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I was wondering if there is a classification for this:

For which $d$ is $D=\mathbb{Z}[\sqrt{d}]$ a UFD, with $d > 1$?

For $d \equiv 1 $ (mod $ 4$), $D$ is not a UFD (proof here).

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When $d \not\equiv 1\; (\textrm{mod}\; 4)$, the ring of integers of $\mathbb{Q}(\sqrt{d})$ is $\mathbb{Z}[\sqrt{d}]$. $\mathbb{Z}[\sqrt{d}]$ is a UFD if and only if it has trivial class group (i.e., the class number of $\mathbb{Q}(\sqrt{d})$ is $1$).

However, it's an open question as to whether or not there are infinitely many $d>0$ with $\mathbb{Q}(\sqrt{d})$ having class number 1, so the answer is not known.

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The first sentence assumes $d$ is square-free. Also, while there's no classification, there is a list of some known UFDs (here). – user236182 Aug 24 '15 at 8:57

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