Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 3 / 1G (Groups, Rings and Modules)

Since $5y^2\le 9$, we have $y\in \{-1,0,1\}$. For $y=0$, $x^2=9$; for $y=\pm 1$, $x^2=4$. Thus the solutions of the Diophantine equation arewhere the two signs in the second family are independent.

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.