What is the Euclidean function for $\mathbb{Z}[\sqrt{14}]$? I've tried a few different pairs of numbers in $\mathbb{Z}[\sqrt{14}]$ and in each case I've been able to find a remainder for which the absolute value of the norm is suitable for the Euclidean algorithm. But the Tooth Fairy tells me that $\mathbb{Z}[\sqrt{14}]$ is not norm-Euclidean. If not the absolute value of the norm, then what is the Euclidean function for $\mathbb{Z}[\sqrt{14}]$? This riddle has me stumped.

It's not the norm function, that much I can tell you for sure.
From an answer to this question Have I found an example of norm-Euclidean failure in $\mathbb Z [\sqrt{14}]$? I have gleaned a pair of numbers that prevent a successful run of the Euclidean algorithm with the absolute value of the norm as the Euclidean function. And it helps that in a comment to yet another question on this topic the answerer gives a specific pair of numbers.
$$\gcd(2, 1 + \sqrt{14}) = 1$$
Now I will work out the specifics of why that specific result can't be obtained by the norm-Euclidean algorithm.
Since $N(1 + \sqrt{14}) = -13$ and $N(2) = 4$, we need to solve $1 + \sqrt{14} = 2q + r$ so that $-4 < N(r) < 4$.
We already know that 2 and $1 + \sqrt{14}$ are coprime, so $r = 0$ is hardly worth mentioning.
The fundamental unit is $\eta = 15 + 4\sqrt{14}$ (note that $N(\eta) = 1$, not $-1$). If $\eta^n = a + b\sqrt{14}$, then it is obvious that $a$ must be odd and $b$ even. But if $2q = c + d\sqrt{14}$ then both $c$ and $d$ are even, and $r = \eta^n$ will not work because we need $b + d = 1$ (though, for what it's worth, $a + c = 1$ is possible).
Let's try now for $N(r) = 2$, which means that $r = \pm(4 \pm \sqrt{14})^n$. If $r = a + b\sqrt{14}$, then $a$ is always even ($n = 0$ doesn't count because that gives us one of the units, which we've already gone over). But $2q = c + d\sqrt{14}$ has $c$ even, and in this case $a + c = 1$ is impossible.
Since $$\left(\frac{14}{3}\right) = -1,$$ 3 is prime in this domain and $N(x) = \pm 3$ is impossible.
Hence the Euclidean algorithm is stymied by this pair of numbers.

I read somewhere that someone has compiled a lot of empirical evidence suggesting that these difficulties can be overcome by "cheating" in $\mathbb Z[\sqrt{14}, \frac{1}{2}]$.
It's not proven, but it does suggest a possible Euclidean function: if the norm is even, then double it. Then we'd have $\tilde N(2) = 8$, and we'd be able to proceed thus: $1 + \sqrt{14} = 2 \times 2 + (-3 + \sqrt{14})$, then $2 = (-3 + \sqrt{14})(-4 - \sqrt{14}) + (4 + \sqrt{14})$ (this still works even though we have to double the norm of $4 + \sqrt{14}$ to 4), and then finally $-3 + \sqrt{14} = (4 + \sqrt{14})(-11 + 4\sqrt{14}) - \eta$. Obtaining a unit as a remainder confirms the two initial numbers are coprime.
But this does not rule out the possibility that someone could find a pair of numbers that similarly stymies this Euclidean function I'm proposing.
