Let
$$
\theta=\frac{1+\sqrt{-p}}2\in\mathcal O_L.
$$
Its minimal polynomial is $X^2-X+m$, and the <algebraic norm> gives
$$
N_{L/\mathbb Q}(n+\theta)=n^2+n+m=:N.
$$
Suppose that $N$ is composite and choose a prime divisor $q\leq\sqrt N<m$. Since $X^2-X+m$ has the root $-n$ modulo $q$, the ideal
$$
\mathfrak q=(q,n+\theta)
$$
is a <prime ideal> of <ideal norm> $q$. The triviality of the <ideal class group> makes $\mathfrak q=(\beta)$ principal, so $|N_{L/\mathbb Q}(\beta)|=q$.

The <ring of integers of a quadratic field> consists of elements
$$
\beta=\frac{a+b\sqrt{-p}}2,
\qquad a\equiv b\pmod2,
$$
whose norm is $(a^2+pb^2)/4$. If $b=0$, this norm is a square and cannot be the prime $q$. If $b\ne0$, then
$$
q=\frac{a^2+pb^2}{4}\geq\frac p4=m-\frac14,
$$
and integrality gives $q\geq m$, again a contradiction. Therefore the <Euler prime-generating quadratic from class number one> yields
$$
\boxed{\ n^2+n+m\text{ is prime}.\ }
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 20G / a / Solution

{c}

Let $m\geq2$, let $p=4m-1$ be prime, and suppose $\mathbb Q(\sqrt{-p})$ has class number one. If $n\geq0$ and $n^2+n+m<m^2$, then $n^2+n+m$ is prime. The key identity is the <algebraic norm>
$$
n^2+n+m=N\left(n+\frac{1+\sqrt{-p}}2\right).
$$
A hypothetical prime divisor below $m$ would split and yield an algebraic integer of that norm, but the positive norm form $(a^2+pb^2)/4$ represents no prime below $m$.

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