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...
 A: 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 $8$.
A: 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 $8$.
A: This is an ordinary linear recurrence, which you can solve using the characteristic equation
$$r+\frac1r=\sqrt 2.$$
The roots are $r=\dfrac{1\pm i}{\sqrt2}$, and the general solution is
$$f(n)=C\left(\dfrac{1+i}{\sqrt2}\right)^n+C^*\left(\dfrac{1-i}{\sqrt2}\right)^n=Ce^{in\pi/4}+C^*e^{-in\pi/4}=A\cos\frac{n\pi}4+B\sin\frac{n\pi}4.$$
The constants $C$ or $A,B$ can be obtained from some given conditions such as $f(0)$ and $f(1)$. This will only solve for integer arguments. For real arguments, you can specify the initial conditions by assigning $f(x)$ arbitrary values for $0\le x < 2$.
This proves that the period is $8$.
