# Show $1+\sqrt{5}$ is irreducible in $\mathbb{Z}[\sqrt{5}]$

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?

• If you really mean $\mathbb Q[\sqrt{5}]$, then $1+\sqrt{5}$ is a unit (as is every other nonzero element). Do you mean $\mathbb Z[\sqrt{5}]$? – Wojowu Jun 19 '17 at 13:25
• You are correct. I am sorry, I fixed it. It should be over $\mathbb{Z}$ of course! – 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 $4$.
• 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? – Jonathan Jun 21 '17 at 12:01
• @Jonathan the remainder for squares modulus 4 are 0,1 so $a^2-5b^2\equiv a^2-b^2$ can't be 2 – Exodd Jun 21 '17 at 18:02