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I have come to know about $\mathcal O_K$ with $K = \mathbb Q(\sqrt{69})$. The norm function needs to be adjusted to absolute value, as is the case with other real rings, but it also needs to be adjusted for primes lying over $23$.

Is there another real quadratic integer ring that is Euclidean but not norm-Euclidean, but for which the norm function can, with two adjustments (absolute value and something else), be used as the Euclidean function? And if so, what is it?

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The answer certainly is not known to be ${\mathbb Z}[\sqrt{14}]$: this ring is Euclidean, as was shown by Harper, and it is not norm-Euclidean, as has been known for a long time. Whether it is Euclidean for a weighted norm is not known (see Prop. 4.11 here), and to the best of my knowledge nobody so far has managed to deal with weighted norms at two prime ideals.

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  • $\begingroup$ That's a paper that I really need to print out, sit down and read beginning to end. There's enough in Proposition 4.11 that's over my head, or it might be explained by what you've written in the lead up to it. One immediate question arises that I will post as a separate question here. $\endgroup$ – Mr. Brooks Jan 3 '17 at 22:31
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Yes, $\mathbb{Z}[\sqrt{14}]$. Absolute value of the norm, then if the norm is not divisible by $14$, square it.

Thus, for example, $\gcd(3, 3 + \sqrt{14}) = 1$ since $3 = 1 \times (3 + \sqrt{14}) + \sqrt{14}$ (now $3 + \sqrt{14}$ has an adjusted norm of $25$ while $\sqrt{14}$ has an adjusted norm of $14$.

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    $\begingroup$ This is probably right in the general idea but I'm skeptic of the specifics. $\endgroup$ – Mr. Brooks Jul 13 '16 at 20:40

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