= 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> gives
$$
\boxed{(2)=P^2,\qquad N(P)=2.}
$$
If $P=(a+bs)$, 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 $P$ gives
$$
(2)I=(\gamma)P.
$$
Norms give $4N(I)=2N(\gamma)$, hence $N(\gamma)=2N(I)$. If $N(I)=p_3\equiv1\pmod{20}$, writing $\gamma=a+bs$ yields
$$
a^2+5b^2=2p_3\equiv2\pmod5,
$$
contradicting the list $0,1,4$ of squares modulo $5$. Therefore \b[no nonprincipal <ideal> has <norm> $p_3$].
As $p_3\equiv1\pmod4$ and $p_3\equiv1\pmod5$, <quadratic reciprocity> gives $(-5/p_3)=1$. Thus $x^2+5$ has two distinct roots modulo $p_3$ and
$$
(p_3)=RR',\qquad N(R)=N(R')=p_3.
$$
The <norm> obstruction shows that $R$ must be principal, say $R=(x+ys)$. Taking its <norm> proves
$$
\boxed{p_3=x^2+5y^2.}
$$
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