The exercise is the following: one should show that $R=\mathbb{Z}[\sqrt{223}]$ has ideal class group which is a cyclic group of order $3$.
I tried to follow the standard path which Marcus uses to determine other ideal class groups: firstly, I calculated $\lambda$ which turns out to be $\sqrt{223}$. Having that, one reduces to check what happens to prime in $R$ which are above $2,3,5,7,11,13$.
I've checked that the prime above $2,5,7,11,13$ are trivial in the ideal class group and using Kummer theorem I've found that $$3R=(3,1+\sqrt{223})(3,1-\sqrt{223}) .$$
Let's call this two primes respectively $p,q$. One can check that $14-\sqrt{223} \in p $ and $14-\sqrt{223} \not \in q$, while the nrm of this element is $-27$, so that one got $p^3=(14-\sqrt{223})=e$ in the ideal class group. With a similar reasoning, one got $q^3=e$ in the ideal class group.
Now, if we manage to demonstrate that $p,q$ are not principal we are done. If one of this ideal were principal, the element which generates it should have norm $\pm 3$. So , we are reduced to show the following :there are no solutions in $\mathbb{Z}$ of $a^2-223b^2=\pm3$.
The one with the $+$ is showed to be impossible using reciprocity, while I've got no idea to end for the other equation.
