Problem 967

[Proof] First, we prove an auxiliary proposition: For any natural number $n$, there exist natural numbers $a, b$, such that
$$\left\{\begin{array}{l} (1-\sqrt{2})^{n}=\sqrt{a^{2}}-\sqrt{2 b^{2}}, \\ a^{2}-2 b^{2}=(-1)^{n} . \end{array}\right.$$

When $n=1$, the conclusion is true. In fact, taking $a=b=1$ suffices. Suppose the conclusion holds for $n=k$. Then for $n=k+1$,
$$\begin{aligned} & (1-\sqrt{2})^{k+1}=(1-\sqrt{2})^{k}(1-\sqrt{2})=\left(\sqrt{a^{2}}-\sqrt{2 b^{2}}\right)(1-\sqrt{2}) \\ = & (a-b \sqrt{2})(1-\sqrt{2})=a+2 b-(a+b) \sqrt{2} \\ = & \sqrt{(a+2 b)^{2}}-\sqrt{2(a+b)^{2}}=\sqrt{a_{1}^{2}}-\sqrt{2 b_{1}^{2}} . \end{aligned}$$

Here $a_{1}, b_{1}$ are natural numbers, and
$$\begin{aligned} a_{1}^{2}-2 b_{1}^{2} & =(a+2 b)^{2}-2(a+b)^{2} \\ & =-a^{2}+2 b^{2} \\ & =-\left(a^{2}-2 b^{2}\right) \\ & =(-1)^{k}(-1) \\ & =(-1)^{k+1} . \end{aligned}$$

This shows that the conclusion also holds for $n=k+1$, thus the auxiliary proposition is true for any natural number $n$.
Using this proposition, it is not difficult to derive the conclusion of this problem. Indeed, if $n$ is even, then
$$(\sqrt{2}-1)^{n}=(1-\sqrt{2})^{n}=\sqrt{a^{2}}-\sqrt{2 b^{2}} .$$

Here $a, b$ are natural numbers, so $a^{2}, 2 b^{2}$ are also natural numbers, and $a^{2}-2 b^{2}=1$.
If $n$ is odd, then
$$(\sqrt{2}-1)^{n}=-(1-\sqrt{2})^{n}=\sqrt{2 b^{2}}-\sqrt{a^{2}},$$

Here $2 b^{2}, a^{2}$ are natural numbers, and
$$2 b^{2}-a^{2}=-\left(a^{2}-2 b^{2}\right)=-(-1)=1 \text {. }$$