TITLE: The units of $\mathbb Z[\sqrt{2}]$
QUESTION [24 upvotes]: How can I show that the units $u$ of $R=\mathbb Z[\sqrt{2}]$ with $u>1$ are $(1+ \sqrt{2})^{n}$ ?

I have proved that the right ones are units because their module is one, and it is said to me to do it by induction on $b$ and multiplication by $-1+\sqrt{2}$. I have already shown that the units of this ring has norm $1$ and all the numbers with norm $1$ are units, this may help.

REPLY [7 votes]: Here is a simpler proof that relies only on basic bounding arguments. Note that if $u$ is a unit then $u, 1/u, -u, -1/u$ are all units also. Now, since once of these is greater than $1$, it suffices to show if $u<1$ then $u$ is a power of $1+\sqrt{2}$. Clearly the nonnegative powers of $1+\sqrt{2}$ monotonically tend to $\infty$ starting from $1$, so we can write $$(1+\sqrt{2})^k\le u <(1+\sqrt{2})^{k+1}$$ for some $k\in\mathbf{Z}^{+}\cup\{0\}$. Dividing by $(1+\sqrt{2})^k$ yields $$1\le u(1+\sqrt{2})^{-k}<1+\sqrt{2}.$$ Note that $u(1+\sqrt{2})^{-k}\in\mathbf{Z}[\sqrt{2}]^{\times}$, and since $1+\sqrt{2}$ is the smallest unit greater than $1$, we must have $u(1+\sqrt{2})^{-k}=1\implies u=(1+\sqrt{2})^k$. Due to norm being multiplicative, all powers of $1+\sqrt{2}$ are units, so we are done.