Prove that $a_n^2 - 2b_n^2 = 1$ if $a_n+b_n\sqrt{2}=(a_{n-1}+b_{n-1}\sqrt{2})^2$ 
Two sequences of positive integers $a_n$ and $b_n$ are defined by $a_1 = b_1 = 1$ and $$a_n+\sqrt{2}b_n=(a_{n-1}+\sqrt{2}b_{n-1})^2$$ for $n \ge 2$. Prove that $(a_n)^2 - 2(b_n)^2 = 1$.

I found that 
$$a_n+b_n\sqrt{2}=(1+\sqrt{2})^{2^{n-1}}$$ 
and I want to prove 
$$a_n-b_n\sqrt{2}=(a_{n-1}-b_{n-1}\sqrt{2})^2$$
but it didn’t work. Please give me some hints. Thanks for your advances.
 A: Hint
$$a_n+\sqrt{2}b_n=(a_{n-1}+\sqrt{2}b_{n-1})^2 \Rightarrow \\
a_n+\sqrt{2}b_n=a_{n-1}^2+b_{n-1}^2+2a_{n-1}b_{n-1}\sqrt{2}\Rightarrow \\
a_n-a_{n-1}^2-b_{n-1}^2=\left(2a_{n-1}b_{n-1}-b_n \right)\sqrt{2}$$
Use that $\sqrt{2}$ is irrational to prove that 
$$a_n-a_{n-1}^2-b_{n-1}^2=2a_{n-1}b_{n-1}-b_n=0$$
From here you can conclude that indeed
$$a_n-b_n\sqrt{2}=(a_{n-1}-b_{n-1}\sqrt{2})^2
$$
A: As an alternate approach, you can directly prove the claim
$$a_n^{\,2}-2b_n^{\,2}=1,\;\text{for all}\;n\ge 2$$
by induction on $n$.

From
$$
a_2+b_2\sqrt{2}
=(a_1+b_1\sqrt{2})^2
=(1+\sqrt{2})^2
=3+2\sqrt{2}
$$
we get $a_2=3$ and $b_2=2$, hence
$$a_2^{\,2}-2b_2^{\,2}=3^2-2{\,\cdot\,}2^2=1$$
which verifies the base case $n=2$.

Next assume $a_n^{\,2}-2b_n^{\,2}=1$, for some $n\ge 2$.

Then from
$$
a_{n+1}+b_{n+1}\sqrt{2}
=(a_n+b_n\sqrt{2})^2
=(a_n^{\,2}+2b_n^{\,2})+(2a_nb_n)\sqrt{2}
$$
we get
$$
\begin{cases}
a_{n+1}=a_n^{\,2}+2b_n^{\,2}\\[4pt]
b_{n+1}=2a_nb_n\\
\end{cases}
\qquad\;\;\;
$$
hence
\begin{align*}
&a_{n+1}^{\,2}-2b_{n+1}^{\,2}\\[4pt]
=\;&(a_n^{\,2}+2b_n^{\,2})^2-2(2a_nb_n)^2\\[4pt]
=\;&a_n^4-4a_n^{\,2}b_n^{\,2}+4b_n^{\,4}\\[4pt]
=\;&(a_n^{\,2}-2b_n^{\,2})^2\\[4pt]
=\;&1\\[4pt]
\end{align*}
which completes the induction.
A: If $a_n+\sqrt{2}b_n=(a_{n-1}+\sqrt{2}b_{n-1})^2=a_{n-1}^2+2b_{n-1}^2+2\sqrt2a_{n-1}b_{n-1}$ 
then $a_{n}=a_{n-1}^2+2b_{n-1}^2$ and $b_n=2a_{n-1}b_{n-1},$ 
so you can prove that $a_n^2-2b_n^2=1$ by induction because
$a_n^2-2b_n^2=(a_{n-1}^2+2b_{n-1}^2)^2-2(2a_{n-1}b_{n-1})^2=a_{n-1}^4+4b_{n-1}^4+4a_{n-1}^2b_{n-1}^2-8a_{n-1}^2b_{n-1}^2$
$=a_{n-1}^4+4b_{n-1}^4-4a_{n-1}^2b_{n-1}^2 =(a_{n-1}^2-2b_{n-1}^2)^2.$
