Find irreducible but not prime element in $\mathbb{Z}[\sqrt{5}]$ I have tried various numbers of the form $a+b\sqrt{5},\ a,b \in \mathbb{Z}$, but cannot find the one needed.
I would appreciate any help.
Update: I have found that $q=1+\sqrt{5}$ is irreducible. Now if I show that 2 is not divisible by $q$ in $\mathbb{Z}[\sqrt{5}]$ then $2\cdot2 = (\sqrt{5}-1)(\sqrt{5}+1)$ and I'm done. Can it be shown without use of norm ?
The last update: I have written the equation $(\sqrt{5}+1)(x\sqrt{5}+y)=2$ and have deduced that the equation has no integer solutions. Thanks to all who helped.
 A: Hint: $(\sqrt{5}+1)(\sqrt{5}-1)=4$
Added: (after the OP's edit) When it comes to showing that $\sqrt{5}+1$ does not divide $2$, note that 
$$\frac{2}{\sqrt{5}+1}=\frac{2(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}.$$  
A: Hint: the norm map is absolutely crucial here. Because it is multiplicative, taking norms preserves the multiplicative structure and transfers divisibility problems from quadratic number fields to simpler integer divisibility problems (in a multiplicative submonoid of $\mathbb Z$). In this case the transfer is faithful, i.e. an element of your quadratic number ring is irreducible iff its norm is irreducible in the monoid of norms. Similarly, in many favorable cases, a quadratic number ring will have unique factorization iff its monoid of norms does.  
For much more on this conceptual viewpoint see this answer. It is crucial to understand this conceptual viewpoint in order to master (algebraic) number theory. Do not settle for ad-hoc proofs when much more enlightening conceptual proofs are easily within grasp.
