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

LetIts minimal polynomial is $X^2-X+m$, and the algebraic norm givesSuppose that $N$ is composite and choose a prime divisor $q\le \sqrt{N}<m$. Since $X^2-X+m$ has the root $-n$ modulo $q$, the idealis 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 elementswhose norm is $(a^2+p{b}^2)/4$. If $b=0$, this norm is a square and cannot be the prime $q$. If $b\ne 0$, thenand integrality gives $q\ge m$, again a contradiction. Therefore the Euler prime-generating quadratic from class number one yields