I want to prove the following result.
If $d$ is a square free integer of the form $ d=n^{2g}+1$,where $ n>4$,then $\ g|h$,where $ h$ is the class number of the field $ \mathbb Q(\sqrt{d})$.
There is a hint which says to look at the ideal $(n)$.
It can be easily seen that the ideal factorizes as $(n)=P\overline{P}$,where $P$ is a prime ideal lying above $(n)$. Then let's say, order of $P$ in the class group is equal to $m$.If I can show that $m=g$, then I will be done.
Say,$P^m=(u+v\sqrt{d})$, then $n^m=u^2-dv^2.$ I can not proceed further.
Any help would be appreciated. Thanks in advance.