$a+b \sqrt{2}=(1+ \sqrt{2})^n$ when $a^2-2b^2=+1 or -1$ I want to prove this statement :
$a+b \sqrt{2}=(1+ \sqrt{2})^n$ when $a^2-2b^2=+1  or -1$
Where a,b,and n are positive integers . 
Usually people consider it true saying $1+ \sqrt{2}$ is fundamental unit of $\mathbb{Z} [ \sqrt{2}]$ , but how can I proof this more rigorously ?? 
 A: Since all elements $(1+\sqrt{2})^n$ are invertible in $\mathbb{Z}[\sqrt{2}]$, because of $-(1+\sqrt{2})(1-\sqrt{2})=1$, the norm of such elements is equal to $\pm 1$. But the norm is $\pm 1=N(a+b\sqrt{2})=a^2-2b^2$.
(Rigorous) Reference: The units of $\mathbb Z[\sqrt{2}]$
A: To prove this is equivalent to proving it's the fundamental unit. Here's a hint for a direct proof, based on understanding how you might show some element is the generator of a cyclic group.
The powers of $\mu = 1 + \sqrt{2}$ form an increasing unbounded sequence of real numbers. Let $\nu$ be a positive unit. Find the largest power of $\mu$ that's less than or equal to $\mu$. Divide $\nu$ by that power to produce a small unit.
A: Define $a_n$ and $b_n$ by 
\begin{eqnarray*}
a_n+b_n \sqrt{2} =(1+\sqrt{2})^n.
\end{eqnarray*}
Their recurrence relation is easily derived by 
\begin{eqnarray*}
a_{n+1}+b_{n+1} \sqrt{2} =(a_n+b_n \sqrt{2}) (1+\sqrt{2}) \\
a_{n+1}=a_n +2 b_n \\
b_{n+1} =a_n+b_n
\end{eqnarray*}
$(a_n,b_n)$ can be shown to satisfy Pell's equation inductively, as follows
\begin{eqnarray*}
a_{n+1}^2-2b_{n+1}^2 =a_n^2+4a_n b_n + 4b_n ^2 -2(a_n^2+2 a_n b_n +b_n^2) =-(a_n^2-2b_n^2) =(-1)^{n+1}.
\end{eqnarray*}
