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.

Ryan's answer is showing how to finish the argument you are making. This answer provides alternative way to solve the equation without splitting into the cases in the first place.
Substituting $y/2$ in place of $y$, we get functional equation $f(x)f(y)=f(x+2y)$, which is little more symmetric. If we now swap $x$ and $y$, we get $f(x+2y)=f(y+2x)$. Put $y=-2x$ and observe that $f(-3x)=f(0)$. Finally, substitute $-x/3$ in place of $x$ and get $f(x)=f(0)$, hence $f(x)$ is a constant function. Let $f(x)=a$, putting into the original equation this gives $a^2=a$, so $a=0$ or $a=1$. Thus $f(x)=0$ and $f(x)=1$ are the only solutions.
