 Solution

Put $s=\sqrt{-5}$ and $P=(2,1+s)$. The polynomial $x^2+5$ reduces to $(x+1{)}^2$ modulo $2$, so the Kummer-Dedekind theorem givesIf $P=(a+bs)$, its norm would imply $a^2+5b^2=2$, impossible for integers $a,b$. Thus $P$ represents the nontrivial ideal class. Since the class number is $2$, every nonprincipal integral ideal $I$ has $[I]=[P]$, and $[IP]=1$. Write $IP=(\gamma )$ with $\gamma \in {\mathcal{O}}_M$; integrality follows from $IP\subseteq {\mathcal{O}}_M$. Multiplying by $P$ givesNorms give $4N(I)=2N(\gamma )$, hence $N(\gamma )=2N(I)$. If $N(I)=p_3\equiv 1(\mathrm{mod}20)$, writing $\gamma =a+bs$ yieldscontradicting the list $0,1,4$ of squares modulo $5$. Therefore no nonprincipal ideal has norm $p_3$.

As $p_3\equiv 1(\mathrm{mod}4)$ and $p_3\equiv 1(\mathrm{mod}5)$, quadratic reciprocity gives $(-5/p_3)=1$. Thus $x^2+5$ has two distinct roots modulo $p_3$ andThe norm obstruction shows that $R$ must be principal, say $R=(x+ys)$. Taking its norm proves