Proof by strong induction of $(a + b\sqrt{2})^n = a_n + b_n \sqrt{2}$ I'm not very familiar with proofs by strong induction. I have a sketch for this one but Iot quite sure about it's validaty.
Let $a$, $b$, $a_n$, $b_n$ be integers such that.
$$(a + b\sqrt{2})^n = a_n + b_n \sqrt{2}$$
where $a$ is the integer closest to $b\sqrt{2}$. Prove that $a_n$ is the integer closest to $b_n\sqrt{2}$.
My sketch:
Suppose the statement holds for all $k\le n  $ we have to show that it holds for $n+1$
$(a + b\sqrt{2})^n = a_n + b_n \sqrt{2} \Rightarrow$ 
$ (a + b\sqrt{2})^{n+1} = (a_n + b_n \sqrt{2})(a + b\sqrt{2}) = a a_n + ab_n\sqrt{2} + \sqrt{2} (a_nb + bb_n\sqrt{2}) $
Since $a_n$ is the integer closest to $b_n \sqrt{2}$, $aa_n$ is the integer closest to $ab_n\sqrt{2} $ and $ba_n$ is the integer closest to $b b_n \sqrt{2}$. 
My intuition tells that  I can prove this with what I wrote, but I can't progress now. 

Hint.
Observe that, if
$$
(a+b\sqrt{2})^n=a_n+b_n\sqrt{2},
$$
then (it can be shown inductively)
$$
(a-b\sqrt{2})^n=a_n-b_n\sqrt{2},
$$
and if $|a-b\sqrt{2}|<1/2$, then clearly, $(a-b\sqrt{2})^n<1/2$.
