I am kinda stuck and am looking for a hint on how to proceed.
Let $m > 2$ be an integer such that $p = 4m − 1$ is prime. Suppose that the ideal class group of $L = Q(\sqrt{−p})$ is trivial. Show that if n > 0 is an integer and $n^2+n+m < m^2$, then $n^2 + n + m$ is prime.
My thoughts so far are:
let $a=(1+\sqrt{(-p)})/2$ the $O_k= \mathbf{Z}[a]$. Also as the class group is trivial $O_k$ is a UFD.
Also $W=n^2+n+m=(a+n)(\bar{a}+n)$ where $\bar{a}$ is the galois conjugate of $a$.
I want to apply dedekind using $a$ which has minimal polymoial $f(X)=X^2-X+m$. We note for all primes $q$ diving $W$ $f(-n)=W=0 \pmod q$ and so as ideals $(q)=(q, n+a)(q,n+\bar{a})=Q_1 \bar{Q_1}$ by dedekind. We note f is inseparable iff $2(-n)-1=0 \pmod q$.
This is where I struggle to make progress. I want to use the $<m^2$ bound somehow and the fact these ideals are principals but I am not sure how. Any hints?
We note our above analysis shows that if W is composite there's an prime ideal with norm $< m$. But $norm(x+y\sqrt{-p})=x^2+py^2$ which is always greater than $m$ if $|y|>1/2$ and we when we have $y=1/2$ we have $|x|\geq 1/2$ and so the norm is greater than m. And if $y=0$ $x$ is an integer and the norm can never be prime. So there are no (principal) ideals of norm a prime less than m and so we have a contradiction.