<s>[INST] 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?
 [/INST] Hint:
$$
\left(\frac{3+\sqrt{17}}2\right)\left(\frac{3-\sqrt{17}}2\right)=\frac{9-17}4=-2.
$$
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