Let $K=\mathbb{Q}(\sqrt{65})$, whose number ring is $\mathcal{O}=\mathbb{Z}\left[\frac{1+\sqrt{65}}{2}\right]$. I know that the number of ideal classes in $\mathbb{Z}\left[\frac{1+\sqrt{65}}{2}\right]$ is $2$, and by Minkowski's bound I know that every ideal is equivalent to one with norm less or equal than $4$. Therefore, if $I\subset\mathcal{O}$ is an ideal, one of the following will happen:
- $||I||=2\Longrightarrow I\mid 2\mathcal{O}=P_1P_2$, with $P_1=(2,\sqrt{65})$ and $P_2=(2,\sqrt{65}+1)$
- $||I||=3\Longrightarrow I\mid 3\mathcal{O}$ which is prime
- $||I||=4\Longrightarrow I\mid 4\mathcal{O}=2\mathcal{O}2\mathcal{O}=P_1P_2P_1P_2$
Therefore, the $P_i$ cannot be principal ideals, whilst $P_i^2$ and $P_1P_2$ are principal (this is the only case where the number of ideal classes is $2$, and I have already proved that both $P_i^2$ and $P_1P_2$ are principal). Now I want to prove that neither $P_1$ or $P_2$ are principal ideals. This is where I am stuck.
If $P_i=(\alpha)$, with $\alpha=a+b\frac{1+\sqrt{65}}{2}$, then $2=||P_i||=||(\alpha)||=|N(\alpha)|$, where $$N(\alpha)=a^2+ab-16b^2$$ So I have to prove that $a^2+ab-16b^2=\pm2$ has no solution in $\mathbb{Z}$. How can I do this? (I will face similar exercises in my final exam so I cannot use a code snippet or wolframalpha or anything like that.)