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Let $K$ be a finite extension of $\mathbb{Q}$. Let $O_K$ be the ring of integers of $K$. Let $O$ be an order of $K$, in short $O$ is a subring of $O_K$ which satisfies $[O_K:O]< \infty$. We say that an ideal $\mathfrak{a}$ of $O$ is proper when $\{x\in K : x\mathfrak{a} \subset \mathfrak{a} \}=O$.

My question is:

Is $\mathfrak{a}$ invertible if $\mathfrak{a}$ is proper? If so, what is the proof?

Remark: When $K$ is imaginary quadratic over $\mathbb{Q}$, it is known that $\mathfrak{a}$ is proper if and only if $\mathfrak{a}$ is invertible. The proof requires the theory of quadratic form. (reference: Cox's book 'Primes of the form $x^2 + ny^2$')

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    Sorry, this holds for quadratic fields; see [here](http://math.uga.edu/~pete/8430notes4.pdf). – 2017-01-30
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    Example 2.3 from [this paper](http://matwbn.icm.edu.pl/ksiazki/aa/aa58/aa5813.pdf) seems to show that this result can't be generalized. – 2017-01-30

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No, this kind of orders are called Gorenstein rings. Every number field of degree greater than 2 has infinitely many orders which are not Gorenstein and also infinitely many orders which are Gorenstein.

This subject is treated in the paper

Christian U. Jensen, Anders Thorup; Gorenstein orders, Journal of Pure and Applied Algebra, Volume 219, Issue 3 (2015), P. 551–562.

where some example is done. Keith Conrad's paper (The conductor ideal) talks about that as well.