Given $d \equiv 1 \pmod 4$, can $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ have a fundamental unit that is not a “half-integer”?

Also $d > 0$, $\mu(d) \neq 0$.

I acknowledge the term "half-integer" is, at best, problematic. What I mean by it is that an integer in $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ is of the form $\frac{a + b \sqrt{d}}{2}$ with $a$ and $b$ both odd.

Please bear with me, I'm very new to this subject. If I've done it correctly, the fundamental units in $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ for $d = 5, 13, 17, 21$ are $$\frac{1 + \sqrt{5}}{2}, \frac{3 + \sqrt{13}}{2}, \frac{5 + 3 \sqrt{17}}{2}, \frac{5 + \sqrt{21}}{2}.$$

Can the fundamental unit in such a domain possibly be of the form $a + b \sqrt{d}$ rather than $\frac{a + b \sqrt{d}}{2}$? And if so, when factoring other integers in that domain, how can you be sure you haven't overlooked a "half-integer" factor?

EDIT: It's been pointed out that I screwed up the calculation in $\mathcal{O}_{\mathbb{Q}(\sqrt{17})}$. Still, I appreciate any insights you all might have regarding fundamental units in these domains.

• It turns out that fundamental units in the rings of integers you describe are always "half-integers". en.wikipedia.org/wiki/… – Greg Martin Jul 14 '15 at 21:25
• @Greg, as much as I'd love to say that Wikipedia led you down the wrong path, it seems that I can't. But I can at least say that that article, technically correct though it may be, is very poorly written and needlessly confusing. – Robert Soupe Jul 15 '15 at 0:36

Yes, it can. An arithmetic mistake on your part has masked an example you would otherwise have found yourself. Verify that $$N\left(\frac{5}{2} + \frac{3\sqrt{17}}{2}\right) = \frac{25}{4} - \frac{153}{4} = -32.$$
In order for $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ to have a half-integer fundamental unit, there has to be a solution to $x^2 - dy^2 = \pm 4$ with $x$ and $y$ both odd, and there isn't when $d = 17$. Consider this modulo 16: the odd squares are 1 and 9, and since $17 \equiv 1 \pmod{16}$, we see that $1 - 1 = 0$, $1 \pm 9 = 8$, $9 - 9 = 0$. But $4 \not\equiv 0$ nor $8 \pmod{16}$. Hence the fundamental unit of $\mathcal{O}_{\mathbb{Q}(\sqrt{17})}$ is $4 + \sqrt{17}$.
At least for primes $p$ what you can do is see whether you can solve $x^2 - dy^2 = \pm 4p$ with $x$ and $y$ odd. For composite numbers, you take this to the constituent prime factors, but if $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ is not UFD then you also have to watch out for potentially multiple distinct factorizations, as well as non-obviously non-distinct factorizations (fun, right?).
Let me give you an example that is neither one more than a power of 2 nor one more than a square: given $d = 41$, we have $32 - 5\sqrt{41}$ as the fundamental unit.