If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, find the period of $f(x)$. If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, then the period of $f(x)$ is?
I tried replacing $x$ with $x-1$ and stuff, but didn't lead me to anything...

Note that $\sqrt{2} = \frac{2}{\sqrt{2}}$. Hence,
$$f(x + 1) + f(x - 1) = \frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}f(x)$$
$$\begin{align}f(x + 1) &= \frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}\left(f(x) - \sqrt{2}f(x - 1)\right)\\
&=\frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}\left(-f(x - 2)\right)\\
&= \frac{1}{\sqrt{2}}\left(f(x) - f(x - 2)\right)\end{align}$$
$$\sqrt{2}f(x + 1) = f(x) - f(x - 2)$$
$$\sqrt{2}f(x) = f(x - 1) - f(x - 3)$$
Hence,
$$f(x - 1) - f(x - 3) = f(x + 1) + f(x - 1)$$
$$f(x + 1) = -f(x - 3)$$
$$f(x) = -f(x - 4)$$
Therefore,
$$f(x) = -f(x - 4) = -(-f(x - 8)) = f(x - 8)$$
thus the period is $8$.
