Forcing the discriminant of an integral basis to be a Carmichael number. I was thinking about the following lemma recently.
Lemma: Let $K=\mathbb{Q}(\theta)$ for some algebraic number $\theta$ and let $n=[K:\mathbb{Q}]$. If $\{\tau_1, \,\dots\,, \tau_n\}$ consists of algebraic integers and the discriminant $\Delta_{K} [\tau_1 ,\, \dots \,, \tau_n]$ is squarefree, then $\{\tau_1,\, \dots\, ,\,\tau_n\}$ is an integral basis for the ring of algebraic integers $\mathcal{O}_{K}$.
[NB: The converse is not true: consider, for example, $\mathbb{Q}(\sqrt{-5})$ with integral basis $\{1, \sqrt{-5}\}$ s.t. $\Delta_{\mathbb{Q}(\sqrt{5})} (1, \sqrt{-5})=-20$.]
It got me thinking about Carmichael numbers (simply because they're squarefree) and so I came up with this question:

How do we restrict $K$ and an integral basis $\mathcal{B}=\{\tau_1,\, \dots\, ,\, \tau_n\}$ for $K$ so that the discriminant $\Delta_{K} (\tau_1,\, \dots \,,\, \tau_n)=C$ is a Carmichael number?

I don't have any coherent ideas, I'm afraid; I hope you don't mind. I've looked around online but I couldn't find anything.

Thoughts: Looking at the definition of $\Delta_{K}$, we might have to restrict the conjugation maps somewhat to make the prime divisors of $C$ pop out.
A Frustrated Sub-question:

Do such discriminants even exist?

I'm looking at $\sqrt{\pm 561}$ and similar numbers to see if I can find one. See my comments below for a description of a similar attempt that might require a computer programme.
Why is this interesting? I don't know: I'm just curious $\ddot\smile$

As you say, by Stickelberger's Theorem, the discriminant of any number field is $0$ or $1$ modulo $4$.
Conversely, if $d \equiv 1 \pmod{4}$ is squarefree, then the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$.  An integral basis for the ring of integers in this case is $\{1$, $\frac{1+\sqrt{d}}{2}\}$.  
It follows that a Carmichael number is a discriminant of a number field if and only if it is $1$ modulo $4$.  Certainly plenty of Carmichael numbers are $1$ modulo $4$ but not all of them: I believe the first counterexample is $8911$.  It is known that there are infinitely many Carmichael numbers (this is one of the more famous results to have come out of my mathematics department), but I am not a Carmichael expert and I don't know whether there are infinitely many which are $1$ modulo $4$.  Is that part of your question?
