A problem on number of solutions of a functional equation: $ f ( x ) = 8 f ( 2 x + 1 ) $ 
Find all functions $ f : \mathbb R \to \mathbb R $ such that $ f ( 0 ) = 1 $ and for all $ x \ne - 1 $,
$$ f ( x ) = 8 f ( 2 x + 1 ) \text . $$

I have found only one solution: $ \frac 1 { ( x + 1 ) ^ 3 } $. The method was by iterated substitution of $ 2 x + 1 $ for $ x $ and finding a pattern in the resulted sequence.
 A: Let $g(x) = f(x-1)$. Then we obtain $g(x) = f(x-1) = 8f(2x-1) = 8g(2x)$, and $g(-1)=1$. Note that values of $g$ for positive $x$ cannot be related to negative $x$. Clearly, $g(0)$ is arbitrary.
To find solutions for positive $x$, we let $h(x) = g(2^x)$, giving $h(x) = 8h(x+1)$. This has solutions $8^{-x} h_1(x)$ for any function $h_1$ with period $1$. Substituting back into $g$ we obtain $g(x) = 8^{-\log_2(x)} h_1(\log_2(x)) = \frac{1}{x^3}h_1(\log_2(x))$, for positive $x$.
The solutions to negative $x$ is handled similarly, now letting $h(x) = g(-2^x)$, and obtaining solutions $h(x) = 8^{-\log_2(-x)} h_2(x)$, for any function $h_2$ with period 1. So we obtain $g(x) = \frac{-1}{x^3}(\log_2(-x))$. We can remove the negative sign by replacing $h_2$ with $-h_2$, since it is an arbitrary function.
So we use $f(x) = g(x+1)$ to find all solutions:
$$f(x) =
\begin{cases}
\frac{1}{(x+1)^3}h_1(\log_2(x+1)) & \text{if $x>-1$}\\
r & \text{if $x=-1$}\\
\frac{1}{(x+1)^3}h_2(\log_2(-x-1)) & \text{if $x<-1$}\\
\end{cases}$$
for all reals $r$, and all functions $h_1, h_2$ with period $1$.
A: A start, laying in bed.
$f(2^z-1) = 8f(2^{z+1}-1)$,
so if $g(z) = f(2^z-1),
g(z)=8g(z+1)$.
Therefore
$g(z) = 8^ng(z+n)$
or
$g(z) = g(z-n)/8^n$.
Putting $z=0$,
since $g(0)=f(0)=1$,
$g(-n)=8^n$.
I'm not sure where to go from here,
and it's late,
so I'll stop.
