 Euler prime-generating quadratic from class number one

Let $m\ge 2$, let $p=4m-1$ be prime, and suppose $\mathbb{Q}(\sqrt{-p})$ has class number one. If $n\ge 0$ and $n^2+n+m<m^2$, then $n^2+n+m$ is prime. The key identity is the algebraic normA hypothetical prime divisor below $m$ would split and yield an algebraic integer of that norm, but the positive norm form $(a^2+p{b}^2)/4$ represents no prime below $m$.