How do I finish solving $f(x)f(2y)=f(x+4y)$? I'm trying to solve this functional equation:
$$f(x)f(2y)=f(x+4y)$$
The first thing I tried was to set $x=y=0$; then I get:
$$f(0)f(0)=f(0)$$
which means that either $f(0)=0$ or we can divide the equation by $f(0)$ and then $f(0)=1$.
Case 1: If $f(0)=0$ then we can try to set $x=0$. Then;
$$f(0)f(2y)=f(4y)$$
$$0=f(4y)$$
Which means that one of the solutions is a constant function $f(x)=0$
Case 2: If $f(0)=1$. This is where I'm stuck.
I tried to set $x=0$, then I get:
$$f(2y)=f(4y)$$
I also tried to set $x=-4y$. Then I get:
$$f(-4y)f(2y)=f(0)=1$$
I think that these two observations could be useful, but I don't know how to continue from here. I have guessed that another solution is $f(x)=1$ but I don't know how to show that there aren't any others as well.

Assuming that $ f(0)=1$, you got
for $ y\in \Bbb R $,
$$f(4y)=f(2y)=f(y)=f(\frac y2)$$
$$=...=f(\frac{y}{2^n})$$
for each $ n\ge 0$.
but by continuity of $ f $ at $ 0$,
$$\lim_{n\to +\infty}f(\frac{y}{2^n})=f(0)=1$$
thus
$$(\forall y\in \Bbb R)\;\; f(y)=1$$
