Showing that $\mathbb Q(\sqrt{17})$ has class number 1 Let $K=\mathbb Q(\sqrt{d})$ with $d=17$. The Minkowski-Bound is $\frac{1}{2}\sqrt{17}<\frac{1}{2}\frac{9}{2}=2.25<3$.
The ideal $(2)$ splits, since $d\equiv 1$ mod $8$. So we get $(2)=(2,\frac{1+\sqrt{d}}{2})(2,\frac{1-\sqrt{d}}{2})$ and $(2,\frac{1\pm\sqrt{d}}{2})$ are two ideals of norm $2$.
Now if we can show that $(2,\frac{1\pm\sqrt{d}}{2})$ are principal ideals, then we know that every ideal class contains a principal ideal, which shows that the class number is $1$.
But how can we show that $(2,\frac{1\pm\sqrt{d}}{2})$ are principal?
 A: Here because of a duplicate. Though it does seem like the original asker does pop in from time to time.
I'm not sure if the duplicate asker is aware of algebraic integers such as $$\theta = \frac{1}{2} + \frac{\sqrt{17}}{2},$$ which is a solution to $x^2 - x - 4$, though the duplicate asker is aware of the Minkowski bound.
So this tells us $N(\theta) = -4$, while obviously $N(2) = 4$. This suggests that $$¿ N\left(\left\langle 2, \frac{1}{2} + \frac{\sqrt{17}}{2} \right\rangle\right) = 4 ?$$ However, if $\mathcal O_{\mathbb Q(\sqrt{17})}$ does indeed have class number 1, that would mean that 16 has one distinct factorization (ignoring multiplication by units) and so $$16 = 2^4 = \left(\frac{1}{2} - \frac{\sqrt{17}}{2}\right)^2 \left(\frac{1}{2} + \frac{\sqrt{17}}{2}\right)^2$$ represents incomplete factorizations, just like $16 = 4^2 = 2 \times 8$ in $\mathbb Z$.
It's not a given that this is a Euclidean domain even if it does have class number 1. However, it wouldn't hurt to try. And so we find by the Euclidean algorithm that $$\gcd\left(2, \frac{1}{2} + \frac{\sqrt{17}}{2}\right) = \frac{5}{2} + \frac{\sqrt{17}}{2},$$ and indeed $$2 + \frac{1}{2} + \frac{\sqrt{17}}{2} = \frac{5}{2} + \frac{\sqrt{17}}{2}.$$
Furthermore, since $$\frac{5}{2} - \frac{\sqrt{17}}{2} \in \left\langle \frac{5}{2} + \frac{\sqrt{17}}{2} \right\rangle,$$ it follows that $$\langle 2 \rangle = \left\langle \frac{5}{2} + \frac{\sqrt{17}}{2} \right\rangle^2.$$ That's a principal ideal after all.
Since $$\left(\frac{17}{3}\right) = -1$$ (that's the Legendre symbol), 3 is prime in this ring. But it's well over the Minkowski bound anyway, so we're done.
EDIT: Jyrki Lahtonen points out a mistake I made regarding $\langle 2 \rangle$. The correct factorization is $$\langle 2 \rangle = \left\langle \frac{5}{2} - \frac{\sqrt{17}}{2} \right\rangle \left\langle \frac{5}{2} + \frac{\sqrt{17}}{2} \right\rangle.$$ This does not detract from the point that these are all principal ideals.
A: Hint:
$$
\left(\frac{3+\sqrt{17}}2\right)\left(\frac{3-\sqrt{17}}2\right)=\frac{9-17}4=-2.
$$
A: We show that $$\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle = \left\langle \frac{5+\sqrt{17}}{2}\right\rangle.$$
Since, $$\frac{5+\sqrt{17}}{2}=2+\frac{1+\sqrt{17}}{2},$$ we have $$\frac{5+\sqrt{17}}{2}\in\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle,$$ thus $$\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle \subseteq \left\langle \frac{5+\sqrt{17}}{2}\right\rangle.$$
Now these two ideals have the same Norm--namely, $2$. Therefore, $$\left\langle 2,\frac{1+\sqrt{17}}{2}\right\rangle = \left\langle \frac{5+\sqrt{17}}{2}\right\rangle.$$
The proof of the principality of $\frac{5+\sqrt{17}}{2}$ is similar.
A: The quadratic integer $n=(1\pm\sqrt{17})/2$ has minimal polynomial $n^2-n-4$, so the ideals
$\left(2,\dfrac{1\pm\sqrt{17}}{4}\right)$
will be principal if natural integers $x,y$ exist such that
$|x^2-xy-4y^2|=2.$
As this condition holds when $x=2,y=1$ we are done.
