather, the ideal cannot generate any more classes than other ideals whose norms are lower. Here the relevant ideals within the Minkowski bound are:
$(3)$, which is inert
$(2,\frac{1+\sqrt{65}}{2})$, which gives a principal ideal when squared.
Thereby the integral domain must be class 2 and have ideal class group $\mathbb Z/2\mathbb Z$, and any ideal with a higher norm must also be principal or give a principal ideal when squared. The latter is what applies to $(5,\sqrt{65})$.