<s>[INST] 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...
 [/INST] 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 $8$.
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