Let $K$ be a real quadratic number field and $\mathcal O_K$ its ring of integers. Let $\mathfrak a \subset K$ be a fractional ideal and $\varepsilon \in \mathcal O_K^\times$.
My question: Can $\varepsilon-1$ always be written as product $xy$ with $x \in \mathfrak a$ and $y \in \mathfrak a^{-1}$?
What I've done so far:
If $\mathfrak a$ is of the form $\alpha \mathcal O_K$ with $\alpha \in K$ it's easy to see that there is such a representation. So in order to find counter-examples we have to look at number fields with class number greater 1.
So I looked a little bit into the first such number field (ordered by discriminant) $K=\mathbb Q(\sqrt{10})$ and haven't found a counter-example there so far. For example if you take $\mathfrak a=(2,\sqrt{10})$ and $\varepsilon = -1$ you have $-2=-2 \cdot 1$ (note that $\mathfrak a^{-1}=\mathfrak a/2$ in this case).
If $\varepsilon-1$ is a unit this works iff $\mathfrak a$ is principal. A short analysis yields that $\varepsilon-1$ can only be a unit in $\mathbb Q(\sqrt{5})$ where we have class number 1.
I just found an counter-example. We consider $K=\mathbb Q(\sqrt{34})$. In $\mathcal O_K$ we define $\mathfrak p:=(6-\sqrt{34})$ and $\mathfrak a:=(3,1+\sqrt{34})$. We have $N(\mathfrak p)=2$ and $N(\mathfrak a)=3$ so both ideals are prime. Clearly $\mathfrak p$ is principal. It can be seen that $\mathfrak a$ isn't. It follows that $\mathfrak p\mathfrak a$ and $\mathfrak p\mathfrak a'$ are the only two ideals of norm $6$ and both can't be principal. This implies that $\mathcal O_K$ doesn't contain elements of norm $\pm 3$ and $\pm 6$.
Now let $x \in \mathfrak a$ and $y \in \mathfrak a^{-1}=\mathfrak a'/3$ with $xy=-2$. Then $3y \in \mathfrak a'$. So $3 \mid N(x)$ and $3 \mid N(3y)$. Furthermore, we have $N(x)N(3y)=36$. Since there are no elements in $\mathcal O_K$ of norm $\pm 3$ and $\pm 6$ this implies that $|N(x)|>6$ and $|N(3y)|>6$. Contradiction.