Norm-Euclidean rings? For which integer $d$ is the ring $\mathbb{Z}[\sqrt{d}]$ norm-Euclidean? 
Here I'm referring to $\mathbb{Z}[\sqrt{d}] = \{a + b\sqrt{d} : a,b \in \mathbb{Z}\}$, not the ring of integers of $\mathbb{Q}[\sqrt{d}]$.
For $d < 0$, it is easy to show that only $d = -1, -2$ suffice; but what about $d>0$?
Thanks.

Your exact question has already been resolved and is today a simple matter of looking it up in the OEIS and sifting out the numbers of the form $4k + 1$. 
$$d = 2, 3, 6, 7, 11 \textrm{ or } 19$$
But it was a long, slow process throughout the 20th century to arrive at this answer, as Ian Stewart and David Tall explain in their excellent book Algebraic Number Theory and Fermat's Last Theorem.

Here I'm referring to $\mathbb{Z}[\sqrt{d}] = \{a + b\sqrt{d} : a, b \in \mathbb{Z}\}$, not the ring of integers of $\mathbb{Q}[\sqrt{d}]$.

Because you said this, it's necessary to sift out the numbers of the form $4k + 1$. Stewart & Tall (and many other authors in other books) show that if a domain is Euclidean then it is a principal ideal domain and a unique factorization domain (the converse doesn't always hold, but that's another story).
So if $d > 5$, $d \equiv 1 \pmod 4$ and $$m = \frac{d - 1}{4},$$ (that's an integer) then $d - 1 = 2^2 m = (-1)(1 - \sqrt d)(1 + \sqrt d)$ represents two distinct factorizations of the same number, which means that $\mathbb Z[\sqrt d]$ is not a unique factorization domain, which in turns means it can't be Euclidean and certainly not norm-Euclidean.
If you had asked about Euclidean in general, you question would be significantly more difficult. The Euclidean function for $\mathcal O_{\mathbb Q(\sqrt{69})}$ was only recently (just a couple of decades ago) discovered: the norm function requires two adjustments and is somewhat frustrating for practical applications.
And it has been proven that $\mathbb Z[\sqrt{14}]$ is Euclidean but not norm-Euclidean, but no one on Math.SE seems to know what the Euclidean function might be (it has been asked).
