= Solution
Write $r=\sqrt{10}$. The ideals are $\mathfrak p_2=(2,r)$, $\mathfrak p_3=(3,r+1)$ and $\overline{\mathfrak p}_3=(3,r-1)$, with residue fields of orders 2, 3 and 3 respectively. Thus their ideal norms are 2, 3 and 3.
The generators of $\mathfrak p_2^2$ are $4,2r,10$. They are multiples of 2, and the integer combination of 4 and 10 giving 2 proves $\mathfrak p_2^2=(2)$. Similarly $\mathfrak p_3\overline{\mathfrak p}_3$ is generated by $9,3r+3,3r-3,r^2-1=9$. The difference of the middle generators is 6; together with 9 this generates 3. Therefore the product is $(3)$.
The ideals $\mathfrak p_2$ and $\mathfrak p_3$ are coprime, so their product equals their intersection. The element $4+r$ belongs to both, hence $(4+r)\subseteq\mathfrak p_2\mathfrak p_3$. Both ideals have norm 6, so equality holds. Likewise $2+r$ belongs to $\mathfrak p_2$ and $\overline{\mathfrak p}_3$, and its norm is $-6$. Thus
$$
\boxed{(2)=\mathfrak p_2^2,\quad(3)=\mathfrak p_3\overline{\mathfrak p}_3,\quad
\mathfrak p_2\mathfrak p_3=(\epsilon+1),\quad
\mathfrak p_2\overline{\mathfrak p}_3=(\epsilon-1).}
$$
The <discriminant> is 40. The <Minkowski bound for ideal classes> in a real quadratic field is $\sqrt{40}/2=\sqrt{10}<4$, so every <ideal class> has an integral representative of norm 1, 2 or 3. The only possible prime ideals with those norms are the ones just listed. Their classes satisfy $[\mathfrak p_2]^2=1$ and $[\mathfrak p_3]=[\overline{\mathfrak p}_3]=[\mathfrak p_2]$. Thus there are at most two classes. The ideal $\mathfrak p_2$ is not principal, since a generator would have norm $\pm2$, already ruled out. Therefore \b[the <class number> is exactly two].
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