$(f(x))^2=f(2x)+2f(x)-2$ Functional Equation

Question

$f$ is a function on real numbers: $$f(x)^2=f(2x)+2f(x)-2$$ and $$f(1)=3$$

What is the value of $f(6)$?

I find a solution $f(x)=2^x+1$. But, I don't know is there more solutions?

Answer 1

Let's look at $g(x)=f(x)-1$. Then $$f(x)^2-2f(x)+1=f(2x)-1\\ g(x)^2=g(2x)$$ There are infinitely many solutions, even continuous ones.

Knowing $f(x)$ gives us $f(2x)$, then $f(4x)$ and so on, but nothing else.

Since $6$ is not a power of $2$ times $1$, you can make $f(6)$ whatever you want.

Answer 2

$f(6)=65$ is not the only solution
Define f as follows:
$f(x)=2^x+1$, for $x=2^n$ or $x=\frac{1}{2^n}$ for $n\in \mathbb{N}$ and 1 otherwise.
In similar way you can define to be 2 otherwise.

Answer 3

The way this question is put, there is a relation to be satisfied between $f(x)$ and $f(y)$ only if $$\frac{x}{y}=2^n$$ for some $n\in\mathbb{Z}$. In particular, the numbers $f(1)$ and $f(6)$ have nothing to do with one another, and you can choose $f(6)$ as you please.

Answer 4

If we write $f(x) = g(x)+1$, the functional equation simplifies to $$ g(2x) = g(x)^2$$ $g(2x)$ must be nonnegative if you want real values, but otherwise $g(x)$ can be arbitrary on, say, $(-2,1] \cup [1,2)$, with $g(2^n x) = g(x)^{2^n}$ for integers $n$. In particular, $f(6)$ is not determined by $f(1)$.