$\mathbb{Q}(\sqrt{23})$ is not a Euclidean number field. The problem I'm facing is that of the tittle: 

Problem. Prove that $\mathbb{Q}(\sqrt{23})$ is not a Euclidean number field.

Since $23\not\equiv 1\pmod{4}$, it must be shown that $\mathbb{Z}[\sqrt{23}]$ is not a Euclidean domain.
I know how to show that it is not norm-Euclidean, but it still could be Euclidean with a different function.
The only way I know how to prove that a domain is not Euclidean is by using Motzkin's theorem:

Theorem. A domain $D$ is Euclidean if and only if
  $\bigcap\limits_{i\in\mathbb{N}} P_0^{(i)} = \emptyset$.

Here $P_0^{(0)} = P_0 := D\setminus\{0\}$ and we define recursively $P^{(i+1)}_0 := (P^{(i)}_0)'$ for all $i\in\mathbb{N}$, where for any $P\subseteq D$ the derived set $P'$ of $P$ is given by $P' := \{b\in P\, ;\, \exists a\in D\text{ such that }a+bD\subseteq P\}$.
It works well for imaginary quadratic fields since they have only $1$ and $-1$ as units: Considering $K := \mathbb{Q}(\sqrt{d})$ for $d < -11$ it can be shown that $P_0^{(i)} = A\setminus\{0, 1, -1\}$ for all integers $i\ge 1$, where $A$ is the ring of integers of $K$, hence by Motzkin theorem $K$ is not Euclidean.
But in $\mathbb{Q}(\sqrt{23})$ we have too many units and I can't even determine $P_0''$.
 A: The ring ${\mathbb Z}[\sqrt{23}]$, like every real quadratic field with class number $1$ and discriminant $< 500$, is Euclidean, as was proved by Harper in his thesis. Other relevant and more recent publications can be found on Ram Murty's homepage; see e.g. here or here.
A: What I'm about to say is based on two results I have not verified for myself. First, I take it on faith that $\mathbb{Z}[\sqrt{23}]$ is a principal ideal domain and therefore a unique factorization domain. Second, that if a ring is Euclidean it has a universal side divisor, and that an element of minimum norm among non-units in an Euclidean ring is a universal side divisor.
Since $N(5 + \sqrt{23}) = 2$ and we're dealing with a unique factorization domain, it follows that any number in this domain with even norm is divisible by $5 + \sqrt{23}$. If a number in this domain has odd norm, then it suffices to add or subtract $1$ to it to make it a number of even norm.
Then for $\gcd(a, b)$ where $a \neq 0$ and $b = 5 + \sqrt{23}$, we can always find a suitable $q \neq 0$ such that $a = qb \pm 1$; obviously $|N(a)| \geq 1$. Therefore $5 + \sqrt{23}$ is a universal side divisor and therefore also $\mathbb{Z}[\sqrt{23}]$ is an Euclidean domain.
I take your word for it that you can prove that this domain is not norm-Euclidean. I'm guessing the minimal example of norm-Euclidean failure is $\gcd(4 + \sqrt{23}, 5)$, you'll let me know if that's wright or wrong. I have no idea what the Euclidean function could be.
