<s>[INST] $a+b \sqrt{2}=(1+ \sqrt{2})^n$ 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 ?? 
 [/INST] 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*}
</s>