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.

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
$a_n^2-2b_n^2=(a_{n-1}^2+2b_{n-1}^2)^2-2(2a_{n-1}b_{n-1})^2=a_{n-1}^4+4b_{n-1}^4+4a_{n-1}^2b_{n-1}^2-8a_{n-1}^2b_{n-1}^2$
$=a_{n-1}^4+4b_{n-1}^4-4a_{n-1}^2b_{n-1}^2 =(a_{n-1}^2-2b_{n-1}^2)^2.$
