definite ring $\mathbb{Z}(\sqrt{5})=\{a+\sqrt{5}b\,|\,a,b\in \mathbb{Z}\}$
show that $4+\sqrt{5}$ is a prime member of $\mathbb{Z}(\sqrt{5})$
definite ring $\mathbb{Z}(\sqrt{5})=\{a+\sqrt{5}b\,|\,a,b\in \mathbb{Z}\}$
show that $4+\sqrt{5}$ is a prime member of $\mathbb{Z}(\sqrt{5})$