Periodic function or not

Question

Given $f(x) : f(x)≠const$ and $\forall x \in \mathbb{R} \space\space f(x-1)+f(x+3)=f(x+1)+f(x+5)$, how do I prove that $f$ is periodic and find what the period is?
All I came up with is that $f(x-1)+f(x+3)=f((x-1)+2)+f((x+3)+2)$.
Assuming $$g(x)=f(x-1)+f(x+3)$$ it implies $$g(x+2)=f(x+1)+f(x+5)=g(x)$$ But how it implies the periodic property of $f(x)$? It seems to be true only if $f(x-1)=f(x+1)$ and $f(x+3)=f(x+5)$. But from initial statement, if $a+b=c+d$ it does not imply $a=c$ and $b=d$.

Answer 1

$f (x-1)+f (x+3)=f (x+1)+f (x+5)=f (x+3)+f (x+7) $

So $f (x-1)=f (x+7) $

So must have period of $8$.

Let $f(x) = 0$ if $0 \le x < 4$ and let $f(x) = 1$ if $4 \le x < 8$. and let $f$ have period $8$ then $f(x-1) + f(x + 3) = 1 =f(x+1) + f(x+5) = f(a) + f(a+4)$. So $f$ need not have any period less than $8$. (For any $0 < d < 4$ then for $-\frac d2 < x < 0$, $f(x)\ne f(x+d)$ so $f$ has no period less than $8$).