Let $K=\mathbb{Q}(\sqrt{10})$ and put $\varepsilon=3+\sqrt{10}$.

(a) Show that 2,3 and $\varepsilon+1$ are irreducible elements in $\mathcal{O}_{K}$. Deduce from the equation

$$6=2 \cdot 3=(\varepsilon+1)(\bar{\varepsilon}+1)$$

that $\mathcal{O}_{K}$ is not a principal ideal domain.

(b) Put $\mathfrak{p}_{2}=[2, \varepsilon+1]$ and $\mathfrak{p}_{3}=[3, \varepsilon+1]$. Show that

$$[2]=\mathfrak{p}_{2}^{2}, \quad[3]=\mathfrak{p}_{3} \overline{\mathfrak{p}}_{3}, \quad \mathfrak{p}_{2} \mathfrak{p}_{3}=[\varepsilon+1], \quad \mathfrak{p}_{2} \overline{\mathfrak{p}}_{3}=[\varepsilon-1] .$$

Deduce that $K$ has class number 2 .

(c) Show that $\varepsilon$ is the fundamental unit of $K$. Hence prove that all solutions in integers $x, y$ of the equation $x^{2}-10 y^{2}=6$ are given by

$$x+\sqrt{10} y=\pm \varepsilon^{n}\left(\varepsilon+(-1)^{n}\right), \quad n=0,1,2, \ldots$$