Show that $11+6\sqrt{-5}$ and $16+3\sqrt{-5}$ are irreducible elements in $\mathbb{Z}[\sqrt{-5}]$. Show that these elements are not associates.
What I did was let $A = \mathbb{Z}[\sqrt{-5}]$. For every element $\alpha$ of $A$ there exist unique $a, b \in \mathbb{Z}$ such that $\alpha = a + b \sqrt{-5}$. Then, I considered the function $N \colon A \to \mathbb{N}$ defined by $N(a + b \sqrt{-5}) = a^2 + 5 b^2$, but I'm not sure how to go from here.
Any tips would be appreciated!
On
HINT. If $\alpha, \beta \in \mathbb{Z}[\sqrt{-5}]$, then $N(\alpha\beta)=N(\alpha)N(\beta)$. If they were reducible then there would be two elements whose product would give the same norm. Can this happen? For the associate, remember elements are associates of each other if there is a unit $u$ with $u\alpha= \beta$. What is the norm of a unit? So can we have $u\alpha= \beta$ for some $u$? [Take norms]
Take $11 + 6\sqrt{-5}$ as example.
Suppose that we write it as the product of two elements of $\mathbb{Z}[\sqrt{-5}]$, i.e. $11 + 6\sqrt{-5} = \alpha \beta$. Taking norm, we have: $$N(11+6\sqrt{-5}) = N(\alpha)N(\beta).$$ But the norm of $11 + 6\sqrt{-5}$ is $301 = 7 \times 43$. This leaves essentially two possibilities: either one of $\alpha$ and $\beta$ has norm $1$, or one of them has norm $7$, since the norms are all positive integers.
However, the second possibility is easily eliminated by writing (say) $\alpha$ as $u + v\sqrt{-5}$ and noting that $u^2 + 5v^2 = 7$ has no solution in integers.
Thus we have (say) $N(\alpha) = 1$, which means that $\alpha$ is a unit in $\mathbb{Z}[\sqrt{-5}]$.
Perhaps you can try solving the rest part.