The units of $\mathbb Z[\sqrt{2}]$ 
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.
 A: 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.
A: First, note that $a+b\sqrt{2}$ is a unit if and only if $a^2-2b^2=\pm1$. Use that to show that if $b\neq 0$ then $|b|\leq |a|< 2|b|$.
Now we first restrict ourselves to $a,b\geq 0$, and prove by induction on $b$. 
If $b=0$ then $a=\pm 1$, and $u>0$ implies $a=1$, so $u=(1+\sqrt 2)^0$.
If $a,b>0$ and $a+b\sqrt{2}$ is a unit, then $$(a+b\sqrt{2})(\sqrt{2}-1) = (2b-a)+(a-b)\sqrt{2}$$
is also a unit.
Since we know that $b\leq a< 2b$ and $b<a$,  we have that $2b-a>0$ and $0<a-b<b$, so by induction:
$$(a+b\sqrt{2})(\sqrt{2}-1) =(1+\sqrt{2})^n$$
But multiplying both sides by $1+\sqrt{2}$ you get:
$$a+b\sqrt{2}=(1+\sqrt{2})^{n+1}$$
Then you have to deal with the case where one of $a,b$ is negative...
