I have an assignment full of questions like these, and I know that using some suitable substitution like replacing $x$ with $(x-1)$ and simplyifying it, I will be able to arrive at a stage where the equations reduce to $f(x)=f(x+T)$ where "T" is the fundamental period.
But I cant think of a smart substitution, and any substitution I try only complicates the question further!
How do I find out the best substitution for a question of this type?
On
I'm going to describe the most general $f$ satisfying the given functional equation $$f(x-1)+f(x+3)=f(x+1)+f(x+5)\ .\tag{1}$$
Let $$g(x):={1\over2}\bigl(f(x)+f(x+4)\bigr)\ .\tag{2}$$ Then from $(1)$ it follows that $$g(x-1)=g(x+1)\ ,$$ whence $g$ has to be periodic of period $2$. Assume now that such a $g$ is given. Then according to $(2)$ the function $f$ has to fulfill the inhomogeneous functional equation $$f(x)+f(x+4)=2g(x)\ .\tag{3}$$ According to general principles of linear algebra the general solution of $(3)$ can be put together from a particular solution $f_p$ of $(3)$ and the general solution of the associated homogeneous equation $$f_h(x)+f_h(x+4)=0\ .\tag{4}$$ Since $g$ is periodic of period $2$ the function $f_p(x):=g(x)$ is obviously a particular solution of (3).
Now $(4)$ says that $f_h$ is periodic of period $4$, up to a sign change. It follows that the function $h(x):=e^{-i\pi x/4}f_h(x)$ is periodic of period $4$.
Putting it all together we see that any solution of $(1)$ is necessarily of the form $$f(x)=g(x)+e^{i\pi x/4}\>h(x)\ ,\tag{5}$$ where $g$ is of period $2$ and $h$ is of period $4$. Conversely it is easy to check that any $f$ given by $(5)$, where $g$ and $h$ are as described, is a solution of $(1)$. In particular all solutions of $(1)$ have period $8$.
With the subst $y=x+1$ we get $$f(y)=f(y+2)+f(y+6)-f(y+4)$$ and if $y=x-1$ $$f(y)=f(y-2)+f(y+2)-f(y+4)$$
Then $f(y+6)=f(y-2)$, that is, the period of $f$ is $8$ (or a divisor).