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.

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    $\begingroup$ Why don't you want to use the norm? $\endgroup$ Commented Feb 7, 2012 at 15:12
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    $\begingroup$ Indeed. Especially since "using the norm" is a bit of an overstatement -- all you're doing is multiplying by $\overline{q}$. $\endgroup$ Commented Feb 7, 2012 at 15:13
  • $\begingroup$ @AlexBecker because then I need to prove that there is a norm on the ring and the ring is integral. $\endgroup$ Commented Feb 7, 2012 at 15:15
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    $\begingroup$ Let's assume that bar means Galois conjugate :). But I think André's suggestion is simpler. $\endgroup$ Commented Feb 7, 2012 at 15:21
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    $\begingroup$ It is actually easier to show that $2$ is irreducible, and that it does not divide $1+\sqrt{5}$ (or $1-\sqrt{5}$)... $\endgroup$ Commented Feb 7, 2012 at 15:59

2 Answers 2


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

  • $\begingroup$ I have shown that by another method. Maybe It's the same as rationalization of denominator. $\endgroup$ Commented Feb 7, 2012 at 15:28
  • $\begingroup$ @Sergey Filkin: Your method works just fine. Both your method and rationalizing solve the problem of finding rationals $x$ and $y$ such that $2/(\sqrt{5}+1)=x+y\sqrt{5}$. $\endgroup$ Commented Feb 7, 2012 at 15:43

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.

  • $\begingroup$ I have solved the problem without use of norm. See my edited post. $\endgroup$ Commented Feb 7, 2012 at 18:28
  • $\begingroup$ Also it's not true that if an element of $\mathbb{Z}[\sqrt{5}]$ is irreducible, then it's norm is irreducible. Take $1+3\sqrt{5}$, it's irreducible, but its norm is not. $\endgroup$ Commented Feb 7, 2012 at 18:33
  • $\begingroup$ But I agree with you that it's enlightening to apply to conceptual proofs instead of low level proofs. $\endgroup$ Commented Feb 7, 2012 at 18:41
  • $\begingroup$ @Sergey The irreducibility of the norm occurs in the monoid of all norms of elements of $\mathbb{Z}[\sqrt{5}]$. Since this is a proper multiplicative submonoid of $\mathbb Z$, the meaning of "irreducible" may differ from that in $\mathbb Z$. Please see the linked post and its references,. $\endgroup$
    – Math Gems
    Commented Feb 7, 2012 at 18:42
  • $\begingroup$ this topic is deeper than I thought. Thank you for the link. $\endgroup$ Commented Feb 7, 2012 at 18:58

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