About finding the period of a function which satisfies $f(x-2)+f(x+2)=f(x)$.

Hint. We have that $$f(x+2)+f(x+6)=f(x+4) $$ and $$ f(x)+f(x+4)=f(x+2) .$$ By adding them, we get $$ f(x)+f(x+6)=0\implies f(x+6)=-f(x).$$
Therefore $$f(x+12)=f((x+6)+6)=-f(x+6)=f(x)$$ for any real $x$ and we may conclude that a period of the function is $12$.

Using this, the characteristic equation of $$f(x+2)-f(x)+f(x-2)=0$$

$t^4-t^2+1=0, t=e^{(2m+1)\pi i/6}$ where $m=0,2,3,5$

$$\implies f(x)=\sum_{m=\{0,2,3,5\}}A_me^{(2m+1)\pi ix/6}$$

Now if $f(x)=f(x+T)\forall x,$ we need $$e^{(2m+1)\pi ix/6}=e^{(2m+1)\pi i(x+T)/6}$$

$\implies\dfrac{(2m+1)\pi T}6\equiv0\pmod{2\pi}$

$\iff(2m+1)T\equiv0\pmod{12}$

As $(2m+1,12)=1$ for $m=0,2,3,5,$ we need $$T\equiv0\pmod{12}$$