If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, find the period of $f(x)$.

Make $x=x-1$

$$f(x)+f(x-2)=\sqrt{2}f(x-1) \tag {1}$$

Make $x=x+1$

$$f(x+2)+f(x)=\sqrt{2}f(x+1) \tag {2}$$

Now sum $(1)$ and $(2)$:

$$2f(x)+f(x+2)+f(x-2)=\sqrt{2}(f(x+1)+f(x-1))=2f(x) $$

$$f(x+2)+f(x-2)=0$$

Now make $x=x+2$ then $f(x+4)=-f(x)$.

Make $x+4$ and get $f(x+8)=-f(x+4)=f(x)$, then the period is $8$.

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 $8$.