= Solution
For $p\equiv1\pmod{20}$, put $\theta=\sqrt{-p}$. Since $-p\equiv3\pmod4$, the <ring of integers of a quadratic field> formula gives $R=\mathbb Z[\theta]$. Modulo five,
$$
X^2+p\equiv X^2+1=(X-2)(X+2).
$$
The two roots are distinct, so the <Dedekind factorization theorem> gives
$$
(5)=P\overline P,\qquad
P=(5,\theta-2),\quad\overline P=(5,\theta+2),\quad
N(P)=N(\overline P)=5.
$$
In particular $P\ne\overline P$.
We prove that $[P]$ has <order of a group element> greater than $n$. Suppose instead that $P^k$ is a <principal ideal> for some $1\leq k\leq n$. Since $P^k\subseteq R$, its generator is in $R$: write $P^k=(a+b\theta)$ with $a,b\in\mathbb Z$. Taking <ideal norms> gives the positive <field norm>
$$
5^k=a^2+pb^2.
$$
But $5^k\leq5^n<p$, so $b=0$. Thus $a^2=5^k$, forcing $k$ to be even and $a=\pm5^{k/2}$. This would give
$$
P^k=(a)=(5)^{k/2}=P^{k/2}\overline P^{k/2},
$$
contrary to <unique factorization of ideals in a number field> and $P\ne\overline P$. Consequently no such $k$ exists. The <finiteness of the ideal class group> ensures that $[P]$ has finite <order of a group element>, so
$$
\boxed{\operatorname{ord}([P])>n}.
$$
For every positive <integer> $n$, the allowed infinitude of <prime numbers> in the progression $1\pmod{20}$ supplies one larger than $5^n$. Applying the construction proves <unbounded element orders in imaginary quadratic ideal class groups>. In particular, this establishes elements of arbitrarily large <order of a group element>, rather than merely unbounded <class numbers>.
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