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? $\endgroup$ – Alex Becker Feb 7 '12 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 $\overline{q}$. $\endgroup$ – Cam McLeman Feb 7 '12 at 15:13
  • $\begingroup$ @AlexBecker because then I need to prove that there is a norm on the ring and the ring is integral. $\endgroup$ – Sergey Filkin Feb 7 '12 at 15:15
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    $\begingroup$ Let's assume that bar means Galois conjugate :). But I think André's suggestion is simpler. $\endgroup$ – Dylan Moreland Feb 7 '12 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 $1-\sqrt{5}$)... $\endgroup$ – Arturo Magidin Feb 7 '12 at 15:59

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

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  • $\begingroup$ I have shown that by another method. Maybe It's the same as rationalization of denominator. $\endgroup$ – Sergey Filkin Feb 7 '12 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 $2/(\sqrt{5}+1)=x+y\sqrt{5}$. $\endgroup$ – André Nicolas Feb 7 '12 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.

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  • $\begingroup$ I have solved the problem without use of norm. See my edited post. $\endgroup$ – Sergey Filkin Feb 7 '12 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. $\endgroup$ – Sergey Filkin Feb 7 '12 at 18:33
  • $\begingroup$ But I agree with you that it's enlightening to apply to conceptual proofs instead of low level proofs. $\endgroup$ – Sergey Filkin Feb 7 '12 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 Feb 7 '12 at 18:42
  • $\begingroup$ this topic is deeper than I thought. Thank you for the link. $\endgroup$ – Sergey Filkin Feb 7 '12 at 18:58

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