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.

The purpose of this community wiki post is to show that $\mathbb Z[\sqrt{14}]$ is does not fail the first step in Motzkin's construction of a Euclidean algorithm, thereby providing some useful background for the other answers.
Every number of even norm is divisible by $4 + \sqrt{14}$, which has a norm of 2.
If $a + b \sqrt{14}$ has even norm and $a$ and $b$ are both even, we can set $$c = \frac{a}{2}, d = \frac{b}{2}.$$ It doesn't matter if $c + d \sqrt{14}$ is itself divisible by $4 + \sqrt{14}$, because 2 is.
Of course $a + b \sqrt{14}$ can still have even norm if $b$ is odd, provided that $a$ is still even. It's only a tiny bit more complicated to show that such a number is also divisible by $4 + \sqrt{14}$. Set $$c = -7b + 2a, d = 2b - \frac{a}{2}$$ and verify that $(4 + \sqrt{14})(c + d \sqrt{14}) = a + b \sqrt{14}$.
Now suppose that $a + b \sqrt{14}$ has odd norm. This means that $a$ must be odd. But if you increment or decrement $a$ by 1, you have a number of even norm which is therefore divisible by $4 + \sqrt{14}$. This means that $4 + \sqrt{14}$ is a universal side divisor in this domain (not everyone likes that terminology, but it seems to be almost standard).
The presence of a universal side divisor means the second subset in Motzkin's construction is non-empty so that Motzkin's construction does not fail at this level.  When the second subset is empty, as for example in $\mathcal O_{\mathbb Q(\sqrt{-19})}$, we can conclude that the ring is not a Euclidean domain.  In this case the second subset is non-empty.  We cannot conclude that $\mathbb Z[\sqrt{14}]$ is Euclidean, but we know that the technique used to prove domains non-Euclidean by failure of Motzkin's construction does not apply to $\mathbb Z[\sqrt{14}]$.  It also gives no guidance as to what the Euclidean function might be.
It should be noted, however, that for many pairs of numbers in $\mathbb Z[\sqrt{14}]$ the absolute value of the norm does properly work as a Euclidean function. For example, using the foregoing, it should be a breeze to find $r = 0$ or 1 for $\gcd(4 + \sqrt{14}, a + b \sqrt{14})$. Certain multiples of $4 + \sqrt{14}$ could be problematic, however, as the other answers will elaborate.
