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...

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 $8$.
