$4+\sqrt{5}$ 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 $\mathbb{Z}(\sqrt{5})$
 A: OK, after giving a wrong solution this morning (I blame it on the lack of coffee...), here is another try:
The standard field norm on $\mathbb{Q}[\sqrt{5}]$ is $N(a+b\sqrt{5}) = |a^2 - 5b^2|$. It is multiplicative, and $N(4+\sqrt{5}) = |4^2 - 5\cdot 1^2| = 11$ is prime, so $4+\sqrt{5}$ is irreducible in $\mathbb{Z}[\sqrt{5}]$, and more generally in the ring of integers of $\mathbb{Q} [\sqrt{5}]$, which is $\mathbb{Z}\left[\frac{1+\sqrt{5}}2\right]$.
Now irreducibility in general does not imply primality, but it does in Euclidean domains. (More general, this statement is true in Unique Factorization Domains, and every Euclidean domain is a UFD.) It is known that $\mathbb{Q}[\sqrt{5}]$ is norm-Euclidean, i.e., that the standard field norm $N$ is Euclidean on the ring of integers $\mathbb{Z}\left[\frac{1+\sqrt{5}}2\right]$. The standard reference for the discussion of the question for which integers $d$ the domain $\mathbb{Z}[\sqrt{d}]$ is Euclidean seems to be the book of Hardy and Wright.
Now if $4+\sqrt{5}$ divides some $a+b\sqrt{5}$ in $\mathbb{Z}\left[\frac{1+\sqrt{5}}2\right]$, then $$  \begin{split} a+b\sqrt{5} &= (4+\sqrt{5})\left(c+d\frac{1+\sqrt{5}}2\right) \\ & = 4c+2d+\frac{5}2d + \left( c+\frac{d}2+ 2d)\right)\sqrt{5}. \end{split} $$
Since the coefficients are integers $a$ and $b$, we get that $d$ is even, and so $c+ d \frac{1+\sqrt{5}}2 \in \mathbb{Z}[\sqrt{5}]$. This implies that $4+\sqrt{5}$ actually divides $a+b\sqrt{5}$ in $\mathbb{Z}[\sqrt{5}]$. Now if $4+\sqrt{5}$ divides a product $xy$ with $x,y \in \mathbb{Z}[\sqrt{5}]$, then it divides $x$ or $y$ in $\mathbb{Z}[\frac{1+\sqrt{5}}2]$, so by the above argument it divides $x$ or $y$ in $\mathbb{Z}[\sqrt{5}]$, showing that it is a prime element of this ring.
