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> 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 $(a\_n)^2 - 2(b\_n)^2 = 1$.
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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.