In this answer and other's that I've read, finding the units in $\mathbb{Z}[\sqrt{2}]$ (elements of the form $a + b\sqrt{2}$ for integers $a$ and $b$) is equivalent to finding the solutions to $a^2-2b^2 = \pm 1$. So from my understanding, for a ring $R$, $u \in R$ is a unit if there exists a $v \in R$ such that $uv = vu = 1_R$, with $1_R$ being the multiplicative identity in the ring.
So I have 2 questions:
- My understanding is that we arrive at this because $(a+b\sqrt{2})(a-b\sqrt{2}) = a^2-2b^2$, so units can only come in conjugate pairs. I could intuitively reason that this is because $(a+b\sqrt{2})x$ would have a $\sqrt{2}$ term if $x$ is not conjugate, but I'm not sure if this is strictly the reason why the units are only conjugates.
- Why can they multiply to $-1$ and still be a unit? Is it because if $(a+b\sqrt{2})(a-b\sqrt{2}) = a^2-2b^2 = -1$, then $(a+b\sqrt{2})^2(a-b\sqrt{2})^2 = (-1)^2 = 1$, and thus the inverse of $(a+b\sqrt{2})$ is $(a+b\sqrt{2})(a-b\sqrt{2})^2$?
Define the norm $N:\mathbb Z[\sqrt 2]\mapsto \mathbb Z$ as follows: $$N(a+b\sqrt 2)=a^2-2b^2\tag{1}$$ or $$N(x)=x\bar{x}\tag{2}$$ Then notice that for all $x,y\in\mathbb Z[\sqrt 2]$, we have that $$N(xy)=N(x)N(y)\tag{3}$$ Thus, suppose that $x\in\mathbb Z[\sqrt 2]$ is a unit. Then let $x^{-1}$ be its inverse. We have by $(3)$ that $$1=N(1)=N(xx^{-1})=N(x)N(x^{-1})$$ and so $N(x)$, which is an integer, is a unit, with multiplicative inverse $N(x^{-1})$. Since the only units in $\mathbb Z$ are $\pm 1$, we have that $N(x)=a^2-2b^2$ must equal $+1$ or $-1$.
This proves that $a^2-2b^2$ must equal $1$ or $-1$ if $x$ is a unit. Now we must prove the converse.
Suppose $N(x)=a^2-2b^2=\pm 1$. Then, by $(2)$, $x\bar{x}=1$ or $x\bar{x}=-1$. If the former is true, then $\bar{x}$ is the inverse of $x$ and $x$ is a unit; if the latter is true, then $-\bar{x}$ is the inverse of $x$ and $x$ is a unit.
Proof complete!
Now I shall address a generalized form of your first question. Suppose that $x,y\in\mathbb Z[\sqrt 2]$, and that $xy\in\mathbb Z$. Write $x=a+b\sqrt 2$ and $y=c+d\sqrt 2$. Then $$xy=(ac+2bd)+(ad+bc)\sqrt{2}$$ For $xy\in\mathbb Z$ to hold, it must be true that the coefficient $ad+bc$ of $\sqrt 2$ vanishes. Thus, we have $$ad+bc=0$$ or $$\frac{a}{b}=-\frac{c}{d}$$ ...which implies that the ratios of the rational and irrational parts of $x$ and $y$ are equal and opposite. This is equivalent to stating the existence of $z\in\mathbb Z[\sqrt 2]$ such that $z|x$ and $\bar{z}|y$.
If you apply this to units, you can see that it is indeed true that all units must come in conjugate pairs.