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?

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.
