Consider the following functional equation over the reals :
$$f(4x) = f(3x) f(x) - f(2x)$$
Apart from $f(x) = 2, f(x) = 0, f(x) = 2 \cos(x), f(x) = 2 \cosh(x)$ are there other continu solutions ?
Are all solutions periodic or constant ?
Notice $f(0) = 2$ for every nonconstant solution of $f(x)$. Ok that needs some explaining and I do not have a full proof but ...
I can almost show that for analytic near $0$ solutions.
$$f(x) = 0 + a x + b x^2 + ...$$
then
$$f(4x) = f(3x)f(x) - f(2x)$$
implies
$$ 4 a x + 4 b x^2 + ... = (3 a x + 3 b x^2 + ...)(a x + b x^2 + ...) - 2 a - 2 b x^2 + ...$$
So $4 a = - 2 a$. so $a$ must be $0$.
Also $4b = 3 a^2 - 2b$ so $b = 0$
I think with some work we can show all coefficients must be $0$.
Second question :
$$g(4x) = g(3x) g(x)^3 - g(2x)$$
What are the continu solutions to this ?
Notice $g(0) = 2^{1/3}$
Is $g$ related to the cube root of the cosine ?
This reminds me of this question :
Solution of $f(3x)=\frac{f(x)f(4x)}{2f(2x)}+\frac{f^2(2x)}{2f(x)}$?
edit : I added the 2cosh example. Thanks to user Gonçalo for pointing it out.
edit
The related generalized equation $$f(n x) = f((n-1) x) f(x) - f((n-2) x)$$
valid for all integer $n$ has as solutions
$f(x) = 2, f(x) = 0, f(x) = 2 \cos(x), f(x) = 2 \cosh(x)$
And I believe these are the only ones. Probably provable by induction and/or pigeonhole.
The original question is thus a variant on this.
Another variant could be for instance the system of equations
$$f(4x) = f(3x) f(x) - f(2x)$$
$$f(7x) = f(6x) f(x) - f(5x)$$
For which again we know at least $4$ solutions and we wonder if the equations force $f$ to be one of those.
Im not sure if this helps but just for background or so.