Euler prime-generating quadratic from class number one
= Euler prime-generating quadratic from class number one
{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 $m$.