# Use the ring ${\bf{O}}[\sqrt5]$ to show that $2$ is irreducible in $\mathbb{Z}[\sqrt5]$

I have this question:

Define

$${\bf{O}}[\sqrt5] = \{c_1 + c_2 \sqrt5 : c_1 + c_2 \in \mathbb{Z} \wedge c_1 - c_2 \in \mathbb{Z}\}.$$

This ring properly contains $\mathbb{Z}[\sqrt5]$. The valuation, $v$, maps ${\bf{O}}[\sqrt5]$ into $\mathbb{N}$ where $c_1 + c_2 \in \mathbb{Z} \wedge c_1 - c_2 \in \mathbb{Z}$,

$$v(c_1 + c_2 \sqrt{5}) = | c_1^2 - 5c_2^2|.$$

You may assume $v$ is multiplicative. $c_1 + c_2\sqrt5 \in {\bf{O}}[\sqrt5]$ is a unit in ${\bf{O}}[\sqrt5]$ if and only if

$$v(c_1 + c_2\sqrt{5}) = c_1^2 - 5c_2^2 = \pm 1.$$

• Prove that there are no integers $a,b$ such that $a^2 - 5b^2 \equiv 2 \pmod 4$. Prove that if $a,b$ are odd then $a^2 - 5b^2 \equiv 1 \pmod 8$.
• Prove $2$ is irreducible in $\mathbb{Z}[\sqrt5]$ and in ${\bf{O}}[\sqrt5]$. HINT: Why is it enough to show that there are no integers $a,b$ with $a^2 - 5b^2 = \pm 2$, and no odd integers $a,b$ with $a^2 - 5b^2 = \pm 8$?
• Find $a,b \in \mathbb{Z}$ such that $a^2 - 5b^2 = \pm 4$. Explain why this gives two essentially factorisations of $4$ into irreducibles in $\mathbb{Z}[\sqrt5]$, but not ${\bf{O}}[\sqrt5]$.

I have done the first question.

I'm stuck on the second one. I can see that I have already shown the bit in the hint, but I don't understand how this comes into play. I think this is because I don't exactly understand how valuations come into play. I get that it is a map which maps the elements in the ring to an integer in $\mathbb{N}$, but how does this help?

For the third one, I get that $a = b = 1$ gives me $a^2 - 5b^2 = -4$. I then though you could say something like, from the ring, can I say that we therefore see that $(1 + \sqrt5)(1 - \sqrt5) = -4$ but I don't get how to carry this on.

Can someone give me any tips or hint on what to do please?

• Hint $\rm\,\ 2 =\alpha \beta \:\Rightarrow\: 4 = v(2) = v(\alpha) v(\beta)\ \$ – Math Gems Apr 19 '13 at 19:41