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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 ?

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*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.

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**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

$$\frac{2}{\sqrt{5}+1}=\frac{2(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}.$$

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.

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