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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 $m$.

The algebraic norm on $\mathbb{Z}[\sqrt{-5}]$ is $N(a+b\sqrt{-5})=a^2+5b^2$ and satisfies multiplicativity of the algebraic norm. The calculation in part (a) therefore gives all elements of norm $9$:If one of these factored into two nonunits, multiplicativity would force both factors to have norm $3$. But the Diophantine equation $a^2+5b^2=3$ has no solution. Hence all six displayed elements are irreducible. This is a standard example in which an irreducible element need not have prime norm.

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