Let $K$ be a quadratic number field, $\mathcal O \subset K$ an order and $I \subset \mathcal O$ an ideal. Denote conjugation by $a \mapsto \bar{a}$. If $I$ is coprime to the conductor of $\mathcal O$, is $I \overline I$ principal?
Without the condition with the conductor, this is clearly false. For example, $K = \mathbb Q(\sqrt 2)$, $\mathcal O = \mathbb Z[2\sqrt 2]$, take $I = 2 \mathbb Z + 2\sqrt 2 \mathbb Z$ equal to the conductor.
For an order $O$ in a quadratic ring of integers let $m = [O_K:O]$.
$mO_K \subset O$.
If $J$ is an ideal of $O$ such that $N(J)=|O/J|$ is coprime with $m$ then $m \in O/J^\times$.
$b\in (JO_K) \cap O$ implies $mb\in mJO_K\subset J$ which implies $b\in J$. Thus $(JO_K) \cap O = J$.
$O/J \subset O_K/(JO_K)$ and $O_K/(JO_K)=mO_K/(J\cap mO_K)\subset O/J$ so that $O/J=O_K/(JO_K)$.
If $I$ is an ideal of $O$ with $N(I)$ coprime with $m$ then $I\overline{I}O_K=(IO_K)(\overline{I}O_K) $, from the factorization of $IO_K$ in prime ideals $(IO_K)(\overline{I}O_K) = N(IO_K)O_K=N(I)O_K$ so that $$I\overline{I} = I\overline{I}O_K \cap O = N(I)O_K \cap O = N(I) O \text{ is principal}$$
In $ O = \mathbb Z[3 \sqrt 2]$, $I = (2,3\sqrt{2}),I\overline{I}=(2,3\sqrt{2})(2,-3\sqrt{2})= (4,-(3\sqrt{2})^2,6\sqrt{2})=(2)$