proof that the pattern exists for all n Explain the pattern 
$$(\sqrt2-1)^1= \sqrt2 - \sqrt1$$
$$(\sqrt2-1)^2 = \sqrt9 - \sqrt8$$    
$$ (\sqrt2-1)^3   = \sqrt{50} - \sqrt{49} $$     
that is $(\sqrt2-1)^n$  is  equal to difference of two consecutive numbers one of which are squares.

We will prove a stronger result that $(\sqrt{2}-1)^{2n}=\sqrt{x_n^2}-\sqrt{2y_n^2}$ for some $x_n, y_n$ where $x_n^2-2y_n^2=1$, and $(\sqrt{2}-1)^{2n-1}=\sqrt{2z_n^2}-\sqrt{w_n^2}$ for some $z_n, w_n$ where $2z_n^2-w_n^2=1$ (here $\{x_n\}, \{y_n\}, \{z_n\}, \{w_n\}$ are all increasing sequences of positive integers such that $x_1=3, y_1=2, z_1=1, w_1=1$.)
$n=1$ case is trivial, so suppose that the result holds for $n$. Then we have $(\sqrt{2}-1)^{2n+1}=(\sqrt{2z_n^2}-\sqrt{w_n^2})(3-2\sqrt{2})=(\sqrt{2(3z_n+2w_n)^2}-\sqrt{(4z_n+3w_n)^2})$. It is enough to prove $2(3z_n+2w_n)^2-(4z_n+3w_n)^2=1$, which is equivalent to $2z_n^2-w_n^2=1$, and this holds due to induction hypothesis. That is, we can set $z_{n+1}=3z_n+2w_n, w_{n+1}=4z_n+3w_n$.
Similarly, we have $(\sqrt{2}-1)^{2n+2}=(\sqrt{x_n^2}-\sqrt{2y_n^2})(3-2\sqrt{2})=\sqrt{(3x_n+4y_n)^2}-\sqrt{2(2x_n+3y_n)^2}$. It is enough to prove $(3x_n+4y_n)^2-2(2x_n+3y_n)^2=1$, which is equivalent to $x_n^2-2y_n^2=1$, and this holds due to induction hypothesis. That is, we can set $x_{n+1}=3x_n+4y_n, y_{n+1}=2x_n+3y_n$. This proves the result.
