Show $1+\sqrt{5}$ is irreducible in $\mathbb{Z}[\sqrt{5}]$. Similar questions use the norm ( see e.g. How to show $1 + \sqrt{ 5 }$ is irreducible ), but I do not know yet about such a mapping, so I can't use it here.

I want to show now that if $$1+\sqrt{5}=(a+b\sqrt{5})(c+d\sqrt{5})$$ at least one of them is a unit i.e. has a multiplicative inverse in $\mathbb{Z}[\sqrt{5}]$. First I tried to characterise the units in $\mathbb{Z}[\sqrt{5}]$ by looking at $$1=(a+b\sqrt{5})(c+d\sqrt{5})$$ and trying to see which properties $a,b,c,d$ need to have to be a unit. But that did not get me very far to be honest. Rearranging the original equation got me nowhere good either.

How can I approach this without using the norm?

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    $\begingroup$ If you really mean $\mathbb Q[\sqrt{5}]$, then $1+\sqrt{5}$ is a unit (as is every other nonzero element). Do you mean $\mathbb Z[\sqrt{5}]$? $\endgroup$ – Wojowu Jun 19 '17 at 13:25
  • $\begingroup$ You are correct. I am sorry, I fixed it. It should be over $\mathbb{Z}$ of course! $\endgroup$ – Jonathan Jun 19 '17 at 13:40

I'm assuming you meant $\mathbb Z[\sqrt 5]$, since $\mathbb Q[\sqrt 5]$ is a field. $$1+\sqrt{5}=(a+b\sqrt{5})(c+d\sqrt{5}) = (ac+5bd) + \sqrt 5(bc+da)$$ $$\implies ac+5bd =1\qquad bc+da=1$$ $$ \implies 1-\sqrt{5}= (ac+5bd) - \sqrt 5(bc+da) =(a-b\sqrt{5})(c-d\sqrt{5}) $$ $$ \implies (1+\sqrt{5})(1-\sqrt{5}) = (a+b\sqrt{5})(c+d\sqrt{5})(a-b\sqrt{5})(c-d\sqrt{5}) $$ $$ \implies -4 = (a^2-5b^2)(c^2-5d^2) $$ Now, if $a^2-5b^2 = \pm 1$, then $(a+b\sqrt{5})(a-b\sqrt{5}) = \pm 1$, so it is a unit. The same holds for $c^2-5d^2$. So If $1+\sqrt{5}$ is reducible, you have $a^2-5b^2=\pm 2$, $c^2-5d^2=\mp 2$ that is impossible modulus $4$.

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    $\begingroup$ Note that this argumentation uses the norm, just not calling it by name. It might thus be easier to first introduce the norm, or at least to use this as an example to motivate such a concept. $\endgroup$ – Dirk Jun 19 '17 at 13:43
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    $\begingroup$ It's worth mentioning that this is the same as the approach using norms, just that you don't mention the word "norm". $\endgroup$ – Wojowu Jun 19 '17 at 13:49
  • $\begingroup$ I believe to understand everything, but your last step. We are saying the only units in $\mathbb{Z}[\sqrt(5)]$ are $\pm 1$, therefore we either want to show that one of the factors is one or that there is a contradiction if we write it as a product of two non units. You showed that the product is $ -4 = (a^2-5b^2)(c^2-5d^2)$, if we assume it can be written as a product of two non units. But how do we reach a contradiction with modular arithmetic here? Why is $a^2-5b^2=\pm 2$, $c^2-5d^2=\mp 2$ impossible? $\endgroup$ – Jonathan Jun 21 '17 at 12:01
  • $\begingroup$ @Jonathan the remainder for squares modulus 4 are 0,1 so $a^2-5b^2\equiv a^2-b^2$ can't be 2 $\endgroup$ – Exodd Jun 21 '17 at 18:02
  • $\begingroup$ I see, excuse me, I'm not familiar with modular arithmetic $\endgroup$ – Jonathan Jun 21 '17 at 18:08

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