Let, $b_{n} = \sqrt a_{n}$
$\therefore b_{n-1} = \sqrt a_{n-1}$
$\therefore b_{n-2} = \sqrt a_{n-2}$
$\therefore \sqrt a_{n} -2 \sqrt a_{n-1} + \sqrt a_{n-2} = 0 \Rightarrow b_{n} -2b_{n-1} + b_{n-2} = 0$
Now, its characteristic equation is $x^{2} -2x +1 = 0 \Rightarrow x = 1, 1$
$\therefore$ General solution is $b_{n} = (A + Bn)x^{n} = (A + Bn)1^{n} = A + Bn$ where $A, B$ are arbitrary constants.
Putting $n=0$ and $n=1$, we get $b_{0} = A \Rightarrow A = 1 \ (\because b_{0} = \sqrt a_{0} = 1)$
Putting $n=0$ and $n=1$, we get $b_{1} = A + B \Rightarrow B = \sqrt {2} - 1 \ (\because b_{1} = \sqrt a_{1} = \sqrt 2 )$
$\therefore b_{n} = 1 + (\sqrt 2 -1)n$
$\therefore a_{n} = [1 + (\sqrt 2 -1)n]^2$